Re: LM78XX imput/output capacitors

On Mon, 17 Sep 2007 02:31:49 +0100, Eeyore
<rabbitsfriendsandrelations@xxxxxxxxxxx> wrote:



John Fields wrote:

Eeyore wrote:
John Fields wrote:
Eeyore wrote:
John Fields wrote:
Eeyore wrote:
John Fields wrote:
Eeyore wrote:
"vincent.thiernesse" wrote:

these are minimum values.

the input capacitor doesn't replace the filter capacitor.

for the filter capacitor, take 1000uF per Ampere for 1 volt input variation.

I take it you mean ripple ? Your numbers are out by about 7-8:1

You need ~ 8000uF for 1V (pk-pk) of ripple @ 1A load and 50Hz.

---
8000µF?

If you want to be pedantic, ***, you might at least try to get
it right.

Let's see your calculation then if you think my number is wrong.

---

Idt 1A * 0.01s
C = ----- = ------------ = 1E-2F = 10000µF
dV 1V

Where the hell do you get 10ms from ? I suggest you go and look at a ripple waveform.

---
LOL, you really _are_ stupid!

What's the period of full-wave rectified 50Hz?

That's not the relevant answer.

What's relevant is the discharge period.

Hint - NOT 10ms. I suggest you either measure or
simulate.

You're not as smart as you claim to be.

---
I've never claimed that I was smart, just less stupid than most.
Including you.

However, in one of your rare moments of lucidity, your "What's
relevant is the discharge period." brought me up short since that's
correct and I hadn't earlier considered that the recharge time was
irrelevant.

Thanks for that. :-)

Fair enough. I'm pleased to hear that.


Moreover, including the recharge time in the waveform's period,
'dt',

Idt
C = -----
dV

makes the solution inaccurate because it doesn't consider the
discharge time boundary created when the rise of voltage from the
rectifier causes the reservoir capacitor's losses to cease.

So what's the right way to do it?

As I see it, first determine the peak DC needed into the regulator.
For a 7824 that would be:

Vin = Vout + Vdo + Vrpl = 24V + 2.5V + 1.0V = 27.5V

The *peak* voltage ? Sure it'll work with that but when you consider low line voltage
conditions, you'll see you need rather more overhead in practice.

---
Brain fart. It's obviously "minimum DC".
---

Where Vin is the peak input voltage to the regulator,
Vout is the output voltage of the regulator,
Vdo is the regulator's dropout voltage, and
Vrpl is the allowable peak-to-peak ripple voltage on the
input of the regulator.

Next, determine the angle which corresponds to the valley voltage of
the ripple:

Vin - Vrpl 26.5V
arcsin = ------------ = arcsin ------- = 74.5°
Vin 27.5V

Then, determine the length of the capacitor's discharge time.

For a 50Hz sinewave, we're dealing with:

1 0.02s
t = ------ = -------
50Hz 360°

which is also:

0.02s
t = ------- ~ 55.5µs/°
360°

Since the cap is fully charged at 90° and discharges to 26.5V in the
time it takes to go from 90° to 0° and then back up to 74.5°, that's
a total of:

n = 90° + 74.5° = 164.5°

To get the discharge time, then, we multiply the total excursion,
164.5° by 55.5µs/°, and wind up with:

164.5° 55.5µs
Td = ------- * -------- ~ 9.13E-3s = 9.13ms
1 °

Now, since we're feeding a fixed voltage (the output of the
regulator) into a fixed resistance, (the load) we can say that the
input current into the regulator will be what the load dissipates
and we can finally say that the value of the reservoir capacitor
should be, at least,

Idt 1A * 9.l3E-3s
C = ----- = -------------- = 9.13E3 ~ 9100µF
dv 1V

Nice analysis. I got the gist of it but didn't pursue it in depth on account of what I know
practically of such things.

---
Well, you started off OK, but you just can't help being obnoxious
with your "I'm better than you are" ***, can you?

LOL, yeah... "precision" audio amps with unregulated rails? Hang it
in your ass, you pretentious jerkoff.
---

I challenge you to find such a short conduction (long discharge)
period in practice. In particular, the transformer's DC resistance (which you haven't
considered) limits the recharge current and this gives rise to a rather longer recharge time.

---
*** you and your challenges. Last time I looked, T still equals RC,
so if you need a shorter recharge time you either get a transformer
with huskier wire or a larger cap.
---

1V of ripple and such a large capacitor value is also unusual btw.

---
In your world, I'm sure that's true. However, if you need to reduce
ripple you basically have two choices: Increase the value of the
reservoir capacitor or reduce the load current.
---

I really do suggest you look at some real world examples. FYI the typical numbers I get for
an 'average' power supply are 2.5ms recharge period and 7.5ms discharge.

---
Who gives a ***? Dumb ass, in the real world I design what's
needed, and the "typical" numbers you see reflect what designers are
doing in your world.
---

A low ripple voltage ought to result in a shorter recharge period you'd think, but the series
resistance counters that rather well.

---
That depends on the value of the series resistance.
---

A very short recharge period would also result in a
particularly lousy power factor btw.

---
So what? If it's a problem you PFC the input.


--
JF
.

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