Why do we use a capacitor of specific value and not an arbitrary value for a full wave rectifier circuit? For example in this circuit diagram below shows a 470uF capacitor so why can't I use a capacitor of 100uF or 1000uF? How can I pick the correct value? Also what is the significance of capacitor C4?

12V DC power supply

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    \$\begingroup\$ there is no circit - post ? \$\endgroup\$ – efox29 Jun 24 '13 at 4:34
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    \$\begingroup\$ Because different parts have different requirements. You can't just stick any part in a circuit and have any expectation that it will work. \$\endgroup\$ – Matt Young Jun 24 '13 at 4:37
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    \$\begingroup\$ I think the best way to answer this question is to recommend you get an education as an electrical engineer if you plan on designing circuitry and need to know how to specify component values and ratings. This question is the equivalent of asking a dentist how to determine if a tooth needs pulled - it takes a little more know how than can be explained in a few minutes. Otherwise, a safe rating for components (such as the 25V on the capacitor) is two times the working value (12V in this case). This part is serving as a bulk capacitor, and other values such as 100uF or 1000uF would work as well. \$\endgroup\$ – Kurt E. Clothier Jun 24 '13 at 5:04
  • \$\begingroup\$ @PeterJ I'm looking at the edit history. How did you get this schematic? \$\endgroup\$ – Nick Alexeev Jun 24 '13 at 5:06
  • \$\begingroup\$ @NickAlexeev, the imgur link was in the markup, just hidden presumably because of the user's rep. \$\endgroup\$ – PeterJ Jun 24 '13 at 5:30

I assume you're talking about C5. This capacitor makes sure that the voltage input of the 7812 is a smooth one. It works like this:

enter image description here

Image Reference: http://www.electronics-tutorials.ws/diode/diode_6.html

The heavier the load (higher current), the faster the capacitor discharges, thus the more ripple there will be. We want an input voltage (to the load) as smooth as possible because sometimes a device acts weird due to a power supply with too large of a ripple.

When the load is so heavy that the ripple is too large, you can use a bigger capacitor, because that smooths the line more. Therefore, it depends on the load what value you need for a capacitor.

You can calculate the capacitance needed for the capacitor for a given mains frequency (not so important) and load current - however, I'd prefer testing around a bit and measure the ripples with a scope.

  • \$\begingroup\$ Good answer. I did not even consider commenting on the purpose of the circuit, rather than on capacitors in general. \$\endgroup\$ – medivh Jun 24 '13 at 8:22
  • \$\begingroup\$ I would add explicitly that source's formula for ripple voltage: V_{ripple} = I{load}/ (f x C), "Where: I is the DC load current in amps, ƒ is the frequency of the ripple or twice the input frequency in Hertz, and C is the capacitance in Farads." So the correct capacitor depends on the acceptable voltage ripple. \$\endgroup\$ – plu Jan 22 at 21:22

Some answers have already explained the ripple but don't go into the calculation. Both are important. You can use the equation:


This describes how current flows through a capacitor. You can simplify it as:


so if at the output of your bridge you draw 100mA, and your C=470uF and you can say:

\$t = {{1}\over{120Hz}} \approx 0.00833s\$

(t=1/f, so f=120Hz since you have 60Hz full wave rectified so 2*60), so rearranging the equation you get:

\$V={{{I}\cdot{t}}\over{C}}={{{0.1}\cdot{0.00833}}\over{470{\mu}F}} \approx 1.7V\$

so your output will charge up to about 12V (neglecting the diode voltage drop) and discharge down to 10.3V and repeat hence you will have a 1.7V ripple on your ouptput.

This method is a good approximation. You also have a regulator after the output which can handle a certain amount of ripple and smooth it out even further. If you reduce your capacitance to say 100uF then your regulator may not be able to handle the increased voltage ripple. If you bump your capacitor up to 1000uF then your ripple will go down even lower, and your design will be better but might cost more. Hope that helps.

