Chip with simple program for Toy

[Dave]

"Dave" <dspear99ca@yahoo.delete.com> wrote in message
news:CUpaf.74209$y_1.11690@edtnps89...

"Pooh Bear" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:436322F0.5BB578F4@hotmail.com...



The other winding is likely 18-0-18 V AC and used for the split (
bipolar )
supply. The centre tap connects to the 'ground' of the supply.

There are two windings, one gives 5VAC single-tapped or two leads, the
second has four leads with varying voltages depending on which leads are
measured. No center tap. Too bad.

Note that 18V AC rectified will be about 24V DC. A +/- 24V DC supply
will
toast
most op-amps.You need voltage regulators to reduce it to typically +/-
15V. The
regulators also remove most of the supply ripple.

Ah, now this is useful information. I have been poring over schematics
for
power supplies and nowhere is it evident (and I obviously lack the
training
to know) that you get higher DC voltage out of a rectifier than the AC
voltage in. Is there a "rule-of-thumb"? I see that above you note about
1.5X multiplier for the DC out of a rectifier. The particular circuit I'm
looking at shows a 32VAC CT transformer which is rectified, filtered,
regulated down to 18VDC, and filtered again. For my particular app, I
don't
necessarily WANT a regulated 18VDC, I am okay with letting it float as the
op-amps are protected by more downstream regulators. 32VAC would give
~48VDC? You'd need one helluva beefy regulator (most of them that I've
seen
can handle up to 30VDC) and heatsink to drop 30VDC!!! If the voltage is
filtered between the rectifier and regulator, why would you need to filter
it again after the rectifier? Could one assume that you'd need smaller
filter caps as you work your way downstream?

In this regard, I have a 12.6VDC CT x-former (which I thought would not be
enough voltage). I really only need 15VDC, so can I expect ~18VDC out of
my
rectifier if I use 12.6VAC in?

I did some more research. Apparently my 12.6VAC transfermer (rated a 3A)
UNLOADED should give me 12.6VAC x 120% x 1.414 =~ 21.4VDC. Perfect for my
app. Max safe loading with 3300uF filter should be about 80% of that or
17VDC... that's unregulated. If I want regulated voltage I shouldn't count
on more than 12.6VDC so I can safely regulate to 12VDC. No? I am using
unoaded voltage because my load will be tiny ~150mA-250mA on a 3A rating.

While I'm at it with this enlightened audience, I have another basic
question: After the rectifier I have a 3300uF 50V electrolytic "filter
cap"
and a 1uF disc "bypass cap". It is my understanding (and, hey, I'm not
batting a thousand here so bear with me) that the filter cap stores up
charge and compensates for voltage drops. Is this right? What does the
bypass cap do?

Thanks in advance for any and all replies.

Dave

Graham

 

[Mxsmanic]

John Doe writes:

But it all makes so much sense having lived through the era being
very interested in personal computing.

I don't understand this statement.

Because it puts the operating system maker's applications at a
significant advantage over the competition.

How?

Are you saying that Microsoft Office is only one application?

Yes.

So why wouldn't they be interoperable?

Because they would all use different file formats, for example.

That's $2,290,000,000 in one quarter.

Yes, and almost all of it is Microsoft Office.

I bet that's more than all other PC software companies combined.

It's not, but it's more than most individual software companies.

--
Transpose mxsmanic and gmail to reach me by e-mail.

 

[Pooh Bear]

Dave wrote:

"Pooh Bear" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:436322F0.5BB578F4@hotmail.com...



The other winding is likely 18-0-18 V AC and used for the split (
bipolar )
supply. The centre tap connects to the 'ground' of the supply.

There are two windings, one gives 5VAC single-tapped or two leads, the
second has four leads with varying voltages depending on which leads are
measured. No center tap. Too bad.

Try looking further.

Remember that a common winding has DC continuity. You should be able to
distinguish those windings that are entirely isolated from each other that way.

Independent windings way still however be combined to make a centre tapped
supply. It's often more convenient for the transformer maker to avoid internal
connections. For example an 18-0-18 supply can be made from 2 x 0-18 windings
by series connecting them.

