BJT

Question 1

Power dissipation of a bjt

Answer 1

Vce x Ic

Question 2

Dc current gain

Answer 2

B = Ic/Ib

Question 3

Describe cutoff region ( 3 points)

Answer 3

1. Ib , Ic , Ie = 0A
2. Ib=0A when Vbe <0.7V
3. collector and emitter behave as an open circuit - turned off switch

Question 4

Describe saturation region ( 4 points)

Answer 4

1.Ib , Ic , Ie > 0A
2. Vbe =0.7V when Ib>0A
3. BJT is operating in saturation region when Ib , Ic , Ie > 0A and Vce ≤ 0.3V
4. collector and emitter behave as an short circuit - turned on switch

Question 5

Describe active region ( 4 points)

Answer 5

1.Ib , Ic , Ie > 0A
2. Vbe =0.7V when Ib>0A
3. BJT is operating in active region when Ib , Ic , Ie > 0A and Vce > 0.3V
4. when bjt is used as an amplifier , it operates in active region

Question 6

Fixed biasing circuits

Answer 6

input loop equation
Vcc - IbRb -0.7V =0

output loop equation
Vc = Vcc - IcRc
Vcc - IcRc - Vce =0

Question 7

Emitter stabilising biasing circuit

Answer 7

input loop equation
Vcc - IbRb -0.7V- IeRe=0

Vcc - IbRb -0.7V - (1+B)IbRe =0

output loop equation
Vc = Vcc - IcRc
Vcc - IcRc - Vce-IeRe =0

Question 8

Voltage divider biasing circuit

Answer 8

1. B x Re ≥ 10 x R2 - use approximate analysis to calculate current and voltage if conditions are true
2. voltage divider rule
   Vb =(R2 / (R1+ R2 ) )Vcc

Question 9

Dc Bias with voltage feedback cricuit