In the circuit to the right, V_tp = -0.4 V and K_p = 20 mA/V^2. If I_0 = 250 uA,
ID: 2080061 • Letter: I
Question
In the circuit to the right, V_tp = -0.4 V and K_p = 20 mA/V^2. If I_0 = 250 uA, how small can the common mode voltage get before M1 and M2 leave saturation. in this part assume R_1 is infinite. in parts (b), (c), (d) assume R_1 is 400K. If I_0 = 250 uA, what is the differential mode gain A_d Se = V_o/(V_g1-V_g2) If I_0 = 250 uA, what is the common mode gain A_cm se = V_o/v_cm. What is v_0(t) if A_dse = 20 V/V, the common mode rejection ratio (CMMR) is 46 dB, v_g1 = 0.3sin(10t) + 2sin(60t), V_g2 = 0.5sin(10t) + 2sin(60t)?Explanation / Answer
Given Vtp =-0.4V and Kp = 20mA/V2
a) I0 = 250uA, R1 = 400K
VG = VGS = (400*10^3(10*10^3 + 10*10^3 )/ 5*10^3)5V
= 250u*300*10^3
VG = 75V
b)Differential Mode gain Ad se = V0 / (Vg1 - Vg2)
Vg1 = (Vcc + Vdd) / 2 => (5 - 5)/2 = 0V
Vg2 = (Vcc -Vdd) / 2 => (5 + 5)/2 = 5V
Ad se = 75 / (0-5) = -15
c) Common mode gain Acm se = V0/ Vcm
= [R3/(R1 + R3)][R4 + R2]/[R2-R4/R2]
= 4.268
Gain = VOUT/(V1-V2)
= VOUT/Vd