Embed Size (px)
Citation preview
Transistor Circuits XVI
Power Amplifiers Part I -
Transformer-Coupled
Basic configuration
Things to know
• 𝑍𝑝 = 𝑍𝑠𝑁𝑝
𝑁𝑠
2
• Zp = Input impedance to the transformer
• Zs = Load or speaker impedance
• Np = Number of turns (windings) in the primary
• Ns = Number of turns (windings) in the secondary – Transformer efficiency ≈ 100%
• 𝑉𝐶𝐸 ≅ 𝑉𝐶𝐶 − 𝑅𝐸𝐼𝐶
• DC resistance of primary turns ≈ 0Ω (negligible).
Circuit for all three examples
68kΩ 8.2kΩ
100Ω
vs
18V
18VVBE = 0.6V
hFE = hfe = 100RL
Transformer
Np/Ns = 20:1
First example problem
• 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
• 𝑉𝐵 =𝑅2
𝑅1+𝑅2𝑉𝐶𝐶 =
8.2kΩ
76.2kΩ18 = 1.937V
• 𝑉𝐸 = 𝑉𝐵 − 𝑉𝐵𝐸 = 1.937 − 0.6 =1.337V
• 𝑉𝐶 = 𝑉𝐶𝐶 = 18V
Second example problem
• If the load resistance is 8Ω, what resistance does the signal current ic in the primary “see” in the circuit shown?
68kΩ 8.2kΩ
100Ω
vs
18V
18VVBE = 0.6V
hFE = hfe = 100RL
Transformer
Np/Ns = 20:1
Second example work
• 𝑍𝑝 = 𝑍𝑠𝑁𝑝
𝑁𝑠
2= 8 20 2 =
3.2kΩ
Third example problem
• 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 8Ω speaker? Hint: The impedance of the primary is like rL of a CE amplifier. Note that the emitter resistance is unbypassed.
68kΩ 8.2kΩ
100Ω
vs
18V
18VVBE = 0.6V
hFE = hfe = 100RL
Transformer
Np/Ns = 20:1
Third example work
• Since the impedance of the primary is like rL, we need to use the formula
𝐴𝑣 =𝑟𝐿
𝑟𝑒′+𝑟𝐸
and then calculate the
value of re’
• Zp = rL = 3.2kΩ; rE = 100Ω
• 𝐼𝐸 =𝑉𝐸
𝑅𝐸=
1.337V
100Ω= 13.37mA
• 𝑟𝑒′ =
25mV
𝐼𝐸=
25mV
13.37mA= 1.87Ω
–(Using 26mV we get 1.945Ω)
• 𝐴𝑣 =𝑟𝐿
𝑟𝑒′+𝑟𝐸
=3.2kΩ
1.87Ω+100Ω=
3.2kΩ
101.87Ω=
31.413
– (Using 1.945Ω for re’ yields a gain of 31.39)
Any questions?
• Contact us at:
– 1-800-243-6446
– 1-216-781-9400
• Email:
