# Help understanging ac signal connected to a charged capacitor (coupling capacitor)

(Assuming ideal elements) My friend is trying to convince me that the voltage across $R_L$ varies between 4 and 6, says the average value doesn't decrease and stays fixed at $5V$ forever. I'm not getting how this is possible - simulate this circuit – Schematic created using CircuitLab

The amplitude of the input ac signal varies between -1 and 1, always less than the capacitor voltage of $5V$. Wouldn't the capacitor see this low voltage and start discharging ? May I ask how the 1V input ac signal can put more charge into the capacitor so that the capacitor's voltage increases, even slighlty ? Makes not much sense to me. Highly appreciate any help. Thanks!

• What does "[4,6]" mean? Ditto elsewhere. – Andy aka Jun 12 '17 at 13:41
• sorry it means any value between 4 and 6 – Hiiii Jun 12 '17 at 13:42
• The say that and don't use obscure references. Your friend is wrong BTW. – Andy aka Jun 12 '17 at 13:44
• There is a DC path around the cap its going to discharge. – sstobbe Jun 12 '17 at 13:55
• Yes if you inserted a precharged cap to 5VDC, RL would see slightly less than 6V to slightly less than 4V. DC would decay as RC time constant to 0VDC. – sstobbe Jun 12 '17 at 14:11

Yes, There is a DC path around the cap its going to discharge. DC response follows RC time-constant.

• Cool plot ! Definitely one picture is better than 1000 words! May I know which tool you're using ? @sstobbe – Hiiii Jun 12 '17 at 14:50
• @Hiiii Its LTSpice linear.com/designtools/software/#LTspice It is Free! – sstobbe Jun 12 '17 at 14:54

How about this, from Signal Wave Explorer: 1MHz DC-block (high pass filter), with 10MHz sin input 1volt peakpeak with 2.5 volt offset • May you comment what is represented by two upper graphs? – MaxMil Dec 18 '17 at 9:07

You can replace the sinewave with the square wave and now the situation will look like this: As you can see the capacitor will have a path for a discharge current to flow. And after some time the capacitor reach steady state and in steady state condition, the average capacitor current is $0A$ over a cycle (I_charge = -I_discharge).
AC Circuit Having Only Capacitor

EDIT

In real amp life the situation will look like this: At DC the Cin capacitor is charged to $2.6V$

For positive half-cycle (+1V peak):

As you can see $5V$ source provide $3.5μA$ of current, but the base current needed to be equal to $9μA$, so this additional current will have to come from the AC signal source.

And the Cin capacitor is discharging (I_discharge = -5.5μA)

As I show in the diagram.

For Negative half-cycle:

This time the base need only $3μA$, but our $5V$ bias source provides $8.5μA$.

This implies that our AC signal source needs to "sink" this excess current $5.5μA$.

And Cin this time is charging (I_charge = +5.5μA)

As you can see the average capacitor current is 0A over a cycle, hence the average (DC) voltage across capacitor stay unchanged (2.6V).

• @Hiiii I update my answer, I hope it will help. – G36 Jun 12 '17 at 17:24