Note that 18V AC rectified will be about 24V DC. A +/- 24V DC supply will
toast
most op-amps.You need voltage regulators to reduce it to typically +/-
15V. The
regulators also remove most of the supply ripple.

Ah, now this is useful information. I have been poring over schematics for
power supplies and nowhere is it evident (and I obviously lack the training
to know) that you get higher DC voltage out of a rectifier than the AC
voltage in. Is there a "rule-of-thumb"?

Yes.

It's the ratio of the rms voltage to the peak voltage. i.e. 1.414. Don't forget
to subtract 1 V typically for the rectifier forward voltage drop though and be
aware that the open circuit voltage is just that. It'll drop on load. The drop
on load is greater for smaller transformers usually btw ( see 'regulation' on
the specs )

I see that above you note about
1.5X multiplier for the DC out of a rectifier. The particular circuit I'm
looking at shows a 32VAC CT transformer which is rectified, filtered,
regulated down to 18VDC, and filtered again. For my particular app, I don't
necessarily WANT a regulated 18VDC, I am okay with letting it float as the
op-amps are protected by more downstream regulators. 32VAC would give
~48VDC?

About that figure certainly.

You'd need one helluva beefy regulator (most of them that I've seen
can handle up to 30VDC) and heatsink to drop 30VDC!!!

Well no.

A regulator only is 'beefy' if it has to dissipate lots of *power*. The voltage
isn't the real problem ( although higher voltage regualtors are somewhat rare ).

Op-amps use mere milliamps and a regulator only dissipates ( Vin - Vout ) *
I_load. Suppose Vin is 30V and Vout is 15V and I_load is 10mA. The regulator
dissipation is 150mW. This is 'nothing'.

If the voltage is
filtered between the rectifier and regulator, why would you need to filter
it again after the rectifier? Could one assume that you'd need smaller
filter caps as you work your way downstream?

You misunderstand.

Once a voltage regulator is in play any further caps aren't there to provide
extra 'filtering'. They are there to ensure circuit stability. See 'decoupling'.

I also think you are confusing filtering and regulating.

In this regard, I have a 12.6VDC CT x-former (which I thought would not be
enough voltage). I really only need 15VDC, so can I expect ~18VDC out of my
rectifier if I use 12.6VAC in?

You mean the transformer is 6.3-0-6.3 ?

You *could* use it to power some op-amp circuitry at alowier than usual voltage.
Depends on your performance criteria.

While I'm at it with this enlightened audience, I have another basic
question: After the rectifier I have a 3300uF 50V electrolytic "filter cap"
and a 1uF disc "bypass cap". It is my understanding (and, hey, I'm not
batting a thousand here so bear with me) that the filter cap stores up
charge and compensates for voltage drops. Is this right? What does the
bypass cap do?

The 'bypass' cap is to do with RF stability. It should be located near to the
active circuitry. The 3300 uF cap does the real work of 'filtering' the supply.
It's more helpful to think of the 3300uF cap as a *storage* cap btw. I feel that
'filtering' is a misuse of the word in this respect.

Graham

 

[Rheilly Phoull]

"Ken Moffett" <KMoffett@mn.rr.com> wrote in message
news:Xns9704E6B68359KMoffettmnrrcom@24.94.170.102...

jh <no@thanks.com> wrote in news:no-96791C.01580504112005@localhost:

Hi,

I was hoping somebody here might be able to help me with a question.
First, some background. A couple of months ago I was trying to record
the sounds of the insides of my computer for an experimental sound
project. I first tried it with a cheap, crappy lapel mic that came
with
a pocket voice recorder. It worked just fine.

Then I borrowed a fairly nice, high quality microphone and tried it
again. Sure enough, this microphone picked up a lot more sounds... in
fact, it recorded all sorts of beeps, buzzes, and hums that weren't
even
there, apparently some sort of electromagnetic interference. I was
amused to find that this high-quality microphone was much more prone
to
picking up this interference than the cheap one I tried earlier.

The thing is, the interference sounds were much more interesting than
the real sounds. Holding the microphone near the graphics card, it
recorded different noises depending on what was being displayed on
screen. The fans sounded like something out of a science fiction
movie.
My personal favorite sound came from the power cord while the computer
was asleep: it made a bizarre sequence of changing pitches that
repeated
every couple of seconds.

