Hey,
i am currently trying to learn and understand analog CMOS design for which I am using Razavis Book "Design of Analog CMOS Integrated Circuits".
Right now, I am reading about Current Mirrors and had a look at the tasks at the end of section 5.
The task is to sketch the voltages Vx and Vy as a function of VDD for the circuits I have attached. They are taken from Razavi - "Design of Analog CMOS Integrated Circuits" page 168.
I would like to not just sketch but to actually calculate the voltages to understand how it works.
Taking the circuit of picture (a), I have the following relation:
$$V_x = V_{DD} - I_{DM2} \cdot R1$$
$$V_x = V_{DD} - k_n \cdot \frac{W}{L} \cdot (V_x - V_{th})^2 (1+ \lambda Vx) \cdot R1$$
Here I solve for Vx and insert different values for VDD. When I compare that to LTSpice simulations it is quite close. I guess small deviations are due to simplifications in the hand calculations.
Can I do the same for the circuit in picture c?
If I understand correctly then my voltage Vx is first governed and set by the voltage divider until it reaches Vx. Then the transistor turns on in pinch-off and draws current from VDD.
Writing down the currents for the case that the voltage Vx already reached Vth (threshold voltage) of M2 (set by the voltage divider of R1 and R3) I get:
$$I_{R1} = I_{DM2} + I_{R3}$$
Now I insert this in the equation from above:
$$V_x = V_{DD} - I_{R1} \cdot R1$$
$$V_x = V_{DD} - (k_n \cdot \frac{W}{L} \cdot (V_x - V_{th})^2 (1+ \lambda Vx) + \frac{Vx}{R3}) \cdot R1$$
Again I solve for Vx and comparing that to LTSpice simulations it seems to be quite close, too. But not as close as the first example.
My question:
Is that the correct way to calculate the operating point for the pictures a and c?
Or do I have to use superposition? Since the voltage divider sets the voltage Vx only until Vx = Vth I do not understand how this would work.
Thank you very much in advance!