Midterm Practice Problems: Solutions Part 2

Problem 4

For minimum Vcm, make sure tail current source stays in saturation

Vcm_min=Vt+von_diff+von_tail

Where von_tail=sqrt(Itail/(Kn’/2*Wtail/Ltail))

And von_diff= sqrt(Itail/(Kn’*Wdiff/Ldiff))

For max vcm, make sure, diffpair device stays in saturation

Vds>=Vgs-Vt

VDD-Vtp-Von_mirror-Vs>Vcm_max-Vs -Vtn

So Vcm_max<VDD-Vtp+Vtn-Von_mirror

Where Von_mirror= sqrt(Itail/(Kp’*Wmirror/Lmirror))

Max output voltage:

Keep mirror device in saturation

Vout<Vdd-Von_mirror

Min output voltage

Keep diffpair device in satuation

Vds>=Vgs-Vt

Vout-vs>Vcm—Vs-Vt

So Vout_min>Vcm_max-Vt

Problem 5

Identify devices as

Bottom left: voltage controlled current source, will produce current gm*vin

Top: constant current source

Top right: cascade device holds voltage at its source constant, passed current

Therefore vout=-gm*R

Conditions for saturation

Label current in bottom left device I1, in right branch I2, so top device carries I1+I2

Voltage at the source of the cascade device

Vs=vbs+Vtp+von_casc where von_casc= sqrt(I2/(Kp’/2*Wcasc/Lcasc))

Vs>von_in

Vdd-Vs>Von_currentsource

To keep cascade device in saturation:

Vb2+Vtp>Vout

Vout=R*I2

Problem 6

Error in the drawing, intention was to have bottom device be a p-device

Then if we neglect gds and gmb we have

Vin-Vb=Vtp+Vtn+Vonp+vonn+R*Iout

Or

Vin-Vb-Vtp-Vtn=R*Iout+sqrt(Iout/(Kn/2’*Wn/Ln))+ sqrt(Iout/(Kp/2’*Wp/Lp)

Which is a quadratic equation for Iout.

Small signal we should get a transconductance Iout/Vin=1/(1/gmn+1/gmp+R)

How would one solve the problem with the devices as drawn?

If the bottom device is in saturation, it sets the current. Iout=Kn’/2*W/L*(Vb-Vs-Vtn)^2

where Vs is the voltage at the source of the bottom device.

In that case, Vin simply sets the voltage at the drain of the bottom FET and thus does not affect the current (neglect gds)

If the bottom device were ohmic, then we could replace it with a resistor, place it in series with the existing resistor and the resulting transfer function would be

Iout/vin=1/(1/gm+R+1/gds_ohmic))