# Coupling Capacitor Values - Emitter-Follower Power Amplifier

#### 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

#### Attachments

• 1.9 KB Views: 14

#### Ylli

Joined Nov 13, 2015
1,063
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.

#### elec_eng_55

Joined May 13, 2018
214
Thanks very much!

David

#### 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.