bridged emitter follower - how does it work?
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



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.