enter image description here For the standard emitter follower you typically need input coupling capacitors and a DC bias of roughly half the supply voltage however what if the input signal itself swings around half the supply voltage?

e.g if the previous stage to an emitter follower is a standard op amp non-inverting amp with a split supply (+12V,-12V) with the input referenced to centre point but the succeeding emitter follower is referenced to -12V (i.e +24V, 0V), so The op amp output is referenced to ''+12v'' relative to the emitter follower? Could this replace the input coupling cap and resistor biasing or would it not work?

Load would be a pair of headphones (50 to 300 ohm)

example of circuit is attached below

enter image description here

Differential emitter follower? differential emitter follower

Inverting amp?

enter image description here

  • \$\begingroup\$ Draw the full idea of the circuit to stop ambiguity and sentences that make no sense. \$\endgroup\$ – Andy aka Apr 26 '20 at 11:03
  • \$\begingroup\$ I did.what sentences make no sense? \$\endgroup\$ – Jay Apr 26 '20 at 11:08
  • \$\begingroup\$ but the preceding emitter follower - what preceding emitter follower? Crop your diagram to fit the page to make it legible, \$\endgroup\$ – Andy aka Apr 26 '20 at 11:12
  • \$\begingroup\$ thanks for correction, it should be succeeding. also fixed size of image... ms paint is not the best editing software \$\endgroup\$ – Jay Apr 26 '20 at 11:15
  • \$\begingroup\$ The board has a built-in schematic editor, which is better to use than paint for schematics. \$\endgroup\$ – Neil_UK Apr 26 '20 at 11:20

Could this replace the input coupling cap and resistor biasing or would it not work?

A more convenient idea would be this: -

enter image description here

It's called a DC coupled amplifier and uses feedback directly from the emitter to set the gain at the output to be: -

$$1 + \dfrac{R2}{R1}$$

Of course, if you really want the return current for your headphones to go via -12 volts (potentially noisier) then you will need an output capacitor.

  • \$\begingroup\$ The issue is I cant really modify the op amp configuration as its part of the output buffer on a DAC PCB. I was thinking there may be another option making use of the differential or ''balanced'' output of this DAC. A bridged amplifier that would recieve this differential signal from the DAC does not reference its output to ground, the headphone/speaker is basically 'floating' between the outputs of each half the bridged amp. I added updated version of schematic to show what I mean. \$\endgroup\$ – Jay Apr 26 '20 at 13:51
  • \$\begingroup\$ This will not work because the NPN is in the loop and adds open loop gain making the circuit unstable. The NPN will have to be a PNP in that case and resistors will have to be added/adjusted. \$\endgroup\$ – le_top Apr 26 '20 at 15:23
  • \$\begingroup\$ @le_top I don't think so - it's an emitter follower and has a voltage gain less than unity and is quite likely to have a frequency response many times greater than the op-amp hence, instability won't happen. If it were a MOSFET source follower, you'd have a more realistic argument. \$\endgroup\$ – Andy aka Apr 26 '20 at 15:33
  • \$\begingroup\$ Ok, my look at the circuit was a bit too fast. Sorry. \$\endgroup\$ – le_top Apr 26 '20 at 16:42
  • \$\begingroup\$ @Andyaka I added a version of your circuit to the OP that using inverting amp, do you see any issues with that? \$\endgroup\$ – Jay Apr 30 '20 at 0:25

If you mean ...


simulate this circuit – Schematic created using CircuitLab

then with suitable op-amp biassing so that its output swings somewhere between the rails, the emitter follower will 'follow' the output at about 0.7 V lower.

  • \$\begingroup\$ yes, thats it just if the op amp is configured as a non-inverting amp/buffer. So the question is would referencing the op amp input to ground take care of biasing? \$\endgroup\$ – Jay Apr 26 '20 at 11:45

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