Okay so from my understanding and looking at the equation for the impedance a capacitor has, the high frequency signals would be relatively easy to be sunk to ground (if the capacitor was grounded) and hence the decoupling caps near ICs. So (I think) from that perspective I'm okay. But it's the big capacitors, say for example on some ESCs, power supplies or super caps for car subwoofers that I don't understand.

From my understanding, the discharge and charge rate are both equivalent assuming the resistance on the charge side is the same as the discharge side. People I ask say it's because it supplies charge when there's like a surge of power needed and eases load from the supply, but surely when there's a say high demand (say bass from the sub), then the second the bass has done, the capacitor would need to equally recharge the same amount (and the same current), hence I can't see what benefit this has to the supply. Even then, my mind would be telling me it's also trying to charge the cap up equally during a power requirement to keep the voltages the same as its parallel to the load (the cap).

I suppose on the case for the car, there would be long wire from the battery to the rear sub and inductance, so it would be advantageous since the charge rate would be slower than the discharge rate if the cap was right next to the speaker, but there's many times when I see the bigger caps on circuitry where I doubt it's the case.

Hope my question makes sense,

  • 2
    \$\begingroup\$ The purpose of a bigger capacitor is ultimately to store a larger amount of charge than a smaller capacitor. If you understand why a decoupling capacitor would be useful then why do you doubt that a bigger capacitor would also be useful if you needed a larger amount of charge, for example for a lower frequency ripple? \$\endgroup\$ Commented Dec 3, 2023 at 19:58
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    \$\begingroup\$ Sorry, I know it's off-topic, but: can you edit your question, writing in proper English? I literally got tired while reading through your question, due to misspellings, miscapitalizations, excessive use of "like" and "say", improper use or lack of ', etc \$\endgroup\$
    – swineone
    Commented Dec 4, 2023 at 5:41
  • \$\begingroup\$ @swineone I appreciate the attempt to read my question, but if you literally got tired if reading simply don’t reply. Why does ‘proper’ English have to exclude regional chances in text? I can understand if it becomes hard to read but many people have been super helpful below. If you want to downvote or get this question cancelled whilst im trying to learn because my English is not very good, then proceed \$\endgroup\$ Commented Dec 4, 2023 at 10:05
  • \$\begingroup\$ @user1850479 hi I understand what your saying, im essentially saying I understand i think mathematically why, just not fully intuitively \$\endgroup\$ Commented Dec 4, 2023 at 10:06
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    \$\begingroup\$ @Georgekirby I will not stretch this discussion beyond what's needed, so consider this my latest reply. You are coming here to ask for help, and you should extend a minimum amount of courtesy to the people who might answer it, which is to make it readable (referring just to use of proper English syntax). A tip: if you copy the contents of your draft question and paste it to ChatGPT with a prompt "Correct English mistakes in the following text:", it'll output a much improved version of what you wrote. \$\endgroup\$
    – swineone
    Commented Dec 4, 2023 at 12:17

4 Answers 4


How to explain concepts

The OP's question is conceptual and requires a conceptual answer. But how do we explain concepts?

Concepts are basic ideas that are generally non-electrical. Therefore, the first means we can use associations with well-known life phenomena (analogies).

The next means by which we can make the transition to specific electrical circuits are equivalent electrical circuits. They are cleared of details that only make it difficult to understand the idea.

"Dynamic self-buffering"

When something (such as money, food, water, etc.) is lacking at some point in life, we need help. It can be external (we find a sponsor) but we can also help ourselves (as we have saved it in advance). In the first case we consume else's funds and in the second our own.

In electronic circuits this is called buffering. Because the circuit buffers itself, and only for a short time, hence this figurative "dynamic self-buffering" name (fabricated by me:-)

"Dynamic self-buffered voltage divider"

The OP's configuration can be thought as a voltage divider where the wire and source resistance form the upper resistor R1, and the load is the bottom resistor R2. Also, a big (decoupling) capacitor is connected in parallel to the divider output (R2). This technique is widely used in circuits.

To understand at the lowest level how the concept is implemented, let's build the circuit and explore its operation by the help of the CircuitLab DC Live Simulation in a few steps. To do this, we must "get rid of the tyranny of time" by replacing (emulating) the charged capacitor with a voltage source.

Imperfect voltage source loaded

R2 = 1 kΩ: Initially imagine a 1 kΩ constant load resistance R2 so the output voltage is half the input voltage.


simulate this circuit – Schematic created using CircuitLab

R2 decreases twice: If, for example, we connect another 1 kΩ resistor (load) in parallel to R2, Vout decreases.


simulate this circuit

R2 increases twice: Conversely, if we connect the 1 kΩ resistor in series to R2, Vout increases.


simulate this circuit

Imperfect voltage source buffered

Slowly changing load: Now let's connect a big "capacitor" (a behavioral voltage source C copying the voltage drop across R2) in parallel to the divider output (R2). Also, connect an ammeter to monitor the current I through the "capacitor". The ammeter has to have some minimum resistance (1 Ω set in its parameters window) to decouple Vc from VR2; otherwise a vicious circle results (Vc is a copy of itself) and CircuitLab reports an error.

If R2 slowly decreases, the "capacitor" slowly discharges, and Vc follows Vout. No current flows through C, and the load is entirely supplied by Vin. This technique is known as "bootstrapping" and is used to virtually increase resistance.


simulate this circuit

If R2 slowly increases, the "capacitor" slowly charges, and Vc follows Vout. No current flows through C again, and the load is entirely supplied by Vin.


simulate this circuit

Rapidly changing load: To imitate it, we have to replace the following behavioral voltage source with a constant voltage source, and repeat the experiments above.

