Question

Answer 1

(a) Circuit1..


For Si , voltage across diode in forward bias = 0.7 V (as shownabove)

Therefore v1 - Vo = Vd = 0.7 V
=> v1 = 0.7V + Vo

Using KCL at v1 ,

10mA = v1/2.2k + Id

=> 10mA = 0.7V/2.2k + Vo/2.2k + Id

=>Vo = 10*2.2 V- 0.7 V - 2.2k *Id

=>Vo = 21.3V - 2.2k*Id--------> (1)

Using KCL at Vo ,

Id = Vo /1.2k

=> 1.2k * Id = 21.3 V -2.2k*Id (by(1) )

=> 3.4k*Id = 21.3 V

Therefore Id = 21.3V/3.4kΩ= 6.265 mA


(b) Circuit 2..........



Onceagain we haveVo - (-5V) =0.7V (For Si , Vd= 0.7 V in forward bias)

=>Vo = 0.7 - 5 = -4.3V

Therefore, Id= (+20V - Vo) / 6.8kΩ
=(+20 V - (-4.3V) ) / 6.8kΩ
=24.3 V / 6.8*1000Ω
=3.57 mA

I Hope this helps you