simulate this circuit – Schematic created using CircuitLab

I was watching this youtube video for an explanation of voltage gain.

Just for verification, is the formula for voltage gain also applicable for circuit below or there is some limitation to it? I was using a simulator and found the gain only is 140. For an input signal of 10mv(peak), the output was only 1.4V(peak.)

I got the following parameters from the simulator:


The coupling capacitor I calculated as below

$$C=\frac{1}{2*\pi*(R_2||\beta*r_e')}=3.79x10^-6 farad$$

My voltage gain (the node between R1 and collector terminal of Q1) should be \$A=\frac{R1}{r_e}= 476\$. Since the emitter is grounded, it can be treated as if the emitter is connected to a load resistance of 0, just like the equation in the video at 1:31.


  • \$\begingroup\$ I don't know what frequency you are using for your source (with a capacitor for coupling, it must be some AC frequency.) And you don't seem to have accounted for the bulk Ohmic impedance of the emitter for the 2N3904, either, which is likely to be close to your value for \$r_e^{'}\$. And there's more. \$\endgroup\$
    – jonk
    Commented May 26, 2021 at 7:31
  • \$\begingroup\$ The formula you used for the coupling capacitor is independent of frequency. Why is that? \$\endgroup\$
    – user16324
    Commented May 26, 2021 at 14:52

2 Answers 2


I'm seeing a gain of about 330 when I simulate that circuit with the default 2N3904 model at 1 kHz. Are you sure you're not dividing by two somewhere?

This is a bad circuit design. Here are some problems I noticed:

  1. You do not give an AC signal frequency. The frequency is also missing from your capacitor calculation. If the frequency is low enough (e.g. 1 kHz), the capacitor could be reducing your gain. You want the capacitor's impedance to be much lower than \$R_2\$ or \$r_e\$.

  2. Your calculation is heavily dependent on beta, which is not a reliable parameter. For example, the default 2N3904 model specifies a beta of 140. Your \$I_C/I_B\$ is not 140 because the simulator uses a more complex and accurate model of the transistor's behavior.

  3. Your DC biasing circuit is unreliable -- there is no emitter resistor, and the base voltage is set with one resistor instead of a voltage divider. The biasing circuit from the second half of your video is much better.

  4. 25 mV is not a good approximation for \$V_T\$. Use 25.9 mV instead.

Going with your \$I_C\$ and \$V_T = 25.9 \mathrm{mV}\$, I get \$r'_e = 0.547 \Omega\$, which suggests a gain of 457. But again, this is a horrible circuit, so the gain being off by ~30% isn't surprising.

Add a 10-ohm emitter resistor and do the calculations again. You'll find that the gain is much closer to what you'd expect.

Leaving out the emitter resistor makes the gain larger at the cost of being less controlled. Real-world amplifiers like op amps use negative feedback to turn a large but uncontrolled gain into a small but precise gain.

EDIT: You should choose the capacitor so that its 3dB point is much less than the signal frequency. If you use the \$1/2\pi f R_{in}\$ formula at 1 kHz, the capacitor will drop ~30% of the input voltage.

Here's a graph showing the voltage division between an 80-ohm resistor and three different capacitors. At 1 kHz, both the 1.5 uF and 4.0 uF capacitors have significant voltage drop. It's best to pick something larger, like 100 uF. You can do AC frequency analysis in your simulator to see the effect on the gain of the amplifier.

What transistor are you using in your simulation? You could try a few different ones.

The Early effect reduces gain, but I don't think it's by that much.

  • \$\begingroup\$ Correct, I made some mistake in the calculation, recheck the output voltage, it swing from 14.877 to 10.135 volt, so Vpeak will be 4.742/2=2.371 so the gain only 237. By the way, I'm using multisim still deviate from your result of 330. Also the capacitor value i wrongly calculated. According to this website electronics-tutorials.ws/amplifier/… it should be 1.86micro farad at 1k frequency? \$\endgroup\$
    – chuackt
    Commented May 26, 2021 at 11:22
  • \$\begingroup\$ Thanks for the suggestion, I added 10ohm and checked the value, the gain much closer to the theoretical value. \$\endgroup\$
    – chuackt
    Commented May 26, 2021 at 11:34
  • \$\begingroup\$ @chuackt I updated my answer. \$\endgroup\$
    – Adam Haun
    Commented May 26, 2021 at 15:07

Here is my simulation and a few notes:



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