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 ≤ π∕ω, 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 ≤ π∕ω) iV CC = 0(π∕ω ≤ t ≤ 2π∕ω)

Using symmetry:

i−V CC = V G ⋅ sin(ωt) RL (π∕ω ≤ t ≤ 2π∕ω) i−V CC = 0(0 ≤ t ≤ π∕ω)

b)


An answer:

PV CC = V CC ⋅ V G 2RLπ∕ω ∫ 0π∕ωsin(ωt)dt = V CC ⋅ V G RL 1 2π∫ 0πsin(𝜃)d𝜃 = V CC ⋅ V G RL 1 π P−V CC = −V CC ⋅ V G 2RLπ∕ω ∫ π∕ω2π∕ωsin(ωt)dt = −V CC ⋅ V G RL 1 2π∫ π2πsin(𝜃)d𝜃 = V CC ⋅ V G RL 1 π

c)


An answer:

PRL = V G2 2RLπ∕ω∫ 02π∕ωsin(ωt)2dt = V G2 RL 1 2π∫ 02πsin(𝜃)2d𝜃 = V G2 2RL

d)


An answer:

Psupply = 2 π≅0.64W PRL = V G2 2RL = 0.5W η ≅0.785

e)


An answer:
This is the power dissipated by the transistors

f)