Chip with simple program for Toy

[Tim Wescott]

On 12/08/2010 01:19 AM, Rich Grise wrote:

panfilero wrote:

Other answers that mentioned sine waves.... I never mentioned sine
waves... I never even mentioned a square wave, I just said a wave that
was high for 50% of the period... basically a pulse waveform with a
duty cycle of 0.5, if you do the math on this, the rms value of a
pulse is it's peak times the square root of the duty cycle, for Vin =
10V, and D = 0.5 you get 7.07V

Vin * sqrt(D) = 7.07V

No, that only applies to a sine wave. For pulses, it's a straight
1-to-1 because the sides are straight up and down. i.e., 50% duty
cycle yields 50% power, and so on. But only for pulses and rectangular
waves. For triangle waves, it's another equation, for sines you use
RMS, and so on.

Do the math again. The root of the mean of the square of
1,0,1,0,1,0,... is the root of the mean of 1,0,1,0,1,0,... which is the
root of 1/2, which is 0.707 (etc.).

Maybe a good way to think of it would be the "effective" value.

A good way to think about it is "effective in what situation". If
you're driving a resistor with a pulsed 12V supply with a 50% duty cycle
(which I think is what Phil was positing), then the current through the
resistor will be I = 12V / R when the thing is on, and I = 0A when the
thing is off. The average power delivered to the resistor will be
1/2((12V)^2 / R + 0), which is (sqrt(1/2) * 12V)^2 / R -- which is the
same power as you'll see dissipated in the resistor if the voltage were
a steady sqrt(1/2) * 12V.

If, on the other hand, you're a resistor on the other side of the output
filter in a switch-mode supply, then the resistor will see the average
voltage -- 6V. In this case the current in the resistor is a steady
(6V) / R, and the power is half as much as if the output filter weren't
there. So in this case the RMS voltage doesn't tell us much, although
the RMS current in the inductor is what you'll need to calculate wiring
losses.

Hope This Helps!

Well...

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Jamie]

Fredxx wrote:

"Rich Grise" <richg@example.net.invalid> wrote in message
news:idnim6$mlm$1@news.eternal-september.org...

panfilero wrote:

Other answers that mentioned sine waves.... I never mentioned sine
waves... I never even mentioned a square wave, I just said a wave that
was high for 50% of the period... basically a pulse waveform with a
duty cycle of 0.5, if you do the math on this, the rms value of a
pulse is it's peak times the square root of the duty cycle, for Vin =
10V, and D = 0.5 you get 7.07V

Vin * sqrt(D) = 7.07V


No, that only applies to a sine wave. For pulses, it's a straight
1-to-1 because the sides are straight up and down. i.e., 50% duty
cycle yields 50% power, and so on. But only for pulses and rectangular
waves. For triangle waves, it's another equation, for sines you use
RMS, and so on.



I regard RMS to mean the equivalent voltage into a resistive load which has
the same heating effect as a constant DC voltage. The principle can apply
to any waveform.

As you say, for a sine wave, the RMS value is 1/root-2 (0.7071) x peak
voltage.

For this square wave, which has a duty cycle of 50% and a "on" value of 10V,
the RMS value is 7.07V. Try using an example 10R load. 100% duty cycle
dissipation is 10W, for 50% it's 5W. (= (7.07V x 7.07V) / 10R)


And all along I've always thought "RMS" was the area of a half

circle where a square can fit! which of course determines the height of
that circle, which is of course .707 of the height of the circle. The
rest is
academic..

Jamie

 

[Rich Grise]

Fredxx wrote:

For this square wave, which has a duty cycle of 50% and a "on" value of
10V,
the RMS value is 7.07V. Try using an example 10R load. 100% duty cycle
dissipation is 10W, for 50% it's 5W. (= (7.07V x 7.07V) / 10R)

I'm sorry, you're going to have to show your work here.

Where did you get the SQRT(2) term, for a square wave?

