Hello,

A long time ago, I read that if you use two capacitors as a voltage divider on an AC input circuit (instead of using a transformer) to supply a low voltage to a bridge rectifier, that the capacitors had to be of equal value.

I have been web searching, and I can't find any information about why the capacitors would have to be of equal value.

Can someone tell why this is true or where to look?

Thanks,
Tammy

 

Using capacitors isn't an efficient method to get a lowered AC voltage. If you need a very small current, then it can save a more costly transformer.

I don't know why they would need to be equal value, unless you wanted to divide by two exactly. You can pick the reactances to give you whatever AC voltage you require.

So, for example, if you have 120VAC and want 12VAC (both at 60Hz),

120VAC- C1 -12VAC -C2 - GND

C1 = 9uF
C2 = 1uF

You can treat the capacitance as resistors, as in the resistor divider formula.

The output current capability is another calculation. As a rule of thumb, you should set the divider current to at least 10X the output current you require. So, if you require 10mA, then you need 100mA to flow through the divider.

So, Vin = 120VAC
Iin = 100mA
Rin = 1200 ohms
Xc = 1/(2*pi*f*C)
C = 1/(2*pi*f*Xc)
C = 1/(2*3.1416*60*1200)
Ct = 2.21uF

So, you'd just set the ratio the same as the above with this total capacitance.

Power dissipation would be significant though, ~12W is wasted. For a total output of 120mW. You can set the 10X requirement a bit lower, but your voltage will droop significantly under load.

Steve

 

The use of such techniques is strongly discouraged here, as there is no positive isolation from mains power.

Use a suitably sized transformer. They are vastly less expensive than a funeral.

 

There is no substitute for a transformer for safety reasons.