In op-amp, how do they remove Quiescent votage (bias voltage of a bjt) from the output?

Question

We use a coupling capacitor to remove the dc component from the output in simple ac amplifier circuits.

In below base biased CE amplifier, the voltage at collector node is not ac, it is pulsating dc with an average value of 15V. After coupling capacitor it becomes ac with average value of 0V.

How do they manage this in an op-amp ? Since op-amp works for dc inputs too, wouldn't a coupling capacitor block both the input and the quiescent bias voltages ?

For ex : As far as this question is concerned, I hope we can assume op-amp is just a two transistor differential amplifier with single ended output and let the gain be 1000; then putting 1mV dc at input gives 1V dc at ouput, but this floats above the already existing quiescent bias voltage of the transistor. If I use a coupling capacitor, it blocks both the quiescent bias voltage and the amplified input voltage. But we want to block only the quiescent bias voltage, not the amplified input voltage. Capacitor blocks all the dc, it doesn't know what is bias and what is input, so I feel we cannot use a capacitor here.

In above circuit \$V_1\$ is dc input and the output is \$V_{out} = V_{CE} + 1000*V_1\$, where \$V_{CE}\$ is the quiescent voltage of Q2 transistor. But we want to remove \$V_{CE}\$ from the output in practical circuits right ?

Answer 1

This can be done with level shifting and requires a bipolar power supply. The simple sketch below illustrates this, showing only the components pertinent to this concept.

The collectors of the first differential pair (Q1 & Q2) are as you described in your question...biased above 0V. The second differential pair (Q3 & Q4) uses PNPs and its output is biased down to just above 0V. This is buffered by the emitter follower, Q5. I showed it fed back to the (-) input of the amplifier simply to illustrate the path for negative feedback.

^{simulate this circuit – Schematic created using CircuitLab}