  • \$\begingroup\$ +1 for the calcs, of course National Semiconductor have had datasheets for these for almost 20 years now. It does not hurt to round up a bit either. Might want to consider how long the run is between C4 and and C5 or C4 and the bridge. Ideally build it and stress test it with a dummy load for real time selection of a good value. NB. many 230-250VAC supplies are in fact 50Hz. \$\endgroup\$ – mckenzm Jan 12 '17 at 0:27
  • \$\begingroup\$ Good answer, except that at peak the capacitor will be charged up to 17.04V (12V RMS => 17V peak), so the ripple will drop around 15V and you have more than 3V dropout on the Linear Regulator. Good enough to make it work (except that on "real" [uk/eu] mains, Vac=230Vrms ±10%). \$\endgroup\$ – HappyCactus Sep 27 '17 at 8:05
  • \$\begingroup\$ Good catch. I edited it just now, It is being reviewed. \$\endgroup\$ – Sajeev Ranasinghe Sep 28 '17 at 11:33

The 7812 voltage regulator requires an input voltage which is several volts higher than 12 in order to function properly (see data sheet). The difference between that minimum voltage and 12V is called "dropout". The 7812 is a fairly high dropout regulator. As long as the input voltage is sufficient (the minimum dropout is met), the regulator can provide a smooth DC, in which the input ripple from the bridge rectifier is reduced by around 80 dB (check the exact ripple reduction decibel figure in the data sheet).

If you measure the voltage on the capacitor you will see that it charges to a higher voltage than 12. The secondary winding of the transformer is 12V, but that's a nominal RMS AC voltage. The peak voltage is actually higher, and the peak voltage is what charges the capacitor. If the secondary windings operate at 12V RMS, then the capacitor will charge to a peak of about 17V. Thus, at the peak, there is 5V of dropout.

On each cycle, the capacitor charges to the peak voltage. Then, it discharges as the regulator draws current from it. The capacitor must be large enough that when the regulator draws current from it between the charge cycles, the voltage will not drop below the minimum voltage specified for that regulator.

This must be ensured under the worst-case load for the regulator when it draws the most current.

Beyond satisfying the worst case current draw, if you further increase the capacitor to a larger value, the only benefit it provides is that it reduces the peak-to-peak ripple. This is a minor benefit, since the regulator is actively reducing that ripple by 80 to 90 decibels already. If the ripple is 0.5V peak to peak at the input of the regulator, and is cut 80 dB, it becomes 50 \$\mu\$V peak to peak at the output. If you reduce the input to 0.3V peak to peak with a larger capacitor, the output ripple goes from 50 \$\mu\$V to 30 \$\mu\$V. Both these values are small and possibly insignificant to the circuit.

If the circuit needs less ripple, by far a better way to get it is to use a better regulator with more decibels of ripple rejection, rather than making the capacitor larger. A regulator that improves rejection from 85 dB to 110 dB will make the same difference as a really huge and impractical capacitor substitution.

A capacitor which is too large stresses the transformer rectifier diodes when power is applied, because the bigger the capacitor, the bigger and more sustained is the inrush current.

  • \$\begingroup\$ By the way, I meant that the LM7812 is a fairly high drop-out regulator. There may be low-drop-out parts which use the 7812 number, to indicate compatibility. \$\endgroup\$ – Kaz Dec 18 '13 at 1:18
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    \$\begingroup\$ Also, reducing ripple might not be just a minor benefit. With lower ripple, you can reduce the input voltage (while maintaining the minimum) and thereby reducing the dissipation in the regulator. \$\endgroup\$ – Kaz Dec 18 '13 at 1:19
  • \$\begingroup\$ Another also: a capacitor value much larger than strictly needed will effectively be a short-circuit on switch-on until it reaches ~some~ level of charge, so (a) your heatsinking of the rectifier diodes might need to be more substantial, and (b) will make the circuit take longer to reach the required output voltage. \$\endgroup\$ – Andrew Morton Jan 27 '16 at 21:53
  • \$\begingroup\$ @AnderMorton I believe that is covered in the very last paragraph, except for the note that it takes longer to reach the operating voltage. \$\endgroup\$ – Kaz Jan 27 '16 at 23:52

The equation I go by is: C=I*(∆V)/(∆T) Where I is the current you want to output ∆V is the Maximum amount of Voltage Ripple (Peak to Peak Ripple of the Capacitor Voltage) that your circuit can safely handle. The minimum peak should be above your voltage regulators Desired input which is usually 3-vokts above your regulated voltage. ∆T can be calculated taking the Minimum Peak To Peak Voltage and dividing it by the maximum Peak Voltage and that value being = X and the solving arcsin(x). The angle you get in degrees must have 90° subtracted from it. 360° occurs in ∆t=1/(50-hertz)=.02 so ∆T=(arcsin(x)-90°)*.02/360°

If I=1Amp, ∆V=5% and Vmax=15-volts then Vmin=14.25 and x=14.25/15=0.95 So ARCSIN(0.95)*.02/360=3,989uF and the nearest standard capacitor value would be 4,700uF for the Filter Capacitor size.


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