LM78XX imput/output capacitors

[jmariano]

Dear All,

Sorry if this is a basic question. I deed search the web for an
answer, but could find one.

I'm designing a power supply with various outputs (+24, +15, +10, -15,
-5), using LM78/79XX line regulators. The datasheet states that a .1
uF capacitor should be placed at the output to improve transient
response, and a .22 uF at the input, if the regulator is far from the
filter capacitor. But when I look at designs published in journal and
such, I found all kind of variations, including different values for
the input cap., even if
the regulator is close to the filter capacitor (in a PCB), and usually
an additional electrolytic capacitor at output.

So I guess my question are: 1 - do I need a capacitor at input and
what is is value? (and is purpose).
2 - Do I need an additional electrolytic capacitor at output, besides
the 0.1uF, and how do I calculate is value.

If anyone knows of a bibliographic reference ware this things are
explained, I would appreciate.

Regards

José Mariano

 

[vincent.thiernesse]

Hello,

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.

for the output, put big value if you want low impedance...you can add a 220
nF plastic capacitor cause it is better at high frequencies.

Regards

Vincent


"jmariano" <[email protected]> a écrit dans le message de

Dear All,

Sorry if this is a basic question. I deed search the web for an
answer, but could find one.

I'm designing a power supply with various outputs (+24, +15, +10, -15,
-5), using LM78/79XX line regulators. The datasheet states that a .1
uF capacitor should be placed at the output to improve transient
response, and a .22 uF at the input, if the regulator is far from the
filter capacitor. But when I look at designs published in journal and
such, I found all kind of variations, including different values for
the input cap., even if
the regulator is close to the filter capacitor (in a PCB), and usually
an additional electrolytic capacitor at output.

So I guess my question are: 1 - do I need a capacitor at input and
what is is value? (and is purpose).
2 - Do I need an additional electrolytic capacitor at output, besides
the 0.1uF, and how do I calculate is value.

If anyone knows of a bibliographic reference ware this things are
explained, I would appreciate.

Regards

José Mariano

 

[Eeyore]

Hello,

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.

Graham

 

[JeffM]

vincent.thiernesse TOP-POSTED [using Outlook Express]:

[...]these are minimum values[...]

For someone who has posted so many times,
http://groups.google.com/groups/sea...gywUTWB72_R7EPUyTwExa7iuCJ4giBMg7Tg&scoring=d
you are STILL posting like a rookie.

Hasn't anyone ever mentioned that you should get a better tool
http://www.google.com/search?q=cach...do-not-*-use-Micro.oft-programs&strip=1#quila
....or learn to configure and use the one you have properly?

 

[MooseFET]

Dear All,

Sorry if this is a basic question. I deed search the web for an
answer, but could find one.

I'm designing a power supply with various outputs (+24, +15, +10, -15,
-5), using LM78/79XX line regulators. The datasheet states that a .1
uF capacitor should be placed at the output to improve transient
response, and a .22 uF at the input, if the regulator is far from the
filter capacitor. But when I look at designs published in journal and
such, I found all kind of variations, including different values for
the input cap., even if
the regulator is close to the filter capacitor (in a PCB), and usually
an additional electrolytic capacitor at output.

So I guess my question are: 1 - do I need a capacitor at input and
what is is value? (and is purpose).

If the LM78XX regulator can see an inductive input line it is likely
to oscillate. You want a low impedance capacitor right at its input
terminals. The values in the data sheets are minimums based on
several assumptions about how the rest of the system looks. They are
usually enough.

Another reason to put a larger capacitor is that logic circuits often
draw current in narrow pulses. A capacitor to ground and some
impedance in the input wire can stop these pulses from making their
way up the power wiring.

One thing that people often overlook is that nearly all capacitors
cost about the same by time they are installed. Using a low number of
different values can often save you more than the price differences.

2 - Do I need an additional electrolytic capacitor at output, besides
the 0.1uF, and how do I calculate is value.

I normally put added capacitance at some small distance from the
output. The purpose is to provide a low impedance path to ground for
the supply. The output impedance of a 78XX rises at high
frequencies. I prefer the impedance to fall.

 

[David L. Jones]

Dear All,

Sorry if this is a basic question. I deed search the web for an
answer, but could find one.

I'm designing a power supply with various outputs (+24, +15, +10, -15,
-5), using LM78/79XX line regulators. The datasheet states that a .1
uF capacitor should be placed at the output to improve transient
response, and a .22 uF at the input, if the regulator is far from the
filter capacitor. But when I look at designs published in journal and
such, I found all kind of variations, including different values for
the input cap., even if
the regulator is close to the filter capacitor (in a PCB), and usually
an additional electrolytic capacitor at output.

