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Transistor Circuits XVI Power Amplifiers Part I - Transformer-Coupled

# Transistor Circuits XIV - Electronicsย ยท Transistor Circuits XVI Power Amplifiers Part I - Transformer-Coupled . Basic configuration. Things to know โข ๐๐=๐ ๐๐ ๐๐

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### Text of Transistor Circuits XIV - Electronicsย ยท Transistor Circuits XVI Power Amplifiers Part I -...

Transistor Circuits XVI

Power Amplifiers Part I -

Transformer-Coupled

Basic configuration

Things to know

โข ๐๐ = ๐๐ ๐๐

๐๐

2

โข Zp = Input impedance to the transformer

โข Zs = Load or speaker impedance

โข Np = Number of turns (windings) in the primary

โข Ns = Number of turns (windings) in the secondary โ Transformer efficiency โ 100%

โข ๐๐ถ๐ธ โ ๐๐ถ๐ถ โ ๐๐ธ๐ผ๐ถ

โข DC resistance of primary turns โ 0ฮฉ (negligible).

Circuit for all three examples

68kฮฉ 8.2kฮฉ

100ฮฉ

vs

18V

18VVBE = 0.6V

hFE = hfe = 100RL

Transformer

Np/Ns = 20:1

First example problem

โข 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

โข ๐๐ต =๐2

๐1+๐2๐๐ถ๐ถ =

8.2kฮฉ

76.2kฮฉ18 = 1.937V

โข ๐๐ธ = ๐๐ต โ ๐๐ต๐ธ = 1.937 โ 0.6 =1.337V

โข ๐๐ถ = ๐๐ถ๐ถ = 18V

Second example problem

โข If the load resistance is 8ฮฉ, what resistance does the signal current ic in the primary โseeโ in the circuit shown?

68kฮฉ 8.2kฮฉ

100ฮฉ

vs

18V

18VVBE = 0.6V

hFE = hfe = 100RL

Transformer

Np/Ns = 20:1

Second example work

โข ๐๐ = ๐๐ ๐๐

๐๐

2= 8 20 2 =

3.2kฮฉ

Third example problem

โข 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 8ฮฉ speaker? Hint: The impedance of the primary is like rL of a CE amplifier. Note that the emitter resistance is unbypassed.

68kฮฉ 8.2kฮฉ

100ฮฉ

vs

18V

18VVBE = 0.6V

hFE = hfe = 100RL

Transformer

Np/Ns = 20:1

Third example work

โข Since the impedance of the primary is like rL, we need to use the formula

๐ด๐ฃ =๐๐ฟ

๐๐โฒ+๐๐ธ

and then calculate the

value of reโ

โข Zp = rL = 3.2kฮฉ; rE = 100ฮฉ

โข ๐ผ๐ธ =๐๐ธ

๐๐ธ=

1.337V

100ฮฉ= 13.37mA

โข ๐๐โฒ =

25mV

๐ผ๐ธ=

25mV

13.37mA= 1.87ฮฉ

โ(Using 26mV we get 1.945ฮฉ)

โข ๐ด๐ฃ =๐๐ฟ

๐๐โฒ+๐๐ธ

=3.2kฮฉ

1.87ฮฉ+100ฮฉ=

3.2kฮฉ

101.87ฮฉ=

31.413

โ (Using 1.945ฮฉ for reโ yields a gain of 31.39)

Any questions?

โ 1-800-243-6446

โ 1-216-781-9400

โข Email:

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