The only problem is, all of these great interference-caused phantom
sounds were almost drowned out by the actual normal sound produced by
the fans, hard drive, etc. in the computer. Needless to say, the
microphone was quite adept at recording these sounds.

So my question is this: is it possible to build a device, or modify a
microphone, so that it picks up ONLY the electromagnetic interference,
but no actual sound?


Thanks,
Josh

p.s.: I hope people don't mind that I'm not including my real email
address. It's probably bad etiquette, but I'm kinda paranoid about
spam.


Any way you can wrap the mic in a "sound proofing" material, so you damp
out the acoustic pickup and leave the EMI sounds?

Forget the mic and use a coil !!

--
Regards ......... Rheilly Phoull

 

[John Fields]

On Fri, 4 Nov 2005 23:10:25 +1300, Jasen Betts
<jasen@clunker.homenet> wrote:

On 2005-11-04, jh <no@thanks.com> wrote:

The only problem is, all of these great interference-caused phantom
sounds were almost drowned out by the actual normal sound produced by
the fans, hard drive, etc. in the computer. Needless to say, the
microphone was quite adept at recording these sounds.

the reason for that is that the dynamic microphone has a voice coil (like in a
loudspeaker) anf that coil picks up all the electromagneic interferance (EMI)
inside your computer.

So my question is this: is it possible to build a device, or modify a
microphone, so that it picks up ONLY the electromagnetic interference,
but no actual sound?

You'll need to shield it against EMI. while not blocking too much sound
that that won't be easy. basically you need to surround it with a conductive
shell, electret mikes (like the "ctrappy lapel mike") are constructed that
way.

---
You've got it backwards. He _wants_ the EMI but _not_ the physical,
mechanical sound.

--
John Fields
Professional Circuit Designer

 

Jasen Betts wrote:

On 2005-11-04, jh <no@thanks.com> wrote:

The only problem is, all of these great interference-caused phantom
sounds were almost drowned out by the actual normal sound produced by
the fans, hard drive, etc. in the computer. Needless to say, the
microphone was quite adept at recording these sounds.

the reason for that is that the dynamic microphone has a voice coil (like in a
loudspeaker) anf that coil picks up all the electromagneic interferance (EMI)
inside your computer.

So my question is this: is it possible to build a device, or modify a
microphone, so that it picks up ONLY the electromagnetic interference,
but no actual sound?

You'll need to shield it against EMI. while not blocking too much sound
that that won't be easy. basically you need to surround it with a conductive
shell, electret mikes (like the "ctrappy lapel mike") are constructed that
way.

Bye.
Jasen

Jasen and OP:

You want to pick up the EMI and not pick up the acoustic sound. The
best way to do that is to replace the microphone with a plain coil of
wire. The coil will pick up the electronic/electromagnetic
interference and not pick up the acoustical noise. You could also do
that by immobilizing the moving part of the microphone so that it only
picks up the EMI and does not respond to sound. That is just the
opposite of what most people want, but having done EMI work for many
years, the sounds of something working are very interesting. If you
have an old LED display pocket calculator, try using that as a noise
source, Listen to the difference between adding two numbers and
getting the logarithm of a number, for example. You can hear EMI noise
also from almost anything electronic.

One other possibility for just a very short time would be to stop the
fan on your computer to stop the acoustic noise, SInce you have the
computer out of the case already, you could probably run without a fan
for a minute of two. Just don't stop the tiny fan on the processor
chip if your computer has one of those.

H. R.(Bob) Hofmann

H. R. Hofmann

 

[Iain Buchanan]

Jasen Betts wrote:

On 2005-11-04, jh <no@thanks.com> wrote:


The only problem is, all of these great interference-caused phantom
sounds were almost drowned out by the actual normal sound produced by
the fans, hard drive, etc. in the computer. Needless to say, the
microphone was quite adept at recording these sounds.


the reason for that is that the dynamic microphone has a voice coil (like in a
loudspeaker) anf that coil picks up all the electromagneic interferance (EMI)
inside your computer.


