I am new to electronics and recently I was learning about npn transistor as an amplifier. I tried to build one in ltspice XVII.

This is my circuit diagram:

enter image description here

but the output was not as expected as shown below:

enter image description here

But in a video on youtube, the person has a resistor in series with the bypass capacitor. When I tried to do the same the result was as shown below:

enter image description here

enter image description here

So I just want to know why this worked? Why it was not working without it? Usually, all the circuit diagrams on the internet don't have this resistor.

Sorry for my bad english. :P

  • \$\begingroup\$ The main problem is not just high gain but small input Vpp range with C1 but no R6. Also Load AC current can never be greater than collector DC current, so make R5>=R1 is a must have. and of course make R6<<R3 to control AC gain R1/R6=Aol \$\endgroup\$ Commented Dec 25, 2018 at 16:37
  • \$\begingroup\$ That resistor reduces the gain, by about 4:1. Thus the Vout is smaller, and shows less distortion (you see nearly a pure sinusoid output). Reactance of 100uF at 1KHz is 1.6 ohms; you added 600 ohms; the 1/gm also is important. \$\endgroup\$ Commented Dec 25, 2018 at 20:54

1 Answer 1


Your first circuit does work. It has very high gain, so that the output signal is very large. It's sinusoidal positive peak is clipped off when it tries to swing above the +30V supply, and its negative peak is clipped off when it tries to swing below ground.

If you reduce the input signal amplitude, you should see an output that is a sine wave as you expect. Try V2: SIN(0 0.01 1000)

The emitter resistor R6 in series with C1 reduces circuit voltage gain to about 18. The load resistor R5 reduces gain even further. The gain reduction of R6 in parallel with R3 is very useful for a few reasons:

  • circuit operation is much more linear

  • gain becomes stable, close to the ratio R1/(R6||R3)

  • circuit input resistance becomes high, and linear.

  • 2
    \$\begingroup\$ R3 is significantly lower than R6, which looks quite uncommon for me. The circuit would behave almost the same without C1 and R6, with a gain that is close to the ratio (R1//R5)/(R3//R6), which is approximately 6. \$\endgroup\$ Commented Dec 25, 2018 at 15:35
  • \$\begingroup\$ Absolutely right. Am too used to circuits where R3>R6. Edited with correction. \$\endgroup\$
    – glen_geek
    Commented Dec 25, 2018 at 15:40
  • 2
    \$\begingroup\$ The described effect is simply : negative feedback ! This principle is applied in nearly each amplifier stage (BJT or opamp) because of the several advantages connected with negative feedback. \$\endgroup\$
    – LvW
    Commented Dec 25, 2018 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.