Transfer Function of a FET Differential Amplifier

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CoolDude420
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Homework Statement


Circuit schematic of a MOSFET differential amplifier with current source I1 feeding output Vout1 at M1 drain, resistor R1 feeding Vout2, input Vin on M1 gate, bias Vb on M2 gate, tail current source Iss to ground, and supply VDD.

[Taken from Razavi's Design of Analog CMOS Integrated Circuits 2nd edition]

Homework Equations

The Attempt at a Solution


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I'm not too sure what to do with this question. Here's what I think of the circuit

This doesn't look like a differential amplifier to me since there's only 1 input. From what, I see, M1 has a constant DC bias current through it of I1. I see the following relationship
$$I_2=I_{SS}-I_1$$
Leading to,
$$V_{out2}=V_{DD}-(I_{SS}-I_1)R_1$$

Note that I2 is the current through R1 and M2

As Vin small signal changes, this I1 is perturbed, which in turn perturbs I_2 which in turns adjusts voltage drop across R1 which leads to Vout2 having a small-signal change. Now, I'm not sure how to get Vout1/Vin or Vout2/Vin. I was thinking of maybe adjusting the equation up there and sub in the Id equation for I1 but that would leave me with a quadratic and squared Vin term.

Also if there's any experienced analog people out there. Is my thought process correct for understanding the circuit or along the right lines?
 
Attachments
  • Circuit schematic of a MOSFET differential amplifier with current source I1 feeding output Vout1 at M1 drain, resistor R1 feeding Vout2, input Vin on M1 gate, bias Vb on M2 gate, tail current source Iss to ground, and supply VDD.
    3OEfR.png
    12.2 KB · Views: 1,339
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CoolDude420 said:
As Vin small signal changes, this I1 is perturbed, which in turn perturbs I_2 ...

But you are given that I1 is an ideal current source. That means it is a constant current regardless of Vin and Vout1, so changes in Vin cannot "perturb" I1.
 
phyzguy said:
But you are given that I1 is an ideal current source. That means it is a constant current regardless of Vin and Vout1, so changes in Vin cannot "perturb" I1.

Hmm good point. In that case then, I have another question. What about like a standard CS source stage with a current source like this
Circuit schematic of a common-source stage with current-source load: VDD at top, current source I1 feeding node Vout, NMOS transistor M1 with gate input Vin and source grounded.


How do these circuits operate then?
 
Attachments
  • Circuit schematic of a common-source stage with current-source load: VDD at top, current source I1 feeding node Vout, NMOS transistor M1 with gate input Vin and source grounded.
    COMMON-SOURCE+STAGE+with+current-source+load.jpg
    20.1 KB · Views: 807
If M1 between D and S appears as a variable resistance controlled by VGS then the voltage across that varying resistance, VDS, can vary even while ID is held fixed.
 
CoolDude420 said:
Hmm good point. In that case then, I have another question. What about like a standard CS source stage with a current source like this()
How do these circuits operate then?

I1 doesn't change, but Vout does change as a function of Vin.

In your original circuit, I didn't say Vout1 doesn't change. It does change as you change Vin.