So my question is this: is it possible to build a device, or modify a
microphone, so that it picks up ONLY the electromagnetic interference,
but no actual sound?


You'll need to shield it against EMI. while not blocking too much sound
that that won't be easy. basically you need to surround it with a conductive
shell, electret mikes (like the "ctrappy lapel mike") are constructed that
way.

Bye.
Jasen
In fact the "crappier" the microphone the more EMI it will pick up!

 

[KruSat]

try to determine the frequency of emi and use a coil to resonate at
that freqn. so that u pick up the max. emi and also with good power....

it also eleminates the need of very high quality amprs.....

also.......

the intresting thing with that sound is that u will actually pick up
the data that is flowing through various cables and wirings of the
PC.....

u can find more info regarding it on the web.......

bye.....

 

[Dave]

"Pooh Bear" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:436AD8B9.5FD46955@hotmail.com...

Remember that a common winding has DC continuity. You should be able to
distinguish those windings that are entirely isolated from each other that
way.

Independent windings way still however be combined to make a centre tapped
supply. It's often more convenient for the transformer maker to avoid
internal
connections. For example an 18-0-18 supply can be made from 2 x 0-18
windings
by series connecting them.


Okay, I'll take a closer look... I know that there are two windings. I know

that one of them has four leads. If I can identify three of those leads,
call them leads (1-2-3) and two combinations give the same voltage, I can
combine for a center-tapped winding? For example, if 1-2 and 2-3 each give
18VAC, that's what I'm looking for as a usable combination? And the common
lead (2 in this case) would be my center tap, no?

In this regard, I have a 12.6VDC CT x-former (which I thought would not
be
enough voltage). I really only need 15VDC, so can I expect ~18VDC out
of my
rectifier if I use 12.6VAC in?

You mean the transformer is 6.3-0-6.3 ?

You *could* use it to power some op-amp circuitry at alowier than usual
voltage.
Depends on your performance criteria.

Another light is flickering in the back of my brain. I am reading what you

wrote, above, and I am thinking that if I use the center tap of a
transformer, I would be effectively cutting the rated output voltage in
half? In my example (12.6VDC CT, 6.3-0-6.3) I thought that if I used the
outer leads to power my rectifier, I'd be getting an effective 12.6VAC input
to the rectifier. Is this not the case?

My outputs from the regulator (~18VDC), then use the center tap as their
ground. Does this cut the voltage in half? As you can tell I'm unclear on
exactly what I need to get my +/-15-18VDC out.

Here's what I want:

120VAC --> transformer --> 12VAC --> rectifier --> 18VDC --> 2 sets of
caps --> two 15-18VDC power supplies, one pos, one neg which use the CT lead
for ground (common).

Do I need to start with a 12.6-0-12.6 transformer? This puts 25VAC into my
rectifier, does it not? That's >35VAC out! It will cook any 30V regulator
UNLESS by using the center tap for gorund you are cutting the voltage in
half.

Geez, just when I was starting to think I was getting a handle on this...

Thanks again for all of your help.

 

[John Popelish]

Dave wrote:

Okay, I'll take a closer look... I know that there are two windings. I know
that one of them has four leads. If I can identify three of those leads,
call them leads (1-2-3) and two combinations give the same voltage, I can
combine for a center-tapped winding? For example, if 1-2 and 2-3 each give
18VAC, that's what I'm looking for as a usable combination? And the common
lead (2 in this case) would be my center tap, no?

Probably. But you should verify that the voltage from 1 to 3 is about
36 volts. It is possible (but not very likely) that there would be
two windings with the same voltage but one of them is phased so that
the total voltage between 1 and 3 adds up to approximately zero.

Another light is flickering in the back of my brain. I am reading what you
wrote, above, and I am thinking that if I use the center tap of a
transformer, I would be effectively cutting the rated output voltage in
half? In my example (12.6VDC CT, 6.3-0-6.3) I thought that if I used the
outer leads to power my rectifier, I'd be getting an effective 12.6VAC input
to the rectifier. Is this not the case?