So I guess my question are: 1 - do I need a capacitor at input and
what is is value? (and is purpose).

Yes, you should have the capacitors. Their purpose is to keep the
regulator stable and stop it from oscillating.

2 - Do I need an additional electrolytic capacitor at output, besides
the 0.1uF, and how do I calculate is value.

Not to keep the regulator stable, no.
You may need additional capacitors on the output located elsewhere on
the board to provide localised decoupling to those parts of the
circuit.

Be aware that "low dropout" voltage regulators are a different beast
and have fairly critical input and output decoupling requirement, this
will be explained in the datasheet for each device. Standard
regulators like the 78xx series are much more tolerant to this.

You might see extra caps in some designs because people tend to get
overzeleous in use of decoupling caps. Nothing wrong with that apart
from extra cost and board space.

If anyone knows of a bibliographic reference ware this things are
explained, I would appreciate.

This might get you started:
http://en.wikipedia.org/wiki/Decoupling_capacitor

Dave.

 

[vincent.thiernesse]

vincent.thiernesse TOP-POSTED [using Outlook Express]:

[...]these are minimum values[...]

For someone who has posted so many times,
http://groups.google.com/groups/sea...gywUTWB72_R7EPUyTwExa7iuCJ4giBMg7Tg&scoring=d
you are STILL posting like a rookie.

What is this, a rookie ?

 

[vincent.thiernesse]

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.

Oh, that's right !!! Sorry, I apologize for my mistaque....

 

[vincent.thiernesse]

vincent.thiernesse TOP-POSTED [using Outlook Express]:

[...]these are minimum values[...]

For someone who has posted so many times,
http://groups.google.com/groups/sea...gywUTWB72_R7EPUyTwExa7iuCJ4giBMg7Tg&scoring=d
you are STILL posting like a rookie.

Hasn't anyone ever mentioned that you should get a better tool ?

No, noone told me....surprised ? Anyway I'll take a look at this, thanks !!!

 

[Eeyore]

I'm designing a power supply with various outputs (+24, +15, +10, -15,
-5), using LM78/79XX line regulators. The datasheet

From which manufacturer ?

states that a .1 uF capacitor should be placed at the output to improve
transient
response, and a .22 uF at the input, if the regulator is far from the
filter capacitor.

More than a couple of inches typically. It's to counter the input inductance
caused by the long pcb trace.

But when I look at designs published in journal and
such, I found all kind of variations, including different values for
the input cap., even if
the regulator is close to the filter capacitor (in a PCB),

The designers are being cautious. An extra cap doesn't cost much. A piece of
returned equipment costs a lot.

and usually an additional electrolytic capacitor at output.

Negative regulators require this IIRC.

So I guess my question are: 1 - do I need a capacitor at input and
what is is value? (and is purpose).
2 - Do I need an additional electrolytic capacitor at output, besides
the 0.1uF, and how do I calculate is value.

If anyone knows of a bibliographic reference ware this things are
explained, I would appreciate.

I think TI or Nat Semi once published a 'Voltage Regulator handbook' - I expect
it would be available as a download these days. The one I'm thinking of is TI -
I have the 1977 copy.

Anyway, IIRC, negative voltage regulators like a slightly larger cap than just
0.1uF on the output pin. I use a 10uF electrolytic in parallel with a 0.1uF
plastic film cap.

Graham

 

[Eeyore]

Not to keep the regulator stable, no.

I think you do with a 79xx especially. I'm sure I once forgot it and it gave
trouble.

Graham

 

[Eeyore]

Oh, that's right !!! Sorry, I apologize for my mistaque....

No problem.

Don't forget to add the transformer loading when calculating 'voltage variation'
too.

Graham

 

[John Fields]

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.

 

[Eeyore]

---
8000µF?

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

Let's see your calculation then if you think my number is wrong. Or did you want me
to suggest an E6 value ?

Graham

 

[John Fields]

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

---

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


For a standard +/- 20% part,


Cnom 10000µF
Cmin = ------- = --------- = 12500µF
0.8 0.8

 

[Eeyore]

---

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.

Graham

 

[John Fields]

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

 

[Eeyore]

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.

Graham

 

[neon]

To begin with if you intend to put a big cap on the output then you will also need a to put a diode across input to output. That is to save the regulator when power is shut off and the output is present because of the cap, For a sytem where there are many boards and circuitry it much beneficial to instal small caps across each device as opposed to a big across the rail,

 

[John Fields]

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.

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