# How can I deduce the power of a capacitor from its datasheet?

I am looking for suitable capacitors for a sound amplifier I have designed. The speaker power should be at least 7 watt (for the sound to be loud enough). (I attach a figure of the design)

So, I think I need capacitors which are suitable for these level of power.

But, in the datasheets, i don't see any specification for the power capability of the capacitors..

Any idea how can I know I have chosen capacitors with high enough power resum capabilities?

Thanks!

• The caps doesn't have power spec. They have capacitnce and ESR, and the voltage. Feb 24, 2019 at 10:09
• Good question. What about maximum current rating? Maybe power isn't an issue, but at the least the terminals must be rated, eg amp caps have huge screw types
– CL22
Feb 24, 2019 at 13:37

Capacitors don't have power ratings because, ideally, they don't dissipate any power. They store energy unlike resistors which consume energy, giving it off as heat.

Instead, you need to consider the following:

• The voltage rating needs to be at least that of the maximum voltage they will see in service.
• For power regulation and loudspeaker connection electrolytics are suitable. Observe polarity.
• The impedance of the loudspeaker decoupling capacitor needs to be low in relation to the speaker impedance. You can calculate the impedance at any frequency from the formula $$\ Z = \frac {1}{2 \pi f C} \$$ where Z is the impedance (ohms), f the frequency (hertz) and C the capacitor value (farads). Choose this for a reasonable bass frequency cut-off point. (Remember that frequencies below this will fall off gradually rather than a sharp cut-off.)
• Up voted this answer. @OP, On a side note, capacitors do have an internal resistance and therefore can dissipate power, but for practical reasons that is not an issue for the application described here (specifically, typical internal resistances are extremely small). Feb 24, 2019 at 10:27
• @Digiproc Even given that, I believe that, generally, if you keep to the voltage rating of the capacitor you will not run into power dissipation issues arising from ESR.
– J...
Feb 24, 2019 at 21:09
• @J...: I suspect that if you run some calculations that you will find that frequency is the problem. ESR won't matter on a capacitor holding steady charge and may not matter much at mains frequencies but at SMPS frequencies it becomes a bigger problem. Feb 24, 2019 at 21:20

Regarding energy storage for the rail, consider that 1 farad when discharged by 1 ampere will sag at 1 volt/second.

Let's assume you have 50 Hz power. Thus you can use a full-wave rectifier and have 100 recharge opportunities per second. Thus your TIME will be 0.01 seconds.

Assume you will accept 0.1 volt sag on the VDD. How big must your capacitor be?

dV/dT = I/C, derived from the derivative of Q = C * V, with C held constant.

Rearrange this, and C = I / (dV/dT) = I * T/ V

In the above case, C = 1 ampere (assumed) * 0.01 second / 0.1 volt = 0.1 farad or

100,000 µF.

================ example of selecting VDD filter for OpAmp =======

Now lets apply what we've learned to a precision opamp circuit, with DC to 100MHz signal (perhaps radar modulation) and we need 12 bit performance (or settling) despite the OpAmp having ZERO dB Power Supply Rejection at high frequencies. Can we design the VDD filtering?

Assume the Opamp must drive 15 picoFarad load at +- 2 volts (4 volts PP) as well as a 250 Ohm feedback resistor.

Summary: we need to provide the surge currents, occurring every 10 nanoSeconds and the VDD capacitor must settle very quickly or NOT SAG more than 1/2^12 or more than 1 Vquanta at the VDD pin.

Assume Vquanta is 1milliVolt (4vpp/4096).

Assume the surge current is I = C * dV/dT + Vout/Rload * Vout/Rfeedback

Isurge = 15pF * 2vPeak * 630Million volts/second + 0 + 2vpeak/250 ohm

Isurge = 15pF * 1.2GigaVolt/sec + 8mA = 18mA + 8mA = 26mA

What size cap (before we worry about VDD ringing, and settling)?

C = I * T/V = 26mA * 10nanoSec / 0.001 = 26mA * 10,000 nano = 260,000 picoF

C = 0.26uF

OK. Now suppose we don't want that cap (plus all the inductance of OpAmp package + PCB traces + PCBvias + 0.26uF inductance) to ring more than 1milliVolt? Can we dampen it? Assume 10nH inductance (to emphasize the seriousness of the challenge).

Fring of 1 uF (round up, to handle temperature deltaCapacitance effects) and 10 nanoHenry? Fring = 0.16 / sqrt(1uF * 10nH) = 0.16 / 1.0e-7 = 1.6MHz.

How to dampen? use resistor of value sqrt(L / C) = sqrt(10nH/ 1uF) = sqrt(0.01) or 0.1 Ohm. Which is an awkward value to find.