LM78XX imput/output capacitors

[Eeyore]

---
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.

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. 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.

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

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.

A low ripple voltage ought to result in a shorter recharge period you'd think, but the series
resistance counters that rather well. A very short recharge period would also result in a
particularly lousy power factor btw.

Graham

 

[John Fields]

Fair enough. I'm pleased to hear that.



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".
---

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" shit, 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 shit? 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.

 

[vincent.thiernesse]

---
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.

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

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

Hello,

You did not take into account the fact that during dischage the wave doesn't
leaves the first sinus at t=0.

So I made the study.

You found 9133 uF for 1 V peak to peak and the corrected value is 9119 uF.

futher investigation shows that for 16V peak to peak we need 374 uF instead
of 397 uF....

so, well, I just did waist my time...

Just don't care...

Vincent

 

[Michael A. Terrell]

Hello,

You did not take into account the fact that during dischage the wave doesn't
leaves the first sinus at t=0.

So I made the study.

Study? Was it published?

You found 9133 uF for 1 V peak to peak and the corrected value is 9119 uF.

futher investigation shows that for 16V peak to peak we need 374 uF instead
of 397 uF....

Make sure to let us know where you buy your 374 uF capacitors.

so, well, I just did waist my time...

Just don't care...

Vincent

Obviously you enjoy 'waisting' time, but dont care about spelling.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

 

[vincent.thiernesse]

Study? Was it published?




Make sure to let us know where you buy your 374 uF capacitors.




Obviously you enjoy 'waisting' time, but dont care about spelling.

Thank you for your time spirit !!!

Please tell me what a DD214 is ? and what a DAV#85 is ?

@+

Vincent

 

[vincent.thiernesse]

Study? Was it published?




Make sure to let us know where you buy your 374 uF capacitors.

OK, I say it cause it seems to be great americain talking: "I buy it in your
ass". This is just a consequence of dollar deflation. Europeans make 375 uF
capacitors to continue being competitive.

Obviously you enjoy 'waisting' time, but dont care about spelling.

First time you talk to me and yet I fell like I should go away....should I ?

 

[John Fields]

Hello,

You did not take into account the fact that during dischage the wave doesn't
leaves the first sinus at t=0.

---
That doesn't matter, since what we're concerned with isn't what
happens during the first cycle after power-up, it's what happens
after the cap manages to charge up as fully as it can when the
rectified sinusoid is at its peak at either 90° or 270°
---

So I made the study.

---
In order to do that properly you'd need to know the circuit details
intimately, including the impedances of the mains, the transformer,
and the reservoir capacitor.

Do you have some numbers?
---

You found 9133 uF for 1 V peak to peak and the corrected value is 9119 uF.

---
Well, you mean _your_ value is 9119µF.


OK, assuming you're right, that's an error, on my part, of:

100 (C1 - C2) 100 * 14µF
Oops% = --------------- = ------------ = 0.154%
C2 9119µF

Not too shabby, considering one can use my method to more or less
choose the minimum capacitance required to assure the
desired/required ripple voltage.
---

futher investigation shows that for 16V peak to peak we need 374 uF instead
of 397 uF....

---
Assuming you did the math right, the difference there is 23µF, which
would make the error:


100 (C1 - C2) 100 * 23µF
Oops% = --------------- = ------------ = 6.15%
C2 374µF

How do you account for the forty-fold discrepancy in error between
the two cases with only a twenty-four fold difference in
capacitance?
---

so, well, I just did waist my time...

 

[vincent.thiernesse]

---
That doesn't matter, since what we're concerned with isn't what
happens during the first cycle after power-up, it's what happens
after the cap manages to charge up as fully as it can when the
rectified sinusoid is at its peak at either 90° or 270°

No, I was not talking about what happens after power-up.

what I mean is: After a every peak of the sine-wave, the signal follows the
(half) sine wave as long as Um w sin(w.t) is lower than I / C.

I was talking about "first sine wave" in the way that each period the signal
leaves a first sine wave to reach a second sinewave....(not clear ?)

---


---
In order to do that properly you'd need to know the circuit details
intimately, including the impedances of the mains, the transformer,
and the reservoir capacitor.

Do you have some numbers?

I took Um=27.5 V...this is enough for my purpose.

yes indeed.

OK, assuming you're right, that's an error, on my part, of:

100 (C1 - C2) 100 * 14µF
Oops% = --------------- = ------------ = 0.154%
C2 9119µF

Not too shabby, considering one can use my method to more or less
choose the minimum capacitance required to assure the
desired/required ripple voltage.

Yes, my conclusion is, your method is enough.

---


---
Assuming you did the math right, the difference there is 23µF, which
would make the error:


100 (C1 - C2) 100 * 23µF
Oops% = --------------- = ------------ = 6.15%
C2 374µF

How do you account for the forty-fold discrepancy in error between
the two cases with only a twenty-four fold difference in
capacitance?

Sorry, my english is limited. I don't understand the question.....

my time is not that precious.

Well, it was just kind of training...

Regards

Vincent

 

[Michael A. Terrell]

Please tell me what a DD214 is ? and what a DAV#85 is ?

Too lazy to use a search engine? DD214 is an honorable discharge
from the US military that you get after completing your six years of
service.

http://www.va.gov/

DAV is Disabled American Veterans, and 85 is the unit number.

http://www.dav.org/


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida