enter image description here

I built a rough test version of the above circuit, it is intended as a bridged (i.e 'balanced') unity gain buffer for headphones. The test circuit seems to work, as in there is sound and it is not blatantly distorted .

The op amp outputs in the circuits are from the balanced XLR output from my DAC, therefore the op amp circuit is internal to the DAC and can not be modified, it's just there for clarity.

If possible my aim is to have distortion no worse than a typical emitter follower (e.g single supply, ac coupled, dc biased) with this circuit, but with a reduced number of parts.

Would this behave like a typical emitter follower or is there any obvious issue?

There are 2 things I'm most interested to know and will be do my best to describe them:

1. The load 'floats' between the 2 outputs it seems, how does the NPN handle negative voltage swings from the op amp?

For example, If the negative side of the load (300 ohm headphone) were connected to -12V supply instead of the output of the negative half as it's shown in circuit then the NPN would be able to source current even during negative voltage swings on the op amp output. So what happens during negative swing when it is connected 'floating' between the negative and postive halves? can the NPN still source current during negative voltage swings from the op amp?

2. The -12V supply is treated as ground for the NPNs, this means Re is connected to the output of the negative voltage regulator, does this matter?

Would it be better/worse to instead connect Re before the regulator, to the negative output of bridge rectifier? This would resemble a more 'normal' single ended supply for the NPN... you would never see a negative regulator on ground for a single supply.

To be clear I am not looking for a 'better' alternative to this circuit, I want to know the problems with this circuit and, if they exist, solutions. If problems exist without solutions then my alternative will simply be to build the standard single supply, AC coupled emitter follower

  • \$\begingroup\$ Fine for mono, For stereo you'd need to configure a custom headphone connector to keep the L and R returns separated, as they are normally commoned and connected to GND. \$\endgroup\$
    – user16324
    May 1, 2020 at 10:26
  • \$\begingroup\$ yeah that is no problem, XLR headphone connectors are used for this purpose \$\endgroup\$
    – Jay
    May 1, 2020 at 10:32

1 Answer 1


For (1), the NPN can't sink current though its emitter, it can only source. R(e) is sinking both the load and transistor current during the low-swing. The way to make a transistor sink current on the low side is to use a push-pull (class B) type driver, which doubles the number of transistors, but is much more efficient.

Anyway, so R(e) needs to be reasonably low enough to sink your load. For 300 ohms it should be about... 300 ohms.

For (2), your input is bipolar, and so is the op-amp output Vout. The amp output will swing above and below ground, with its gain set by R1 and R2. For the NPN emitter-follower to track that, it needs to be able to follow the op-amp down to Vout - V(be).

For example, if op-amp Vout is -1V, then V(be) will need to be able to follow down to -1V - 0.7V = -1.7V. So you need the negative Ve supply.

(n.b, typical NPN V(be) is 0.7V)

If you want to be able to use ground for V(e), that is, for the amp to be uni-polar, you need to add an offset the input signal to bring it such that Vout swings between V(be) and +12V. You can do this by AC coupling it and using a voltage divider to set the bias at the op-amp input to at least 6V+Vbe, or +6.7V. That's what's shown in this link.

Anyway, I did a quick Falstad sim of your setup, with a small change. Try it here: link

enter image description here

Two more things. First, if you take the op-amp feedback from the NPN's emitter instead of the op-amp Vo, you will get better accuracy: V(e) will be a copy of the input owing to the negative feedback to the op-amp. This is in the sim; the op-amp gain is set to 1.

Second, it's not strictly necessary to have the DC blocking caps to the load, since in theory the two emitters should be at the same voltage when the input is at zero and so no current will flow through the speaker. At the very least you could get away with just one DC blocker. You can edit the sim and try it yourself.

MORE: If you poke around in Falstad, you'll find in the 'Transistors' pull-down there's a sample push-pull driver. You can enhance that one using the op-amp feedback, like I showed you for the emitter-follower version. The op-amp will help mask the crossover distortion inherent in the push-pull.

  • \$\begingroup\$ OK I think I understand why we consider the input to be bipolar in this example, the output of the negative half of the circuit is not actually ''floating'' like I thought, it sits at GND minus Vbe. One question, you explain here how my test circuit is able to make sound with bipolar input: ''R(e) is sinking both the load and transistor current during the low-swing.'' Im assuming there is a downside to this compared to correctly biasing the NPN to swing only between ground and positive supply? higher output impedance during negative swing? \$\endgroup\$
    – Jay
    May 1, 2020 at 10:15
  • \$\begingroup\$ There’s nothing ‘incorrect’ about having Ve being negative vs. ground. If you want to have Ve be ground, then you need to set the idle point to be midway between ground and +12V using the same technique as the example emitter follower you show. You can try this in the sim. \$\endgroup\$ May 1, 2020 at 13:34

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