Thanks,
Rich

 

[Phil Allison]

"Rich Grise is a MORON "

Fredxx wrote:

For this square wave, which has a duty cycle of 50% and a "on" value of
10V,
the RMS value is 7.07V. Try using an example 10R load. 100% duty cycle
dissipation is 10W, for 50% it's 5W. (= (7.07V x 7.07V) / 10R)

I'm sorry, you're going to have to show your work here.

** Wot a fucking idiot.

Where did you get the SQRT(2) term, for a square wave?

** Err - calculating power dissipation involves squaring a voltage or
current...



...... Phil

 

[John Fields]

On Thu, 09 Dec 2010 13:02:47 -0800, Rich Grise
<richg@example.net.invalid> wrote:

Fredxx wrote:

For this square wave, which has a duty cycle of 50% and a "on" value of
10V,
the RMS value is 7.07V. Try using an example 10R load. 100% duty cycle
dissipation is 10W, for 50% it's 5W. (= (7.07V x 7.07V) / 10R)

I'm sorry, you're going to have to show your work here.

Where did you get the SQRT(2) term, for a square wave?

---

Assuming a voltage source, 10VDC into a 10 ohm load will dissipate, in
the load:

E˛d 100 * 1
P = ----- = --------- = 10 watts
R 10R

from the same source, 10VDC into a 10 ohm load at 50% duty cycle will
dissipate, in the load:

E˛d 100 * 0.5
P = ----- = ----------- = 5 watts
R 10R

Now, since the RMS value of a waveform is that value which will create
the same amount of heat as DC would into an equivalent load, in order
to create 5 watts of heat in a 10 ohm resistor, using DC, we can
rearrange:

E˛d
P = -----
R
into:

Et = sqrt(PR) = sqrt (5W * 10R) = 7.07VDC.

crosschecking:

E˛d 7.07˛ * 1 50 * 1
P = ----- = ----------- = -------- = 5 watts
R 10R 10R

So, a 7.07VDC supply would supply as much power as a 10V square wave
into the same load resistance, which would cause the terminal
temperature of the load to be equal, in either case, making the RMS
voltage of a square wave equal to 0.707 times that of its peak
voltage.

---
JF

 

[Baron]

Winston Inscribed thus:

Michael A. Terrell wrote:

Winston wrote:

(...)

Use a respirator and have a vacuum ready. This is dusty business.


For drywall?

Draw the outline and use a utility knife to cut a deep grove along
the
lines. Then cut groves to opposite corners. Whack with a hammer and
peel out the scrap. VERY little dust created.

The paper backing would tear out the back side
of the drywall in the corners. It wouldn't be
visible but it would be a weakness in the 'rock
where a crack could form.

Pop a 5mm drill through in each corner !
FWIW there is a plastic template available for single and double gang
drywall boxes.

Filleted corners
reduce the chances of cracking.

--Winston <-- In Earthquake country

--
Best Regards:
Baron.

 

[Winston]

Baron wrote:

Winston Inscribed thus:

(...)

The paper backing would tear out the back side
of the drywall in the corners. It wouldn't be
visible but it would be a weakness in the 'rock
where a crack could form.

Pop a 5mm drill through in each corner !
FWIW there is a plastic template available for single and double gang
drywall boxes.

So, the filleted inside corner created by the drill bit
limits the stress rise as compared to a sharp corner
or a corner that has a chipped - out back side?

I suppose this would reduce the likelihood of a crack
forming, as during an earthquake for example.

I see your point, Baron. Good Idea! Thanks!



--Winston

 

[Michael A. Terrell]

Baron wrote:

Winston Inscribed thus:

Michael A. Terrell wrote:

Winston wrote:

(...)

Use a respirator and have a vacuum ready. This is dusty business.


For drywall?

Draw the outline and use a utility knife to cut a deep grove along
the
lines. Then cut groves to opposite corners. Whack with a hammer and
peel out the scrap. VERY little dust created.

