current mirror connection. which way round?

[surreyian]

hello,

How do we know which way is the correct connection?

here is my understanding why circuit B doesnt work. But i cannot put my head around why circuit A works.
in circuit B. Break loop at A and inject a signal. If A increase then causing M1 to go Low, giving more current in M0 and M2. The negative feedback resistor R0 will reduc the over gain, hence over time, the circuit will stuck at zero. Is this understaning correct?

Cirucuit A has the correct connection is the working circuit, can someone help me to understand why we have to connect this way.

 

[godfreyl]

I'll stick my neck out and say they both look like bad circuits. What are they supposed to be for?

In circuit B, R0 increases the loop gain so if some current is injected to start with, I would expect the current to increase and stay high after the external current injection is removed.

In circuit A, R1 decreases the loop gain below unity, so I would expect the current to reduce to zero when the external current injection is removed.

If no current is injected, I don't think circuit A will ever switch on by itself. Circuit B would probably start due to leakage current, but the current would quickly rise to a high (and badly defined) level.

 

[erikl]

Simulation shows that actually both circuits stabilize at equal currents and node voltages -- even with a resistance of 1kΩ instead of the given 1Ω, the current and drain voltage deviations are ≦ 1‰ .

 

[godfreyl]

I simulated it with the circuit below, setting either R1 or R2 to zero. The results were very different.