I'm trying to understand the operation of a class A common-emitter amplifier, but I'm struggling to understand how the equivalent resistance for the cutoff frequency of C1 is determined. I also can't see how the equivalent circuit at the bottom left is correct.

Since the circuit splits in 3 after C1, it makes sense that there would be three parallel resistances. My first instinct would be to follow each of these to ground, and sum resistance along the way. Which would result in R1 | R2+RC+RE | RE, which is clearly wrong. I also thought to sum only the resistors before reaching the NPN, which left me with R1 | R2+RC.

So my question is, is there a rule for which resistors to include in the equivalent resistance of this circuit?

  • \$\begingroup\$ Have a look here: electronics-tutorials.ws/amplifier/… \$\endgroup\$ – Oldfart Oct 20 '19 at 6:43
  • \$\begingroup\$ @Oldfart I wish that web page was accurate in computing Zin. But it isn't. At least, not for the schematic they show. \$\endgroup\$ – jonk Oct 20 '19 at 9:41

enter image description here

How about now?

Assuming hFE Re*Rc >= R12C1 for same break point, let Re =0 for AC

  • \$\begingroup\$ Thanks! So CE provides an AC ground which makes RE effectively 0. I'm still a little unsure about the R2 path, though. Is there an AC short circuit from R2 to ground? \$\endgroup\$ – user10257574 Oct 20 '19 at 6:48
  • \$\begingroup\$ A supply rail is assumed to be like an ideal voltage source: infinitely strong. So it acts just like a ground when looking at the impedance. (Or think of it this way: any current going into it will go through the supply to ground) \$\endgroup\$ – Oldfart Oct 20 '19 at 7:51
  • \$\begingroup\$ for AC , ideal is all rails are 0V \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 20 '19 at 7:53

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