# How to bias the Q55, Q54 transistors in my design (what's the right voltage value needed for it)?

I want to make a simple differential amplifier and when I bias the pnp transistor that's connected to Vcc I notice in simulation that the application is too small (less than 1) and it go around 4.96v with a peak to peak voltage less than a milli volt I'm using the 2N2222 npn transistor and the 2N3905 pnp transistor (I set it's Vaf to be 53v) I'm biasing the circuit using a branch of a ptat circuit I used a base2 net lable to bias the right transistor and a base3 to bias the left one The two input signals is sine wave with only 1 milivolt different from each other (which is the right V8 voltage source) V9 and V10 are the noise (symmetrical signals) The first two branches is the ptat circuit (proportional to absolute temperature) the transistors (Q67, Q86, Q90, Q94, Q98) are for current amplification in order not to lower the branch current when biasing (not to lower the derived current from the branch which needed to pass through transistors NPN bases when biasing it) The next branch is for biasing using the it's transistors bases (and this branch is biased from the ptat circuit through the Q63 transistor) And the other two branches is the amplifier (differential stage of common emiter but using a PNP transistor instead of Rc "in order to get higher gain") Now this is the explanation of the circuit does anyone knows why I have that problem I tried to solve with no success

• Is this strictly a simulation question? Or do you intend on constructing some device out of real parts? Also, can you describe your design in more detail? Break it down into sections and talk about each? Commented Sep 18, 2023 at 16:17
• Both actually i intend to make an amplifier out of transistors and I don't understand why the simulation goes like that (it only drops 40milivolt on the pnp Q55 transistor i will edit my question and add more details Commented Sep 18, 2023 at 18:09
• Please run your text through a word processor. It’s currently one long sentence. Commented Sep 18, 2023 at 18:28
• @winny Do you use "word processor" in analogy to "food processor"? Commented Sep 18, 2023 at 18:38
• @user107063 No, it’s a term. Look it up. Commented Sep 18, 2023 at 18:41

# What the OP's circuit is

This looks like an NPN differential amplifier (Q95-Q96) with a PNP current mirror (Q54-Q55) acting as a dynamic load...

# What it should be

But to be so it is necessary that one of the top PNP transistors be diode-connected. Such a configuration is biased by an emitter current source (Q105) whose current splits into two parts flowing accordingly through the two (left and right) legs. So, you can properly bias Q95-Q96 by adjusting the emitter current produced by Q105.

# Understanding the circuit

The best way to understand this seemingly weird circuit is to trace its evolution step by step. I have illustrated this approach using simple CircuitLab simulations where you can manually change the quantities and observe the result.

## The idea of dynamic load

... i tried to (in another schematic) make a common emitter amplifier with a pnp transistor and an npn current source

This OP's experiment can serve us as an excellent starting point for building the differential circuit.

Fighting true current sources: I will even allow myself to go back a step by showing what the concept is here; we can figuratively call it "fighting current sources". At first glance, the idea is absurd - to connect two current sources in series and start changing their currents in the opposite direction (one, the other or both at the same time).

simulate this circuit – Schematic created using CircuitLab

A conflict arises (as in the games of "tug of war" and "arm wrestling") and the voltage at their midpoint vigorously changes... and that means amplification. As you can see, it is enough to slightly change the current(s) (in milliamps) and the voltage changes a lot (up to kilovolts, because CircuitLab current sources are ideal). The voltage drops across the sources change because this way they try to restore the desired current. So, we conclude, this way we can make amplifiers with gigantic gain.

