Homework Help.

[Daniel Pitts]

Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have
one particular homework problem that I think I've gotten correct, but
I'd like to have someone with more experience confirm. Unfortunately
most of my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:


..--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-


My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer. Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction, which
cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for the
..7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server flaked
out on the first attempt (as it does far too often), but I'm not always
sure.

 

[Phil Hobbs]

On 01/28/2013 12:08 PM, Daniel Pitts wrote:

Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have one
particular homework problem that I think I've gotten correct, but I'd
like to have someone with more experience confirm. Unfortunately most of
my classmates finish the homework on the due-date, so I can't compare
notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:


.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-


My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer. Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction, which
cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for the
.7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server flaked
out on the first attempt (as it does far too often), but I'm not always
sure.

RMS is a time average based on the power dissipated. An ideal diode in
the secondary side conducts on the positive half-cycle and not on the
negative half-cycle. On the positive, the instantaneous current is just
the same as it would be if the diode weren't there, and therefore the
instantaneous power is also unchanged. There's no power dissipated on
the negative half cycle, so the average power is half of what it would
be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes down
by a factor of.....?

Cheers

Phil Hobbs





--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[Daniel Pitts]

On 1/28/13 9:14 AM, Phil Hobbs wrote:

On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have one
particular homework problem that I think I've gotten correct, but I'd
like to have someone with more experience confirm. Unfortunately most of
my classmates finish the homework on the due-date, so I can't compare
notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:


.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-


My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer. Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction, which
cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for the
.7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server flaked
out on the first attempt (as it does far too often), but I'm not always
sure.

RMS is a time average based on the power dissipated.An ideal diode in
the secondary side conducts on the positive half-cycle and not on the
negative half-cycle. On the positive, the instantaneous current is just
the same as it would be if the diode weren't there, and therefore the
instantaneous power is also unchanged. There's no power dissipated on
the negative half cycle, so the average power is half of what it would
be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes down
by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it was
the square root of the average of the squares of all the values - a sort
of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4 when
the diode is in the circuit. This would lead to a Pavg = 15v^2/220ohm ~=
1watt.

Is that the proper approach?

Thanks,
Daniel.

 

[Phil Hobbs]

On 01/28/2013 12:45 PM, Daniel Pitts wrote:

On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have one
particular homework problem that I think I've gotten correct, but I'd
like to have someone with more experience confirm. Unfortunately most of
my classmates finish the homework on the due-date, so I can't compare
notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:


.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-


My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer. Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction, which
cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for the
.7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server flaked
out on the first attempt (as it does far too often), but I'm not always
sure.

RMS is a time average based on the power dissipated.An ideal diode in
the secondary side conducts on the positive half-cycle and not on the
negative half-cycle. On the positive, the instantaneous current is just
the same as it would be if the diode weren't there, and therefore the
instantaneous power is also unchanged. There's no power dissipated on
the negative half cycle, so the average power is half of what it would
be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes down
by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it was
the square root of the average of the squares of all the values - a sort
of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4 when
the diode is in the circuit. This would lead to a Pavg = 15v^2/220ohm ~=
1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate
the two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root, which
is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[Daniel Pitts]

On 1/28/13 10:05 AM, Phil Hobbs wrote:

On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have
one
particular homework problem that I think I've gotten correct, but I'd
like to have someone with more experience confirm. Unfortunately
most of
my classmates finish the homework on the due-date, so I can't compare
notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:


.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-


My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction, which
cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for the
.7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server flaked
out on the first attempt (as it does far too often), but I'm not always
sure.

RMS is a time average based on the power dissipated.An ideal diode in
the secondary side conducts on the positive half-cycle and not on the
negative half-cycle. On the positive, the instantaneous current is just
the same as it would be if the diode weren't there, and therefore the
instantaneous power is also unchanged. There's no power dissipated on
the negative half cycle, so the average power is half of what it would
be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes down
by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it was
the square root of the average of the squares of all the values - a sort
of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4 when
the diode is in the circuit. This would lead to a Pavg = 15v^2/220ohm ~=
1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate
the two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root, which
is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

 

[Tim Wescott]

On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have
one
particular homework problem that I think I've gotten correct, but
I'd like to have someone with more experience confirm. Unfortunately
most of
my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:


.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-


My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction,
which cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for
the .7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server
flaked out on the first attempt (as it does far too often), but I'm
not always sure.

RMS is a time average based on the power dissipated.An ideal diode in
the secondary side conducts on the positive half-cycle and not on the
negative half-cycle. On the positive, the instantaneous current is
just the same as it would be if the diode weren't there, and
therefore the instantaneous power is also unchanged. There's no power
dissipated on the negative half cycle, so the average power is half
of what it would be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes
down by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it was
the square root of the average of the squares of all the values - a
sort of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4
when the diode is in the circuit. This would lead to a Pavg =
15v^2/220ohm ~= 1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate the
two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root,
which is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the RMS
voltage goes down by what factor?

Cheers

Phil Hobbs


Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

It's probably close enough.

I would strongly suggest that you slap some graph paper down and sketch
the voltage waveform seen by the resistor. Even if the instructor
doesn't need to see it, it should help your understanding immensely.

If you're on a track to be an EE and not just a technician, and if you've
taken calculus, and if you have the time, you may want to take your
sketch of the waveform at the resistor and calculate the RMS voltage.

In this case, RMS would be defined as the square root of the time average
of the square of the voltage. Finding the time average of the square of
the voltage can either be done intuitively (you know what the time
average is for a whole cycle, so you can easily do it for a half cycle),
but grinding through the exercise of integrating the voltage-squared over
one cycle, then dividing by the cycle time, will be good for your soul.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

 

[Daniel Pitts]

On 1/28/13 12:03 PM, Tim Wescott wrote:

On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have
one
particular homework problem that I think I've gotten correct, but
I'd like to have someone with more experience confirm. Unfortunately
most of
my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:


.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-


My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction,
which cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for
the .7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server
flaked out on the first attempt (as it does far too often), but I'm
not always sure.

RMS is a time average based on the power dissipated.An ideal diode in
the secondary side conducts on the positive half-cycle and not on the
negative half-cycle. On the positive, the instantaneous current is
just the same as it would be if the diode weren't there, and
therefore the instantaneous power is also unchanged. There's no power
dissipated on the negative half cycle, so the average power is half
of what it would be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes
down by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it was
the square root of the average of the squares of all the values - a
sort of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4
when the diode is in the circuit. This would lead to a Pavg =
15v^2/220ohm ~= 1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate the
two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root,
which is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the RMS
voltage goes down by what factor?

Cheers

Phil Hobbs


Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

It's probably close enough.

I would strongly suggest that you slap some graph paper down and sketch
the voltage waveform seen by the resistor. Even if the instructor
doesn't need to see it, it should help your understanding immensely.
I've seen, and sketched, enough half-waves that I'm not that concerned

about seeing it. I do understand that much.

