Fesch
Newbie
Hi,
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 following circuits:
Both pictures 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), can I calculate Vx by using following formula and solving for Vx?
When I try this and compare my results with LTSpice it is very close. Not exactly but I guess this is due to some simplifications in my equations.
How do I do this 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 of M2 (set by the voltage divider of R1 and R3) I get:
Now I insert this in the equation from above:
Solving for Vx I get results that are not so close to my LTSpice simulation anymore.
They are not super far off but I have the feeling that I miss something here.
E.g. For VDD = 3 V, I get Vx = 1.11 V (hand calculation) and Vx = 0.929 V from LTSpice.
Using the same parameters for LTSpice as for my hand calculations.
Or do I have to calculate Vx by using 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!
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 following circuits:
Both pictures 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), can I calculate Vx by using following formula and solving for Vx?
When I try this and compare my results with LTSpice it is very close. Not exactly but I guess this is due to some simplifications in my equations.
How do I do this 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 of M2 (set by the voltage divider of R1 and R3) I get:
Now I insert this in the equation from above:
Solving for Vx I get results that are not so close to my LTSpice simulation anymore.
They are not super far off but I have the feeling that I miss something here.
E.g. For VDD = 3 V, I get Vx = 1.11 V (hand calculation) and Vx = 0.929 V from LTSpice.
Using the same parameters for LTSpice as for my hand calculations.
Or do I have to calculate Vx by using 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!