Boost Converters v Voltage Doublers

Question

Can you explain the basic reason why boost converters seem to be more popular than voltage doubler circuits for load bearing output?

I have built these two circuits in the simulator and they both basically achieve the same result of charging an 80 V battery from a 48 V supply.

I have built the boost converter on a circuit board and wasted many hours blowing up MOSFETs and making the circuit work on a real-world circuit board with reasonable success.

I have also built a voltage doubler circuit that achieves producing the desired voltage output but only when not under any real load. Yet in the simulator it works as well as the boost converter.

What are practicality / lifespan / cost reasons why boost converters are used for high current voltage doubling? Is there any point in me persevering with making my voltage doubler circuit work properly?

It is all a learning curve really but it would be nice to be able to actually charge my 80 V battery at the end of it.

So far I can charge the battery at 1 A with my home made boost converter as long as I remember to unplug the input power before the output power, it blows the MOSFET if I disconnect the output while its running, but I think that's a microcontroller issue not responding to the rise in voltage fast enough when the load is disconnected and the PWM duty becomes to high, that I should be able to overcome if I think about it.

Answer 1

Both topologies are used. In very low power applications you can find charge pump circuits like the capacitive voltage doubler, but the more power must be converted, the more likely you see inductors in the solution.
There are big differences to deal with:

An inductor based converter applies a voltage across the inductor and let the current rise linear. This voltage must be turned off before the inductor saturates or the current rises excessive and destroyes the switch. You seem to have experience with this effect. The amount of energy stored in the inductor is controlled by the time the voltage is applied.

In a charge pump converter you have a very large instant current when the switch connects the pump capacitor. In low power converters with capacitors below some uF the switch cost is affordable, but even in your example circuit this current is above 100 A using a good capacitor. However the amount of enery stored in the capacitor can not easily be controlled, there is no linear dependency between charge time and energy.

The regulator options are also very different:
When the inductor current is turned off, the magnetic field collapses and this produces a voltage across the inductor. If there is no load, this voltage is very high, in practice typically 20 to 100 times the supply voltage used. You seem to have some experience with this effect too, you mentioned a MOSFET breakdown at the moment of load disconnect. So a boost converter must know the output voltage and calculate the next needed active time of the switch. That is close to zero if the target voltage is reached.

Since the charge pump doesn't know the energy transferred per cycle, it can only control the frequency of the cycles and monitor the result. The maximum output voltage is twice the input voltage, so there is no dangerous high voltage using a free running frequency. This is a typical use case, where split supplies for OpAmps or line drivers must be created from a single supply and only a few mA current are required.

My last point is the choice between useful components for the two topologies:
In the mW range an inductor is bigger and more expensive than the equivalent capacitor needed, and if no fast regulation is required, a charge pump is used with the benefit of better EMI values. But if you need capacitors above 100 uF, the charge pump solution becomes expensive. All electrolytic capacitors have a short lifetime compared to inductors and ceramics are not available in this range. Inductors however exist in almost any power range.

Answer 2

A boost converter regulates its output, while a doubler only (tries to) doubles. If the doubler is loaded, then of course the output voltage will fall (or if the input voltage falls), but any time the loading decreases the output voltage, then the efficiency drops. This is because while the input current is always twice the output current, the output voltage is not necessarily twice the input voltage.

Conversely, a boost converter regulates the output voltage the desired value, irrespective (within reason) of the input voltage or output loading. This type of converter may be over 90 % efficient.

Answer 3

Voltage doublers are usually used for sine wave applications, Your voltage doubler will cause the capacitor to charge through the MOSFETs without any current limiting. If you observe these currents in your simulation, you will see currents that are far above the pulse peak current rating of the MOSFETs. Also there may be a finite amount of time where both MOSFETs are conducting. I'll do a simulation and come back later with more definitive results and perhaps an answer with a way to design a reliable voltage doubler that will work on a square wave. But it may wind up as more of a boost converter with inductance.

OK, I ran a simulation with a 5 kHz square wave and it does not look all that bad, except for the initial transition which causes a current spike of about 140A. After that the spikes are about 30A, and power dissipation in the MOSFETs is less than 5W, for an output of about 84 VDC into 100 ohms - about 72 watts. The MOSFETs have about 500 mOhm RdsON, and that accounts for the initial charging of C1 and C2.

Adding a 100 ohm gate resistor and 2 ohms in series with C1 tames the current surges to about 15 amps. The output is now about 80 VDC or 65 watts, with input of 77 watts and R3 dissipating 7 watts.