A friend (who knows even less about electronics than me!) asked me if I could build a circuit that can convert 8 V 50 Hz AC into a DC power supply so he can power a small board that is normally driven by batteries. The load will have a low current and also the circuit will not need to run for more then a couple of minutes at a time.

The obvious solution is to make a full wave rectifier and to smooth the voltage out with a smoothing capacitor. I'm not quite sure what voltage he wants so I'm going to allow it to be tuneable via a trimming potentiometer. I also want to add an LED to show it works. What I've come up with is the following circuit:


simulate this circuit – Schematic created using CircuitLab

The problem I've got is that I don't have access to the power supply he wants to use to drive it (the joys of lockdown). Which means I'll need to build it with out being able to test it at 50 Hz. I've built something on breadboard and tested it with a DC power supply. Based on that I'm happy with the values for R1, R2 and R3 and the diodes are spec'ed to switch fast enough.

What I can't work out is the value of the smoothing capacitor. If it's too small I think I'll get 50 Hz drops in the output voltage which will upset the down stream circuit. Too high and could I get trouble charging it?

Is there a formula or even a rule of thumb that will give me the right value?

Edit: I've updated the diagram. As people rightly pointed out I've had the LED in backwards. Fortunately this was just a mistake in the diagram. The breadboard prototype was the right way round - it did light up.

  • \$\begingroup\$ This LM317 datasheet suggests good examples \$\endgroup\$
    – jay
    Commented Aug 22, 2021 at 17:09
  • 1
    \$\begingroup\$ Your LED is wired backwards and you should find out what voltage he wants and also not use a pot for retuning that voltage. Find out what he wants and do the job properly. \$\endgroup\$
    – Andy aka
    Commented Aug 22, 2021 at 18:38
  • \$\begingroup\$ @jay Which part of the datasheet suggests good example values for the input smoothing/bulk capacitor? I think it does not. \$\endgroup\$
    – Justme
    Commented Aug 22, 2021 at 18:45
  • \$\begingroup\$ @Justme Sorry, and thanks for letting me answer something better. My opinion was, instead of telling Stomcloud that his design may not work, I wanted to suggest looking into using a adjustable voltage regulators. Once he starts reading it, he would sense many factors apply the filtering selections. Since he wants to know that very answer, the cap selection, I will give him an answer right below. \$\endgroup\$
    – jay
    Commented Aug 22, 2021 at 19:13
  • 1
    \$\begingroup\$ A rule of thumb is 2,200 µF per 1 A current. \$\endgroup\$
    – WalterH
    Commented Aug 24, 2021 at 16:07

3 Answers 3


What I can't work out is the value of the smoothing capacitor I'll need. If it's too small I think I'll get 50Hz drops in the output voltage which will upset the down stream circuit.

That trouble is called "Ripple". When it cause problem to down stream, than let's name it "Ripple Effect", a new terminology now on.

Too high and could I get trouble charging it?

Yes, that trouble is called "surge current" through diode. What happens is that the "Voltage Source" tries to charge the capacitor with all possible force, and only limits are hidden (from the schematics) impedance inherent to the components. When that current goes over the "rated maximum" of the parts involved in the current path (transformer, rectifier, capacitors), that will harm the corresponding parts. These are all in the datasheet.

Is there a formulae or even a rule of thumb that will give me the right value?

Yes. The formulae is "ripple voltage calculation formula", let's name it, if it did not have name before. The idea is finding the degree of smoothing effect of the capacitor by specifying tolerable amount of the voltage variation on the load.

The input variables to the formula are: a. Input AC voltage, b. Load current, c. The ripple tolerance on the load.
The formula, to approximate the cap, is: dV = 1/C x I x dt

Trans former output = 9V AC
Rectifier drop = 1V
Load resistance = 100 ohm
Tolerable ripple = 0.5V

1) Peak voltage at the load side of the bridge rectifier
= Vac x root(2) - 2 x Vrect = 9V * 1.4 - 2 x 1V = 10.6V
2) The load current, Iload = Vpeak / Rload = 10.6V / 100 = 106mA
3) C = {Iload x [1 / (line_frequency x 2_for_full_bridge)]} / Vripple
= {106mA x [1 / (50Hz x 2)]} / 0.5V = 0.00212F | 10.6V => 2,200uF | 15V

The same formula is used as well to calculate the input capacitor value between rectified AC and a voltage regulator.


What you have drawn, at least the first phase, is common practice, however, using the pot to change output voltage is not - wastes un-necessary power and other problems. What you can do is buying a Low Drop Out (LDO) voltage regulator, it's less efficient than SMPS but matches your requirement here (presumably! it depends on some other factors but you have not mentioned about the load).

About the cap, the higher the cap the higher the time constant, thus lower ripple induced on power lines. go too high and that's gonna cause problems too. It really depends on your specific application. But my experience thus far, in cases like this, is that they are usually in the range of uF's. These designs usually include a way of stepping down mains voltage, then putting it through the process that is rectification and smoothing, so most stuff you search for online already assume the frequency of these kinds of circuits to be 50-60 Hz.

PS1: I think you got your LED the wrong way, watch out!

PS2: I have given an answer here before to a similar question.

PS3: If efficiency is not that important, you can even use a half-wave rectifier and then feed it into a regulator.


The first problem is that if the load draws a non-constant current, the output voltage of the resistive divider is not constant either. Generally you don't supply power via resistor divider but a regulator. And the LED is upside down as well. Also, 8 VAC rectified has 11V peak voltage which gets charged to the capacitor.

Otherwise there is a simple formula for calculating the capcitor value. Q=I×t and Q=C×U. You just need to know the maximum current and allowable ripple voltage and mains ripple frequency after full bridge rectifier is 100 Hz.


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