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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?

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  • \$\begingroup\$ Have a look here: electronics-tutorials.ws/amplifier/… \$\endgroup\$
    – Oldfart
    Commented Oct 20, 2019 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
    Commented Oct 20, 2019 at 9:41

2 Answers 2

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There are 3 independant high pass filters in your circuit formed by the 3 capacitors acting with their associated resistances.

The cut off frequency of each filter is:-

fc = 1/(2 * pi * R * C) where R is each filter's associated resistance.

For the output coupling capacitor C2, R = (Rc + RL)

For the input coupling capacitor C1, R = Rs + Rin where Rs is the signal source resistance which could be the output resistance of a previous stage and Rin = R1//R2//(hFE*re) where re = 25 mV/Ic

If we assume Rs = 0 ohms then the input filter's associated resistance is just R1//R2//(hFE*re) but this assumes that the emitter bypass capacitor is large enough to make the impedance of the parallel combination of CE and RE very small at frequency fc.

For the emitter by-pass capacitor, R = RE//(re+((R1//R2//Rs)/hFE))

If as before, we assume that Rs = 0 ohms then the filter's associated resistance, R becomes just RE//re where re = 25 mV/Ic

That isn't the whole story though because those 3 high pass filter are effectively cascaded. If all 3 filters had the same cut-off frequency fc (-3 dB frequency) then the overall frequency response of the amplifier would be 9 dB down at fc.

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  • \$\begingroup\$ When your amplifier contains three independent high pass effects there is a simple method to find a good approximation for the resulting (combined) cut-off frequency fo based on the three cut-offs for each filter: fo=SQRT(fo1²+fo2²+fo3²). Each cut-off frequency fo,n is the inverse of the corresponding time constant T: fo,n=1/Tn \$\endgroup\$
    – LvW
    Commented Dec 23, 2022 at 9:04
  • \$\begingroup\$ Comment: The maximum error of this square root approximation is app. 6% for three equal cut-offs. This error becomes smaller for unequal cut-off frequencies. As we can see, the largest value of fo,n willl dominate. \$\endgroup\$
    – LvW
    Commented Dec 23, 2022 at 10:13
  • \$\begingroup\$ @LvW Thanks for the info, very interesting. I wasn't aware of that formula. I was just about to get back to you and question the formula after running some simulations and finding that for three independent equal cut-off filters the overall -3 dB frequency is exactly twice the cut-off frequency of each individual filter where as your formula gives an overall -3 dB frequency at about 1.7X the individual cut-offs. I make that an error of about 13% (for an individual cut-off of 1 kHz the formula gives 1732 Hz compared to an actual overall of 2 kHz which is an error of (2000-1732)/2000 = 13.4%) \$\endgroup\$
    – user173271
    Commented Dec 23, 2022 at 10:46
  • \$\begingroup\$ Thanks for your response. Yes - you are right, there was an error in my second comment: For TWO equal decoupled first-order circuits the max error of the SQRT-approximation is app. 12.5%. For three equal cut-offs the error is even somewhat larger. \$\endgroup\$
    – LvW
    Commented Dec 23, 2022 at 11:39
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enter image description here

How about now?

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

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  • \$\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
    Commented Oct 20, 2019 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
    Commented Oct 20, 2019 at 7:51
  • \$\begingroup\$ for AC , ideal is all rails are 0V \$\endgroup\$
    – D.A.S.
    Commented Oct 20, 2019 at 7:53

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