Fighting current "sources": The "true current sources" above have inside an internal resistor and a voltage source that varies its voltage to maintain the desired current regardless of the load. In the practical implementation, the voltage source is taken out (so that it can be used by other devices) and the current "source" is just a dynamic resistor that changes its resistance to keep the current constant. Now the voltage drop across the current "source" is limited to the supply voltage (aka compliance voltage). I cannot show it here because there are only ideal current sources in CircuitLab...

simulate this circuit

Fighting resistors: ... but I will demonstrate it with simple variable resistors (voltage divider). You just want to change their resistance significantly to mimic a dynamic resistor.

simulate this circuit

## Implementation

Fighting transistors: In differential circuits, this configuration of two fighting current sources is implemented by two transistors (one PNP and the other NPN) connected by their collectors (the cited OP's idea above). By carefully changing in opposite directions their input (base-emitter) voltages in the simulation below, you can control their collector-emitter "resistances". Note that for simplicity, in this conceptual circuit, I have referenced Vin1 to V+ and Vin2 to V- (not as usual to ground).

simulate this circuit

Cloned Q1: In the above differential configuration, the two transistors are stacked on top of each other, but in practical implementation it is convenient for both to be NPN and grounded. Here we come up with a clever trick - using a current mirror to clone Q1 and put the clone Q4 on Q2. Now both input voltages Vin1 and Vin2 are grounded.

simulate this circuit

As above, carefully bias the two transistors so that to obtain zero output voltage. I did it by guesswork starting with Vin1 and Vin2 values around 650 mV. This requires some dexterity and patience but "the end justifies the means" as the ancient Romans said :-) To do this, open the properties window of one input voltage and change it in the appropriate direction so that Vout approaches 0 V. Then do the same with the other voltage (you can having to repeat this iterative procedure several times). In short, increasing Vin1 increases the current of Q1, Q3 and Q4 respectively, and Vout increases. Conversely, increasing Vin2 increases the current of Q2 and Vout decreases.

After you have adjusted the quiescent output voltage, change the input voltages in opposite directions (differentially). By the help of Q3, Q4 copies the Q1's current and contrasts it with the Q2's current; as a result, Vout vigorously changes.

Emitter side biasing: The problem with the above circuit solution, however, is that the emitters of the input transistors Q1 and Q2 are connected to a fixed ground and no common-mode input voltage can be applied. The circuit will amplify both common mode and differential input signal equally; it will not differentiate between them.

To solve the problem, we can apply another clever trick - insert a current source between the emitters and the negative power supply. By changing its current, we adjust the quiescent collector currents so that to obtain a quiescent 5 V output voltage Vout. Now the emitters "move" with the input common voltage (when both input voltages simultaneously change) and the collector currents, accordingly Vout, do not change. You can try it in the simulation below.

simulate this circuit

As you can see from the graph below, only a 10 mV amplitude of the input voltage (Vin1) is sufficient to obtain the maximum swing of the output voltage.

# Saga of biasing

## The motive

Since this word is often used mechanically but OP is interested in what exactly it is, I have shown below the actual idea behind it.

## The problem

Compared to an op-amp, a transistor is a "crappy" device because it does not start working as soon as the input voltage starts to increase above zero but later, somewhere around 0.6V.

## The remedy

So the input voltage must somehow "jump" this "deadband". This can be done in two ways:

1. An offset voltage is contained in Vin: This is what I meant in Schematic 5 above when I said, "carefully bias the two transistors so that to obtain zero output voltage". I said "carefully" because after about 650 mV the collector current starts to change significantly with a small change in Vin. In this case, we see only one AC + DC input voltage source.

simulate this circuit

For example, in the schematic above, AC = 10 mV and DC = 641.5 mV; so Vin wiggles between 631.5 mV and 651.5 mV (too small to see it in the graph below). As a result, Vout wiggles around 5 V.

2. An offset voltage is added to Vin: Usually, the input voltage source is a "pure" AC source so we need to add a DC offset (aka "bias voltage") to it. The simplest way to do this is by connecting, according to KVL, a bias source Vbias in series with Vin. Now there two separate voltage sources - AC = 10 mV and DC = 641.5 mV

simulate this circuit

The result of all this is exactly the same as above.

So the so-called biasing means simply to add a constant (bias) voltage to the input voltage.

## Setting the bias voltage

But what should its value be? It is logical to choose it so that the quiescent output voltage is in the middle (5 V) of the output range (10 V). You can do it with calculations, graphically or experimentally as I have done above. The final result is important - the output voltage should be in the middle so that it can then, under the influence of the input voltage, change from end to end.