If you're on a track to be an EE and not just a technician, and if you've
taken calculus, and if you have the time, you may want to take your
sketch of the waveform at the resistor and calculate the RMS voltage.
I'm on track to be a hobbyist, nothing more ;-). However, I think

you're correct that it my help me to do that work. Its been a while
since I've done any real calculus.

In this case, RMS would be defined as the square root of the time average
of the square of the voltage. Finding the time average of the square of
the voltage can either be done intuitively (you know what the time
average is for a whole cycle, so you can easily do it for a half cycle),
but grinding through the exercise of integrating the voltage-squared over
one cycle, then dividing by the cycle time, will be good for your soul.
Yes, probably true. Deriving the formulas has always been a useful

exercise for me. I just didn't really have the time to sit down with a
pencil and paper to do so. Maybe tonight.

Was my final answer of 8 watts for the Pavg correct?

 

[George Herold]

On Jan 28, 4:06 pm, Daniel Pitts
<newsgroup.nos...@virtualinfinity.net> wrote:

On 1/28/13 12:03 PM, Tim Wescott wrote:



On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have
one
particular homework problem that I think I've gotten correct, but
I'd like to have someone with more experience confirm. Unfortunately
most of
my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:

.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-

My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction,
which cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for
the .7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server
flaked out on the first attempt (as it does far too often), but I'm
not always sure.

RMS is a time average based on the power dissipated.An ideal diode in
the secondary side conducts on the positive half-cycle and not on the
negative half-cycle. On the positive, the instantaneous current is
just the same as it would be if the diode weren't there, and
therefore the instantaneous power is also unchanged. There's no power
dissipated on the negative half cycle, so the average power is half
of what it would be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes
down by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it was
the square root of the average of the squares of all the values - a
sort of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4
when the diode is in the circuit. This would lead to a Pavg > >>>> 15v^2/220ohm ~= 1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

                = 1/R * (time average of (instantaneous voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate the
two expressions for average power:

1/R * Vrms^2  = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root,
which is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

It's probably close enough.

I would strongly suggest that you slap some graph paper down and sketch
the voltage waveform seen by the resistor.  Even if the instructor
doesn't need to see it, it should help your understanding immensely.

I've seen, and sketched, enough half-waves that I'm not that concerned
about seeing it.  I do understand that much.

If you're on a track to be an EE and not just a technician, and if you've
taken calculus, and if you have the time, you may want to take your
sketch of the waveform at the resistor and calculate the RMS voltage.

I'm on track to be a hobbyist, nothing more ;-).  However, I think
you're correct that it my help me to do that work. Its been a while
since I've done any real calculus.

In this case, RMS would be defined as the square root of the time average
of the square of the voltage.  Finding the time average of the square of
the voltage can either be done intuitively (you know what the time
average is for a whole cycle, so you can easily do it for a half cycle),
but grinding through the exercise of integrating the voltage-squared over
one cycle, then dividing by the cycle time, will be good for your soul.

Yes, probably true.  Deriving the formulas has always been a useful
exercise for me. I just didn't really have the time to sit down with a
pencil and paper to do so.  Maybe tonight.

Was my final answer of 8 watts for the Pavg correct?-

Grin, Here's another way to get to the answer. (Just to check
yourself.)

Without the diode what's the average power?

Then the diode blocks the signal for 1/2 the time so your answer
should be 1/2 the above number.... does that check?

George H.



Hide quoted text -

- Show quoted text -

 

[John Larkin]

On Mon, 28 Jan 2013 13:23:11 -0800 (PST), George Herold
<gherold@teachspin.com> wrote:

On Jan 28, 4:06 pm, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
On 1/28/13 12:03 PM, Tim Wescott wrote:



On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have
one
particular homework problem that I think I've gotten correct, but
I'd like to have someone with more experience confirm. Unfortunately
most of
my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:

.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-

My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction,
which cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for
the .7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server
flaked out on the first attempt (as it does far too often), but I'm
not always sure.

RMS is a time average based on the power dissipated.An ideal diode in
the secondary side conducts on the positive half-cycle and not on the
negative half-cycle. On the positive, the instantaneous current is
just the same as it would be if the diode weren't there, and
therefore the instantaneous power is also unchanged. There's no power
dissipated on the negative half cycle, so the average power is half
of what it would be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes
down by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it was
the square root of the average of the squares of all the values - a
sort of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4
when the diode is in the circuit. This would lead to a Pavg =
15v^2/220ohm ~= 1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

                = 1/R * (time average of (instantaneous voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate the
two expressions for average power:

1/R * Vrms^2  = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root,
which is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

It's probably close enough.

I would strongly suggest that you slap some graph paper down and sketch
the voltage waveform seen by the resistor.  Even if the instructor
doesn't need to see it, it should help your understanding immensely.

I've seen, and sketched, enough half-waves that I'm not that concerned
about seeing it.  I do understand that much.

If you're on a track to be an EE and not just a technician, and if you've
taken calculus, and if you have the time, you may want to take your
sketch of the waveform at the resistor and calculate the RMS voltage.

I'm on track to be a hobbyist, nothing more ;-).  However, I think
you're correct that it my help me to do that work. Its been a while
since I've done any real calculus.

In this case, RMS would be defined as the square root of the time average
of the square of the voltage.  Finding the time average of the square of
the voltage can either be done intuitively (you know what the time
average is for a whole cycle, so you can easily do it for a half cycle),
but grinding through the exercise of integrating the voltage-squared over
one cycle, then dividing by the cycle time, will be good for your soul.

Yes, probably true.  Deriving the formulas has always been a useful
exercise for me. I just didn't really have the time to sit down with a
pencil and paper to do so.  Maybe tonight.

Was my final answer of 8 watts for the Pavg correct?-

Grin, Here's another way to get to the answer. (Just to check
yourself.)

Without the diode what's the average power?

Then the diode blocks the signal for 1/2 the time so your answer
should be 1/2 the above number.... does that check?

That's a nice intuitive simplification.

You can imagine the transformer driving a resistor, no diodes, and
calculate the power, P = Erms^2/R.

Now split the resistor into two equal ones of resistance R, and drive
them from diodes in opposite directions. They conduct on opposite half
cycles. The transformer still sees R, so the power delivered from the
transformer is the same, but by conservation of energy and by symmetry
each resistor burns half of the power P. Now back-calculate the RMS
voltage across one of the resistors.

You can do similar intuitive thinking about complex signals by
splitting them up with ideal filters and dumping into equal
terminating resistors, then apply conservation of energy.


--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro acquisition and simulation

 

[Daniel Pitts]

On 1/28/13 4:37 PM, John Larkin wrote:

On Mon, 28 Jan 2013 13:23:11 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Jan 28, 4:06 pm, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
On 1/28/13 12:03 PM, Tim Wescott wrote:



On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have
one
particular homework problem that I think I've gotten correct, but
I'd like to have someone with more experience confirm. Unfortunately
most of
my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:

.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-

My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction,
which cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for
the .7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server
flaked out on the first attempt (as it does far too often), but I'm
not always sure.

