Understanding Minimum Output Voltage of Cascode Current Mirror

Question

I guess that I have some serious trouble understanding how to get the saturation equation for the cascode current mirror right which is known from all the textbooks as

$$ V_{out} = 2 V_{DS, sat} + V_{th} $$

It's quite intuitive to get to the point that the gate voltage at Q4 needs to be

$$ V_{G, Q4} = 2 V_{DS, sat} + 2 V_{th} $$

Now when looking at how a MOSFET is saturated we use \$ V_{DS} \geq V_{GS} - V_{th} \$. So applying this to Q4, why do we just simply do

$$ V_{D, Q4} = V_{out} = V_{G, Q4} - V_{th} = 2 V_{DS, sat} +V_{th} $$

? That doesn't work for me because we look at \$ V_{G, Q4} \$ with respect to ground, so for me it makes sense to write

$$ V_{out} = V_{G, Q4} - V_{S, Q4} - V_{th} = V_{G, Q4} - V_{D, Q2} - V_{th} $$

But we don't know \$V_{D, Q2}\$. When assuming \$V_{D, Q2} = V_{D, Q1}\$ we get

$$ V_{out} = 2 V_{DS, sat} + 2 V_{th} - V_{DS, sat} - V_{th} - V_{th} = V_{DS, sat} $$

which obviously is nonsense... So where did I get stuck or where did I go seriously wrong? Any help is greatly appreciated!

Answer 1

so for me it makes sense to write

$$ V_{out} = V_{G, Q4} - V_{S, Q4} - V_{th} = V_{G, Q4} - V_{D, Q2} - V_{th} $$

But that does not make sense as $$ V_{G, Q4} - V_{S, Q4} = V_{GS, Q4} $$

So you're calculating Vds of Q4 and not Vout.

You should add the Vds of Q2 into the equation.

Answer 2

Here is my stepwise thought process to find the Minimum Output Voltage of Cascode Current Mirror.