You would get 12.6 *1.414 - 1.2 = 16.6 volts between the positive
supply output (+8.3 volts) and the negative supply output (-8.3
volts). The center tap just gives you a middle voltage between them
to call zero, so the ends can be called a positive and a negative supply.

My outputs from the regulator (~18VDC), then use the center tap as their
ground. Does this cut the voltage in half? As you can tell I'm unclear on
exactly what I need to get my +/-15-18VDC out.

Here's what I want:

120VAC --> transformer --> 12VAC --> rectifier --> 18VDC --> 2 sets of
caps --> two 15-18VDC power supplies, one pos, one neg which use the CT lead
for ground (common).

Do I need to start with a 12.6-0-12.6 transformer? This puts 25VAC into my
rectifier, does it not? That's >35VAC out! It will cook any 30V regulator
UNLESS by using the center tap for gorund you are cutting the voltage in
half.

This is what you need, and each regulator will see half of the 25 *
1.414 = 35 or about 17 or 18 volts between one end of that 35 volts
and the middle voltage of the center tap.

Geez, just when I was starting to think I was getting a handle on this...

Thanks again for all of your help.

 

[jh]

Thanks for everyone for your ideas. I'll try some stuff (when I get a
chance; it might be a while) and post anything interesting. I'm hoping
to eventually (maybe after I learn more about electronics) build a
reasonably sensitive EMI-only "microphone" to use as a sort of computer
stethoscope.

Here's my idea: could a computer technician, with this tool and some
practice, hear and immediately recognize the EMI signature of a dying
power supply or other bad component, the same way an expert pilot can
instantly diagnose engine troubles just from their noises? Certainly,
many technicians can recognize the distinctive sound made by some dying
hard drives; but most components don't make any noise that we can hear
naturally.

-- Josh


In article <1131113675.779200.106750@g49g2000cwa.googlegroups.com>,
hrhofmann@att.net wrote:

Jasen Betts wrote:
On 2005-11-04, jh <no@thanks.com> wrote:

The only problem is, all of these great interference-caused phantom
sounds were almost drowned out by the actual normal sound produced by
the fans, hard drive, etc. in the computer. Needless to say, the
microphone was quite adept at recording these sounds.

the reason for that is that the dynamic microphone has a voice coil (like
in a
loudspeaker) anf that coil picks up all the electromagneic interferance
(EMI)
inside your computer.

So my question is this: is it possible to build a device, or modify a
microphone, so that it picks up ONLY the electromagnetic interference,
but no actual sound?

You'll need to shield it against EMI. while not blocking too much sound
that that won't be easy. basically you need to surround it with a
conductive
shell, electret mikes (like the "ctrappy lapel mike") are constructed that
way.

Bye.
Jasen


Jasen and OP:

You want to pick up the EMI and not pick up the acoustic sound. The
best way to do that is to replace the microphone with a plain coil of
wire. The coil will pick up the electronic/electromagnetic
interference and not pick up the acoustical noise. You could also do
that by immobilizing the moving part of the microphone so that it only
picks up the EMI and does not respond to sound. That is just the
opposite of what most people want, but having done EMI work for many
years, the sounds of something working are very interesting. If you
have an old LED display pocket calculator, try using that as a noise
source, Listen to the difference between adding two numbers and
getting the logarithm of a number, for example. You can hear EMI noise
also from almost anything electronic.

One other possibility for just a very short time would be to stop the
fan on your computer to stop the acoustic noise, SInce you have the
computer out of the case already, you could probably run without a fan
for a minute of two. Just don't stop the tiny fan on the processor
chip if your computer has one of those.

H. R.(Bob) Hofmann

H. R. Hofmann

 

[kell]

jalbers@bsu.edu wrote:

I know that the work required to fully charge a capacitor is given as:
1/2 C V^2

How much work is required to charge a capacitor to approximately 63% or
2 RC time constants? I know that it is going to be the area under the
curve: Vsource [1-exp(-t/RC)] from zero to 2RC , but it has been a long
time since I have done any integration. I am looking for a ball park
estimation if the exact answer is too difficult to derive. Any help
would be greatly appreciated. Thanks

Wait, isn't 63% just one time constant? I'm pretty sure it is. Plug
it into a calculator, or even just do a napkin calculation using 2.7 as
an approximation for e. The quantity inside the brackets of the
exponential formula is about 17/27, or 63% for RC=1.
At any rate, the amount of work would equal the amount of energy
stored, which as you said is 1/2 C V^2, where V is just the voltage on
the cap. You don't have to do any calculus if you can calculate the
voltage on the cap. But calculate the voltage as a function of time
correctly.

