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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.

Conditions:

  • 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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Parents
  • 1. Given the conditions, the transistor should be in its linear region. Assuming small signal voltage gain is what we mean by gain "Av= vout / vs", then we need the Rc value to evaluate Av= beta * Rc / hie (hie is the input impedance in the BJT hybrid parameters model)

    2. Av= beta * Rc / hie ( Rs considered negligeable)

    3. Very sensitive to temperature variations

    4. You can add a resistor between emitter and ground (that will introduce a negative feedback for stabilization) and maybe a capacitor in parallel. 

Reply
  • 1. Given the conditions, the transistor should be in its linear region. Assuming small signal voltage gain is what we mean by gain "Av= vout / vs", then we need the Rc value to evaluate Av= beta * Rc / hie (hie is the input impedance in the BJT hybrid parameters model)

    2. Av= beta * Rc / hie ( Rs considered negligeable)

    3. Very sensitive to temperature variations

    4. You can add a resistor between emitter and ground (that will introduce a negative feedback for stabilization) and maybe a capacitor in parallel. 

Children
  • Merci Gaetano! Answers 1, 3 and 4 are fully correct! On question 2, may be if you elaborate further you can come to a gain expression that gives Ac independent of beta (if beta is large, above 80; which is the case with standard BJTs like BC107, BC108 or BC109 having beta of several hundreds. But let's see our coming other feedback!

  • 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.