Problem 1:

Source follower controls the current, diode connected load device will always be in saturation, but when the current is too high, vout will go too low and push the input device ohmic

Always: vout=Vdd-Vtp-Vonload

where Vonload=sqrt(Id/(Kp’/2*Wp/Lp))

Id=Kn’/2*Wn/Ln*(vin-Vtn) if vout>vin-Vtn so just plug this equation into the first

If vout<vin-Vtn, Id=Kn’*Wn/Ln*((vin-Vtn)*vout-vout²/2), so plug in above and solve the quadratic equation.

Problem 2:

This is a basic source follower

Gain: vout/vin=-gm/gds/(1+sCl/gds)

Input referred noise: vn²=8/3 kT 1/gm df

UGBW=gm/(2*pi*C)

Von=sqrt(Id/(K’/2*W/L)<voutmin

The situation is this: gm needs to meet a certain minimum value in order to meet the noise and bandwidth specs. One of the two is more stringent. Find that one.

Work with the given current, calculate W/L. Check von, if Von is too large, simply increase W/L, Higher W/L makes von smaller, bandwidth wider and noise lower.

Problem 3:

When simplifying this without gds and Cgd, we end up with two poles. One where the input current gets converted to a gate voltage – the pole is at

P0=gmn/2*pi*C1

And one where the output current from the first mirror is converted to a gate voltage in the p-mirror.

P1=gmp/2*pi*C2

Where C1 and C2 are composed of Cgs(1x), Cgs(10x) plus Cdb on the input side (and for C2 also the Cdb of the n mirror output).

Cgs dominates and is a function of W*L. Choose L=Lmin. Now gm is a function of sqrt(W) and Cgs is a function of W, so a smaller W will give a higher bandwidth.

Set the pole value to 100MHz, then find the W value for which PO is ok, then repeat for the p – side.

Once Wp is determined, Id is known and Lp is known, we can calculate voutmax as

Vdd-vonp