3.5 Analyzing transistor circuits

Same approach as for diode circuits here. You have to do part of the work.

Transistors - Level 2

Here is the circuit that you will analyze. You have to calculate the currents and voltages for all components.

BJT-ex2

Transistors - Level 3

Now you will analyze a circuit with a somewhat higher complexity.

BJT-ex4

Transistors - Level 4

The circuit below contains two transistors. The emitter current of the first transistor is also the base current of the second one. This configuration is called a Darlington pair, and allows for a higher current gain. At the start of this level you may assume that β is ∞. This means that you can ignore the base current, since IB = IC / β = 0 A. Note however, that this does not imply that the transistor is in cut-off. In practice, the base current is still non-zero, so the transistor is not in cut-off mode. It’s just small enough so that you can ignore it in your calculations.

BJT-ex5

Transistors - Level 5

Another popular transistor configuration is the current mirror. In the circuit below, the two transistors are identical (typically implemented on the same chip), so they have identical values for β, and the relationship between base current and base-emitter voltage is also identical. Since both bases are connected, and also both emitters, the base-emitter voltages are also the same. As a result, both transistors have the same base current, and since they have the same β, they also have the same collector current. Hence the name “current mirror”.

BJT-ex6

Transistors - Level 6

In the previous challenges, the transistor was used in the linear mode. As mentioned in the introduction, transistors are very often used as controllable switches, so in cut-off mode or saturation mode. The circuit below is an example.

BJT-ex7

40 thoughts on “3.5 Analyzing transistor circuits

    • Because the zener diode is fed with a 6V power supply, and the breakdown voltage is lower than 6V.
      Note that this is still an assumption that should be checked afterwards.
      Suppose e.g. that the base current is very large (which is usually NOT the case), then the voltage across RB might so large that the breakdown voltage is not reached.

  1. Level 4, question 1: You asked 2x the power of the second transistor, i guess this is has to be 1x power of the first, and 1x of the second.

  2. level 4 question 1: bèta is infinite so Ib=0 in both transistors. This means they are both in cut-off. How do we know the voltage Vbe in this case? I assumed 0 but this turned out to be wrong.

    • Probably you took 0.7 V for the voltage VCE across the transistor.
      This is a common mistake.
      The voltage VBE is 0.7 V, but VCE can approach 0 V as indicated in the question.

    • It is an assumption, so you are not sure yet.
      If e.g. the voltage across the transistor would be negative, the assumption would be false, and the transistor would be in saturation (see level 2).
      You do know that the transistor is not in cut-off though. You know why?

  3. Level 3 – Question 3:
    The explanation says the Ic will drop when the transistor goes in saturation mode, which I understand, but then it says the Zener diode will no longer break down, why is that?

    • That is indeed a tricky one.
      If Ic drops, the voltage across Re drops.
      This voltage plus 0.7 V for Vbe will be less than the breakdown voltage, so the zener will no longer break down.

  4. How do I calculate the current Ie2? What I did is detract the diode voltages from Vth, so it looks like this:
    6-0,7-0,7= 4,6V
    and then you just have a voltage divider left between the 2 resistors of 500 and 500k Ohm, so I calculated the voltage for the 500 Ohm resistor:
    (500/500500)*4,6= 0,0045954V
    and divided by 500 Ohm that becomes 0,00000919 A (9,19 µA).

    I feel like there’s another way to do it, using the equation with beta, but I can’t think of it.

    • Well, your way of calculating makes no sense. Sorry.
      I have never seen a voltage division formula like yours (500/500500 ????).
      When you use the Thévenin equivalent circuit, there is no voltage divider left.
      What you can do, is use KVL in the loop consisting of Vth Rth Vbe1 Vbe2 and Re.

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