I want to know if there is a rough any ready way of calculating the capacitance required to decouple the power supply used in an amplifier. (Rough and ready because I'll be adding 30% to the value anyway!)

Questions: 1) Does this depend on the type of amplifier used (in my case, common emitter stage going into an AB pushpull)?

2) Does this depend on the signal to be amplified, (in my case just a simple sinewave).

3) How is this linked to the power rating of the amplifier (in my case only a few watts.)

4) Does the voltage effect it? (apart from the voltage rating of the cap being high enough for the supply! (in my case 25v))

5) Does the frequency effect this, (in my case 50kHz to 5MHz)

In short, what steps/calculations should I make to calculate the values of the bypass caps needed to decouple the supply and prevent distortion due to supply ripple voltage?

  • \$\begingroup\$ Your bypass capacitor should do these following tasks. >>1. Transient current demand of your amplifier( in driving any loads) >>2. You didn't mentioned the application of amplifier, what u r driving(rise time of driving and load details), any specific freq region you want to bypass. Baseb on this facts, cap value wil depnd. 1. it depends on load driving 2. No i think 3. Voltage of cap hould be derated based on ceramic or tantalum or aluminium etc and see is it ok for application 5. This frequency most of the capacitors are OK \$\endgroup\$
    – user19579
    Aug 16, 2013 at 11:25
  • \$\begingroup\$ The amplifier I have designed uses a +25V/-25V power supply and drived a 50Ohm load. \$\endgroup\$
    – Tim M
    Aug 16, 2013 at 11:53
  • \$\begingroup\$ You've told us nothing about your power supply and the ripple is primarily based on the power supply type, how good it is and what load conditions it will see. \$\endgroup\$
    – Andy aka
    Aug 16, 2013 at 14:19

3 Answers 3


I think that the general rule of thumb for decoupling caps is "the bigger, the better!".

The best way to figure out whether your decoupling is good enough is to build the circuit and measure the ripple. The second best way is to simulate the circuit in Spice and measure the ripple.

However, if you do want to have a rough estimation of the order of magnitude, you need to take into account the following parameters:

  • Output impedance of your power supply - \$R\$
  • Average switching current and its duration - \$I_A\$ and \$T\$
  • The maximal allowed ripple in the supply's voltage - \$V_{r_{max}}\$

It is clear that the ripple of the supply's voltage is due to the voltage drop on its internal output impedance, and that this voltage drop is equal to voltage drop on the cap:

$$V_{r}(t)=I_{R}(t)R=\frac{1}{C}\Delta Q(t)$$

Note that \$I_{R}\$ in the above equation is not the total current drawn by the load, but the fraction of this current which is drawn from the power supply (the other part is drawn from the capacitor).

If you do the algebra and substitutions, you'll get to the following equation:


Where \$I_C(t')\$ is the current drawn from the capacitor.

In order to find the maximal voltage drop you need to find the maximum of the above function. This requires the differentiation with respect to \$t\$ and finding the value of \$t\$ for which the derivative is equal to zero. Due to the fact that the current drawn from the capacitor depends on the voltage on the capacitor and the current drawn by the load, the above differentiation is not simple and requires an exact characterization of the switching current profile.

However, you do not want an exact solutions, but just estimations, therefore we can make several assumptions which will simplify the problem:

  • The current drawn from the power supply when the ripple is at maximum is \$\frac{V_{r_{max}}}{R}\$. We can assume linear ripple, which means that the average current drawn from the supply is \$I_{R_{A}}=\frac{V_{r_{max}}}{2R}\$
  • The average current drawn from the capacitor is then \$I_{C_{A}}=I_A-I_{R_{A}}=I_A-\frac{V_{r_{max}}}{2R}\$.
  • The total voltage drop on the capacitor due to the above average current which flows during time period of \$T\$ (switching time) is \$\Delta V_C=\frac{1}{C}I_{C_A}T=\frac{T}{C}*(I_A-\frac{V_{r_{max}}}{2R})\$

Accepting all the above assumptions and requiring \$\Delta V_C=V_{r_{max}}\$ leads to the following capacitance value:

$$C=T\left ( \frac{I_A}{v_{r_{max}}}-\frac{1}{2R} \right )$$


I've derived the above equation just now. It may be completely wrong. However, I see that the dependency of the required capacitance on the parameters of the problem is intuitively correct:

  • The higher the switching current the bigger the capacitance you need
  • The lower the desired ripple the bigger the capacitance you need
  • The higher the internal output resistance of the supply the bigger the capacitance you need
  • The switching time dependence is a bit tricky: it has no affect on the first term in parentheses due to the averaging of the current. Therefore, the shorter the switching time, the bigger capacitance you need.

It will be wise to test this model, and, as you said, anyway take the capacitor which is bigger than predicted by this equation.

I'll be glad to get a feedback on this model.

