Potentiometer w/ Capacitor connected to the middle

Thread Starter

*Neon23

Joined Mar 19, 2018
5
I'm currently studying about transistors and I would like some guidance on this part
where an amplifier has this kind of circuit



For the DC Equivalent Circuit
The potentiometer would be a fixed resistor and the capacitor would be open and would result to a circuit of





On an AC Equivalent Circuit

the capacitor would be shorted. However I can't seem to understand fully how would the potentiometer act on this circuit.

From what I assume, it would act just like the DC equivalent circuit that is shown previously. Or it would split into two variable resistors shown below
upload_2018-3-19_18-36-4.png
 

DickCappels

Joined Aug 21, 2008
10,179
That sounds like a reasonable assumption. Splitting the resistor into top and bottom parts is a nice way to make the circuit easier to analyze.
 

WBahn

Joined Mar 31, 2012
30,058
From what I assume, it would act just like the DC equivalent circuit that is shown previously. Or it would split into two variable resistors shown below
It would behave like the split circuit, not the DC circuit. As the frequency goes up, the impedance of the capacitor goes down and it progressively "shorts out" that portion of the potentiometer that it is in parallel with.
 
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