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KCC's quizzes AQQ235 about a bipolar common emitter amplifier - a kind proposal from our colleague Martin Walker

Apologize for our non-technical audience since this quiz is more for our FAEs...

A kind proposal from our colleague Martin Walker, ADI Product Marketing Engineer, UK:

Most of our electronic engineers have seen this sort of circuit in their first year study time using a bipolar transistor in its 3 famous basic configurations: common emitter, common collector and common base.

Here above is a common emitter configuration.


  • Vcc = 6V
  • Vout = 3V
  • Tc = 25°C

Q1 is a BJT NPN with current gain β of several hundreds.

RB1 and RB2 are large compared to RC and RS

 Questions :

  1. Is there enough information to work out the gain of this circuit?
  2. If so, what is the voltage gain of the circuit?
  3. Is it a good amplifier?
  4. If not, what would you do to mitigate for its limitations?

Again, many thanks Martin!

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[edited by: emassa at 2:19 PM (GMT -4) on 22 May 2023]
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  • Thanks Bernhard for your feedback.

    Not sure about your statement on question 1: indeed amplification is somehow influenced by the gain beta of the BJT, but when it is large, the AC voltage gain is less dependent ! It is like the open gain of an opamp: when approaching infinite, the close loop gains become totally indendent. 

  • That circuit / question is a classic New College candidate interview question going back to when I interviewed 46 years ago here in Wilmington. It gets to the heart of if the student was paying attention in Electronics I class. They always get bogged down in large signal vs small signal models and they insist that the resistor values must be known and what's beta (as other respondents here seem to have) etc. Beta (and the base current) has nothing to do with small signal analysis. Yes, the small signal gain varies with DC operating point Ic and of course temperature due to Vt = kT/q.

  • Hi Kuo-Chang, you can write down Av= -Rc / re (with re equal to the input resistance looking into the emitter terminal with base grounded) but you still have a strong dependency of the voltage gain from beta. Starting from the definition of re:= delta Vbe / delta Ib, you can prove re= hfe / hie or beta / hie. So variations of beta reflects on re and on Av. Different situation with a resistor Re added between the emitter terminal and ground. In this case, with high values of beta, you have Av= - Rc / Re and and a relative independency from beta thanks to the negative feedback. 

  • OK, it's time to publish the official (and correct) answer (event though, here some subjectivity can be made! Again, many thanks to our colleague Martin Walker for having proposed this original topic

    1. Although we don’t have any component values, we can actually work out the circuit gain. Effectively, the collector load and biasing network have been set to give a half-rail output (3V/6V).
    2. The gain is -116 or 41dB

    In effect:

    Gain of grounded common emitter is -gmRc

    Transconductance gm=  IC/(kT⁄q)

    RL is set to give VCC / 2 so Rc=((VCC-VOUT))/IC

    So, voltage gain A :


    As = 0.0258 at 25 °C, A=-116 (~41 dB)

    This leads to the rule of thumb that the internal resistance of the emitter re = 25 / IC where IC is in mA

    1. No; the gain is highly temperature dependent because VT (the thermal voltage, kT/q changes with temperature and. if the input signal is large enough, it will substantially modulate the collector current Ic, making the gain non-linear.

     4. Adding some negative feedback, usually achieved by adding an emitter resistor Re (much larger than the internal resistance of the emitter terminal, re) will improve the amplifier. This will reduce the gain to -Rc/Re but it will be more stable with temperature, more linear with applied signal input and will increase the bandwidth of the amplifier.

    For a full explanation, see The Art of Electronics (Third Edition) Chapter 2 – 2.3.4. The Common Emitter Revisited.

     You can even simulate it to see the gain is 41dB in the ac analysis and see the non-linearity in the transient simulation.  Martin had attached the LTspice file but it should show the same results in EESim.

    Big applause to our 4 winners (based also on reasoning and explanation provided):

     ,  ,  and  

    And be ready for the next coming challenge!

  • I am indebted to Paul Brokaw for the question.  Here's his talk on ADITube:

    Sheerluck Ohms and the 33dB Solution

    Hope you had fun with this.


  • Yes, but only for the start after power-up :-)

    Interpreting Vout as output bias voltage, Ic=(Vcc-Vout)/Rc=3V/Rc

    gm=Ic/Ut with Ut=kT/e=25.7mV

    Small signal input resistance is rin=(beta*1/gm)||Rb1||Rb2


    Is it a good amplifier: No, the bias point is not stable. The easiest way would be, to add an emitter resistor with a value 5 to 10 * 1/gm

  • Right answers Robert! Congrats!