Coupling Capacitor Values - Emitter-Follower Power Amplifier

Thread Starter

elec_eng_55

Joined May 13, 2018
214
Hi:

I have simulated an Emitter-Follower Power Amplifier in LTSpice. I am confused
about the required value for C1 and C2.

The computed values are as follows:
C1 = 1/[2 * π * 20Hz * 152Ω * 0.1] = 524 μF
C2 = 1/[2 * π * 20Hz * 8Ω * 0.1] = 10 μF

However, in order for the output voltage to start at zero in transient graph,
C1 = 1,000 μF and C2 = 2,500 μF.

Why would this be?

David
 

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Ylli

Joined Nov 13, 2015
1,086
C1 = 1/[2 * π * 20Hz * 152Ω * 0.1] = 524 μF
C2 = 1/[2 * π * 20Hz * 8Ω * 0.1] = 9947 μF

The starting at less than zero is simple the RC phase shift.
 

Audioguru

Joined Dec 20, 2007
11,248
Nobody makes an audio power amplifier like that. Instead it is a room heater that heats with 9.42W all the time even when it is not playing sounds. The output clips when the power in the 8 ohm speaker is only 0.76W.

Your amplifier operates in class-A, but real modern audio power amplifiers are cool class-AB or cold class-D.
With a 12V supply, a class-AB amplifier would heat with about 0.2W without sounds and produce an output at clipping of about 1.56W.
A bridged class-D amplifier would produce almost no heat and produce 8.3W into the 8 ohm speaker.
 
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