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How do you know what value of smoothing capacitor to use in front of a voltage regulator?

I am using a 7805 to power a circuit that draws 30 mA at 5v.  I feed the 7805 with a 14v AC supply, full-wave rectified.

I can find the formulas for calculating ripple from the rectifier, but does the voltage regulator care about ripple on its input if the voltage stays above the dropout?
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but does the voltage regulator care about ripple on its input if the voltage stays above the dropout?

No it does not. The input and output filter caps are more for requlator stablity and oscillation prevention. The specific regulator datasheet will give the best information on size, need and purpose of the input and output caps.


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Most 7805 specs show 0.33uF on the input and 0.1uF on the output.
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The datasheet for the reg (http://focus.ti.com/lit/ds/symlink/tl780-05.pdf) states tthat a 0.33 uF cap on the input is required "when the regulator is far from the power-supply filter."

This seems to imply some sort of additional power supply filter is required, but it doesn't say anything about the nature of that filter.
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I think the essential nature of the question has been missed. The power is coming from a rectified AC supply so the capacitor has to do more than just a little extra filtering of a DC supply, it has to supply the bulk of the current to the circuit during the "valleys" of the AC waveform.

As a first approximation, say the 14VAC supply is RMS, so the peaks are 14*sqrt(2) = 19.8V, minus two diode drops for the bridge (0.7V each) gives 18.4V. Say 18V to be conservative. So this input cap charges to 18V during the peaks and can't drop to any lower than 7V so the 7805 doesn't drop out.

This means the cap's voltage goes from 18V to 7V over an 8.33ms period (one cycle of the rectified waveform). Not exactly, but it's conservative. Given the current draw of 30mA, with a little extra 5mA for the 7805 we run the calculation C=i*dt/dV=0.035*0.00833/(18-7)=26.5 microfarads.

I would start with a 33uF capacitor.

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....and completely forgetting to answer the OP's question, no the regulator does not care about 120 Hz ripple on the input (60 Hz full-wave rectified) as long as the voltage stays above the dropout. In fact the 7805 datasheet generally states a ripple rejection parameter at 120 Hz, at least 62 dB for the ON Semi MC7805. This means that an 11V ripple at the input will look like 11*10^(-62/20) = 8.7mV ripple at the output, to a first approximation (and will probably be much better). Adding even a small 0.1uF capacitor on the output will help with that too, though no digital circuit will care about 8.7mV of ripple.

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I'd start with 100uf 50v (available 20 for $3.00 with free shipping on ebay). It's axiomatic that you'll always draw more current from a power supply than you think you will (add-ons, enhancements, etc). And 35V is the minimum voltage rating I'd use for a 14vrms input. 50V is even better.
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Thank you for that explanation!  The connection between the unregulated and regulated supply seems to be missing in a lot of discussions I have found.

One more question:  the calculation seems to assume that the output current @ 5V is the same as the input current @ 18V.  Is this the case?
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The output current of the 7805 is the same as the input current, less ~5mA for the 7805 regulator itself. So I guess the answer to your question is yes, it's the same as the input current to the 7805. But it's not the same as the current provided by your 14VAC supply. This AC supply provides current in bursts to the 33uF (or 100uF) capacitor, recharging it quickly at the "peaks" of the rectified waveform, then providing no current when the AC voltage goes into a valley and below whatever voltage is on the capacitor.

Hopefully that is not overly confusing matters.

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Some interesting thoughts here and I don't want to confuse matters but an 18 volt feed to the 7805 can mean it is going to have to dissipate a lot of heat if the output current is more than about 100mA. I would use a lower voltage say around 9 volts and compensate with a higher value capacitor myself. However I am often wrong smiley
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I would use a lower voltage say around 9 volts and compensate with a higher value capacitor myself. However I am often wrong


Your not wrong in this case.  smiley-wink It's always good to lessen the dissipation burden on the voltage regulator as much as possible. That aside, I powered my RS-232 serial arduino clone board for over a year with a regulated 2 amp +15vdc laptop power module. The on board regulator did run pretty warm but I never had a problem with it. Keep in mind even the simplest 7805 regulator as self-protection modes where it shuts off output if too much current is drawn from it or if it overheats.

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an 18 volt feed to the 7805 can mean it is going to have to dissipate a lot of heat if the output current is more than about 100mA

Agreed - but the 14V AC bus was the closest source of power, and I only needed 30 mA.
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My advice is to use a switching (buck) regulator. Let's assume 35V input and you need 5V@30mA so you have a power dissipation of 900mW... The effectivity factor is then 20%. Switchers offer atleast 85%! And you will need much less space for that. An easy way is to use Micromodules, e.g. from Linear Technology. So you haven't to suffer for a correct (optimal) layout.
I myself built many switchers ant it isn't that complicated...
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Power supply filtering is a well-understood science.  There's a reasonable explanation of ripple on wikipedia: http://en.wikipedia.org/wiki/Ripple_%28electrical%29

If I didn't make mistakes, the relevant equation works out to
C = I / (2 * f * Vripple)
So if you wanted the ripple going into the regulator to be about 5V (a sawtoothy waveform of about 14V +/-2V, all nicely above 5V and etc), you'd need a capacitor of about 50uF.
Bigger is better; conventional power supply design is sorta "stick as big a cap in there as you can fit."
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Actually no one make in the real world calculation for this, use the 1000uf/25 volts, is ok for your application
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