where we left off...
Io/Iref = area of EBJ Q1/area of EBJ Q2 = 1/(1+2/beta), want beta large
if beta very high,  Io~~Iref
ex: beta = 100 -> Io/Iref = 1/1.02 = .9803
	beta = 300 -> Io/Iref = 1/1.007 = .993

Io/Iref = m != 1

Iref = Ic1 + Ib1 + Ib2 = Ic1(1 + (1+m)/beta)

Io = m*Iref/(1+(1+m)/beta) -> Io/Iref = m/(1+(1+m)/beta)

Ro2 = delVa/delIo = Va2/Io

try to derive correction term

exercise 6.6:
recall Iref = 2Ic1/beta + Ic1 = Ic1(1+2/beta) -> Ic1 = Iref/(1+2/beta) = 9.80392e-4
beta = 100
Iref = 1mA
Note: Vbe - Vce1 = 0 -> Vce1 = Vbe, ditto for Vce2, so Vce1 = Vce2
Vbe = Vtln(Ic1/1e-15) = .026ln(98) = 0.7199V
Ic1 = Is1 e^(Vbe/Vthresh) = 1e-15*98e10 = 98e-5 = 0.98mA
Ro = Ro2 = Va2/Io = 100/1e-3 ~~ 100Kohms

Io = Iref*m/(1+(1+m)/beta) * (1+(Vo-Vbe)/Va2) = .001*1/(1+2/beta) * (1+ (5-.7179)/100) = 1.02mA

frequency response for amplifiers:
parasitic capacitances: gate/drain, gate/source, gate/body, etc

good range for audio is 20Hz to 20KHz

open circuit time constant metnod (OCTC)
2 port amplifier w/ Vin and Vout
not easy: 
	1. derive the input/ouput voltage transfer ratio using node or mesh eqs: Vo/Vin = 
	2. set s = jw
	3. mag(Vo(jw)/Vin(jw)) = M
	4. set M = 1/sqrt(2)
	5. solve for w, then f = w/2pi

easy:
methodology: example 3 pole transfer function
T(s) = Vo(s)/Vi(s) = a0/((tau1s+1)(tau2s+1)(tau3s+1))
multiply out denominator
b3 = tau1tau2tau3
b2 = tau1tau2 + tau1tau3 + tau2tau3
b1 = tau1 + tau2 + tau3
b0 = 1
T(s) = a0/(b3sss + b2ss + b1s + b0)

assertion: near the 3db point, the 1st order term typically dominates all other terms
T(s) ~~ a0/(b1s + 1) = a0/(sum<i=1,n>(taui*s + 1))

OCTC method
	1. compute the thevenin resistance facing each capacitor with all other capacitors killed (open)
	2. form the product tauj = RjCj
	3. sum all time constants (b1)
	4. wv(estimated) = 1/(sum<j=1,n>(RjCj))


look at example 6.6, we'll work it thursday