looking at a Vo/Io graph
gives us a conductance curve same as 1/Ro
Vov is where the conductance curve starts to reasonably flatten
have to find lowest possible value of Vo to maintain mostly flat curve region
Va' = early voltage per length, multiple by 1 micron to get early voltage
find change in output current resulting from change in Vo

assuming Q1,Q2 matched perfectly
must find Vgs
Iref = Id1 = 1/2 * kn' * (W/L)(Vgs - Vthresh)^2
100uA = 1e-4 * 10 *(Vgs - .7)^2 -> Vgs = 1.0162V
then, Vov = Vgs - Vthresh = .3162V
R = (Vdd - Vgs)/Iref = 19840 ohms
Va' = Va/L
Va = Va'L = 20V
Ro2 = Va/Io = 20/1e-4 = 200Kohms = deltaVo/deltaIo -> deltaIo = 1/200000 = 5uA

desire: A's voltage out changes by 1V
we want Io to vary no more than 1%

given 1% of Io = 1uA
So: Ro2 = delVo/delIo = 1uA
Ro2 = Va'L/Io = 1uA = 10L/1e-4

W/L equal between parts
10/1 = W2/5 -> W2 = 10um


current mirror setup can be used to create multiple mirrors off of the gate node


solution for figure 6.7 (a current steering circuit):
I2 = Iref*(W2/L2)/(W1/L1) = 60uA -> W2/W1 = 6
I3 = Iref*(W3/L3)/(W1/L1) = 20uA -> W3/W1 = 2
I4 = I3
I5 = I4*(W5/L5)/(W4/L4) = 80uA -> W5/W4 = 4

Vov2 = .2V
Id2 = 1/2 * kn' * (W2/L2) * Vov*Vov -> W2 = 15um -> W1 = 2.5um -> W3 = 5um
I5 = 1/2 * kn' * (W5/L5) * Vov5*Vov5 -> W5 = 50um -> W4 = 12.5um



current mirror with BJTs, almost exactly the same setup
problem is that gates don't transfer 0 current
Iref = Ic + 2Ic/beta = (1+2/beta)Ic, assume Io = Ic
Io/Iref = Ic/(Ic(1+2/beta)) = 1/(1+2/beta)
high beta drives 2/beta term toward 0 -> good