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There are several ways to bias the quiescent point of transistor, when used as an amplifier. Among which voltage divider is best known for its improvements (improved temperature drift and beta variation). But as we use different transistors, even voltage divider cannot make amplifier beta independent.

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I am specially interested in the biasing of differential transistor pair (amplifier), where biasing is done only by replacing emitter resistor with active device known as constant current source, or should be done; as stated in Wikipedia and many other sites:

"...the differential pair is directly biased from the side of the emitters by sinking/injecting the total quiescent current. The series negative feedback (the emitter degeneration) makes the transistors act as voltage stabilizers; it forces them to adjust their VBE voltages (base currents) to pass the quiescent current through their collector-emitter junctions..."

I somehow still don't understand how a transistor could conduct any current without Vbe connected between the base and emitter of transistor. But by forcing the Vbe to "occur" and conduct current, whereas Vbe is forced by the current source...

Can anyone explain me what is actually happening here? Or at least show a practical example, where this kind of biasing is put into use.

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  • \$\begingroup\$ You can read some articles from Bob Cordell: Designing Audio Power Amplifiers. The degeneration resistors are put in series with emitters even with current source as depicted. \$\endgroup\$ – Marko Buršič Nov 26 '17 at 21:36
  • \$\begingroup\$ Saying it forces Vbe is not a good explanation. Having the current source connected to the emitter produces whatever emitter voltage is needed to allow that current through. For example if the voltage at the base is 0V, the emitter will be at -0.6v, if 5V, then 4.4V, etc. \$\endgroup\$ – τεκ Nov 26 '17 at 21:37
  • \$\begingroup\$ @τεκ If the constant current source is placed instead of emitter (tail) resistor: If there is no voltage across the base-emitter diode of diff. pair then no current flows through it and the constant current source collector sees the diff pair as huge resistor (as none of them conducts), and therefore no current should be flowing through both. That is how I see it... \$\endgroup\$ – Keno Nov 27 '17 at 9:40
  • \$\begingroup\$ Any diff. pair amplifier needs a DC current path for the base current. Is this so hard to understand? \$\endgroup\$ – G36 Nov 27 '17 at 13:45
  • \$\begingroup\$ @G36 As wikipedia says (if I understood correctly), there should be no voltage divider but only current source at emitter to bias the base of transistor.. \$\endgroup\$ – Keno Nov 27 '17 at 13:48
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Look at the example when you can see that indeed the \$I_{C1}\$ and \$I_{C2}\$ are almost beta independent. There is a slightly \$\beta\$ influence because \$Ic = I_E\cdot \frac{\beta}{\beta + 1}\$

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In this circuits we have:

$$I_{C1} = I_{C2} = 0.5\cdot I_{EE}\cdot\frac{\beta}{\beta + 1} $$

And the base current is:

$$I_B = \frac{I_C}{\beta}$$

The voltage at the base is:

$$V_B = -I_B\cdot R_B$$

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Source: Bob Cordell, Designing Audio Power Amplifiers

The design of Figure 3.2 differs from that of Figure 3.1 by the addition of emitter degeneration resistors R15 and R16 and by reducing C1 from 300 pF to 30 pF. The pair of 234-Ω emitter degeneration resistors implements 10:1 degeneration of the input differential pair by increasing the total emitter-to-emitter resistance RLTP from 52 Ω to 520 Ω. This reduces its transconductance by a factor of 10. In order to keep the negative feedback gain crossover frequency fc at the same 500 kHz for equivalent stability, C1 must be reduced by that same 10:1 factor.

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  • \$\begingroup\$ Ok, I get the degeneration thing but why isn't the base of Q1 biased to set Q point of operation? If it is not set then the crossover distortion occurs and amplifier wouldn't work properly. \$\endgroup\$ – Keno Nov 27 '17 at 13:57
  • \$\begingroup\$ If you disconnect the Q1 base or remove the input signal source. Q1 will be cut-off and Q2 will be fully ON. Hence Q4 is OFF. And Q7 will be in saturation and the output voltage will reach positive saturation voltage Vout_max = Vcc - 0.2V - 2Vbe. No crossover distortion will occur. \$\endgroup\$ – G36 Nov 27 '17 at 14:15

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