For a simple MOSFET amplifer with active load (shown in Figure 1033 attached), the transistor parameters are: Vtn=1v, Vtp1=Vtp2=-1V, Kp1=Kp2=Kn=100 uamps/V^(2), and lambda1=lambda 2=lambda n=.02V^(-1). Let V+=10V and Iref=100 uAmps.
(a) Find Vsg, the control voltage.
(b) What value of voltage, Vi, will produce Vds0=Vsd2 (saturated voltage)?
(c) Determine the open circuit small signal gain.
After some work, I determined that Vsg=Vtn+sqrt(Iref/Kn)=2V. However, after that, I am stuck. Please tell me how you solve for b and c.
Thank you.© BrainMass Inc. brainmass.com August 15, 2018, 7:13 am ad1c9bdddf
Your answer for Vsg is almost correct. You neglected lambda though. You also left out the 1/2 factor in the IDS saturation equation for a MOSFET (unless your Kp1 constant includes that 1/2 factor). Some textbooks do this different than others.
Some write it as:
IDS_sat = 1/2 * kp * (Vgs - Vtp)^2 * (1 + lambda p Vsd)
others write it as
IDS_sat = kp * (Vgs - Vtp)^2 * (1 + lambda p Vsd), where the 1/2 is inside the kp constant. Since you have done it this way, I'll assume you used this equation.
Your equation will change if you use lambda2, but it will be harder to solve:
Iref = kn * (Vgs - Vtp)^2 * (1 + lambda p*Vsd)
The solution below answers what the control voltage, value of voltage, and the gain in the open circuit small signal is with respect to the MOSFET amplifier with an active load. Transistor parameters, lambdas, and levels of voltage are given to answer these questions.