When designing an amplifier stage, like this, how does it work in a linear fashion? enter image description here

Suppose you're applying some feedback, the output may well be stabilized to zero, but the more "wrong" \$R_C\$ is (in terms of base-emitter-voltage of \$Q_3\$ and quiescent collector current of \$Q_2\$), the more the differential pair is pushed away from its linear region. How is this dealt with properly? Secondly, it is said, that the voltage \$V_o\$ is dependent on \$gm\$ of the differential pair, how is this meant? Imagine there is a Capacitor for single-pole compensation between Collector and Base of \$Q_3\$ how does the \$gm\$ of the differential pair affect slew rate? This is all a bit blurry in my head, please can someone help me clearing it up. Thanks

Edit: You can assume that the differential input is ideally 0, tied to gnd. At perfect conditions of \$R_C\$ the base voltage of \$Q_3\$ is exactly as needed to make \$V_o\$ 0V. But the slightest variation of either \$I_{C2}\$,\$R_C\$ or \$V_{BE}\$ of \$Q_3\$ make that impossible. So you could feedback \$V_o\$ to base of \$Q_2\$. But then you little by little leave the linear area of the differential pair with \$R_C\$ going "wronger". How can this problem be dealt with properly. And, as asked above, how is \$V_o\$ dependent on \$gm\$ of the differential pair?

Edit2: The amplification of the difference voltage of a diff-pair is as follows:enter image description here

When I chose \$R_C\$ slightly wrong, I have a constant bias (\$v_{diff}\$) to achieve that \$v_o\$ is 0 (by feedback). Now when a signal is applied to the base of the input transistor, \$v_{diff}\$ is already offset and thus pushed further into non-linear region. How can this be avoided?

Edit 3: The problem could also be described this way: To get zero output through feedback, there will be \$v_{id} \neq 0\$ which would decrease linearity of the diff-pair since the incoming voltage to be amplified would rather alternate around a point left or right of \$x = 0\$ and therefore be in a more non-linear region. The worse the collector resistor of Q2 is aligned, the more linearity suffers. How can this be possibly avoided without having to look into the transisitors specification with a magnifying glass, but rather have it work for various transistors?

Another main question is how the compensation capacitor is dependent on the \$gm\$ of the diff-pair. How do they interrelate? Actually these should be bjts as opposed to FETs

The FETs are actually BJTs

I only know that this has something to do with slew rate, but I can't seem to find out what exactly.

  • 2
    \$\begingroup\$ Where's the input? \$\endgroup\$
    – Andy aka
    Jul 30, 2014 at 13:06
  • \$\begingroup\$ Usually the input is Base of \$Q_1\$ but for this matter this is not important. See edit. \$\endgroup\$
    – jjstcool
    Jul 30, 2014 at 13:10
  • 2
    \$\begingroup\$ If both bases are tied to ground, feedback would seem (more or less by definition) impossible, so I assume the application of feedback will involve removing the ground connection. Unless I misunderstand the circuit, applying feedback from Vo to the base of Q2 will result in positive feedback, and your description is correct. So the obvious way to fix the problem is to provide feedback to Q1, instead. \$\endgroup\$ Jul 30, 2014 at 13:59
  • \$\begingroup\$ Oh, yes of course feedback to the other base. However, linearity suffers since the two input bases are biased out slightly out of balance, means at 0V \$V_o\$ vdiff is not equal to 0, so further signal differences are not as linearly amplified as if there were no bias. How is this best avoided? \$\endgroup\$
    – jjstcool
    Jul 30, 2014 at 14:43
  • \$\begingroup\$ @JJstcool: out-of-balance (asymmetry) is NOT identical to non-linearity. What is your primary goal? Amplification of voltage differences (both inputs used)? Large one-sided differential amplification? Or DC bias with Vo=0V? \$\endgroup\$
    – LvW
    Jul 30, 2014 at 15:06

1 Answer 1


At first, there is a certain asymmetry caused by the base current for Q3. As a result, Vc1 and Vc2 cannot be equal. Of course, Rc must be chosen properly to allow the correct bias point for Q3.

The one-sided differential gain (referred to the base of Q1) of the main amplifier is Ad=+gm*Rc/2. Thus, the transconductance gm - set by the current source I1 - plays a main role.

It is no surprise that any change of Ic and/or Rc influences the bias point for Q3 and the quiescent voltage Vo. This is a classical "offset" effect. But you should not try to feed back a voltage from the Vo node to the base of Q2. This would result in POSITIVE feedback (instability). But a feedback path to the base of Q1 should work. However, the gain of the whole circuit will be affected, of course.

  • \$\begingroup\$ Yes, as I said above, I mistakenly chose the wrong input for feedback. However, do you guys see my problem? See edit 2. \$\endgroup\$
    – jjstcool
    Jul 30, 2014 at 15:10

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