If R2 rapidly decreases, the "capacitor" does not change its voltage and supplies the load (in cooperation with Vin).


simulate this circuit

If R2 rapidly increases, the "capacitor" does not change its voltage as above and sinks the load current (like a Zener diode).


simulate this circuit

Practical circuits

Let's finally investigate two practical circuits of imperfect 10 V voltage sources with 10 Ω internal resistance R (including the wire resistance) - first without, and then with backup capacitor C. They are loaded by a 100 mA pulse current source IL that "pulls down" for one second the output at the 50th and 100th second (it is implemented as a programmable so-called CSV current source from the CircuitLab library).

Unbuffered imperfect voltage source

First let's disconnect the capacitor. The result is obvious...


simulate this circuit

... the output voltage drops to ground...

STEP 4.1a

... when the load current "jumps" to 100 mA.

STEP 4.1b

Buffered imperfect voltage source

Then let's restore the capacitor.


simulate this circuit

Now the output voltage slightly drops...

STEP 4.2a

when the load current "jumps" to 100 mA.

STEP 4.2b


  • A capacitor connected in parallel to the load acts as a "backup battery" that can dump short load changes.

  • The larger the capacitance, the longer the load voltage drops can be, but the time to recover the charge (the time between two adjacent drops) will be longer.


For your bass example, we can devise a following scenario.

If you have a 1A power supply, it can't provide a peak current of 2A which an amplifier may need under heavy transient.

But if you put a capacitor in between the power supply and amplifier, the power supply can charge the capacitance at 1A when there is no peak demand, and during 2A peak demand, the capacitor can provide the missing 1A while discharging.

Of course real world scenarios are far more complex than that.

Wires have resistance which limits how much peak current is available, and wires have also inductance which limits how quickly the peak current is available.

So larger caps are just for storing and releasing energy at low frequencies, and small caps that can't store much capacitance, can work with higher frequencies, able to release the energy faster than the slow larger caps.

  • 1
    \$\begingroup\$ Thankyou for the response, so I guess it would be only useful if the (say amplifier here) had quick demands, if it needed constant 2A it would be pointless? \$\endgroup\$ Commented Dec 4, 2023 at 10:10
  • 1
    \$\begingroup\$ @Georgekirby: Correct. The power supply has to supply the average demand. The capacitor temporarily supplies the peak demand. \$\endgroup\$
    – MSalters
    Commented Dec 4, 2023 at 11:36

One of the uses of capacitors is short term energy storage. For example single phase AC power goes away 100-120 times per second, so in order to make constant DC energy needs to be stored on the high parts of the voltage wave, and consumed on the low parts.

The big caps in older non-switching power supplies are often called "smoothing capacitors" but what they actually do is charge up and store energy at the peak of the wave. During the rest of the mains period, the rectifiers are off and the device is powered just by the caps. Even in a more modern switching supply using PFC, power from the mains is pulsed, and power output is DC, so there is still a need to store and release energy. For a solar inverter it's the same in reverse: constant DC power in, pulsed AC power out.

In the energy storage use case, the impedance of the cap dropping with frequency isn't actually useful. I mean it would be better if impedance was flat at the low end, but then it would be a battery.

Now the huge cap for car audio doesn't just do energy storage, it's also a decoupling cap. Consider a power supply or a battery on the left, it has impedance, voltage will sag under load and it doesn't react instantly. Now add more resistance and inductance with wires, and put a load at the end.

enter image description here

Whether it's an amp in a car, or a chip on a motherboard, it's the same. In this case the cap will do some energy storage, but it has two other important parts to play:

  • Lower the supply impedance seen by the load, especially at high frequency when the wiring inductance becomes a problem.

This keeps voltage at the load stable and avoids voltage drops which may cause the load to malfunction or shut down.

Supply impedance seen by the load is the parallel combination of the local cap and the sum of wiring plus power supply impedance. The former drops at high frequency, the latter rises at high frequency.

  • Act as a current divider

Because the cap is in parallel with the supply+wires, load current will be shared between the two in inverse proportion of their impedance. This means high frequency current will mostly go through the cap.

In other words the supply current through the wires will be a lowpass filtered, averaged version of the load current while the cap provides the high frequency part.

So for example if the load is a motor ESC that chops current at high frequency, it's drawing current that looks like a square wave. The cap charges and discharges accordingly, and the power supply/battery only has to provide the average current.

The cap's value can be huge or small, depending on what frequency you want the crossover between cap and power supply to be. Long wires mean lower frequency means a bigger cap.


In addition to the responses, I would like to mention something, perhaps trivial but not obvious:

As mentioned:

  1. You have your battery. You have your load (speakers).
  2. Ideally, the speakers consume the energy provided by the battery.
  3. In practice, the battery may not be able to provide the required energy quickly, perhaps for powerful beats
  4. Adding capacitor stores energy when the load demand is low (speaker not needing too much power)

Now, regarding your question:

Yes, you can always find a system which when played very loud, will drain the capacitor and will not give the capacitor time to charge.

This is then a design issue. Either the battery cannot provide enough power, the capacitor is too small, the load requires too much power, or a combination off.

One should take into consideration the load, battery specifications and the capacitor simultaneously when designing the system.

  • \$\begingroup\$ Well explained... \$\endgroup\$ Commented Dec 5, 2023 at 14:58

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