Emitter follower o/p coupling cap question

Thread Starter

samy555

Joined May 24, 2010
116
I've designed an Emitter Follower stage working on a 3-volt battery and IC = 3.25 mA
I do the design so that the output Impedance = 8 ohms because I want to run a 8 ohm load. I do not intend to actually run the circuit, I do it for study and analysis.​
Her is the circuit and some of my calculations.​
The input signal was 300mvolt 1kHz, and the problem was that the output signal was distorted as the following picture​

But when I removed the coupling output capacitor C1 I got a clean output signal like this:​

How can I get a clean output signal in the presence of C1??​
Thanks​
 

crutschow

Joined Mar 14, 2008
25,686
The problem is that you calculated values for the small signal linear operation of the amp but it is actually operating in the large signal region where the dynamic range of the circuit comes into play. As long as you keep the peak AC signal at less that 20mV or so you stay in the linear region and your calculations work. Above that you need to look at the circuit biasing which affects the large signal operation.

R3 must supply (sink) the current through the capacitor to the load for the negative portion of the waveform, which limits the negative current. You have biased the base of the transistor at about 2.4V meaning the emitter is biased at about 1.8V. This voltage will be divided by the ratio between the 8 ohm and 470 ohm resistor on the negative half cycle. The maximum negative voltage is thus 1.8V * (8 /(8+470)) = 30mV, which is about what your simulation shows.

To get a larger negative voltage you need to bias the base at a higher voltage and/or reduce the value of R3. Typically to drive a low impedance load such as a speaker a push-pull complimentary (class B or AB) stage is used where a NPN provides the positive current and a PNP provides the negative current. That way you don't have to waste so much bias current to get a reasonable output level.
 

crutschow

Joined Mar 14, 2008
25,686
The Zo applies to a small signal (less than 20mV peak as I previously stated) but not for a large signal. For a negative going large signal the current can't be supplied by the transistor since the transistor can only provide positive current, thus it has to be supplied by R3 to ground and its impedance is 470Ω. The small signal equations you show don't apply to a large signal.
 

studiot

Joined Nov 9, 2007
4,998
You have biased the base of the transistor at about 2.4V meaning the emitter is biased at about 1.8V.
Can you explain this further?

3*(20/26.2) = 2.29 and this will be reduced further by any shunting effect of the transistor base current.

Also if samy is right about Ic = 3.25mA I make Ve to be around (3.25)*(.47) = 1.53 volts

Of course 2.29 - 1.53 = 0.76, which is a tad on the high side for a well biased 2N3904
 

crutschow

Joined Mar 14, 2008
25,686
Can you explain this further?

3*(20/26.2) = 2.29 and this will be reduced further by any shunting effect of the transistor base current.

Also if samy is right about Ic = 3.25mA I make Ve to be around (3.25)*(.47) = 1.53 volts

Of course 2.29 - 1.53 = 0.76, which is a tad on the high side for a well biased 2N3904
I slipped a digit. The approximated base voltage should have been 2.3V, not 2.4V giving about 1.6V at the emitter (give or take). This would give a bias current of 1.6V / 470 = 3.4mA. But my calculations were rough just to illustrate my point. A simulation gives a base voltage of 2.24V and an emitter voltage of 1.55V, giving an emitter current of 3.3mA.
 
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