The problem is quite articulated so I will start from a more generic description and get more specific as I go on.
I need help to design a braking resistor system for a fairly uncommon application. The whole traction system looks something like this:
Power line are rectified with a 3 phase passive bridge. The DC Bus voltage (540V) is then divided by a capacitive voltage divider giving
2 virtual DC busses of 270V each. Each motor inverter is supplied by these artificial 270 V. The motors itself are fairly small and
have approximately 40 W nominal power. There are approximately 300 of these motors and motor inverters supplied by the same DC bus. Half of
them on fed by the "upper" side and the other half by the "lower" side.
Now the problem I am trying to solve is the shifting of the "virtual" DC middle point (I call it virtual because it's only given by the
capacitive voltage divider).
In a normal operation all the motors are equally loaded and turn with the same RPM, but it can happen that some motor have to stop. If many motors fed by the same DC "side" need to be stopped, then the dc middle point shifts and the voltage on the dc bus is no longer equally divided. This can lead to get, for example 350V on the high side and 190V on the lower side.
To solve the issue and keep the middle point in the middle I have come up with a braking resistor system that helps dumping the excessive
power of one or the other DC side. The circuit looks like this:
The idea is pretty clear: if, for example the upper voltage gets dangerously high, turn the high side on and dump the energy with the
braking resistor. In the same way if the lower voltage gets too high the low side MOSFET is turned on to burn the excess energy.
In the schematic there is 1 switch but in reality I am paralleling 4 MOSFETs to distribute the current. an important point is also that
I am not driving the MOSFETs with a PWM. Either on or off. It might happen that the unbalance persists for minutes and maybe even hours
so the gate needs to be kept continuously on. This is my question:
Is this the right circuit for such application?
I tried answering the question by building the circuit and testing it. I added a couple of fuses in series with the MOSFETs to prevent
blowing up 8 MOSFETs in case a shoot-through happens. I than turned the low side MOSFET on and the fuses turned into nice lamps
for a brief moment. I assume the high side also turned on but I can't figure out why and how I can prevent that from happening.
is there an alternative/safer method to drive the MOSFETs "always on"?
Best Answer
1) Shoot-through.
Looking at IR2110 block diagram, it appears the chip does not have any circuit to prevent shoot-through. If HIN and LIN are active at the same time, both FETs will turn ON. So you will have to either design the circuit which generates those signals to make sure it does not happen, or use a MOSFET driver which is protected against shoot-through.
Don't implement that kind of stuff in software.
2) This is a bootstrapped driver
The driver for the high-side FET is powered from the bootstrap cap (between Vb and Vs) which is recharged via the diode when the top FET turns OFF and the bottom FET turns ON. You can't use this kind of driver to hold the top FET in ON state for a long time, as the driver consumes some current which will eventually discharge the cap:
With a 100nF cap and 230µA, voltage will drop at dv/dt=i/C=2.3V/ms so you can expect it to hold the top FET ON for a few milliseconds at most and then it will either detect the undervoltage and turn the FET OFF, or keep it on with reduced gate voltage which will increase RdsON until it catches fire.
A common gotcha is you can't even turn the top FET ON if the bootstrap cap is not charged, which will be the case if the output has been sitting at midsupply like in your circuit. You'd have to first pulse the low-side FET ON to charge the bootstrap cap, then turn the top FET ON, and repeat the process at least every millisecond to recharge the cap.
Also your STP5NK50 FET has max Id of 4.4A, 1.5 ohms RdsON and as you would expect for a low-current FET with a tiny chip it has high RthJC of 1.78°C/W, all of which is inadequate.
Now...
No 😁
You talk about 100kW power, like 3000 x 40W motors... About 0.15A per motor... So if all the motors one one side are at max power and all the motors on the other side are off, that's (roughly) 200 amps going through your braking resistor and it will dissipate 50kW. That's going to be a big resistor. It's an impractical solution.
My preferred solutions (best first):
If you order 3000 of them, surely the manufacturer is going to be interested in meeting your needs... since the inverter is probably going to use a IRF2110 anyway and that takes 600V, maybe all you need for 540V rating is a change of capacitors and MOSFETs. And maybe they already sell an inverter rated for 600V...
As for the motors, if the insulation withstands the voltage, you can keep your existing ones if the inverters support PWM output to vary the voltage...
You don't have to use just one huge transformer, you can use several smaller ones, whichever is cheaper.
That's supposing you keep your existing capacitive divider arrangement... It's a bad idea IMHO but well.
So, in your schematic...
Besides fixing all the problems outlined above, I'd just replace the braking resistor R15 with an inductor and feed a 50% PWM signal to the MOSFET driver. Size up the FETs and inductor just as you would do for a buck DC-DC converter, since this is essentially a buck converter...
Note a synchronous buck is inherently bidirectional. Current in the inductor can go both ways. It will output half the input voltage and keep your two capacitors at identical voltages. What it will actually do is discharge the capacitor which has higher voltage into the one with lower voltage while keeping losses low.
Also you don't have to design one huge 100kW buck converter. You can make it multichannel or just make several ones. Since resistive losses scale as RI^2, twenty units handling 10A each are potentially a lot less annoying to make than one single unit handling 200A. This will probably end up cheaper than the electricity wasted in the 50kW resistor.
Also having several separate units makes it less of a single point of failure. Useful since, if this blows up, all the inverters will be over-volted too. Maybe a comparator which shuts down the whole system in case the voltages get dangerously high would be a good idea.
I would also recommend an overcurrent protection on the buck but I'm afraid that would spoil the spirit of running 270V rated ESCs in situations where you get "350V on the high side and 190V on the lower side" ...