1970s Design Indulgence

Immediately past the crossover the current through the upper emitter resistor comprises that required by the load, plus a non-constant 7mA for the NFB (which has been "pre-paid" by unequal Vq's), and the recharging current to replenish the output capacitor (which has not been "pre-paid" as of yet).

It's not that the output capacitor has been discharged. It has been partially discharged because it was pulling the load negative under instruction of the lower output transistor. The law of conservation of energy always applies, and so electrical charge energy has been converted into mechanical movement of the speaker cone (sound energy - a pressure wave), and heat.

Now, this isn't taught (except by some prof. in Sofia - and me), the simple college and university explanation of what a DC blocking capacitor does is a crime, which leads to false accusation by the "all capacitors are bad brigade" - because they are obviously under-educated. If you are reading this, you are now more advanced than many.

The charge replenishment - a kind of inrush once the voltage rises through signal zero - must flow through the upper emitter resistor. Regardless of what current it is, Ohms law states that it will result in a voltage, and that voltage will be higher than we expected.

"I" being the symbol for current, then the charge lost must be -di/dt, and is being replaced by +di/dt. It was lost in time period dt, so the recharge must be the same dt, but dt is the inverse of frequency. Will that matter? It is far beyond the -3dB frequency. What we need is some values to plug into the charge equation CV=IT (what, that again? Yes!).

Actually, we can forget frequency and thus time "t" because it's not at the -3dB cut. We can forget C because it's the same in both directions. So it's either V or I. I is current or the flow of charges, so -di and +di cancel. We are left with V. So, as we lost going down, we must add what we lost going up.

Maybe we're back at gm-doubling? We set Vq. Earlier, we arrived at Vq = 18mV, but to obtain some equilibrium on test, that had to be dropped to 10.5 mV. What will happen if we halve the upper emitter resistance?

By Ohms law, for the same current, the voltage will halve. As we are replacing charge, then the current must match what was lost when positive conduction happens, so the voltage on the upper emitter resistor will match that on the lower emitter resistor - when conducting. But it will be half when quiescent.

So, is the answer to place an equal value emitter resistor in parallel? Then we adjust the lower emitter resistor voltage to 18mV?

We shall see.