RMS is a time average based on the power dissipated.An ideal diode in
the secondary side conducts on the positive half-cycle and not on the
negative half-cycle. On the positive, the instantaneous current is
just the same as it would be if the diode weren't there, and
therefore the instantaneous power is also unchanged. There's no power
dissipated on the negative half cycle, so the average power is half
of what it would be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes
down by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it was
the square root of the average of the squares of all the values - a
sort of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4
when the diode is in the circuit. This would lead to a Pavg =
15v^2/220ohm ~= 1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate the
two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root,
which is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

It's probably close enough.

I would strongly suggest that you slap some graph paper down and sketch
the voltage waveform seen by the resistor. Even if the instructor
doesn't need to see it, it should help your understanding immensely.

I've seen, and sketched, enough half-waves that I'm not that concerned
about seeing it. I do understand that much.

If you're on a track to be an EE and not just a technician, and if you've
taken calculus, and if you have the time, you may want to take your
sketch of the waveform at the resistor and calculate the RMS voltage.

I'm on track to be a hobbyist, nothing more ;-). However, I think
you're correct that it my help me to do that work. Its been a while
since I've done any real calculus.

In this case, RMS would be defined as the square root of the time average
of the square of the voltage. Finding the time average of the square of
the voltage can either be done intuitively (you know what the time
average is for a whole cycle, so you can easily do it for a half cycle),
but grinding through the exercise of integrating the voltage-squared over
one cycle, then dividing by the cycle time, will be good for your soul.

Yes, probably true. Deriving the formulas has always been a useful
exercise for me. I just didn't really have the time to sit down with a
pencil and paper to do so. Maybe tonight.

Was my final answer of 8 watts for the Pavg correct?-

Grin, Here's another way to get to the answer. (Just to check
yourself.)

Without the diode what's the average power?

Then the diode blocks the signal for 1/2 the time so your answer
should be 1/2 the above number.... does that check?

That's a nice intuitive simplification.

You can imagine the transformer driving a resistor, no diodes, and
calculate the power, P = Erms^2/R.

Now split the resistor into two equal ones of resistance R, and drive
them from diodes in opposite directions. They conduct on opposite half
cycles. The transformer still sees R, so the power delivered from the
transformer is the same, but by conservation of energy and by symmetry
each resistor burns half of the power P. Now back-calculate the RMS
voltage across one of the resistors.

You can do similar intuitive thinking about complex signals by
splitting them up with ideal filters and dumping into equal
terminating resistors, then apply conservation of energy.

This is exactly the analysis I did originally, but according to Phil
Hobbs the analysis is wrong.

So, who is correct?

 

[Tim Wescott]

On Mon, 28 Jan 2013 18:34:05 -0800, Daniel Pitts wrote:

On 1/28/13 4:37 PM, John Larkin wrote:
On Mon, 28 Jan 2013 13:23:11 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Jan 28, 4:06 pm, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
On 1/28/13 12:03 PM, Tim Wescott wrote:



On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due
to scheduling constraints I skipped the "AC" electronics
course. I have one particular homework problem that I think
I've gotten correct, but I'd like to have someone with more
experience confirm. Unfortunately most of my classmates finish
the homework on the due-date, so I can't compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:

.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-

My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a sine waveform, the peak voltage is 60v*sqrt(2),
approx 85v.

Using the "ideal" model, the diode only conducts one direction,
which cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I
am missing something, a pointer to what I'm missing should be
enough. I know I could get a slightly more accurate result if I
account for the .7v drop, but I don't think the instructor
cares, as long as I understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server
flaked out on the first attempt (as it does far too often), but
I'm not always sure.

RMS is a time average based on the power dissipated.An ideal
diode in the secondary side conducts on the positive half-cycle
and not on the negative half-cycle. On the positive, the
instantaneous current is just the same as it would be if the
diode weren't there, and therefore the instantaneous power is
also unchanged. There's no power dissipated on the negative half
cycle, so the average power is half of what it would be without
the diode.

Power is proportional to voltage squared, so the RMS voltage
goes down by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it
was the square root of the average of the squares of all the
values - a sort of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by
4 when the diode is in the circuit. This would lead to a Pavg =
15v^2/220ohm ~= 1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous
voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate
the two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root,
which is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the
RMS voltage goes down by what factor?

Cheers

Phil Hobbs

Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or
42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

It's probably close enough.

I would strongly suggest that you slap some graph paper down and
sketch the voltage waveform seen by the resistor. Even if the
instructor doesn't need to see it, it should help your understanding
immensely.

I've seen, and sketched, enough half-waves that I'm not that
concerned about seeing it. I do understand that much.

If you're on a track to be an EE and not just a technician, and if
you've taken calculus, and if you have the time, you may want to
take your sketch of the waveform at the resistor and calculate the
RMS voltage.

I'm on track to be a hobbyist, nothing more ;-). However, I think
you're correct that it my help me to do that work. Its been a while
since I've done any real calculus.

In this case, RMS would be defined as the square root of the time
average of the square of the voltage. Finding the time average of
the square of the voltage can either be done intuitively (you know
what the time average is for a whole cycle, so you can easily do it
for a half cycle), but grinding through the exercise of integrating
the voltage-squared over one cycle, then dividing by the cycle time,
will be good for your soul.

Yes, probably true. Deriving the formulas has always been a useful
exercise for me. I just didn't really have the time to sit down with
a pencil and paper to do so. Maybe tonight.

Was my final answer of 8 watts for the Pavg correct?-

Grin, Here's another way to get to the answer. (Just to check
yourself.)

Without the diode what's the average power?

Then the diode blocks the signal for 1/2 the time so your answer
should be 1/2 the above number.... does that check?

That's a nice intuitive simplification.

You can imagine the transformer driving a resistor, no diodes, and
calculate the power, P = Erms^2/R.

Now split the resistor into two equal ones of resistance R, and drive
them from diodes in opposite directions. They conduct on opposite half
cycles. The transformer still sees R, so the power delivered from the
transformer is the same, but by conservation of energy and by symmetry
each resistor burns half of the power P. Now back-calculate the RMS
voltage across one of the resistors.

You can do similar intuitive thinking about complex signals by
splitting them up with ideal filters and dumping into equal terminating
resistors, then apply conservation of energy.

This is exactly the analysis I did originally, but according to Phil
Hobbs the analysis is wrong.

So, who is correct?

John and Phil. John's is _not_ the analysis that you did that Phil
didn't like. His method will get you the right answer, if you mind your
P's and Q's.

(I ain't gonna say more until you go back and look at what John said
_really closely_, then revisit what RMS means, then come back. Sorry --
but you did say it's homework).

