0

This probably has a really obvious answer, but it's bugging me, so here goes.

Why do we need two transistors for a differential amplifier? Instead of of using a basic differential pair, can't I just use something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Can't we just use something like this to cancel out the common mode signal and get just the difference (Since V_bias is in the order of Volts and difference is usually in mV/uV)?

Thanks in advance!

user_9
  • 257
  • 2
  • 9
  • 3
    You can use a single BJT as a differential amplifier. Examine [Q4 here](https://electronics.stackexchange.com/a/503581/38098) or [Q2 here](https://electronics.stackexchange.com/a/257324/38098). – jonk Oct 24 '20 at 17:17
  • 2
    If you add Re (low resistance) between emitter and ground on Q1 it turns into a differential amplifier with different input impedances, of course. One input is to the base and the other to the emitter. This circuit is used in some older power amplifiers for input and negative feedback. – Peter MP Oct 24 '20 at 17:38
  • A differential pair is just a type of differential amplifier. Of course, the pair implies "two transistors", but there are many mathematical benefits (e.g. hyperbolic tangent transfer characteristic) to using this specific structure which is why it's ubiquitous. It requires matched transistors, which is why it's mainly used in ICs since it's easier to control in that environment. You need to buy matched pair packages (DMMT3904W) if you want to use them as effectively in discrete design. – Ste Kulov Oct 24 '20 at 18:07

2 Answers2

3

A differential amplifier generally has two transistors for symmetry, because in many applications, it is desirable for both inputs to have similar characteristics, and for them both to be referenced from the same node (e.g., ground).

You can create a differential amplifier from a single transistor by applying the signals to the base and the emitter, but the input impedance seen by the two sources will be very different — and there's also a DC offset voltage between them. In some applications, this doesn't matter.

In your diagram, you're simply adding voltages by placing them in series, which is not a "differential amplifier" at all. In many applications, both sources must be ground-referenced, so this configuration is not suitable. It also means that the source impedances are all in series, which will cause problems unless they are very small — close to ideal voltage sources.

Dave Tweed
  • 168,369
  • 17
  • 228
  • 393
2

Your signal sources are never ideal, nor are they always floating. Connecting them in series will cause them to distort each other at best, and short each other out at worst. The transistors are there so the sources can drive their own higher impedance input so that they do not affect each other, nor are loaded down by the input as much. It also makes it so they can both use the same reference voltage (i.e. ground).

DKNguyen
  • 54,733
  • 4
  • 67
  • 153