 

[Jon Elson]

Jasen Betts wrote:

On 2005-11-04, jh <no@thanks.com> wrote:



The only problem is, all of these great interference-caused phantom
sounds were almost drowned out by the actual normal sound produced by
the fans, hard drive, etc. in the computer. Needless to say, the
microphone was quite adept at recording these sounds.



the reason for that is that the dynamic microphone has a voice coil (like in a
loudspeaker) anf that coil picks up all the electromagneic interferance (EMI)
inside your computer.



So my question is this: is it possible to build a device, or modify a
microphone, so that it picks up ONLY the electromagnetic interference,
but no actual sound?



You'll need to shield it against EMI. while not blocking too much sound
that that won't be easy. basically you need to surround it with a conductive
shell, electret mikes (like the "ctrappy lapel mike") are constructed that
way.


No, that's the opposite of what he wants. He WANTS the EMI. A coil of fine

wire, perhaps half an inch in diameter, with a hundred turns of wire or
so should
do it. You might be able to salvage such a coil out of an old speaker
or headphone,
Carefully remove the coil from the rest of the speaker structure, and
connect the two
wires to a microphone cable.

Jon

 

[kell]

kell wrote:

jalbers@bsu.edu wrote:
I know that the work required to fully charge a capacitor is given as:
1/2 C V^2

How much work is required to charge a capacitor to approximately 63% or
2 RC time constants? I know that it is going to be the area under the
curve: Vsource [1-exp(-t/RC)] from zero to 2RC , but it has been a long
time since I have done any integration. I am looking for a ball park
estimation if the exact answer is too difficult to derive. Any help
would be greatly appreciated. Thanks

Wait, isn't 63% just one time constant? I'm pretty sure it is. Plug
it into a calculator, or even just do a napkin calculation using 2.7 as
an approximation for e. The quantity inside the brackets of the
exponential formula is about 17/27, or 63% for RC=1.
At any rate, the amount of work would equal the amount of energy
stored, which as you said is 1/2 C V^2, where V is just the voltage on
the cap. You don't have to do any calculus if you can calculate the
voltage on the cap. But calculate the voltage as a function of time
correctly.

Correction: I meant to say, "where t=RC," not "where RC=1"

 

[Rich Grise]

On Fri, 04 Nov 2005 14:23:00 -0800, Mr. Wizard wrote:

What is meant by additive & multiplicative noise in electronic
amplifier circuits?

Have you tried the other side of google at all?

http://www.google.com/search?q=%22additive+noise%22+electronic+circuits
http://www.google.com/search?q=%22multiplicative+noise%22+electronic+circuits

Have Fun!
Rich

 

[John Larkin]

On 4 Nov 2005 13:32:03 -0800, "jalbers@bsu.edu" <jalbers@bsu.edu>
wrote:

I know that the work required to fully charge a capacitor is given as:
1/2 C V^2

How much work is required to charge a capacitor to approximately 63% or
2 RC time constants? I know that it is going to be the area under the
curve: Vsource [1-exp(-t/RC)] from zero to 2RC , but it has been a long
time since I have done any integration. I am looking for a ball park
estimation if the exact answer is too difficult to derive. Any help
would be greatly appreciated. Thanks

If V is the final charged voltage, the energy to get the cap to 0.632
of V is just

e = 1/2 * C * (0.632*V)^2.

which is 0.4 times as much energy as the whole shebang.

Of course, that doesn't count the energy lost in a charging resistor,
if any.