  • \$\begingroup\$ There are actually two rules of thumb: Bigger is better. And, low inductance is better. Unfortunately these rules often conflict. \$\endgroup\$
    – The Photon
    Aug 16, 2013 at 16:17

Go to http://www.duncanamps.com/psud2/index.html and download their FREE "PSU Deisgner". It allows you to construct your power supply schematically and see output voltage, output current, and ripple results. You can easily edit component values (such as filter cap) and re-simulate as often as you like.

You'll need to know how much current your amp draws and its rail (power supply) voltage to set the load resistor (using ohm's law, R=E/I) or you can set your load as a "current sink" (fixed current, regardless of voltage).

Then, simulate the power supply and look at the "difference" value (between max and min voltage) of your load. This will be your ripple voltage. If it's no more than 0.0032% of the supply voltage than it's 90dB down from your power supply voltage, and should satisfy even audiophile standards.

You can also try different caps and see how much ripple voltage (or current) you get. Take the log of your ripple voltage over your supply voltage (or ripple current over supply current) and multipy by twenty to see your ripple in dB.

Good luck!

P.S.: ...or were you referring to an OUTPUT bypass cap? In which case, use

f = 1/(2π x r x c), where:

f is the lowest frequency you want to pass in Hz (typically 10 or 20 for an audio amp),

r is your load resistance in ohms (typically 6-8Ω for your average speaker + maybe an ohm for the impedance of your output stage), and

c is your capacitance (in farads - i.e. 1µF would be .000001)

  • 1
    \$\begingroup\$ Just FYI: "Audiophile Standards" is more like -140 dB on the power rails. Frequently these guys don't use negative feedback, so their PSRR is around 0 dB-- and even with that they have noise floors better than -135 dB. \$\endgroup\$
    – user3624
    Aug 28, 2013 at 4:09

Questions 1 - 3 are all linked to the same issue: Power consumed from the supply rails.

1: An AB Push-Pull amplifier has a 75% efficency.

2: A sinewave will use more power than a squarewave with a 10% PWM cycle, for example.

3: Take your RMS ouptut rating in watts and multiply by 1.414 to get the peak watts and divide that by .75 to get the wattage that needs to go in the supply rails. Use Power Law and the supply voltage to determine the effective resistance of the amplifier.

There is a very simple equation that will allow you to get the capacitance answer you want. It is: C=T/R


C = capacitance in Farads

T = time in seconds

R = resistance in Ohms.

Calculate the peak current draw from your amplifier and use Ohm's Law and Power Law to find the equivalent resistance in Ohms.

T is the amount of time it takes for the voltage in the capacitor to drop by 1 time constant, or 63% of full voltage. The voltage decline is not linear, but sinusoidal. For your purposes you can think of it as linear. Want to have a voltage drop of 6.3% instead of 63%? Multiply the capacitance by 10X.

One thing that big capacitors are good at is supplying lots of power. What they are not good at is reacting quickly. If you want to filter MHz frequencies from your amplifier you might want to consider putting a .1uF decoupling capacitor in parallel with your monster decoupling caps or putting an inductor in series between VCC and the filtering caps. The formula for that is:



XL = the inductive reactance (or AC resistance) of the inductor

F = the frequency you want to block, in Hz

L = the inductor value, in Henrys

The wire gauge should be large enough to pass the peak desired current without heating and the XL value should be 100X+ higher than the equivalent resistance of the amplifier at the frequency you want to block.

  • \$\begingroup\$ Your method doesn't really work correctly. Your formula for caps, C=T/R, only work for caps that will be discharged to about 66% of their initial voltage. If a real amp discharged to that level there would probably be serious audio distortion or other issues. There are other problems with the answer, but mostly semantic issues with how you've described things. \$\endgroup\$
    – user3624
    Aug 29, 2013 at 21:20
  • \$\begingroup\$ The C=T/R calculation can be modified as described in the answer to achieve a lower voltage drop. \$\endgroup\$ Aug 30, 2013 at 13:23
  • \$\begingroup\$ It's good that you added that to the answer. Now, say that AB amps vary widely in efficiency depending on the signal being amplified, but is typically in the 50 to 80% range. And explain that dividing the power rating by 0.75 is to compensate for the 75% efficiency and the real number should be substituted in there. And finally, you might put in something about the power supply rejection ratio (PSRR) of the amp. Do that and I'll change my -1 into a +1. \$\endgroup\$
    – user3624
    Aug 30, 2013 at 14:09
  • \$\begingroup\$ This article: en.wikipedia.org/wiki/Amplifier#Class_B indicates the theoretical maximum efficiency of an AB push-pull amplifier is below the 78.5% theoretical maximum of a class B amplifier. While your points are well taken, please don't forget the OP is looking for a rough-and-ready set of calculations. I have provided that. \$\endgroup\$ Aug 30, 2013 at 14:29
  • \$\begingroup\$ For a full-scale square wave the efficiency is much higher than 78%. You also have to remember that you are writing an answer for more than just the OP. If your answer appealed to a wider audience then you'll get more upvotes. \$\endgroup\$
    – user3624
    Aug 30, 2013 at 14:46

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