This is the circuit I'm referring to:

enter image description here

This stage is often used as input stage of audio amplifier.

I am wondering, how can the input signal be normally amplified since there is no DC bias across the base of both transistors?

In my book, it says that we have to drive this stage with "line-level" amplitude signal (that is around 1V). But still, if the amplitude swings for 1V from its reference point which is probably 0V, then only a bit of a positive signal wave would be amplified, since transistor starts conducting at Vbe of approx. 0.7V!

Without base bias voltage, the output signal would look like this, right?

enter image description here

Instead of this, right?

enter image description here

  • \$\begingroup\$ since there is no DC bias across the base of both transistors? Not across both but that is not needed, the individual transistors need to be DC biased and that is done by the current source \$I_0\$ \$\endgroup\$ – Bimpelrekkie Nov 1 '17 at 14:26
  • \$\begingroup\$ But this circuit in 90% of the application is used together with split supply (symmetrical supply). electronics.stackexchange.com/questions/335930/… Resistor R2 and R6 bias the input stage. And with single supply, we always use the voltage divider to bias the circuit. \$\endgroup\$ – G36 Nov 1 '17 at 14:29
  • \$\begingroup\$ @Bimpelrekkie I know but if the resistance seen by the collector of current source is too high, then the constant current source will saturate (V_ce approx. 0V) and no current will flow through, right? \$\endgroup\$ – Keno Nov 1 '17 at 14:59
  • \$\begingroup\$ Do you mean, if the current source \$I_0\$ (possibly implemented with a transistor) has no headroom (voltage to work with) then the transistors are not biased. Yes that's true but the circuit will only work if the transistors (Q1, Q2) are biased. Same as that they need a supply voltage. The designer has to make sure that \$I_0\$ does have enough headroom under all circumstances where the circuit is supposed to work. \$\endgroup\$ – Bimpelrekkie Nov 1 '17 at 15:02
  • \$\begingroup\$ @Bimpelrekkie So, if they are biased (but not shown here) then G36 is right (he's referring to my previous question about diff-amp) ? \$\endgroup\$ – Keno Nov 1 '17 at 16:11

Since Io is a current source the emitters of the transistors will actually be at approximately -0.7V so the transistors will be in their active region.

If the emitter resistors are too low however the amplifier will saturate if the signal goes to high or too low though. Normally this type of stage is used where Vi2 is derived from the output of the following amplifier and provides negative feedback.

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  • \$\begingroup\$ But if there is no base voltage then the collector resistance seen by the collector of the current source (lets say, it is a bjt) is to high and constant current source will saturate, therefore no current will flow. \$\endgroup\$ – Keno Nov 1 '17 at 14:55
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    \$\begingroup\$ If one of the base nodes is not used it must be grounded (DC=0). More than that, any input signal must be referenced to ground. Hence, any DC base current must go through the signal source. If the signal source is coupled via a capacitor a separate resistor to ground is needed. \$\endgroup\$ – LvW Nov 1 '17 at 14:59

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