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How to get a differential gain > 40V/V?

How to get a differential gain > 40V/V with a diferential pair with PMOS input stage and resistive loads with Vdd=2.5V, Vss=-2.5V and Vo=0V by changing any value of bias current, transistor size and resistance?

Differential Amplifier with Resistive Loads

* Set options

.option post probe

* Element Statements

vdd vdd 0 2.5

vss vss 0 -2.5

m1 vout- vin+ 1 vss cmosn w=100u l=1.0u

m2 vout+ vin- 1 vss cmosn w=100u l=1.0u

r1 vdd vout- 50k

r2 vdd vout+ 50k

i1 1 vss 100u

e1 vin+ 2 3 vss 0.5

e2 vin- 2 3 vss -0.5

vcm 2 vss 2.5

vs 3 vss dc=0

* Control statements

.dc vs -0.5 0.5 0.01

.probe v(vout+,vout-)

.tf v(vout+,vout-) vs

.op

* Include device model with filename "cmos05.mod"

.include cmos05.mod

.end

1 Answer

Relevance
  • 1 decade ago
    Favorite Answer

    Since you're not going to change the transconductance of the devices about the only thing you -can- to is increase the load impedance (since load impedance times transconductance equals stage gain). And you'll also have to increase the supply voltage(s) to keep the currents the same (because the transconductance is a function of source-drain current).

    Doug

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