# How can I bias the two transistors (Q59, Q58) two main transistors in my common emitter pnp amplifier? [duplicate] [ I tried in my design to make a common emitter stage amplifier using a pnp transistor and a npn transistor as a current source but I didn't know to bias either of them (well as for Q54 I think I've biased correctly I'm not sure) the simulation showed an operating point way lower than I expected (as you notice it's 1.178v) so the signal came out clipped. How can I increase my operating point value so it doesn't clip my signal?

• You mean Q64 & Q80 in the title isn't it?
– sai
Sep 21 at 12:25
• The transistors Q80, Q64 are for biasing (to produce voltage can be applied to the bases of Q58, Q59) Sep 21 at 15:09
• Yes it did I will add the answer there👍👍 Sep 21 at 15:15
• @Alghaith, Let's clarify some things. First, by saying Q59, Q58 in the title, you actually mean Q54, Q55, right? Then, you want to oppose Q54 to Q55 by connecting their collectors and applying the input voltage to Q54 and a constant input voltage to Q55, right? Such a simpler configuration (than the discussed in your previous question) is a common-emitter amplifier (Q54) with a simple active load (Q55) acting as a constant current source. Yesterday we were talking about a more complex configuration where the input voltage of Q55 varies oppositely (differentially) to Q54's input voltage... Sep 21 at 19:20
• ... If all this is so, there is only one main problem left to solve - how to apply the grounded input source voltage to the base of Q54 whose emitter is connected to Vcc and not to ground. This problem is usually solved with a current mirror (see, for example, current-feedback amplifier in Google). Sep 21 at 19:32

In the place of Q55, if you had a resistor, you can easily fix the operating point by adjusting the value of the resistor but, it will reduce the gain. If you want to use Q55, you'll need to have a feedback loop to set the DC operating point and the bandwidth of that loop should be much lower than the signal frequency.

• .Do i need this loop in every amplifier circuit to set the operating point or this method applies only on this circuit? Sep 21 at 14:05
• Any high gain amplifier cannot be used in open loop because of the exact problem that you are facing. The output operating point will not be possible to be set optimally because of the high gain. Small mismatches will move the output too high or too low. High gain amplifiers in open loop are basically comparators. Even if you somehow bias the amp, it is not really useful to use an open loop amplifier because the gain would vary a lot across PVT.
– sai
Sep 21 at 14:58
• This was extremely useful to me sir it has answered alot of questions thank you 👍👍 Sep 21 at 15:11

By saying Q59, Q58 in the title, you probably mean Q54, Q55 (Q1, Q2 in the schematics below). It seems you want to oppose Q1 to Q2 by connecting their collectors and applying the input voltage to Q1 and a constant input voltage to Q2. Such a simpler configuration (than the discussed in your previous question) is a common-emitter amplifier (Q1) with a simple active load (Q2) acting as a constant current source.

## Floating input source, grounded reference simulate this circuit – Schematic created using CircuitLab

## How do we ground the input source?

There is only one main problem left to solve - how to apply the grounded input source voltage to the base of Q1 whose emitter is connected to Vcc and not to ground. simulate this circuit

## Grounded input source, grounded reference

This problem is usually solved with a current mirror (see, for example, "current-feedback amplifier" in Google). simulate this circuit

In the more complex configuration the input voltage of Q2 varies oppositely (differentially) to Q1's input voltage. It can be used as another input; thus we obtain a differential stage with dynamic load. simulate this circuit

# Biasing

To bias a transistor means to add a constant voltage Vbias about 0.65 V. Figuratively speaking, we can do it in two ways:

• by "lifting" the base with Vbias

• by "lowering" the emitter with Vbias

I have already tacitly used the first option above by initially increasing the input voltages with Vbias. However, this assumes that the emitters are of fixed voltage (grounded). This means that the amplifier is "purely differential" - its two input voltages cannot change simultaneously within wide limits (the so-called common mode). So we have to "pull down" the common emitters with some "soft" element.

## Emitter resistor

Collector-resistor stage: The humble resistor Re can serve as such an element as it faithfully served 100 years ago in the classic long-tailed pair. From one side, it is useful that by changing its resistance, the voltage drop across Rc varies, and this way we can set the quiescent output voltage; but from the other side, this is a disadvantage because the output voltage changes when both input voltages change (common mode). simulate this circuit

By sweeping Re, we can precisely see how it determines Vout. Dynamic-load stage: Let's now check if this is also the case with our dynamic-load stage. Surprisingly for us, Vout hardly changes when we change Re! simulate this circuit

We can see it very well if we sweep Re. Aha... the explanation is simple. As we change Re, both branch currents change. In the right branch, transistors Q1 and Q2 change their "resistances" in opposite directions; as a result, their midpoint does not change its Vout voltage (sort of like a virtual ground).

## Emitter current source

Collector-resistor stage: To improve the classic long-tailed pair, they replaced the emitter resistor with a constant current source Ie. simulate this circuit

The results are the same. Dynamic-load stage: As above, when changing Ie, Vout hardly changes for the same reasons. simulate this circuit This gives us reason to conclude that the emitter current source is not absolutely necessary in an amplifier stage with a dynamic load.