I am studying analogue electronics. In specific BJT transistors. There is this question that kind of confuses me:


Somehow the first question managed to confuse me really easily, and I am still stuck on that one. For the second request I calculated \$I_B\$ as follows: $$I_B = \frac{I_C}{\beta} = \frac{20mA}{200} = 100\mu A$$ Then considering of wanting \$V_{CE} = 5V\$. I calculated as follows: $$R_B = \frac{V_{CC}}{I_B} = \frac{20V}{100\mu A} = 200k\Omega$$ $$R_C = \frac{V_{CE}}{I_C} = \frac{5V}{20mA} = 250\Omega$$ As maximum input signal I suppose it is the maximum value that will eventually saturate \$V_{CC}\$, but at the moment I am confused about this as well. I suppose this is not a good circuit, as well, if it is not, why?

Thank you in advance for any clarifications given!


  • \$\begingroup\$ It's not very clear just what you're asking. Your analysis of the base current looks alright, but I'd recommend using something more like 10V for the resting voltage, as it gives the most headroom for your signal. \$\endgroup\$
    – C_Elegans
    Commented Apr 1, 2018 at 16:33
  • \$\begingroup\$ I'm basically asking to possibly helping me giving an answer to the questions and an eventual explanation of the answers... \$\endgroup\$
    – Sevenarth
    Commented Apr 1, 2018 at 16:35
  • \$\begingroup\$ @C_Elegans what do you mean by giving the most headroom? This is basically new stuff for me, so I am completely new to transistors \$\endgroup\$
    – Sevenarth
    Commented Apr 1, 2018 at 16:35

1 Answer 1


What would be an appropriate value for \$V_{CE}\$ in the absence of a signal?

The ideal value of \$V_{CE}\$ is one that gives the maximum amount of room for the output signal to swing. So for instance, if \$V_{CE}= 18V\$, then the voltage can only go up by 2V before it reaches the supply rail and begins to clip. A similar thing happens if \$V_{CE}=2V\$, except that the voltage now can only go down by \$2V-V_{CEsat} \approx 1.5V\$. The ideal voltage for \$V_{CE}\$ is halfway between the positive and negative supplies, as that gives the maximum amount of room for the amplified signal to be.

Calculate the base current \$I_B\$ and the base resistor \$R_B\$

Your calculation for \$I_B\$ looks good, but your calculation for \$R_B\$ could add the a diode drop to the base voltage, though in a real design this wouldn't matter that much.

Calculate the value of \$R_C\$ to give your chosen value of \$V_{CE}\$

This looks fine, as long as you change \$V_{CE}\$ as described above

What is the maximum input signal

The problem doesn't give any description of the input source, which is rather important if you want to use a constant diode drop model for \$V_{BE}\$. Assuming an answer in amps is what is wanted, calculate the base current needed to make\$V_{CE} = V_{CEsat} \approx 0.3V\$.

Is this a good amplifier?

No. Realistically, you need emitter degeneration on a practical amplifier, as it makes the gain of the transistor depend more on \$\frac{R_C}{R_E}\$ and less on \$\beta\$. Additionally, you want a voltage divider type biasing network instead of the single resistor, again so it is easier to bias with a variable \$\beta\$. The stage will work fine in the ideal case presented here, but with the variations in beta among transistors in the real world, the circuit becomes much less practical.


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.