BJT Biasing and Amplification | Analog Circuits | EE/EC | ESE Previous Year Questions | BYJU'S GATE

[@saurabhjha5401]

Please take a session on Small signal Analysis of BJT and MOSFET.

[@mdsakhwathossain8161]

in last problem, as VE=0 and also 5K ohm resistor is grounded, bjt supposed to be cutoff. Then vo=5 volt, assming zener will not break down

[@syedfardeen1630]

Sir the question which you mentioned wrong at 35:23 was actually not wrong. The answer to this question is (B) -15V. Here is the solution-: Ve= -8 + 0.7 Ve= -7.3 Icq= 3 mA (given) Vc= 2Ic + Vcc= 6 + Vcc Vb= -8 For action region, Vc

[@KumarPattapu-ym2wu]

32:53 Yes you, are right , but my solution in another method, assume that bjt in saturation and lets take Vec =0.2 volt and it gives Vc=-7.5 volt, then (-7.5-Vcc)/2k = 3mA it gives Vcc =-13.5 volt , and for active region Vec greater than 0.2 it gives Vcc less than -13.5, then option B) is correct -15volt