As you can see I have designed a simple buffer circuit with a current source. The output is nice and steady as you can see (\$V = 2.6+1.6\sin(2\cdot\pi\cdot 1000\cdot t)\$ is the input, frequency of 1kHz). The current source provides 1mA.
However, when I add a capacitor between the output node and the 8-ohm resistor to function as a dc offset remover, the output gets defaced. The value of the capacitor I added was 1 microfarad.
At first I thought maybe the capacitor and resistor form a high-pass filter on the output node and that's why my signal got distorted, but changing the value of capacitor absolutely did not help and my output was still distorted. This is the output for a 2microfarad value.
So I wanted to 1) ask about the reason for this, since I've been using coupling capacitors at the output of my circuits and never had this problem before, and 2) is there any fix to it? That 8-ohm resistor is actually a speaker, and thus no dc offset should be applied to it, is there any alternative dc-offset removing methods I can use in my circuit?
Best Answer
Without the output capacitor the top transistor supplies all the current to the 8 ohm load. It is an emitter follower and this configuration is ideal for this type of drive. The lower transistor hardly plays any role in this circuit and you could actually remove it and get virtually identical results.
When you use a capacitor in series with the 8 ohm load, things change considerably. Now the lower transistor is expected to remove charge from said capacitor when the input signal is at its trough or most negative peak. However, the current sinking capabilities of the lower transistor (configured as a weak current sink) means that it can't do this to any degree comparable with the top emitter follower transistor's ability to source current.