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I thought that capacitors will only act as a conductor while is is being charged and once fully charged it will not conduct any more. I know that current does not actually pass through the capacitor due to the plate between the insulator between the two plates but I use the word 'conduct' here for simplicity.

At low frequencies and DC currents the capacitor gets fully charged very early at the beginning of the cycle and acting afterward as an open circuit until the voltage drops in the second half cycle below the capacitor voltage when the capacitor starts to discharge. As a result the capacitor in the HP filter will will block low frequencies.

Also at high frequencies there is less electricity in the cycle so the capacitor does not get fully charged and it is always either charging and conducting or discharging so the current is not blocked at all and the capacitor is acting as a short circuit. As a result in HP filter high frequencies will pass.

Now if the capacitor receives two signals at the same time, the first has very low frequency and the other has very high frequency, the first will cause the capacitor to be fully charged so not allowing any more current to pass.

How the high frequency signal will pass now?

Will it actually pass?

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2 Answers 2

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And at low frequencies and DC currents the cappacitor gets fully charged very early at the beginning of the cycle and acting afterward as an open circuit until the voltage drops in the second half cycle below the capacitor voltage when the capacitor starts to discharge.

No, you have been given a completely unhelpful picture of how a capacitor behaves.

If you apply DC to a capacitor, then it takes an initial charging current while its voltage is changing. Once the capacitor has charged to the supply voltage, no further current is drawn. Both of these states are summarised by the equation I = C*dV/dt, the capacitor draws a current proportional to its capacitance, and the rate of change of voltage across it. Initially, there's a large dV/dt, so a high current. At the end dV/dt is zero, so the current is zero.

Exactly the same equation can be used to predict its behaviour with AC.

High frequency AC has a large dV/dt, so the capacitor draws a large current, and appears as a relatively low impedance. Low frequency AC has a lower dV/dt, so the capacitor behaves as a higher impedance.

In a linear circuit, voltages and currents can be superposed without interference. This means you can analyse the response of the filter components separately to any given input signal. If you supply all of those signals at the same time, its response will just be the sum of those responses. In a highpass filter with a series capacitor, high frequency signals will find a low series impedance and be coupled to the output well, low frequency signals will find a high impedance and be poorly coupled to the output.

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  • \$\begingroup\$ Your explanation is way better then mine, let me tidy up my answer. \$\endgroup\$ Apr 21, 2021 at 12:54
  • \$\begingroup\$ this changed my thoughts about capacitors ! another question : if the capacitor will conduct as far as there is change in voltage and and the resultant current will depend upon the rate of change, then in low frequency is the capacitor conduct until the voltage begins to drop but the current will decrease along with voltage rate decrease ? \$\endgroup\$
    – dev65
    Apr 21, 2021 at 13:27
  • \$\begingroup\$ @dev65 is the capacitor conduct until the voltage begins to drop- no, the capacitor does not 'conduct until', it simply draws current when the voltage across it is changing. Conversely, it's the current flowing at its terminals that changes its voltage. the current will decrease along with voltage rate decrease, not sure what you're trying to say here - certainly the lower the rate of change of voltage, the lower the current. \$\endgroup\$
    – Neil_UK
    Apr 21, 2021 at 14:11
  • \$\begingroup\$ @Neil_UK I mean in low frequency the capacitor will remain active for the first half of the first positive cycle but it may sound as a short circuit since the change in voltage will be very small \$\endgroup\$
    – dev65
    Apr 21, 2021 at 14:54
  • \$\begingroup\$ @dev65 No, the capacitor is not 'active' for some of the cycle, it is doing its 'capacitor thing' for 100% of every cycle. Its 'capacitor thing' is drawing current proportional to the rate of change of voltage across it. It does not become a short-circuit at any time. However, if you connect an uncharged capacitor to a voltage source, it will draw a very large current while it charges up to the voltage of the source, because you are impressing a very large dV/dt on it. \$\endgroup\$
    – Neil_UK
    Apr 21, 2021 at 17:00
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TLDR: a capacitor is a linear device. If you apply the sum of two signals (current or voltage) to it, its response (voltage or current) will be the sum of what the responses for the respective signals on their own would have been.

So any DC charge on the capacitor will not make a difference to how an AC signal gets through: the DC charge just acts as an additional voltage source in the circuit.

You can consider a capacitor as an ideal voltage source at any given point of time. The difference to an ideal voltage source is that the voltage across a capacitor changes over time when there is a current across the capacitor: in that case the voltage rises at a rate inversely proportional to the capacitor's capacitance.

At high frequencies, the current only has very little time to effect a change in the voltage across the capacitor, so the capacitor voltage changes rather little, making it pass AC voltages with comparatively little change.

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  • \$\begingroup\$ Actually it depends on capacitor. Capacitors are not ideal and certain types of chip capacitors can lose capacitance over DC bias. \$\endgroup\$
    – Justme
    Oct 8, 2023 at 22:17
  • \$\begingroup\$ @Justme Sure: ceramic capacitors in particular tend to saturate pretty badly, taking rather little charge at the end of their voltage range. A better controlled gradation is available with capacitance diodes. But those are deviations from the idealised device, llike leakage and resistance are. \$\endgroup\$
    – user107063
    Oct 8, 2023 at 22:25

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