I have a 9V AC source which will have a max load of 100mA. I'm trying to figure the value of a smoothing capacitor with less than a 10% ripple. I don't have a scope or else I'd just try a bunch out.

I'm just a hobbyist, but I'm trying to understand this. Since I'm not sure if I'm doing this right, I'll just step through what I've done so far.

First I found this formula: $$V_{ripple}={V_p \over R_LC} \times \Delta T$$

I'm using 60Hz 9VAC that is full wave rectified and I want a 10% ripple so: $$(9V \times .1)={9V \over R_LC} \times {1 \over 60Hz}$$

I'm calculating my resistive load with Ohm's law.
$$R = {9V \over .1A} \quad R = 90 \Omega$$

So plugging that in I get: $$.9V={9V \over 90\Omega \; C} \times {1 \over 60Hz}$$

If I solve for C (admittedly using wolfram alpha), I get: $$C={1\over 540}F \quad C=1851 \mu f$$

1850µf seems pretty high to me... And if I lower the ripple percent it get crazy high.

Is this the right way to find this? Or did I screw up somewhere?

Also, my circuit has the capacitor between the full bridge rectifier's positive and negative outputs. I've only seen this type of smoothing circuit, but maybe my circuit needs to change?

  • 3
    \$\begingroup\$ After full wave rectification, ripple is at 120Hz, which will halve the required capacitance. \$\endgroup\$
    – user16324
    Commented Jun 22, 2015 at 9:38

2 Answers 2


Your calculations are good but if you are using a full bridge rectifier (diode bridge, I think?), your frequency is the double of the input, so here 120Hz.

Moreover, there is the voltage drop of the diodes, so with a full bridge it's roughly 1.2V. Hence Vp = 9-1.2 = 7.8V.

Notice that a ripple of 10% is very good. Usually we add a regulator after the capacitor to have a better constant voltage supply.


1000 microfarad is your recommended filter capacitance preffered value and yes it seems high but it must store energy between the rectified AC crests if the cap wasn't there you would get 48% ripple which is unworklable for most applications These days most powersupplies are switchmode which gives the designer the option of storing energy at the rectified mains level which gives a physicaly smaller capatitor in fact a 10microfarad would do better than the 1000microfarad if it was a SMPS


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.