I don't understand, why the voltage V2 is approximately Vb. Can't V2 be 0, if Vb is enough high, so that the emmiter-current of Q2 is Ie2?

The assistent only told me, that because both transistors are aproximately the same and because the two resistors R2 are also aproximately the same, therefor the Voltage V2 has to be the same as Vb. The point, where I get confused is, that the Current through the left R2 resistor is dependet of the Voltage V2. Correct me if I'm wrong.

I hope someone can help me.

Have a nice day:)

Question a)

  • \$\begingroup\$ You are completely ignoring the feedback via R3 in your reasoning. The feedback will take care that the base of Q2' will have the same level as the base of Q2. \$\endgroup\$ – Curd Jun 22 '15 at 15:41

Here is my explanation (calculation): V3=Vout+0.7V with Vout=V1=0.7V

V3=(0.7+0.7)V=1.4V .

This result must be in accordance with Ohms law for R2: V3=3-I2R2=3-0.5mA*3.2kohms=3-1.6=1.4V.

This calculation shows that both results for V3 are equal for Ic=0.5mA only. Because this is half of the emitter current source both transistors draw the same current. Hence, both base-emitter voltages must be equal.


To answer the question finally (the task was to find the relationshipo between V2 and VB) - here is a corresponding formula:

V2=VB + VT*ln{[IE2*R2/(Vcc-2*VBE)-1]}

Rough calculation (with VT=26mV, IE2=1mA, R2=3.2k, VBE=0.7V):

V2=VB + VT*ln[(3.2/1.6)-1)

V2=VB .


If you recognize this to be intended to be an amplifier (like an op-amp) with negative feedback, then you can instantly see that the voltages V2 and Vb have to be about the same (iff it's working properly).

If you look at what would happen if the voltages were imbalanced, then there are two situations:

  1. If the voltage at the base of Q2' is higher then it grabs most of the current and the output goes to Vcc- Vbe from the follower. So, Q1 turns on and sucks current through R1 and reduces the voltage at the base of Q2' and balance is achieved.

  2. If the voltage at the base of Q2 is higher then the output goes 'low', but it cannot go any lower than Vb-Vbe (where the transistor is saturated). The output goes to Vb - 2Vbe, and through R3 to Q1. If Vb > 3Vbe then Q1 cannot turn off, and balance won't be achieved, but if it can reduce current through Q1 then the voltage at the base of Q2' rises and balance is achieved.

  • \$\begingroup\$ Thanks for the answer. How do I see that this is an negative feedbacked opamp? \$\endgroup\$ – blakebaxley Jun 22 '15 at 15:17
  • \$\begingroup\$ Assuming you recognize the differential pair and output circuit, you can see the feedback from the differential pair back to an input. You could ass-u-me that it's negative feedback (risky, but on an exam question probably a decent bet), or simply run through the simple analysis I went through above to assure yourself that it is negative (not positive) feedback. \$\endgroup\$ – Spehro Pefhany Jun 22 '15 at 15:33
  • \$\begingroup\$ okei thanks now I see it. I would be very happy if you eventually could explain me, why the output goes to Vb-2Vbe? is it because the base-voltage at Q3 is Vb-Vbe and then there is a new voltage drop of Vbe? Thank you \$\endgroup\$ – blakebaxley Jun 22 '15 at 16:02
  • \$\begingroup\$ Regarding negative feedback: The whole feedback path contains one phase inversion only (Q1). Moreover, it is well known that the long-tailed pair produces no further phase inversion between V2 and V3 (but we have 180 deg phase shift between v2 (base) and the collector voltage of Q2´). \$\endgroup\$ – LvW Jun 22 '15 at 16:58
  • \$\begingroup\$ @blakebaxley At point circled-1, the voltage will have to be Vbe for Q1 to be in the linear region. Base current is less than 30uA (from beta, Vcc and 1k load) so that adds 0.9V through R3 so the output can't be more than 1.6V or so. \$\endgroup\$ – Spehro Pefhany Jun 22 '15 at 17:45

Have a look at the big picture (ignoring the details): The bases of Q2 and Q2' are the negative and positive inputs of a differential amplifier (so called "long tailed pair") in a configuration with negative feedback. One input is held at \$V_B\$. The other input is regulated to be \$V_B\$ as well because of the negative feedback.

You can see it as following circuit: enter image description here

  • \$\begingroup\$ Curd, I am not quite sure if the above circuit is a "good" analogy (which can help to understand/solve the given task). At first, the opamp gain is (nearly) infinite in contrast to the gain of the long-tailed pair. Secondly, the voltage Vout in the opamp-based circuit results from a small difference (V2-VB), which means that the opamp provides signal gain. This also is in contrast to the original circuit where - at least - theoretically V2=VB is possible (pure common-mode operation) without any differential gain. \$\endgroup\$ – LvW Jun 23 '15 at 13:54
  • \$\begingroup\$ @LvW: It is not just an analogy. The subcircuit consisting of Q2, Q2' and Q3 actually is an OpAmp (not a particularly good one, it's far from ideal, but it works as an OpAmp). Also I think you contradict yourself in your argument: first you say an OpAmp would have infinite gain (which is only true for an ideal OpAmp). Then you say an OpAmp would require a non-infinitesimal difference between V2 and VB (which is only true for an non-ideal OpAmp with finite gain). Also note that the original question only asked for an approximate statement about the relationship of Vb and V2. \$\endgroup\$ – Curd Jun 23 '15 at 21:20
  • \$\begingroup\$ @Lvw: Also note: it is a well known basic fact that the voltage difference between inverting and non-inverting inputs of an OpAmp is negligibly small (=0 for an ideal OpAmp) if it is working in linear (i.e. non-saturated) mode (see e.g. en.wikipedia.org/wiki/Operational_amplifier#Closed_loop) \$\endgroup\$ – Curd Jun 23 '15 at 21:25
  • \$\begingroup\$ Curd, please note that I spoke about a gain that is "nearly" infinite. I know about the difference between real and ideal. My only point was that I am not sure if the analogy can help to understand the given circuit (and the claim that V2=VB). The diff. voltage between the opamp inputs is nearly (!!) zero because of the very large open-loop gain. However, this argument does not hold for the given circuit. Therefore, I think, it is appropriate to look for another explanation. That`s all. \$\endgroup\$ – LvW Jun 24 '15 at 7:05
  • \$\begingroup\$ @LvW: I just wanted to point out the inconsistency of your argument: on one hand you critizized the fact that the gain of the actual circuit is not "(nearly) infinite" as oppposed to an ideal OpAmp. On the other hand you critizized that a non-ideal OpAmp would have V+ != V-. \$\endgroup\$ – Curd Jun 24 '15 at 8:33

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