Differential Amplifier - Active Load Current MIrror Logic

Question

I'm trying to wrap my head around the workings of a differential amplifier when a current mirror is used as an 'active load'

With reference to this Youtube video leecture: here

My understanding so far (when using resistors as loads rather than current mirror)

• A constant current sourrce is sinking say 10mA in the tail
• This means each leg has 5 mA when both inputs are matched
• When one input increase, its transistor allows more current, increasing the voltage drop across its load resistor thus lowering the output voltage that side
• This also means that less current is 'available' for the opposite transistor so the current through its load resistor decreases, its voltage drop decreases and so the output voltage increases.

If we now replace the load resistors with the current mirror (where I think I'm misunderstanding some steps):

• When the input voltage increases (left side) the current through the transistor increases
• The voltage on the base of the mirror transistor (left side) changes [how and which way?]
• This is mirrored to the base of the other current mirror transistor, causing the same current to flow through it

Now, with resistor loads the current was split such that it would add to the 10mA of the tail, with more on one side and less on the other. With the mirror, if left side goes up to 8mA and it's mirrored across - both sides now have 8ma (where 16mA isn't obtainable because the tail is constant at 10mA)

So what is the behaviour/action happening that allows for the increased gain on the output? The video talks about 'reversing the polarity' of the current on the right hand side but i'm not sure i understand how that can work - Can current flow both down through the mirror and up through the tail? But then how is this aligned to the constant 10mA?

Many thanks in advance for any pointers you can provide.

Answer 1

Fundamentally, in the small_Signal model, in a transistor_driving_R_load stage, the gain will be

• gm * ( Rload || R_early_effect)

whereas in transistor_driving_activeLoad stage, the gain will be

• gm * (R_early_effect || R_early_effect)

With R_early_effect able to be >> R_load , the gain with the active_load can be >> gain of the transistor_driving_R_load.

(R_load is a physical resistor, whereas R_early_effect is a measured parameter of the semiconductor).

Also, I recall using the 3_transistor current mirror topology (THREE Transistors), at 1 milliamp, with actual 2N2905 transistors, with emitter swamping resistors to make up for mismatched Vbe.

On a Fairchild curve tracer (perhaps one reason Schlumberger bought Fairchild Instrument), that 3_transistor current mirror measured

• 5,000,000 ohms

thus equivalent Early Effect voltage was ...... 5,000 volts.