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

 

[John Larkin]

On Mon, 28 Jan 2013 18:34:05 -0800, Daniel Pitts
<newsgroup.nospam@virtualinfinity.net> wrote:

On 1/28/13 4:37 PM, John Larkin wrote:
On Mon, 28 Jan 2013 13:23:11 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Jan 28, 4:06 pm, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
On 1/28/13 12:03 PM, Tim Wescott wrote:



On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have
one
particular homework problem that I think I've gotten correct, but
I'd like to have someone with more experience confirm. Unfortunately
most of
my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:

.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-

My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction,
which cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for
the .7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server
flaked out on the first attempt (as it does far too often), but I'm
not always sure.

RMS is a time average based on the power dissipated.An ideal diode in
the secondary side conducts on the positive half-cycle and not on the
negative half-cycle. On the positive, the instantaneous current is
just the same as it would be if the diode weren't there, and
therefore the instantaneous power is also unchanged. There's no power
dissipated on the negative half cycle, so the average power is half
of what it would be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes
down by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it was
the square root of the average of the squares of all the values - a
sort of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4
when the diode is in the circuit. This would lead to a Pavg =
15v^2/220ohm ~= 1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate the
two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root,
which is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

It's probably close enough.

I would strongly suggest that you slap some graph paper down and sketch
the voltage waveform seen by the resistor. Even if the instructor
doesn't need to see it, it should help your understanding immensely.

I've seen, and sketched, enough half-waves that I'm not that concerned
about seeing it. I do understand that much.

If you're on a track to be an EE and not just a technician, and if you've
taken calculus, and if you have the time, you may want to take your
sketch of the waveform at the resistor and calculate the RMS voltage.

I'm on track to be a hobbyist, nothing more ;-). However, I think
you're correct that it my help me to do that work. Its been a while
since I've done any real calculus.

In this case, RMS would be defined as the square root of the time average
of the square of the voltage. Finding the time average of the square of
the voltage can either be done intuitively (you know what the time
average is for a whole cycle, so you can easily do it for a half cycle),
but grinding through the exercise of integrating the voltage-squared over
one cycle, then dividing by the cycle time, will be good for your soul.

Yes, probably true. Deriving the formulas has always been a useful
exercise for me. I just didn't really have the time to sit down with a
pencil and paper to do so. Maybe tonight.

Was my final answer of 8 watts for the Pavg correct?-

Grin, Here's another way to get to the answer. (Just to check
yourself.)

Without the diode what's the average power?

Then the diode blocks the signal for 1/2 the time so your answer
should be 1/2 the above number.... does that check?

That's a nice intuitive simplification.

You can imagine the transformer driving a resistor, no diodes, and
calculate the power, P = Erms^2/R.

Now split the resistor into two equal ones of resistance R, and drive
them from diodes in opposite directions. They conduct on opposite half
cycles. The transformer still sees R, so the power delivered from the
transformer is the same, but by conservation of energy and by symmetry
each resistor burns half of the power P. Now back-calculate the RMS
voltage across one of the resistors.

You can do similar intuitive thinking about complex signals by
splitting them up with ideal filters and dumping into equal
terminating resistors, then apply conservation of energy.

This is exactly the analysis I did originally, but according to Phil
Hobbs the analysis is wrong.

So, who is correct?

The diode doesn't (as you originally posted) cut VRMS in half, it cuts the power
dissipated by the resistor in half. I think that's the difference. I think you
corrected the VRMS thing in a later post.

Never bet against Phil on stuff like this.

Incidentally, his book on electro-optics is worth having for anyone serious
about electronics. The other necessary book is The Art Of Electronics by H+H.



--

John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators

 

[Daniel Pitts]

On 1/28/13 8:25 PM, John Larkin wrote:

On Mon, 28 Jan 2013 18:34:05 -0800, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 1/28/13 4:37 PM, John Larkin wrote:
On Mon, 28 Jan 2013 13:23:11 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Jan 28, 4:06 pm, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
On 1/28/13 12:03 PM, Tim Wescott wrote:



On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I have
one
particular homework problem that I think I've gotten correct, but
I'd like to have someone with more experience confirm. Unfortunately
most of
my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:

.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-

My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction,
which cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be enough. I
know I could get a slightly more accurate result if I account for
the .7v drop, but I don't think the instructor cares, as long as I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server
flaked out on the first attempt (as it does far too often), but I'm
not always sure.

RMS is a time average based on the power dissipated.An ideal diode in
the secondary side conducts on the positive half-cycle and not on the
negative half-cycle. On the positive, the instantaneous current is
just the same as it would be if the diode weren't there, and
therefore the instantaneous power is also unchanged. There's no power
dissipated on the negative half cycle, so the average power is half
of what it would be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes
down by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it was
the square root of the average of the squares of all the values - a
sort of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4
when the diode is in the circuit. This would lead to a Pavg =
15v^2/220ohm ~= 1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate the
two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root,
which is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

It's probably close enough.

I would strongly suggest that you slap some graph paper down and sketch
the voltage waveform seen by the resistor. Even if the instructor
doesn't need to see it, it should help your understanding immensely.

I've seen, and sketched, enough half-waves that I'm not that concerned
about seeing it. I do understand that much.

If you're on a track to be an EE and not just a technician, and if you've
taken calculus, and if you have the time, you may want to take your
sketch of the waveform at the resistor and calculate the RMS voltage.

I'm on track to be a hobbyist, nothing more ;-). However, I think
you're correct that it my help me to do that work. Its been a while
since I've done any real calculus.

In this case, RMS would be defined as the square root of the time average
of the square of the voltage. Finding the time average of the square of
the voltage can either be done intuitively (you know what the time
average is for a whole cycle, so you can easily do it for a half cycle),
but grinding through the exercise of integrating the voltage-squared over
one cycle, then dividing by the cycle time, will be good for your soul.

Yes, probably true. Deriving the formulas has always been a useful
exercise for me. I just didn't really have the time to sit down with a
pencil and paper to do so. Maybe tonight.

Was my final answer of 8 watts for the Pavg correct?-

Grin, Here's another way to get to the answer. (Just to check
yourself.)

Without the diode what's the average power?

Then the diode blocks the signal for 1/2 the time so your answer
should be 1/2 the above number.... does that check?

That's a nice intuitive simplification.

You can imagine the transformer driving a resistor, no diodes, and
calculate the power, P = Erms^2/R.

Now split the resistor into two equal ones of resistance R, and drive
them from diodes in opposite directions. They conduct on opposite half
cycles. The transformer still sees R, so the power delivered from the
transformer is the same, but by conservation of energy and by symmetry
each resistor burns half of the power P. Now back-calculate the RMS
voltage across one of the resistors.

You can do similar intuitive thinking about complex signals by
splitting them up with ideal filters and dumping into equal
terminating resistors, then apply conservation of energy.

This is exactly the analysis I did originally, but according to Phil
Hobbs the analysis is wrong.

So, who is correct?



The diode doesn't (as you originally posted) cut VRMS in half, it cuts the power
dissipated by the resistor in half. I think that's the difference. I think you
corrected the VRMS thing in a later post.
Ah, okay. so I think I get it now. That makes sense.

