I have been following a lecture on net about a two stage amplifier example. It is basically a BJT amplifier with a PNP emitter follower as shown below. I named some of the nodes for clarity:

enter image description here

The design restrictions in the example of the video lecture are Ie2 = 2mA and Ic = 1mA and the absolute gain is 5 which means |Rc/Re1| = 5.

The lecturer starts designing this amplifier from the very right side setting Ve2 to 5V for max swing.

So for 2mA he sets Re2 to 2.5k.

So far so good, then he tries to set Ve1 for maximum swing as well and he sets Ve1 to 5V as well.

I'm stuck at that point, If Ve1 and Ve2 is both 5V how come the PNP transistor goes to active region?

Basically my question is: Is Ve1 dependent on Ve2? Or is Ve2 dependent to Ve1? Will one of them fix the other one's voltage? Which one controls the other?

Because otherwise setting both to 5V doesn't make sense in design process to me. I'm very confused about it, would appreciate any clarification.

The video lecture is a bit long but in case here is the link:

  • 1
    \$\begingroup\$ If they're both in the active region, Ve1 should be about a diode drop below Ve2. This is a fairly small difference given how unpredictable designing with transistors is. I haven't watched the video, but I suspect the lecture was just assuming they are roughly the same to make the design process easier. \$\endgroup\$
    – jramsay42
    Commented Nov 1, 2018 at 1:53
  • \$\begingroup\$ The lecturer made a mistake. For a gain of 5.0000X, with Ic1 = 1mA, the 'reac1' is 26 ohms. If Rc1 = 5Kohm, then the total of Re1 + 'reac1' must be 1K, and Re1 is 1,000 - 26 = 974 ohms. The Rb1 will be 974 mV + ~~ 0.6 volts = 1.574 volts. Assume very high beta, and pick Rb1 and Rb2 to produce 1.574 volts on Base1. \$\endgroup\$ Commented Nov 1, 2018 at 4:10

1 Answer 1


The PNP is a follower, as you mentioned, so Ve2 approximately follows Ve1 with an offset of about a diode drop, as the previous comment mentioned. The collector emitter current of a transistor is an exponential function of the base emitter voltage. That is to say that a small change in emitter voltage will cause a large change in emitter current. If you decrease Ve1, the emitter current in the follower with increase, resulting in a fall in emitter voltage. This then results in a decrease in emitter current until everything balances again. So you have negative feedback. This is how it follows, with a gain slightly less than 1.

  • \$\begingroup\$ Ve2 follows Ve1 OR Ve1 follows Ve2? which one is the independent? \$\endgroup\$
    – GNZ
    Commented Nov 1, 2018 at 3:20
  • \$\begingroup\$ @Genzo each is dependent on the other to a lesser or greater degree but the convention is that with an emitter follower, the emitter follows the base. In this circuit the signal flows from left to right. I can't think of an application where one would apply a signal to the emitter and take an output from the base. \$\endgroup\$ Commented Nov 1, 2018 at 4:51
  • \$\begingroup\$ "application where one would apply a signal to the emitter and take an output from the base" ... a current mirror, in a way. (I_ce is the input, V_be is output) \$\endgroup\$
    – Pete W
    Commented Jan 7, 2021 at 18:33

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