The paper backing would tear out the back side
of the drywall in the corners. It wouldn't be
visible but it would be a weakness in the 'rock
where a crack could form.

Pop a 5mm drill through in each corner !
FWIW there is a plastic template available for single and double gang
drywall boxes.

Filleted corners
reduce the chances of cracking.

If he ever tried what I'm talking about he would find that you can
cut more than half way through drywall with a utility knife. Done
properly, it continues in a straight line, leaving nothing but paper to
cut but he can waste as much time as he wants.


--
You can't fix stupid. You can't even put a band-aid on it, because it's
Teflon coated.

 

[Baron]

Winston Inscribed thus:

Baron wrote:
Winston Inscribed thus:

(...)

The paper backing would tear out the back side
of the drywall in the corners. It wouldn't be
visible but it would be a weakness in the 'rock
where a crack could form.

Pop a 5mm drill through in each corner !
FWIW there is a plastic template available for single and double gang
drywall boxes.

So, the filleted inside corner created by the drill bit
limits the stress rise as compared to a sharp corner
or a corner that has a chipped - out back side?

Thats the idea.
Those templates I mentioned. I just looked at mine. Its actually an
8mm drill (supplied) and has a series of guide holes (imagine figure of
eight but square sides) every few mm. It also has four very small pips
or pins on the back to stop it sliding about whilst drilling. A quick
tap with a hammer pops out the blank (sometimes). The biggest problem
I find, is the cardboard honeycomb in studding, sticks to the waste and
makes it awkward to remove.

I suppose this would reduce the likelihood of a crack
forming, as during an earthquake for example.

I see your point, Baron. Good Idea! Thanks!



--Winston

--
Best Regards:
Baron.

 

[Winston]

Michael A. Terrell wrote:

(...)

If he ever tried what I'm talking about he would find that you can
cut more than half way through drywall with a utility knife. Done
properly, it continues in a straight line, leaving nothing but paper to
cut but he can waste as much time as he wants.

That's how I cut 'repair panels' to size.
Score, snap and cut the backing through to
separate the pieces. It works great and results
in straight cuts. No 'inside' corners, no issues.

--Winston <-- deeply appreciates your permission

 

[Rich Grise]

Michael A. Terrell wrote:

Baron wrote:

Filleted corners
reduce the chances of cracking.

If he ever tried what I'm talking about he would find that you can
cut more than half way through drywall with a utility knife. Done
properly, it continues in a straight line, leaving nothing but paper to
cut but he can waste as much time as he wants.

On the "Home Improvement" TeeVee shows, I've seen them use a "drywall saw,"

which looks like a small keyhole saw.

Cheers!
Rich

 

[DJ Delorie]

Rich Grise <richg@example.net.invalid> writes:

On the "Home Improvement" TeeVee shows, I've seen them use a "drywall saw,"
which looks like a small keyhole saw.

I saw a "drywall saw" once, but it looked just like a hammer...

 

[Michael A. Terrell]

Rich Grise wrote:

Michael A. Terrell wrote:
Baron wrote:

Filleted corners
reduce the chances of cracking.

If he ever tried what I'm talking about he would find that you can
cut more than half way through drywall with a utility knife. Done
properly, it continues in a straight line, leaving nothing but paper to
cut but he can waste as much time as he wants.

On the "Home Improvement" TeeVee shows, I've seen them use a "drywall saw,"
which looks like a small keyhole saw.

Yse, they have tto use every gadget. They also have a nice wet/dry
shop vac to clean up the plaster dust.


--
You can't fix stupid. You can't even put a band-aid on it, because it's
Teflon coated.

 

[Michael A. Terrell]

DJ Delorie wrote:

Rich Grise <richg@example.net.invalid> writes:
On the "Home Improvement" TeeVee shows, I've seen them use a "drywall saw,"
which looks like a small keyhole saw.

I saw a "drywall saw" once, but it looked just like a hammer...

I see your puny hammer & raise you a 20 LB sledge!