Never bet against Phil on stuff like this.
Good to know ;-)

Incidentally, his book on electro-optics is worth having for anyone serious
about electronics. The other necessary book is The Art Of Electronics by H+H.

 

[Daniel Pitts]

On 1/28/13 7:34 PM, Tim Wescott wrote:

This is exactly the analysis I did originally, but according to Phil
Hobbs the analysis is wrong.

So, who is correct?

John and Phil. John's is _not_ the analysis that you did that Phil
didn't like. His method will get you the right answer, if you mind your
P's and Q's.
I see that now. Thanks.

(I ain't gonna say more until you go back and look at what John said
_really closely_, then revisit what RMS means, then come back. Sorry --
but you did say it's homework).

I think I've got it now. Thanks everyone for your help and suggestions!

 

[Mark Storkamp]

In article <fgJNs.86314$LS5.49792@newsfe10.iad>,
Daniel Pitts <newsgroup.nospam@virtualinfinity.net> wrote:

On 1/28/13 8:25 PM, John Larkin wrote:
On Mon, 28 Jan 2013 18:34:05 -0800, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 1/28/13 4:37 PM, John Larkin wrote:
On Mon, 28 Jan 2013 13:23:11 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Jan 28, 4:06 pm, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
On 1/28/13 12:03 PM, Tim Wescott wrote:



On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I
have
one
particular homework problem that I think I've gotten correct, but
I'd like to have someone with more experience confirm.
Unfortunately
most of
my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:

.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-

My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction,
which cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be
enough. I
know I could get a slightly more accurate result if I account for
the .7v drop, but I don't think the instructor cares, as long as
I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server
flaked out on the first attempt (as it does far too often), but
I'm
not always sure.

RMS is a time average based on the power dissipated.An ideal diode
in
the secondary side conducts on the positive half-cycle and not on
the
negative half-cycle. On the positive, the instantaneous current is
just the same as it would be if the diode weren't there, and
therefore the instantaneous power is also unchanged. There's no
power
dissipated on the negative half cycle, so the average power is
half
of what it would be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes
down by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it
was
the square root of the average of the squares of all the values - a
sort of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4
when the diode is in the circuit. This would lead to a Pavg =
15v^2/220ohm ~= 1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous
voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate the
two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root,
which is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the
RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

It's probably close enough.

I would strongly suggest that you slap some graph paper down and
sketch
the voltage waveform seen by the resistor. Even if the instructor
doesn't need to see it, it should help your understanding immensely.

I've seen, and sketched, enough half-waves that I'm not that concerned
about seeing it. I do understand that much.

If you're on a track to be an EE and not just a technician, and if
you've
taken calculus, and if you have the time, you may want to take your
sketch of the waveform at the resistor and calculate the RMS voltage.

I'm on track to be a hobbyist, nothing more ;-). However, I think
you're correct that it my help me to do that work. Its been a while
since I've done any real calculus.

In this case, RMS would be defined as the square root of the time
average
of the square of the voltage. Finding the time average of the square
of
the voltage can either be done intuitively (you know what the time
average is for a whole cycle, so you can easily do it for a half
cycle),
but grinding through the exercise of integrating the voltage-squared
over
one cycle, then dividing by the cycle time, will be good for your
soul.

Yes, probably true. Deriving the formulas has always been a useful
exercise for me. I just didn't really have the time to sit down with a
pencil and paper to do so. Maybe tonight.

Was my final answer of 8 watts for the Pavg correct?-

Grin, Here's another way to get to the answer. (Just to check
yourself.)

Without the diode what's the average power?

Then the diode blocks the signal for 1/2 the time so your answer
should be 1/2 the above number.... does that check?

That's a nice intuitive simplification.

You can imagine the transformer driving a resistor, no diodes, and
calculate the power, P = Erms^2/R.

Now split the resistor into two equal ones of resistance R, and drive
them from diodes in opposite directions. They conduct on opposite half
cycles. The transformer still sees R, so the power delivered from the
transformer is the same, but by conservation of energy and by symmetry
each resistor burns half of the power P. Now back-calculate the RMS
voltage across one of the resistors.

You can do similar intuitive thinking about complex signals by
splitting them up with ideal filters and dumping into equal
terminating resistors, then apply conservation of energy.

This is exactly the analysis I did originally, but according to Phil
Hobbs the analysis is wrong.

So, who is correct?



The diode doesn't (as you originally posted) cut VRMS in half, it cuts the
power
dissipated by the resistor in half. I think that's the difference. I think
you
corrected the VRMS thing in a later post.
Ah, okay. so I think I get it now. That makes sense.

Never bet against Phil on stuff like this.
Good to know ;-)

Incidentally, his book on electro-optics is worth having for anyone serious
about electronics. The other necessary book is The Art Of Electronics by
H+H.

Now that he's worked out the answer, it would be interesting to know
what his instructor expects the answer to be. I once took a community ed
night class at a local college. During the day the room was used for a
vocational electronics program. They had this same basic problems worked
out on the blackboard, but had the OP's original wrong answer at the end.

 

[Tom Biasi]

On 1/29/2013 12:40 PM, Mark Storkamp wrote:

In article <fgJNs.86314$LS5.49792@newsfe10.iad>,
Daniel Pitts <newsgroup.nospam@virtualinfinity.net> wrote:

On 1/28/13 8:25 PM, John Larkin wrote:
On Mon, 28 Jan 2013 18:34:05 -0800, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 1/28/13 4:37 PM, John Larkin wrote:
On Mon, 28 Jan 2013 13:23:11 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Jan 28, 4:06 pm, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
On 1/28/13 12:03 PM, Tim Wescott wrote:



On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I
have
one
particular homework problem that I think I've gotten correct, but
I'd like to have someone with more experience confirm.
Unfortunately
most of
my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:

.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-

My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction,
which cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be
enough. I
know I could get a slightly more accurate result if I account for
the .7v drop, but I don't think the instructor cares, as long as
I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server
flaked out on the first attempt (as it does far too often), but
I'm
not always sure.

RMS is a time average based on the power dissipated.An ideal diode
in
the secondary side conducts on the positive half-cycle and not on
the
negative half-cycle. On the positive, the instantaneous current is
just the same as it would be if the diode weren't there, and
therefore the instantaneous power is also unchanged. There's no
power
dissipated on the negative half cycle, so the average power is
half
of what it would be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes
down by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it
was
the square root of the average of the squares of all the values - a
sort of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4
when the diode is in the circuit. This would lead to a Pavg =
15v^2/220ohm ~= 1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous
voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate the
two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root,
which is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the
RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

It's probably close enough.

I would strongly suggest that you slap some graph paper down and
sketch
the voltage waveform seen by the resistor. Even if the instructor
doesn't need to see it, it should help your understanding immensely.

I've seen, and sketched, enough half-waves that I'm not that concerned
about seeing it. I do understand that much.

If you're on a track to be an EE and not just a technician, and if
you've
taken calculus, and if you have the time, you may want to take your
sketch of the waveform at the resistor and calculate the RMS voltage.

