Speedlight battery circuit?

I have no idea what charging circuit is in the speed lights of the last couple of decades. I’m thinking of the ones that use 4-AA batteries. Do they depend on voltage or enough current to charge the capacitors? For instance, the AAs produce 6 volts. Could the capacitors charge with the 4.5-volts but more capacity of three D cells?

The voltage of the capacitors is typically around 350V. The voltage of the batteries must be transformed to that level. The lesser the voltage of the batteries the higher currency they must deliver, the sooner they are exhausted. Thus when choosing batteries of lesser voltage, their capacity (size) must be chosen larger. (And of course the transformer stage adapted too.)

If exchanging batteries for larger ones of same voltage, typically nothing has to be changed, but the bigger ones yield more charging cycles. (The inner resistance of the batteries though may play a role as in charging speed, up to even overstressing the transforming circuit.)

Voltage multiplication is frequently via an oscillator which converts the DC to AC. The AC is then stepped-up via a transformer to charge the capacitor. You can hear the oscillator as it charges.

The other odd thing about xenon flash is that the electrodes on the flashtube are connected directly to the high voltage capacitor. There is no on/off 'switch' per say. An 'exciter' triggers the flashtube to discharge. That is why you should not fool with a flash tube if the capacitor is charged.

Most on-camera flash units can take rechargeables or alkaline...four AA alkaline total 6V and as the batteries get consumed they eventually are about 4.8V and a longer charge time.
Rechargeables start at about 1.35V each so four start at 5.4V (no load) but about 1.25V each when under load, and remain about that level 5V until the slope falls off the cliff when the cell is discharged.