John

 

[Rich Webb]

On Fri, 04 Nov 2005 19:35:04 -0600,
droors_20@hotmail-dot-com.no-spam.invalid (Voshempa) wrote:

I've been looking at Plasma and LCD tvs and have noticed that there
are variations in the contrast ratio (1000:1, 3000:1, etc.) and the
resolution. I understand how resolution numbers work with computer
monitors but wasn't sure if it was much of the same with a TV. I'm
guessing that the better the resolution on the TV, then the more
crisp the image will be?

Also, what is factored in when looking at a contrast ratio on a
certain TV? How do the variations affect the screen image? The higher
the better?

Any information would be helpful. Thanks.

Unless there is some fine print that says exactly how the contrast ratio
was determined, consider it just a "marketing number" that somebody
thought sounded good. There is probably some value in the number when
used to compare similar products from the same manufacturer.

There is an ANSI standard method of determining contrast. There's also
one for measuring brightness. Don't assume that a gadget that's "3000
ANSI lumens, 2000:1 contrast ratio" has measured both the brightness and
the contrast using ANSI methods.

--
Rich Webb Norfolk, VA

 

[kell]

kell wrote:

kell wrote:
jalbers@bsu.edu wrote:
I know that the work required to fully charge a capacitor is given as:
1/2 C V^2

How much work is required to charge a capacitor to approximately 63% or
2 RC time constants? I know that it is going to be the area under the
curve: Vsource [1-exp(-t/RC)] from zero to 2RC , but it has been a long
time since I have done any integration. I am looking for a ball park
estimation if the exact answer is too difficult to derive. Any help
would be greatly appreciated. Thanks

Wait, isn't 63% just one time constant? I'm pretty sure it is. Plug
it into a calculator, or even just do a napkin calculation using 2.7 as
an approximation for e. The quantity inside the brackets of the
exponential formula is about 17/27, or 63% for RC=1.
At any rate, the amount of work would equal the amount of energy
stored, which as you said is 1/2 C V^2, where V is just the voltage on
the cap. You don't have to do any calculus if you can calculate the
voltage on the cap. But calculate the voltage as a function of time
correctly.

Correction: I meant to say, "where t=RC," not "where RC=1"

Okay, I just now did a calculation to give the amount of energy
required
INCLUDING heat lost in the charging resistor.
Current through the rc combo is I = (V/R)e^(-t/RC)
Power = VI = (V^2/R)e^(-t/RC)
Energy = integral of Power with respect to time = -V^2 C e^(-t/RC) + K
set the constant K so that the integral is zero when t = 0
E = V^2 C (1 - e^(-t/RC))

Hope I did that right.

 

[Art]

These numbers are supplied by the manufacturers and normally are used by the
consumers to compare relative items before purchase. However, as stated, the
measurement procedures implemented by each company may very. Best
recommendation is for the prospective purchaser to physically compare items,
preferably without the intrusion of a high pressure salesperson being
involved. Then making an educated selection before dropping the quid on the
telly. Trying to do this via the internet, in sales brochures, etc normally
ends in dissatisfaction in the opinion of the end user.
"Rich Webb" <bbew.ar@mapson.nozirev.ten> wrote in message
news:ec7om1h75s3tn8jd1g1hvbf2stb281andq@4ax.com...

On Fri, 04 Nov 2005 19:35:04 -0600,
droors_20@hotmail-dot-com.no-spam.invalid (Voshempa) wrote:

I've been looking at Plasma and LCD tvs and have noticed that there
are variations in the contrast ratio (1000:1, 3000:1, etc.) and the
resolution. I understand how resolution numbers work with computer
monitors but wasn't sure if it was much of the same with a TV. I'm
guessing that the better the resolution on the TV, then the more
crisp the image will be?

Also, what is factored in when looking at a contrast ratio on a
certain TV? How do the variations affect the screen image? The higher
the better?

Any information would be helpful. Thanks.

Unless there is some fine print that says exactly how the contrast ratio
was determined, consider it just a "marketing number" that somebody
thought sounded good. There is probably some value in the number when
used to compare similar products from the same manufacturer.

There is an ANSI standard method of determining contrast. There's also
one for measuring brightness. Don't assume that a gadget that's "3000
ANSI lumens, 2000:1 contrast ratio" has measured both the brightness and
the contrast using ANSI methods.

--
Rich Webb Norfolk, VA

 

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