I'm on track to be a hobbyist, nothing more ;-). However, I think
you're correct that it my help me to do that work. Its been a while
since I've done any real calculus.

In this case, RMS would be defined as the square root of the time
average
of the square of the voltage. Finding the time average of the square
of
the voltage can either be done intuitively (you know what the time
average is for a whole cycle, so you can easily do it for a half
cycle),
but grinding through the exercise of integrating the voltage-squared
over
one cycle, then dividing by the cycle time, will be good for your
soul.

Yes, probably true. Deriving the formulas has always been a useful
exercise for me. I just didn't really have the time to sit down with a
pencil and paper to do so. Maybe tonight.

Was my final answer of 8 watts for the Pavg correct?-

Grin, Here's another way to get to the answer. (Just to check
yourself.)

Without the diode what's the average power?

Then the diode blocks the signal for 1/2 the time so your answer
should be 1/2 the above number.... does that check?

That's a nice intuitive simplification.

You can imagine the transformer driving a resistor, no diodes, and
calculate the power, P = Erms^2/R.

Now split the resistor into two equal ones of resistance R, and drive
them from diodes in opposite directions. They conduct on opposite half
cycles. The transformer still sees R, so the power delivered from the
transformer is the same, but by conservation of energy and by symmetry
each resistor burns half of the power P. Now back-calculate the RMS
voltage across one of the resistors.

You can do similar intuitive thinking about complex signals by
splitting them up with ideal filters and dumping into equal
terminating resistors, then apply conservation of energy.

This is exactly the analysis I did originally, but according to Phil
Hobbs the analysis is wrong.

So, who is correct?



The diode doesn't (as you originally posted) cut VRMS in half, it cuts the
power
dissipated by the resistor in half. I think that's the difference. I think
you
corrected the VRMS thing in a later post.
Ah, okay. so I think I get it now. That makes sense.

Never bet against Phil on stuff like this.
Good to know ;-)

Incidentally, his book on electro-optics is worth having for anyone serious
about electronics. The other necessary book is The Art Of Electronics by
H+H.

Now that he's worked out the answer, it would be interesting to know
what his instructor expects the answer to be. I once took a community ed
night class at a local college. During the day the room was used for a
vocational electronics program. They had this same basic problems worked
out on the blackboard, but had the OP's original wrong answer at the end.

Just because the instructor displayed a workout doesn't mean he or she

said it was correct.

 

[Daniel Pitts]

On 1/29/13 9:40 AM, Mark Storkamp wrote:

In article <fgJNs.86314$LS5.49792@newsfe10.iad>,
Daniel Pitts <newsgroup.nospam@virtualinfinity.net> wrote:

On 1/28/13 8:25 PM, John Larkin wrote:
On Mon, 28 Jan 2013 18:34:05 -0800, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 1/28/13 4:37 PM, John Larkin wrote:
On Mon, 28 Jan 2013 13:23:11 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Jan 28, 4:06 pm, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
On 1/28/13 12:03 PM, Tim Wescott wrote:



On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but due to
scheduling constraints I skipped the "AC" electronics course. I
have
one
particular homework problem that I think I've gotten correct, but
I'd like to have someone with more experience confirm.
Unfortunately
most of
my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the figure:

.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-

My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one direction,
which cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something? If I am
missing something, a pointer to what I'm missing should be
enough. I
know I could get a slightly more accurate result if I account for
the .7v drop, but I don't think the instructor cares, as long as
I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP server
flaked out on the first attempt (as it does far too often), but
I'm
not always sure.

RMS is a time average based on the power dissipated.An ideal diode
in
the secondary side conducts on the positive half-cycle and not on
the
negative half-cycle. On the positive, the instantaneous current is
just the same as it would be if the diode weren't there, and
therefore the instantaneous power is also unchanged. There's no
power
dissipated on the negative half cycle, so the average power is
half
of what it would be without the diode.

Power is proportional to voltage squared, so the RMS voltage goes
down by a factor of.....?

I guess I don't fully understand what RMS is then... I thought it
was
the square root of the average of the squares of all the values - a
sort of Euclidean distance.

If I understand what you're saying, the Vrms would be divided by 4
when the diode is in the circuit. This would lead to a Pavg =
15v^2/220ohm ~= 1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the resistance.
That's the whole point of RMS--you can use AC or DC or some nasty
switching waveform, but the RMS value will tell you how much power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous
voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage, equate the
two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square root,
which is where the name comes from: root( mean( square voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the same, the
RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2), or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

It's probably close enough.

I would strongly suggest that you slap some graph paper down and
sketch
the voltage waveform seen by the resistor. Even if the instructor
doesn't need to see it, it should help your understanding immensely.

I've seen, and sketched, enough half-waves that I'm not that concerned
about seeing it. I do understand that much.

If you're on a track to be an EE and not just a technician, and if
you've
taken calculus, and if you have the time, you may want to take your
sketch of the waveform at the resistor and calculate the RMS voltage.

I'm on track to be a hobbyist, nothing more ;-). However, I think
you're correct that it my help me to do that work. Its been a while
since I've done any real calculus.

In this case, RMS would be defined as the square root of the time
average
of the square of the voltage. Finding the time average of the square
of
the voltage can either be done intuitively (you know what the time
average is for a whole cycle, so you can easily do it for a half
cycle),
but grinding through the exercise of integrating the voltage-squared
over
one cycle, then dividing by the cycle time, will be good for your
soul.

Yes, probably true. Deriving the formulas has always been a useful
exercise for me. I just didn't really have the time to sit down with a
pencil and paper to do so. Maybe tonight.

Was my final answer of 8 watts for the Pavg correct?-

Grin, Here's another way to get to the answer. (Just to check
yourself.)

Without the diode what's the average power?

Then the diode blocks the signal for 1/2 the time so your answer
should be 1/2 the above number.... does that check?

That's a nice intuitive simplification.

You can imagine the transformer driving a resistor, no diodes, and
calculate the power, P = Erms^2/R.

Now split the resistor into two equal ones of resistance R, and drive
them from diodes in opposite directions. They conduct on opposite half
cycles. The transformer still sees R, so the power delivered from the
transformer is the same, but by conservation of energy and by symmetry
each resistor burns half of the power P. Now back-calculate the RMS
voltage across one of the resistors.

You can do similar intuitive thinking about complex signals by
splitting them up with ideal filters and dumping into equal
terminating resistors, then apply conservation of energy.

This is exactly the analysis I did originally, but according to Phil
Hobbs the analysis is wrong.

So, who is correct?



The diode doesn't (as you originally posted) cut VRMS in half, it cuts the
power
dissipated by the resistor in half. I think that's the difference. I think
you
corrected the VRMS thing in a later post.
Ah, okay. so I think I get it now. That makes sense.

Never bet against Phil on stuff like this.
Good to know ;-)

Incidentally, his book on electro-optics is worth having for anyone serious
about electronics. The other necessary book is The Art Of Electronics by
H+H.

