# Low Voltage AC Smoothing Capacitor Calculation

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?

• After full wave rectification, ripple is at 120Hz, which will halve the required capacitance. – Brian Drummond Jun 22 '15 at 9:38