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Transistor Circuits XVI Power Amplifiers Part I - Transformer-Coupled

# Transistor Circuits XIV - Electronics · Transistor Circuits XVI Power Amplifiers Part I - Transformer-Coupled . Basic configuration. Things to know • 𝑍𝑝=𝑍 𝑁𝑝 𝑁𝑠

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### Text of Transistor Circuits XIV - Electronics · Transistor Circuits XVI Power Amplifiers Part I -... 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?

– 1-800-243-6446

– 1-216-781-9400

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