Question

2) Design a BJT differential amplifier to amplify a differential input signal of 0.2 V and provide a differential output signal of 4 V. To ensure adequate linearity, it is required to limit the signal amplitude across each base-emitter junction to a maximum of 5 mV, note this means emitter resistors. Another design requirement is that the differential input resistance be at least 80 k22. The BJTs available are specified to have ß > 200. Give the circuit configuration and specify the values of all components.

Answer 1

Answer :

Given that :BJT differential amplifier to amplify a differential input signal of 0.2V and provide a diffeential output signal of 0.2V

جسے Vs = y 05m) +45mv + - m 5 نے وار 0.2y IELM NN RE IELE 2

Rin
Σ 80KΩ ;β Σ 200

ta24 Az(diff) = 0.2 20

Re = 20(RE+r.)...

VRE e RE 95 Fromdiagram oriere = 19 5

RE = 19re........ ..............2

Pid = 2(8+1) Retre) > 80K......

Substitute 2 in 1,we get

2(201)(19r_e+r_e)

> 80K

80 x 103 Tez 2 x 20 2 x 201 x 20 = 9.9522

re>9.9512..........

Substitute 4 & 5in 1,we get

R_c=20(189.05+9.95)

R_c=3.98K$\Omega$

Te2 = lei = = gm = gm

11 Pe 9.95 = 0.1005

VI 9.95

I 25 x 10-3 Ica2 = Ical = VT 9.95 = 9.95 - = 2.51m A

I=I_E1+I_E2

I=(2.51+2.51)mA

I=5.02mA

Modified Diagram :

Rc 3.98k2 Rc 3.98ke - v. 4g + $t q kt TIE, my +95 my IE2 +950V- M RE = 189.05.12 189.05 B> 200 1=5.02MA Rin 80K IE, - IEZ - 2.51mA