I have a transformer which outputs 0.33 A at 24 VAC. I would like to rectify this down to 5 VDC and minimal current, 50 mA being the maximum. I came across the following schematic which seems to be what I'm looking for.

Capacitor rectification

The only difference is that the output current in this schematic is ten times what I need.

It would be all too easy to simply divide the capacitors by ten. Can anyone explain to me the formula for calculating the capacitance needed to achieve a final result of 5 V, 30 mA coming from 24 VAC, 0.33 A? By the way, D2 is a 5.1 V, 1.3 W Zener diode.

  • \$\begingroup\$ This circuit is often called a "capacitive dropper". The voltage depends on the current - it's not a constant voltage source. You can tweak C1 according to your voltage and current needs. If your current usage is variable, you might want to put a zener diode in parallel with the output. \$\endgroup\$
    – user253751
    Sep 26 at 21:13
  • 2
    \$\begingroup\$ In general, you will be better off rectifying the 24VAC into about 34VDC and then dropping to 5VDC with a separate DC-DC regulator. The output voltage will be more stable, the idle power usage will be lower and the parts will be smaller. \$\endgroup\$
    – jpa
    Sep 27 at 9:24

5 Answers 5


Unless you have a reason to limit the current it makes no difference if a power supply is capable of more current than you need at a given voltage, if a device draws 50 mA at 5 V it will draw that much at 5 V whether the supply is capable of 50 mA or 50 A.

This goes back to Ohm's Law $$ I=\frac{E}{R}$$

The supply shown uses a capacitor to drop voltage. The output voltage will be partially dependent on the current drawn, so the voltage will vary with load, will not be well regulated, and will tend to have a lot of ripple. It's also not very efficient, somewhere around 10%. This is okay for something where the load is relatively constant and you don't need good regulation or low ripple.

One thing you could do is modify the circuit for better regulation by first dropping to a higher voltage and then regulating from that, here I've done it with a simple zener and resistor but a 5V LDO regulator should work as well and give you closer to 5 V. I've reduced C1 to 15 uF to lower the dissipation of the 10 V Zener.

enter image description here

You could also use a a more conventional rectifier, such as half wave, full wave or bridge, and a voltage regulator. This can be a simple zener, transistor type, three terminal integrated regulator, or a switching regulator.

If you still need to limit the current to a certain maximum you can add a current limiting circuit.



simulate this circuit – Schematic created using CircuitLab

Here's what you get with the existing circuit and a nominal 50mA approximate load (I've changed D1 to a 5.6V Zener and C2 to 1000uF).

enter image description here

As you can see the quality of the '5V', even with 1000uF filter, is not very good. There is also the matter of the 68uF capacitor, which needs a fairly high voltage and ripple current rating, a bit inconvenient.

Unfortunately the peak voltage of a half-wave rectified 24VAC is a bit high for the cheapest alternative simple regulator solutions- it could easily exceed 40V with mains a touch high and poor transformer regulation, which rules out many parts without some form of voltage dropper or pre-regulation. You could consider a module such as this or design one.

  • \$\begingroup\$ I'm using a LM2591HV module to run the 12V timer I added to my soldering station a similar situation. \$\endgroup\$
    – Jasen
    Sep 28 at 6:40

The circuit looks as if it should work as it stands. C1 is acting as a series impedance, similar to a series resistor, but without the dissipation. D2 clips the sine waves at 5.1 V, the current through D2 is limited by C1. The 1N4002 rectifies the resultant clipped sine wave and C2 is a smoothing (reservoir) capacitor to reduce output ripple. C1 and C2 are not a capacitive divider.

Since you only require 1/10th the current you could reduce C1 by a factor of 10 to 6.8 uF.

A back of the envelope calculation suggests that, in the original circuit, the zener diode is dissipating around 1.2W worse case, so reducing the value of C1 reduces the power dissipation in the zener diode to a 1/10th of what it was. An ordinary 1/4 W would do. However the forward drop (0.7V) of the 1N4002 needs to be taken into account and this will reduce the output voltage to about 4.4V so maybe you should use a 5.6V zener. Leave C2 as it is.


Can anyone explain the formula for calculating the capacitance needed to achieve a final result of 5 V, 30 mA coming from 24 VAC, 0.33 A?

\$X_C\$ is the impedance of a capacitor (in Ohms) at some frequency, and is defined as \$X_C = \dfrac{1}{2\pi f C}\$.

Assuming the frequency is 60Hz, we can solve for it in the schematic:

$$ \require{cancel} X_C = \dfrac{1}{2\pi \cdot 60 \cdot 68µ} $$

$$ X_C = \dfrac{1}{ 6.28318 \cdot 60 \cdot 68µ} $$

$$ X_C = \dfrac{1}{0.0256353744} $$

$$ X_C \approx 39\Omega $$

So C1 is effectively like a 39Ω resistor at this frequency.

To rearrange and solve for C given a random chosen Xc of 67Ω:

$$ X_C = \dfrac{1}{2\pi f C} $$

$$ 67Ω \cdot 2\pi 60 = \dfrac{1 \cdot \cancel{2\pi 60}}{\cancel{2\pi 60} C} $$

$$ \dfrac{67Ω \cdot 2\pi 60}{1} = \dfrac{1}{C} $$

$$ \dfrac{1}{67Ω \cdot 2\pi 60} = \dfrac{C}{1} $$

$$ \dfrac{1}{67Ω \cdot 2\pi 60} = C $$

$$ C = 3.959\text{e-5} \approx 40µF $$

Remember that the source is AC and you're only doing a single-diode rectification, so 50% of the time the AC will be "useless" for making DC.

As the voltage in C2 increases from zero, the "useless" time will grow even larger, as the voltage must not only be positive of zero, but greater than the volts on C2 + diode drop of the 1N4002 to put any electricity into C2. Anyways, this equates to a low duty-cycle, meaning that 24V is not going constantly through a 39Ω Xc to D2. This low duty cycle will decrease the average power dissipated in D2 linearly (and power available for the load.)

As Xc gets larger, expect the ripple voltage to get larger as well, even if drawing less current. For this reason, the Xc is usually chosen to give more than adequate current for the given load. Could increase C2 to combat this, but then you get slower start-up time...


I take it you realise that it is ok to draw less current from that power supply than the 300mA maximum that it is capable of supplying and the reason you want to redesign it to have a smaller current supplying capacity is so that you can use a smaller capacitor.

The ac voltage source is only producing usable current during its swing from near its negative peak to its positive peak. During the ac voltage source's swing from near its positive peak to its negative peak the capacitor's charging current is being sunk through the forward biased diode D2.

So you are effectively dealing with a half-wave rectified waveform which has an rms voltage value of Vpeak/2 = 24*sqrt(2)/2 = 17 V.

Now the current available from the output is 17/Xc where Xc = 1/(2 * pi * f * C1).

So, available output current for C1 = 47uF is equal to 17/56.43 = 300 mA.

Transposing the formula to make the capacitance the subject gives:-

C1 = I/(17 * 2 * pi * f)

I make it that you require a 4.7uF capacitor for a 30 mA output capability. This makes sense because the capacitor value is reduced by a factor of 10 for a 10 fold reduction in current capability. You may want to increase the value of C1 above 4.7uF to give yourself some headroom.


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