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

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Transistor Circuits XVI Power Amplifiers Part I - Transformer-Coupled

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

Page 1: 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‘ 

Transistor Circuits XVI

Power Amplifiers Part I -

Transformer-Coupled

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Basic configuration

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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‰ˆ 100%

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โ€ข ๐‘‰๐ถ๐ธ โ‰… ๐‘‰๐ถ๐ถ โˆ’ ๐‘…๐ธ๐ผ๐ถ

โ€ข DC resistance of primary turns โ‰ˆ 0ฮฉ (negligible).

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Circuit for all three examples

68kฮฉ 8.2kฮฉ

100ฮฉ

vs

18V

18VVBE = 0.6V

hFE = hfe = 100RL

Transformer

Np/Ns = 20:1

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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.

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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 โˆ’ ๐‘‰๐ต๐ธ = 1.937 โˆ’ 0.6 =1.337V

โ€ข ๐‘‰๐ถ = ๐‘‰๐ถ๐ถ = 18V

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Second example problem

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

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68kฮฉ 8.2kฮฉ

100ฮฉ

vs

18V

18VVBE = 0.6V

hFE = hfe = 100RL

Transformer

Np/Ns = 20:1

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Second example work

โ€ข ๐‘๐‘ = ๐‘๐‘ ๐‘๐‘

๐‘๐‘ 

2= 8 20 2 =

3.2kฮฉ

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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.

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68kฮฉ 8.2kฮฉ

100ฮฉ

vs

18V

18VVBE = 0.6V

hFE = hfe = 100RL

Transformer

Np/Ns = 20:1

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Third example work

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

๐ด๐‘ฃ =๐‘Ÿ๐ฟ

๐‘Ÿ๐‘’โ€ฒ+๐‘Ÿ๐ธ

and then calculate the

value of reโ€™

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โ€ข Zp = rL = 3.2kฮฉ; rE = 100ฮฉ

โ€ข ๐ผ๐ธ =๐‘‰๐ธ

๐‘…๐ธ=

1.337V

100ฮฉ= 13.37mA

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โ€ข ๐‘Ÿ๐‘’โ€ฒ =

25mV

๐ผ๐ธ=

25mV

13.37mA= 1.87ฮฉ

โ€“(Using 26mV we get 1.945ฮฉ)

โ€ข ๐ด๐‘ฃ =๐‘Ÿ๐ฟ

๐‘Ÿ๐‘’โ€ฒ+๐‘Ÿ๐ธ

=3.2kฮฉ

1.87ฮฉ+100ฮฉ=

3.2kฮฉ

101.87ฮฉ=

31.413

โ€“ (Using 1.945ฮฉ for reโ€™ yields a gain of 31.39)

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Any questions?

โ€ข Contact us at:

โ€“ 1-800-243-6446

โ€“ 1-216-781-9400

โ€ข Email:

โ€“ [email protected]

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