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When I design a basic power supply that uses a full wave rectifier, The smoothing capacitor is very large. The output of power supply is 5V and 1A.

The ripple voltage equation is: V = I / (f*C)

f = 100 Hz and I assume that ripple voltage are 10 % (0.5V).

The capacitor value is 20 mF. I think that's too much and the cap is not available practically. I saw many previously designed power supplies and their capacitors ranged from 470 uF to 2200 uF. What Am I missing?

Thank you very much,

enter image description here

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    \$\begingroup\$ "20 mF" = ???? Do you mean 20,000 uF (micro farads)? "Too much" meaning too large of a capacitance value? The equation you give is basically correct, though I believe it is missing a scaling term ( 2 * pi = 6.26). However, more importantly, the "I" in that equation is the capacitor current, not the load current being supplied to the resistor (presumably the "1 A" you mention in the first paragraph). Ripple voltage is determined in large part by the output load resistor. The equation you are using does not have an R term. \$\endgroup\$
    – FiddyOhm
    Oct 21, 2016 at 17:50
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    \$\begingroup\$ @FiddyOhm: Get with the times! We've all switched to using SI units for capacitance, and that includes "millifarads" instead of "thousands of microfarads". Also, the 2*pi correction is incorrect -- we're talking about whole cycles here, not radian frequency. And the I is indeed the load current, which is the capacitor current between the charging pulses. I don't know who upvoted your comment, but it's essentially wrong on every point. \$\endgroup\$
    – Dave Tweed
    Oct 21, 2016 at 17:57
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    \$\begingroup\$ @David: Please don't get your wool panties in a bunch over this! If you go onto DigiKey or Mouser and look up a few large value capacitors (e.g. 47,000 mfd), you will find they are described exactly like "47000 uF", not "47 mF". And besides that's how Snoop Dog always specifies electrolytic caps when he draws them on his schematics. Capacitive reactance = Xc = 1 / 2*pifc where c is expressed in farads, f in Hertz and Xc in Ohms. "Wrong on every point" - get with it man!This is the InterNet, idiots become geniuses in a micro-second (uS = 0.001 mS) simply by writing words they don't understand \$\endgroup\$
    – FiddyOhm
    Oct 21, 2016 at 18:43
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    \$\begingroup\$ This unregulated supply will only make exactly 5V at certain mains voltages and certain loads .Mains volt tolerances and relatively poor load regulation mean that all credible 5V supplies are regulated . \$\endgroup\$
    – Autistic
    Oct 21, 2016 at 23:11
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    \$\begingroup\$ @FiddyOhm: Clearly, it's taking you a lot longer than a microsecond to get there. I still don't understand the point you're trying to make. The capacitive reactance at a single frequency is not very helpful here, where we're trying to determine the peak-to-peak ripple voltage. Because the output of the rectifier is not a sine wave, you would have to combine the results for an infinite series of frequencies in order to get a useful result. It's much simpler to consider the instantaneous voltages and currents directly. \$\endgroup\$
    – Dave Tweed
    Oct 22, 2016 at 10:49

2 Answers 2

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The equation is simple enough:

$$\Delta V = \frac{\Delta Q}{C} = \frac{I \Delta t}{C}$$

Solve for C and plug in the numbers:

$$C = \frac{I \Delta t}{\Delta V} = \frac{1 \text{A} \cdot 10 \text{ms}}{0.5 \text{V}} = 20 \text{mF}$$

If that's the amount of droop you want to have at that current, then you need the 20 mF capacitor.

Supplies that use smaller capacitors are:

  • using less current
  • allowing more droop (more headroom on the input side of a voltage regulator)
  • operating at a higher frequency (60 Hz vs. 50 Hz)
  • overly optimistic(?)
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    \$\begingroup\$ ...have a voltage regulator? \$\endgroup\$
    – gbarry
    Oct 21, 2016 at 17:57
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    \$\begingroup\$ @gbarry it's a good idea to minimize the ripple on the output of a rectifier like this, because the down-stream Vregs might not have good powersupply rejection ratio or load/line regulation specs (so the ripple will be seen in a large extent on the output of the Vregs) \$\endgroup\$
    – KyranF
    Oct 21, 2016 at 17:59
  • \$\begingroup\$ @gbarry so, What would be the value of the cap when I use a voltage regulator? \$\endgroup\$ Oct 21, 2016 at 18:14
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    \$\begingroup\$ That would depend on the specific regulator and transformer you choose. Just as an example, let's suppose you have an LM7805 regulator (2V dropout) and a transformer secondary that's 8Vrms. This gives you a peak voltage of 11.3V before the rectifier, or 10V after the rectifier. This leaves 10V - 7V = 3V range for the droop or ripple voltage. Plug the numbers into the equation above, and you get 3.3 mF absolute minimum; 4.7 mF would be a better choice. \$\endgroup\$
    – Dave Tweed
    Oct 21, 2016 at 18:24
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Your equation is correct. As to what you are missing.. 20,000uF is not unreasonable for a linear supply of that size, but most 5W supplies these days will be switching, so the output filter capacitor can be much smaller. The input filter capacitor on a single phase supply has to operate at mains frequency so it is larger, but the switching regulator can tolerate a lot more ripple voltage without the output voltage being affected.

Also note that your ripple is peak-to-peak. If you look at in terms of peak ripple (5V +/- 0.5V) you can use a 10,000uF capacitor which is pretty compact.

At 1A linear regulators are not very attractive- you're going to end up throwing away close to half the power to deal with diodes drops and mains voltage tolerance, so you've got a 5W heatsink to deal with. At a couple hundred mA your capacitor size is proportionally lower and heatsinks may not be required.

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