Exercise 8.6 A bipolar output stage

a)


An answer:
With the transistor characteristics and with the circuit in the previous figure, vOUT (t) = vG(t).

For 0 t πω, vG(t) 0 and the NPN conducts. Hence the current from the positive supply voltage during one period of vG is:

iV CC = V G sin(ωt) RL (0 t πω) iV CC = 0(πω t 2πω)

Using symmetry:

iV CC = V G sin(ωt) RL (πω t 2πω) iV CC = 0(0 t πω)

b)


An answer:

PV CC = V CC V G 2RLπω 0πωsin(ωt)dt = V CC V G RL 1 2π0πsin(𝜃)d𝜃 = V CC V G RL 1 π PV CC = V CC V G 2RLπω πω2πωsin(ωt)dt = V CC V G RL 1 2ππ2πsin(𝜃)d𝜃 = V CC V G RL 1 π

c)


An answer:

PRL = V G2 2RLπω02πωsin(ωt)2dt = V G2 RL 1 2π02πsin(𝜃)2d𝜃 = V G2 2RL

d)


An answer:

Psupply = 2 π0.64W PRL = V G2 2RL = 0.5W η 0.785

e)


An answer:
This is the power dissipated by the transistors

f)