I've been trying to cascade BJT amplifier to get a gain of 800. My input is 10 mV, and I want an output of 8 V. My question is, how does the gain of common emitter amplifiers work? I've created a BJT that gives a gain (Av) of 7, so the output is 70 mV. When I cascaded it to another emitter amplifier that was built the EXACT same as the previous, the output it gave was 2 V, meaning a gain of 200. I couldn't find any calculations or logical reasons why it does this. If anyone knows of a formula or something that explains this it would be much appreciated! Even a short explanation will do. This is my circuit. The first stage is the buffer stage with a gain of around 0.8. It thus gives a 10*0.8 = 8 mV peak-to-peak output.

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Adding the common emitter amplifier stage, the output is 600 mV peak-to-peak as seen below:

enter image description here

I cannot seem to apply the formula below to get even close to similar results for the output at the second stage. This is the formula I've learned and tried to apply but it doesn't work for me or explain the 8 mV to 600 mV gain.

enter image description here

  • 2
    \$\begingroup\$ Can you show us the circuit diagram with the component values? \$\endgroup\$
    – G36
    Nov 6, 2021 at 7:31
  • \$\begingroup\$ BJT's have a current gain, not a voltage gain. The voltage gain depends on the choice of the load resistors as well. \$\endgroup\$
    – Janka
    Nov 6, 2021 at 7:47
  • \$\begingroup\$ I would love to give a simulated circuit now, but my simulation software is working against me. Will post as soon as I come right. \$\endgroup\$
    – Bottle
    Nov 6, 2021 at 7:49
  • \$\begingroup\$ I've updated the question, the circuit doesn't represent the example I gave but I think if I can understand this I should be good with cascading BJT's. Thanks again in advance! In this case, the Beta is 200 for both transistors. \$\endgroup\$
    – Bottle
    Nov 6, 2021 at 8:06
  • \$\begingroup\$ Your schematic does not cascade two common emitter amplifiers and therefore it contradicts your text. Please fix this up. \$\endgroup\$
    – Andy aka
    Nov 6, 2021 at 8:45

1 Answer 1


Gain of second stage is \$R_{C2}/(R_{E2}+r_e)\$ where \$r_e = 25\text{ mV}/I_c\$.

So, gain of second stage = approx 75

The overall gain of the whole circuit is then reduced below a value of 75 because you have a fairly large source resistance (10k) forming a potential divider with the input resistance of the rest of the circuit.

The drop in gain due to the source resistance acting in conjunction with the input resistance of the rest of the circuit can be calculated from the common potential divider equation which is $$R_{\text{in}}/(R_s+R_{\text{in}})$$

The result of that would then be multiplied by the other cascaded gains of the circuit to calculate the overall gain.

So, to use that equation, you need to calculate the input resistance of the circuit.

To calculate the input resistance of the circuit start at the output and work backwards towards the input.

Firstly, calculate the resistance seen looking into the base of Q2 which is $$(R_{\text{E2}}+r_e) \times (\beta+1)$$

The resistance seen looking into the base of Q2 is seen from the base of Q1 as being in parallel with RE1, R3 and R4 and then multiplied by \$\beta+1\$.

So, the resistance seen looking into the base of Q1 is:

$$((R_{E2}+r_e)(\beta+1) \parallel R_{E1} \parallel R_3 \parallel R_4)(\beta+1)$$

This result is then in parallel with R1 and R2. Once that result is calculated you should have a value for Rin which can be used in the first equation I gave, the potential divider equation, to calculate the gain reduction factor.

Using a value of 100 for beta and calculating re as 7.7 ohms I get a value for Rin of 36137 ohms giving a gain reduction factor of 0.78

75 * 0.78 is equal to approx 59.

See if you get the same!

  • \$\begingroup\$ Thank you, is there any formula you know of that includes the source resistance for the output? \$\endgroup\$
    – Bottle
    Nov 6, 2021 at 11:11
  • \$\begingroup\$ @Bottle See edit \$\endgroup\$
    – user173271
    Nov 6, 2021 at 11:47
  • \$\begingroup\$ Thank you very very much! I got the same answer yes. You've saved me a lot of work trying to understand all this work and I appreciate it a lot! My final question is, if I were to add another transistor containing R5, R6, RE3, would the equation then be: ((RE3+re)(beta+1)//(RE2//R5//R6//RE1//R3//R4))(beta+1)? Just wanna make sure I completely understand. \$\endgroup\$
    – Bottle
    Nov 6, 2021 at 12:48
  • \$\begingroup\$ @bottle No, if you were to add a second common emitter amplifier after the second stage, biased by R5 & R6 with an emitter resistor RE3 then the resistance seen looking into the base of Q2 would be unchanged. Instead you would get a gain increase due to the gain of the added stage but a new potential divider has now been formed between the output resistance of the 2nd stage, which is equal to RC2, and the input resistance of the added third stage. So you would now have 2 more factors to multiply by to calculate the overall gain. \$\endgroup\$
    – user173271
    Nov 6, 2021 at 13:26
  • \$\begingroup\$ Ok, I think I understand, just not sure yet, so the resistance seen looking in to the base of Q2 stays the same (RE2+re)(beta+1). And the resistance looking into the base of Q3 is then (RE3+re)(beta+1). So the resistance becomes: ((RE3+re)(beta+1)//(RE2+re)(beta+1)//RE2//R5//R6//RE1//R3//R4) = G. and the gain would then be: G/(G+RC2)? Thanks for the help so far and sorry for bothering so much. \$\endgroup\$
    – Bottle
    Nov 6, 2021 at 14:12

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