(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!

  • \$\begingroup\$ What does "[4,6]" mean? Ditto elsewhere. \$\endgroup\$ – Andy aka Jun 12 '17 at 13:41
  • \$\begingroup\$ sorry it means any value between 4 and 6 \$\endgroup\$ – Hiiii Jun 12 '17 at 13:42
  • \$\begingroup\$ The say that and don't use obscure references. Your friend is wrong BTW. \$\endgroup\$ – Andy aka Jun 12 '17 at 13:44
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    \$\begingroup\$ There is a DC path around the cap its going to discharge. \$\endgroup\$ – sstobbe Jun 12 '17 at 13:55
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    \$\begingroup\$ 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. \$\endgroup\$ – 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.

enter image description here enter image description here

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

enter image description here

  • \$\begingroup\$ May you comment what is represented by two upper graphs? \$\endgroup\$ – MaxMil Dec 18 '17 at 9:07

You can replace the sinewave with the square wave and now the situation will look like this:

enter image description here

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


In real amp life the situation will look like this:

enter image description here

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).

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

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