--
You can't fix stupid. You can't even put a band-aid on it, because it's
Teflon coated.

 

[Baron]

Rich Grise Inscribed thus:

Michael A. Terrell wrote:
Baron wrote:

Filleted corners
reduce the chances of cracking.

If he ever tried what I'm talking about he would find that you can
cut more than half way through drywall with a utility knife. Done
properly, it continues in a straight line, leaving nothing but paper
to cut but he can waste as much time as he wants.

On the "Home Improvement" TeeVee shows, I've seen them use a "drywall
saw," which looks like a small keyhole saw.

Cheers!
Rich

Their fine for shaped holes even though they rag the edges of the hole.

--
Best Regards:
Baron.

 

[Baron]

Michael A. Terrell Inscribed thus:

DJ Delorie wrote:

Rich Grise <richg@example.net.invalid> writes:
On the "Home Improvement" TeeVee shows, I've seen them use a
"drywall saw," which looks like a small keyhole saw.

I saw a "drywall saw" once, but it looked just like a hammer...


I see your puny hammer & raise you a 20 LB sledge!

Aw ! In that case, wana borrow my JCB

--
Best Regards:
Baron.

 

[Michael A. Terrell]

Baron wrote:

Michael A. Terrell Inscribed thus:


DJ Delorie wrote:

Rich Grise <richg@example.net.invalid> writes:
On the "Home Improvement" TeeVee shows, I've seen them use a
"drywall saw," which looks like a small keyhole saw.

I saw a "drywall saw" once, but it looked just like a hammer...


I see your puny hammer & raise you a 20 LB sledge!

Aw ! In that case, wana borrow my JCB

I broke up two dump trucks full of old concrete with that hammer in a
couple days when I was 30. it took the county longer to load the trucks
than it took me to bust it apart. Adrenaline can do wondrous things, if
you don't let it kill you.


Using a backhoe is cheating, BTW.


--
You can't fix stupid. You can't even put a band-aid on it, because it's
Teflon coated.

 

[Michael A. Terrell]

Baron wrote:

Rich Grise Inscribed thus:

Michael A. Terrell wrote:
Baron wrote:

Filleted corners
reduce the chances of cracking.

If he ever tried what I'm talking about he would find that you can
cut more than half way through drywall with a utility knife. Done
properly, it continues in a straight line, leaving nothing but paper
to cut but he can waste as much time as he wants.

On the "Home Improvement" TeeVee shows, I've seen them use a "drywall
saw," which looks like a small keyhole saw.

Cheers!
Rich

Their fine for shaped holes even though they rag the edges of the hole.

And stir up enough dust to choke a horse.

--
You can't fix stupid. You can't even put a band-aid on it, because it's
Teflon coated.

 

[Bill Bowden]

On Jan 14, 6:15 pm, Rich Grise <ri...@example.net.invalid> wrote:

John Fields wrote:

Other power supplies, notably high voltage, have what's called a
"bleeder" resistor tied across their outputs so that when the supply
is powered down, any energy stored in its output is dissipated safely,
and quickly, so that any inadvertant human/machine contact won't
result in harm to the human.

Remember the desktops of the olden days, where if you didn't have a hard
drive yet, you had to put a dummy load on the PS, just so it could maintain
regulation?

Cheers!
Rich

I used to use a 12 volt automotive tail light as a load on the 5 volt
output to get the thing to turn on.

-Bill

 

[John Fields]

On Fri, 14 Jan 2011 18:15:49 -0800, Rich Grise
<richg@example.net.invalid> wrote:

John Fields wrote:

Other power supplies, notably high voltage, have what's called a
"bleeder" resistor tied across their outputs so that when the supply
is powered down, any energy stored in its output is dissipated safely,
and quickly, so that any inadvertant human/machine contact won't
result in harm to the human.

Remember the desktops of the olden days, where if you didn't have a hard
drive yet, you had to put a dummy load on the PS, just so it could maintain
regulation?

Cheers!
Rich

---
Kinda...

---
JF