new lecture!
differential amp, common sources, resistors on the drains, constant current source on the common source
Vt = 0.5V
given kn' * W/L = 4mA/vv
find value of Vid = Vg1-Vg2 that causes all I to flow through Q1 and corresponding Vd1 and Vd2.
KVL: Vg1-Vgs1+Vgs2-Vg2 = 0 
if all I through Q1, Id1 = Wkn'/2L * (Vgs1-Vt)^2 -> Vgs1 = sqrt(2IL/Wkn') + Vt = 0.9472V
KVL: Vd1+Id1Rd-1.5 = 0 -> Vd1 = 1.5-Id1Rd = 0.5V
KVL: Vd2+Id2Rd-1.5 = 0 -> Vd2 = 1.5

so....Vgs2 = Vt

KVL: Vs+Vgs2-Vg2=0 -> Vs+Vt-Vg2=0 -> Vs = Vg2-Vt
Vs+Vgs1-Vg1=0 -> Vs = Vg1-Vgs1
Vg2-Vt=Vg1-Vgs1 -> Vgs1-Vt = Vg1-Vg2 = Vid = 0.9472 - 0.5 = 0.4472V

ex7.3, p696
bias current=.4mA
uCox=.2mA/vv
find W/L ratios and gm if the mosfets are operated at Vov=0.2V
for each value of Vov, give |Vid|max for which (Vid/2Vov)^2 = 0.1 -> Vid = 0.06324V
I=.4mA
from 7.23, id1=I/2+I/Vov * Vid/2 * sqrt(1-(Vid/2Vov)^2) = 3.2e-4A
id1 = Wkn'/2L * Vov*Vov -> W/L = 2id1/VovVovkn' = 80
gm1 = 2Id1/Vov = 2mA/V
always trading gain for linearity
A = -gmRo

common mode rejection - we apparently need a tattoo of this

first, we want to utilize the diff amp in a balanced mode
start off with Vg1=Vg2=Vcm
Vcm must be large enough to keep Q1/Q2 in saturation
mathematically: Vg1 = DC + sig component = Vcm+Vid/2, Vg2 = Vcm-Vid/2

small sig analysis:
Vcm is shorted
Q1/Q2 biased by I/2 by Vcm
id1=gm(Vid/2) = gmVgs1
id1=gm(-Vid/2) = -gmVgs1
also, Vo1+id1Rd=0 -> Vo1=-gmVidRd/2
also, Vo2+id2Rd=0 -> Vo1=gmVidRd/2

Vo1/Vid=-gmRd/2
Vo2/Vid=gmRd/2

define differential gain as Ad = Vo/Vid = (Vo2-Vo1)/Vid = gmRd = I/Vov