This is a question which I had asked earlier.

Tinkering around, I found out that if I connect a capacitor to the load resistor, i.e. if I pass the output through a RC circuit, the voltage retains it original value, and the frequency too, if the capacitance of this capacitor is the same as the previous ones in the circuit. However, the square wave's upper edges are somewhat rounded, and somehow touch the V_init. Can someone give a possible explanation to this, as to how to apply Thevenin's theorem here, if it's actually feasible here, and how does this configuration relate to the initial circuit without a load resistor connected to the output?

  • \$\begingroup\$ It's unclear what load resistor you are talking about and how you might connect the capacitor you talk about. Also, what is V_init??? What does "this configuration" actually mean and what is the initial circuit you mention? \$\endgroup\$
    – Andy aka
    Aug 4, 2021 at 13:00
  • \$\begingroup\$ @Andyaka you can check out my initial question, the link of which has been mentioned above. This is just a follow-up question. \$\endgroup\$ Aug 4, 2021 at 17:12

1 Answer 1


This is a common design fault. Each stage is Common Emitter or active “pull” only , while the “push” is the collector R.

When an AC coupled Load is connected, that load R MUST be greater than Rc to avoid overloading the collector R pullup as although the series C is a HPF, the collector sees it as a partial LPF additional load.

  • \$\begingroup\$ Can you explain it a bit more, I didn't really get you. \$\endgroup\$ Aug 4, 2021 at 10:00
  • \$\begingroup\$ I understand that the output is an oscillating waveform, and I am somewhat using a high pass filter connected to the ground. That would mean it would block the low frequencies, and not the high ones. How does it become a low pass filter though? \$\endgroup\$ Aug 4, 2021 at 10:05
  • \$\begingroup\$ OK, thanks a lot, I've understood now how it works. Just can you tell me, why is it that as I bring down my cutoff frequency lower, the waveform turns into more and more square shaped, instead of being rounded? \$\endgroup\$ Aug 4, 2021 at 10:26
  • \$\begingroup\$ Load voltage regulation error of any peak is just the inverse of impedance for a constant drive current. e.g. for 1% voltage error, source R must be ~1% of load. \$\endgroup\$ Aug 4, 2021 at 10:39
  • \$\begingroup\$ I don't really get you... Also, I was wrong, as the cutoff frequency increases, the voltage error decreases. It does act as a low pass filter, I don't know why. Can you please explain? \$\endgroup\$ Aug 4, 2021 at 12:32

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