Now that he's worked out the answer, it would be interesting to know
what his instructor expects the answer to be. I once took a community ed
night class at a local college. During the day the room was used for a
vocational electronics program. They had this same basic problems worked
out on the blackboard, but had the OP's original wrong answer at the end.

Some of my classmates were comparing answers, and my answer was the "odd
man out". Others had calculated the average voltage. When I explained
how I got my answer (both the "intuitive" way and by integration), they
seemed very confused. One of them asked the instruct about average
power and average voltage. I'm not sure the instructor was paying full
attention but they said yes that would work.

Mathematically I can see it not working, so I'm sticking with my answer
and current understanding (Thanks Phil Hobbs), even if the instructor
does say it is wrong when he finally corrects the homework.

Anyway, thanks everyone for all your help. I appreciate it very much,
and I feel like I have a better grasp on this concept than many of my
classmates.

Ironic since most of them have taken an "AC" course, and I skipped that
one.

 

[amdx]

On 1/28/2013 10:25 PM, John Larkin wrote:

On Mon, 28 Jan 2013 18:34:05 -0800, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 1/28/13 4:37 PM, John Larkin wrote:
On Mon, 28 Jan 2013 13:23:11 -0800 (PST), George Herold
gherold@teachspin.com> wrote:


The diode doesn't (as you originally posted) cut VRMS in half, it cuts the power
dissipated by the resistor in half. I think that's the difference. I think you
corrected the VRMS thing in a later post.

Never bet against Phil on stuff like this.

Incidentally, his book on electro-optics is worth having for anyone serious
about electronics. The other necessary book is The Art Of Electronics by H+H.

Mikek
And don't wait for the next edition of, Art of Electronics. I've been
waiting 6 or 7 years, maybe 10!
Mikek

PS. I found this,

"UPDATE ( 01 / 24 / 2012 ): Sadly, the schedule for the 3rd edition of
The Art of Electronics has slipped. This post quotes a mail by Winfield
Hill in which he explains the current state of the book and says that it
"should be out by late 2012 or early 2013”. Considering the history of
delays of the 3rd edition, I think that early 2013 is more probable, and
I would not be surprised at all if it will be much later. It seems
obvious that both authors would rather delay the 3rd edition even
further than to rush their work and deliver a book of lower quality.
Anyway, as soon as I learn of a new schedule I will add it here."

Here:
http://www.wisewarthog.com/electronics/horowitz-hill-the-art-of-electronics.html

And just for giggles, this from 2004.


I have found in a german group, that the 3rd editon of the book
"Horowitz, Hill - art of electronics" shall be released in march 2004.
This means, next month. Does anybody know something about this?
Paul

urban myth

Thanks,
- Win


<Bugs Bunny mode ON>

Ah... whadda you know, buddy?

<Bugs Bunny mode OFF>
Ralph

 

[Phil Hobbs]

On 02/01/2013 05:39 PM, amdx wrote:

On 1/28/2013 10:25 PM, John Larkin wrote:
On Mon, 28 Jan 2013 18:34:05 -0800, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 1/28/13 4:37 PM, John Larkin wrote:
On Mon, 28 Jan 2013 13:23:11 -0800 (PST), George Herold
gherold@teachspin.com> wrote:


The diode doesn't (as you originally posted) cut VRMS in half, it cuts
the power
dissipated by the resistor in half. I think that's the difference. I
think you
corrected the VRMS thing in a later post.

Never bet against Phil on stuff like this.

Incidentally, his book on electro-optics is worth having for anyone
serious
about electronics. The other necessary book is The Art Of Electronics
by H+H.



Mikek
And don't wait for the next edition of, Art of Electronics. I've been
waiting 6 or 7 years, maybe 10!
Mikek

PS. I found this,

"UPDATE ( 01 / 24 / 2012 ): Sadly, the schedule for the 3rd edition of
The Art of Electronics has slipped. This post quotes a mail by Winfield
Hill in which he explains the current state of the book and says that it
"should be out by late 2012 or early 2013”. Considering the history of
delays of the 3rd edition, I think that early 2013 is more probable, and
I would not be surprised at all if it will be much later. It seems
obvious that both authors would rather delay the 3rd edition even
further than to rush their work and deliver a book of lower quality.
Anyway, as soon as I learn of a new schedule I will add it here."

Here:
http://www.wisewarthog.com/electronics/horowitz-hill-the-art-of-electronics.html


And just for giggles, this from 2004.


I have found in a german group, that the 3rd editon of the book
"Horowitz, Hill - art of electronics" shall be released in march 2004.
This means, next month. Does anybody know something about this?
Paul

urban myth

Thanks,
- Win


Bugs Bunny mode ON

Ah... whadda you know, buddy?

Bugs Bunny mode OFF
Ralph

Win and Paul have sent a bunch of chapters off to the publisher already,
so it looks a bit more real this time.

Cheers

Phil Hobbs

(Off home to an early bedtime--my plane to SF leaves at 7:00 tomorrow
from JFK, so I'll be leaving home about 4:15, blech.)

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[Daniel Pitts]

On 1/30/13 8:42 AM, Daniel Pitts wrote:

On 1/29/13 9:40 AM, Mark Storkamp wrote:
In article <fgJNs.86314$LS5.49792@newsfe10.iad>,
Daniel Pitts <newsgroup.nospam@virtualinfinity.net> wrote:

On 1/28/13 8:25 PM, John Larkin wrote:
On Mon, 28 Jan 2013 18:34:05 -0800, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

On 1/28/13 4:37 PM, John Larkin wrote:
On Mon, 28 Jan 2013 13:23:11 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Jan 28, 4:06 pm, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:
On 1/28/13 12:03 PM, Tim Wescott wrote:



On Mon, 28 Jan 2013 10:48:59 -0800, Daniel Pitts wrote:

On 1/28/13 10:05 AM, Phil Hobbs wrote:
On 01/28/2013 12:45 PM, Daniel Pitts wrote:
On 1/28/13 9:14 AM, Phil Hobbs wrote:
On 01/28/2013 12:08 PM, Daniel Pitts wrote:
Hello Electronics Wizards,

So, I'm taking a Linear Circuits class at a local JC, but
due to
scheduling constraints I skipped the "AC" electronics
course. I
have
one
particular homework problem that I think I've gotten
correct, but
I'd like to have someone with more experience confirm.
Unfortunately
most of
my classmates finish the homework on the due-date, so I can't
compare notes.

The problem as stated:

Find the peak and average power delivered to Rl in the
figure:

.--o^-._-^-* -. 2:1 .---->|---o--.
| Fuse ^ | | Diode^ |
o S* || *S \
S || S /
120V rms S || S \ 220 ohm
S || S /
o | | \
| | | |
`-------------. .------.---o--.
|
-----
---
-

My approach to solving this is:

120v rms becomes 60v rms on the secondary of the transformer.
Assuming a
sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.

