Question

10v R3 R4 20k 20K Vo R2 38.6k M1 M2 V2 R6 R5 500K 500K Q2 Q1 -10v

For this differential amplifier determine:
a) The differential gain
b) The CMRR

Given:
MOSFET: K=1mA/V^2, Vth=1V, r0=infinity
BJT: Beta=200, Va=100

Answer 1

Soli Data given in the question is as follows K = 1mA/p Vin=lv, rozdo B= 200 VA=100 tlov Rz=20k • Ry= 20K vo R₂ & 38.6k V2 IFY R6= 500k - БъК I Qy Q2 - lov The left side of the abore figure is a Bút Current Mirror Honce Applying Kr Meget- 10-38.6K I - 0.7 to=0 → 20- Oritz 38.6K I E Transistor , and Q2 19.3=38.6KI TT Perfectly matched. 0.5mA Iz 19.3 38.6K Iz 0.5mA

id = id 2 = I = 0.25mA idi= id= 0.25mA Now We know that formula for Trans conductance in a Mosfet is given by - I am Jan Bunge la 2L Kna inconco IMA/ 9 id=0-25mA Subistating & the above Dota in equation Veget- Im= 2x2 x lxo 3x0.25x103 Gm= 1xło Honce, the Diffential gain is given by (Adl= - gm Ro gmro -Gm Ro 2 | Ad) lxo 3x 20x103 Ad 20 Similarly the Differential Mode gain is given by Common Ac = .Ru RE - Ro Rc 20 N 500 25 Ac=

(11) CMRR: It is the ability of a Circuit to Reject the Common Mode gain, it is given by- CMRR) = Ad AC How subsiteting the Values of Ad and Ac in equation ② Weeget- 20 (CMRP) = (1/25) (CMRR) = 500 In decibels hee Con expressed as- (CMRR) dB = AC = 53.67 dB 20log | Ad (COMPR) aga 20log (500) (CMRR do 2 53.67 dB Hence, the value of CMRR is 53.67 dB.