Electronic – Operational Amplifier Output Stage Biasing

amplifierbjtltspiceoperational-amplifiertransistors

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I am trying to bias my op-amp output stage using diodes. I am using a current source for the bias current but later I will replace it with BJT current source.

However, it doesn't seem to work. Whatever the value of the Ibias, Q7 is never in active mode. I am not sure what I am doing wrong here.

Any help is appreciated. Thanks!

Best Answer

Your amplifier can be a fair bit simpler (and more correct.)

  1. Keep \$Q_2\$, \$Q_3\$, \$Q_8\$, \$Q_9\$, \$Q_{11}\$, and \$C_1\$. That topology is fine.
  2. Keep \$D_1\$, \$D_2\$, \$Q_7\$, \$Q_{10}\$, \$R_7\$, and \$R_9\$. But delete \$I_1\$ and instead connect \$Q_7\$'s base to \$Q_{11}\$'s collector.
  3. Dump \$Q_5\$, \$Q_6\$, \$R_1\$, and \$R_\text{ref1}\$ and replace the entire mess with an ideal current sink, for now. I don't even want to think about it. So just stick a current sink in there and worry about it, later. Set the tail current to something reasonable.
  4. Dump \$Q_1\$, \$Q_{12}\$, \$Q_{13}\$, \$R_2\$, and \$R_5\$. Replace all that with another current sink. (Make the sink value the same as in step #3?) Tie this current sink to the base of \$Q_{10}\$.

That should about do it. Haul some parts out to the trash, for now. Get this thing working. Then worry about those two current sinks.

If you get the above going okay then you can replace both current sinks with a single current mirror. The diode-connected side feeding the diff-amp pair and the other side feeding your \$Q_{11}\$ VAS. Stick a resistor (call it \$R_x\$ for now) between the diode-connected side's collector and your diff-pair emitter node to set the current. Then test the whole thing again.

If that works, then you can get fancy and remove \$R_x\$ as the current setting method and come up with something better to set the mirrored currents.

But at least this should get you closer.


If you start with this:

schematic

simulate this circuit – Schematic created using CircuitLab

It's inputs have no source of recombination current so \$I_\text{CBO}\$ is what's making things happen.

Now, set it up as a simple voltage follower:

schematic

simulate this circuit

And it will do what you expect, I think.

Now you can replace those current sinks. I'll use something simple:

schematic

simulate this circuit

And if you like hook that up as a follower and add a signal to the (+) input and try it out.

I was confused by the other stuff you added. You wrote no explanation for it.

By the way, the circuit I was thinking about with \$R_x\$ was this:

schematic

simulate this circuit

That's not so good. But it is passable for purposes of an example.