How to increase load current for capacitive AC voltage regulator circuit?

Question

I am studying fans and their working while visiting a factory, and I came across one problem that they're trying to solve. For an AC capacitive voltage regulator, how can we increase the amount of current that could reach from the divider for a given voltage drop? This made me read more about AC voltage regulators but I couldn't find much apart from the output current point of view, the most I could find was this thread here which talks about the disadvantages of capacitive dividers. Is there a way this can be done?

Basically what they're trying to do is, they can control BLDC fans using either a manual regulator or a remote-based regulator, but they want to use both. And that's where the Ic current issue is occurring.

I'll try to explain what I could understand more on the problem, the way they control the BLDC motor is simple when they're using a remote-based regulator as they can change the duty cycle according to the signal data received. For manual regulators, they use the Iccurrent and the phase difference obtained(because of the capacitors) from the regulators to analyse what is the state of the regulator and accordingly set the duty cycle. But the problem when both remote-based regulators and normal regulators are used is that when the regulator is at a lower step, say step 1, the current obtained from it is around 600mA, which is way more than what they need(50-60mA).

Answer 1

For an AC capacitive voltage regulator, how can we increase the amount of current that could reach from the divider for a given voltage drop?

You can't achieve anything better than 1:1 i.e. current in equals current out.

Theoretically the "dropper" needn't be a capacitor; it could be an inductor too. An inductor can be made that has the same impedance as the capacitor dropper (for a given operating frequency) and it would not waste power in the delivery of a 1:1 current to the load but, we don't do that because inductors (for a given impedance) are so much more bigger than the equivalent capacitor.

However, the great advantage of using a wound component is that you can add a secondary winding and gain much more output current but, it's not a simple reactance dropper any more. However, it is much, much more safer than an impedance dropper because the windings can use insulation from each other and protect the user of potentially lethal voltages.

Answer 2

how can we increase the amount of current

Capacitance Transformers work exactly the same as the impedance ratio of resistors in a voltage divider where the loop impedance draws the current.

The downside is the very low impedance of a switched voltage of an impulse or step risetime conducts more current than the sine wave frequency risetime. (Ic=CdV/dt)

So other current limiting elements are necessary tradeoffs to maximum power conversion or current limiting.

e.g. if a 50:1 voltage divider is needed from AC, one might have 1nF series to 49 nF shunt and with more current scale up both values. But other elements both in series and shunt must be considered to limit the current with a 4kV impulse or 350V abrupt turn on which may affect transient stresses in lossy devices or arcing in the highest impedance element.

For 120V , 60Hz it is rare to see C dividers generate more than 5W to a 5V load.

What is a Capacitive transformer?

Answer: https://www.wikiwand.com/en/Voltage_transformer

No solution may be suggested until the load impedance and all the specs: are sufficiently defined.

An inductance transformer is a far better AC solution above a certain low current threshold, but the cost of copper windings and core is much higher than metal film capacitors.