Will capacitor in high pass filter pass high frequencies while it is blocking low frequencies?

Question

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?

Answer 1

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.

Answer 2

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.