Transistor Circuits XVI Power Amplifiers Part I - Transformer-Coupled
Transistor Circuits XVI
Power Amplifiers Part I -
Transformer-Coupled
Basic configuration
Things to know
β’ ππ = ππ ππ
ππ
2
β’ Zp = Input impedance to the transformer
β’ Zs = Load or speaker impedance
β’ Np = Number of turns (windings) in the primary
β’ Ns = Number of turns (windings) in the secondary β Transformer efficiency β 100%
β’ ππΆπΈ β ππΆπΆ β π πΈπΌπΆ
β’ DC resistance of primary turns β 0Ξ© (negligible).
Circuit for all three examples
68kΞ© 8.2kΞ©
100Ξ©
vs
18V
18VVBE = 0.6V
hFE = hfe = 100RL
Transformer
Np/Ns = 20:1
First example problem
β’ Find the dc base-to-ground voltage VB, collector-to-ground voltage VC, and emitter-to-ground voltage on the circuit shown. Assume that the dc resistance of the primary windings is negligible.
First example work
β’ ππ΅ =π 2
π 1+π 2ππΆπΆ =
8.2kΞ©
76.2kΞ©18 = 1.937V
β’ ππΈ = ππ΅ β ππ΅πΈ = 1.937 β 0.6 =1.337V
β’ ππΆ = ππΆπΆ = 18V
Second example problem
β’ If the load resistance is 8Ξ©, what resistance does the signal current ic in the primary βseeβ in the circuit shown?
68kΞ© 8.2kΞ©
100Ξ©
vs
18V
18VVBE = 0.6V
hFE = hfe = 100RL
Transformer
Np/Ns = 20:1
Second example work
β’ ππ = ππ ππ
ππ
2= 8 20 2 =
3.2kΞ©
Third example problem
β’ In the circuit shown, let vo be the signal voltage across the primary turns of the transformer. What is the ratio vo/vs if the load on the secondary is an 8Ξ© speaker? Hint: The impedance of the primary is like rL of a CE amplifier. Note that the emitter resistance is unbypassed.
68kΞ© 8.2kΞ©
100Ξ©
vs
18V
18VVBE = 0.6V
hFE = hfe = 100RL
Transformer
Np/Ns = 20:1
Third example work
β’ Since the impedance of the primary is like rL, we need to use the formula
π΄π£ =ππΏ
ππβ²+ππΈ
and then calculate the
value of reβ
β’ Zp = rL = 3.2kΞ©; rE = 100Ξ©
β’ πΌπΈ =ππΈ
π πΈ=
1.337V
100Ξ©= 13.37mA
β’ ππβ² =
25mV
πΌπΈ=
25mV
13.37mA= 1.87Ξ©
β(Using 26mV we get 1.945Ξ©)
β’ π΄π£ =ππΏ
ππβ²+ππΈ
=3.2kΞ©
1.87Ξ©+100Ξ©=
3.2kΞ©
101.87Ξ©=
31.413
β (Using 1.945Ξ© for reβ yields a gain of 31.39)
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