Using the "ideal" model, the diode only conducts one
direction,
which cuts the Vrms in half.

Pavg = 30v^2/220ohm ~= 4watts
Ppeak = 85v^2/220ohm ~= 33watts

Does my approach look correct, or am I missing something?
If I am
missing something, a pointer to what I'm missing should be
enough. I
know I could get a slightly more accurate result if I
account for
the .7v drop, but I don't think the instructor cares, as
long as
I
understand that I could.

Thanks,
Daniel.

P.S. Sorry if this is a duplicate post, I think my NNTP
server
flaked out on the first attempt (as it does far too
often), but
I'm
not always sure.

RMS is a time average based on the power dissipated.An
ideal diode
in
the secondary side conducts on the positive half-cycle and
not on
the
negative half-cycle. On the positive, the instantaneous
current is
just the same as it would be if the diode weren't there, and
therefore the instantaneous power is also unchanged.
There's no
power
dissipated on the negative half cycle, so the average power is
half
of what it would be without the diode.

Power is proportional to voltage squared, so the RMS
voltage goes
down by a factor of.....?

I guess I don't fully understand what RMS is then... I
thought it
was
the square root of the average of the squares of all the
values - a
sort of Euclidean distance.

If I understand what you're saying, the Vrms would be
divided by 4
when the diode is in the circuit. This would lead to a Pavg =
15v^2/220ohm ~= 1watt.

Is that the proper approach?

Thanks,
Daniel.

Power dissipated is RMS voltage squared divided by the
resistance.
That's the whole point of RMS--you can use AC or DC or some
nasty
switching waveform, but the RMS value will tell you how much
power
you'll deliver to the load, _on_average_.

Instantaneous power = (instantaneous voltage)^2/R

Average power = time average of instantaneous power

= 1/R * (time average of (instantaneous
voltage)^2)

(note the order of the operations--square first, then average).

Then to get back to the right way to express the voltage,
equate the
two expressions for average power:

1/R * Vrms^2 = 1/R * (time average of (instantaneous
voltage)^2)

A little algebra gives

Vrms = sqrt(time average of (instantaneous voltage)^2)

i.e. you square, then time average, and then take the square
root,
which is where the name comes from: root( mean( square
voltage)).

Solving the equation for Vrms,

Vrms = sqrt(Pavg * R)

So if the average power is down by half, and R stays the
same, the
RMS
voltage goes down by what factor?

Cheers

Phil Hobbs

Ah, that would be a factor of sqrt(2). I think I've got it now.

So, 60v rms input across a diode would result in 60v/sqrt(2),
or 42v.

That gives (60/sqrt(2))^2/220ohm = 1800/220 ~= 8 watts.

Is my analysis on Ppeak correct? 60v rms * sqrt(2) ~= 85v, and
85v^2/220ohm ~= 33watts?

Actually, is my analysis of what the transformer does correct?

It's probably close enough.

I would strongly suggest that you slap some graph paper down and
sketch
the voltage waveform seen by the resistor. Even if the instructor
doesn't need to see it, it should help your understanding
immensely.

I've seen, and sketched, enough half-waves that I'm not that
concerned
about seeing it. I do understand that much.

If you're on a track to be an EE and not just a technician, and if
you've
taken calculus, and if you have the time, you may want to take
your
sketch of the waveform at the resistor and calculate the RMS
voltage.

I'm on track to be a hobbyist, nothing more ;-). However, I think
you're correct that it my help me to do that work. Its been a while
since I've done any real calculus.

In this case, RMS would be defined as the square root of the time
average
of the square of the voltage. Finding the time average of the
square
of
the voltage can either be done intuitively (you know what the time
average is for a whole cycle, so you can easily do it for a half
cycle),
but grinding through the exercise of integrating the
voltage-squared
over
one cycle, then dividing by the cycle time, will be good for your
soul.

Yes, probably true. Deriving the formulas has always been a useful
exercise for me. I just didn't really have the time to sit down
with a
pencil and paper to do so. Maybe tonight.

Was my final answer of 8 watts for the Pavg correct?-

Grin, Here's another way to get to the answer. (Just to check
yourself.)

Without the diode what's the average power?

Then the diode blocks the signal for 1/2 the time so your answer
should be 1/2 the above number.... does that check?

That's a nice intuitive simplification.

You can imagine the transformer driving a resistor, no diodes, and
calculate the power, P = Erms^2/R.

Now split the resistor into two equal ones of resistance R, and drive
them from diodes in opposite directions. They conduct on opposite
half
cycles. The transformer still sees R, so the power delivered from the
transformer is the same, but by conservation of energy and by
symmetry
each resistor burns half of the power P. Now back-calculate the RMS
voltage across one of the resistors.

You can do similar intuitive thinking about complex signals by
splitting them up with ideal filters and dumping into equal
terminating resistors, then apply conservation of energy.

This is exactly the analysis I did originally, but according to Phil
Hobbs the analysis is wrong.

So, who is correct?



The diode doesn't (as you originally posted) cut VRMS in half, it
cuts the
power
dissipated by the resistor in half. I think that's the difference. I
think
you
corrected the VRMS thing in a later post.
Ah, okay. so I think I get it now. That makes sense.

Never bet against Phil on stuff like this.
Good to know ;-)

Incidentally, his book on electro-optics is worth having for anyone
serious
about electronics. The other necessary book is The Art Of
Electronics by
H+H.

Now that he's worked out the answer, it would be interesting to know
what his instructor expects the answer to be. I once took a community ed
night class at a local college. During the day the room was used for a
vocational electronics program. They had this same basic problems worked
out on the blackboard, but had the OP's original wrong answer at the end.

Some of my classmates were comparing answers, and my answer was the "odd
man out". Others had calculated the average voltage. When I explained
how I got my answer (both the "intuitive" way and by integration), they
seemed very confused. One of them asked the instruct about average
power and average voltage. I'm not sure the instructor was paying full
attention but they said yes that would work.

Mathematically I can see it not working, so I'm sticking with my answer
and current understanding (Thanks Phil Hobbs), even if the instructor
does say it is wrong when he finally corrects the homework.

Anyway, thanks everyone for all your help. I appreciate it very much,
and I feel like I have a better grasp on this concept than many of my
classmates.

Ironic since most of them have taken an "AC" course, and I skipped that
one.

Yep, so the instructor has considered this approach and answer wrong...
His approach was to find the average voltage and use that in v^2/r. Of
course that isn't the answer I came up with... I used wolframalpha.com
to calculate my result:

http://www.wolframalpha.com/input/?i=r%3D220%2C+v%3D60+sqrt%282%29%2C+%28integrate+%28v*sinx%29%5E2%2Fr+from+0+to+%28pi%29%29%2F%282pi%29

{r == 220, v == 60 Sqrt[2], Integrate[(v Sin[x])^2/r, {x, 0, Pi}]/(2 Pi)}

Unless I don't understand the definition of "Average" (sum over count,
or integral over range) or "Power" (current times voltage).