How to bias a common-source amplifier with a current-source load?

This is my answer to the SE EE question about the famous 'dynamic load'.

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Basic idea

It is quite paradoxical because in this arrangement, two current sources are connected in series. According to the laws of electrical circuits, this is incorrect because even with a minimal difference in their currents, each of them will struggle to establish its current in the circuit. As a result of this "tug of war", the voltage at their midpoint will increase in the corresponding direction to infinity. But this "negative" phenomenon is used in modern integrated circuits to obtain something very useful - extremely high gain.

Implementation

How to set Vin?

Here are three techniques for this purpose:

Simply set Vin by its parameters window in the simulator
Use the FET transfer characteristic
Introduce negative feedback and calculate Vin needed from the resistor ratio

For the purposes of my intuitive explanation, I will use the first option. To show the idea in depth, I have explored below in the same step-by-step way three possible implementations - using "ideal" current sources, dynamic resistors and FETs.

"Ideal" current sources

Strictly speaking, we cannot connect two ideal current sources in series if only because CircuitLab will not let us. To "cheat" it, we need to degrade the sources a bit by connecting very high resistances in parallel. It turns out that the internal resistance of the voltmeters (1 GΩ) can play this role.

I1 = I2 = 1 mA: If two 1 mA current sources are exactly the same (this is only possible here in the simulator), the voltage at their midpoint is exactly half of the supply.

Simulate this circuit – Schematic created using CircuitLab

I1 = 1.000004 mA, I2 = 1 mA: It is enough to increase I1 by only 4 nA, and the output voltage V2 increases with 2 V...

Simulate this circuit

I1 = 1 mA, I2 = 1.000004 mA: ... or increase I2 by only 4 nA, and the output voltage V2 decreases with 2 V.

Simulate this circuit

I1 = 1.000004 mA, I2 = 0,999996 mA: If we simultaneously and oppositely change the two currents in one...

Simulate this circuit

I1 = 0,999996 mA, I2 = 1.000004 mA: ... or the other direction...

Simulate this circuit

... the output voltage V2 changes with 4 V in the according direction.

Dynamic resistors

In most cases, constant current sources are implemented as constant-current dynamic resistors. They are passive devices that do not contain internal power sources. Since there are no such elements in the CircuitLab library, we can simulate them with manually controlled variable resistors R1 and R2. To show that the two implementations are equivalent, I have adjusted the resistance to give the same output voltage as above.

R1 = R2 = 5 kΩ: The benefit of this trick is immediately visible - the initial resistance of the resistors is only 5 kΩ instead of "infinitely high".

Simulate this circuit

R1 = 1 kΩ, R2 = 9 kΩ: If we decrease R1 with 4 kΩ, R2 will increase with 4 kΩ, and V2 will increase with 4 V.

Simulate this circuit

R1 = 9 kΩ, R2 = 1 kΩ: Conversely, if we increase R1 with 4 kΩ, R2 will decrease with 4 kΩ, and V2 will decrease with 4 V.

Simulate this circuit

MOSFET

Transistors (MOSFET here) behave like the dynamic resistors above - they change its static resistance to keep the current constant. The only difference is that they are voltage-controlled dynamic resistors. So, to set the output current and accordingly, the output voltage, we have to adjust its input Vgs voltage. This is not so easy because the drain current changes drastically for very small changes in Vin around the threshold, but it is worth doing to see exactly what the transistor is doing. Now let's repeat the above Constant Vin1 and Vin2 experiments obtaining equivalent output voltages.

Constant Vin1 and Vin2