[SOLVED] Why is the DC level of the output not well-defined?

[alexyangfox]

The book says ,
the DC output level is not well-defined.
When the DC voltages of Vb,Vin and Vdd are defined,
I think if I have the equations for Id of the two MOSFETS equal,
I can have a solution for Vout.
If so, Vout shouldn't be not well-defined.
could anyone explain this to me?
Thank you in advance.
yours sincerly ,
alex.

 

[big2]

You are right,for this open circuit Vout is not well-defined,
so as usual we must add a Feedback circuit to get we want.

 

[pavel_adameyko]

Hello
in open circuit variation gain A=Vout/Vin is rather high, so output is very sensitive to variations of Vb, Vin, temperature, mismatch and so on

 

[alexyangfox]

Hello
in open circuit variation gain A=Vout/Vin is rather high, so output is very sensitive to variations of Vb, Vin, temperature, mismatch and so on

Hello,pavel_adameyk
Could you explain it in more details?

Is the key point the HIGH GAIN?

if the upper MOSFET is diode connected,will the output be well-defined?

Maybe I don't quite understand what "well-defined" means.

thank you very much.
Best wishes to you.

 

[pavel_adameyko]

Is the key point the HIGH GAIN?

if the upper MOSFET is diode connected,will the output be well-defined?

Maybe I don't quite understand what "well-defined" means.

thank you very much.
Best wishes to you.

Hi
Yes you are right - if the upper PMOS is diode-connected, the output is well-defined. It is equal to VDD - Vgs,pmos. If we assume VDD is constant then all variations of output is due to Vgs,pmos. This quantity has rather little spread over temperature, corners and mismatch. For example, if typical value of Vgs,pmos is near 600 mV, sigma would be near tens milliVolts. Also, as the gain of amplifier equal to Gm,nmos/Gm,pmos (rather low) little variations of Vin wouldn't lead to large deviations at output.

In your case gain is equal to Gm,nmos*(Ro,pmos//Ro,nmos) which can be as much as 100, so if the input differ for let's say 20 mV from it original value, the output would deviate for 2V. In such circuit due to temperature, parameters mismatch the output can be any value between GND and VDD. So you need to use negative feedback to stabilize it (like when you work with OPAMPs)

Regards.

 

[alexyangfox]

Thank you very much.
Very helpful!
Best wishes to you.