I have already made a DC-DC converter which takes 12V or 24V from batteries and boosts it to about 300 VDC for use as the DC link voltage for a standard 240 Volt VF drive and three phase motor. This was actually the result of a long process of various trial and error attempts, simulations, and tests, and now I am trying to complete the process by making observations about my results and implementing changes which will result in a reliable unit that will do what I need. And then I may try a new design which should be more efficient, less expensive, smaller, and more practical. So I'd like to use this thread to explain the design process from my perspective, and perhaps be of use to others who may want to attempt something similar. And as I proceed (since there is yet more work to be done), I'd appreciate any suggestions on better ways to do this.
To begin, I'll provide the basic specifications. I want to be able to power and control a 2HP three phase AC motor, on a small utility vehicle or lawn tractor, using batteries of 12 volts to 48 volts total. The vehicle weighs roughly 100 pounds and I would expect a fully loaded weight of about 500 pounds including batteries, motor, controller, and rider. For this project it will be used only as a utility vehicle, maybe pulling a small cart, and it should have an operating time of 1/2 to 1 hour. Speed will be no more than 6 MPH.
So here's a rough calculation of what is needed. This should actually be done before selection of the motor, but I'll just verify that it's adequate. I'll start with the maximum grade, which I'll say might be 20%. So for a 500 pound vehicle that requires a force of 100 pounds exerted by the tires. They are roughly 16" diameter, for a radius of 8". Thus I will need a torque of 100*8/12 = 67 Lb-Ft. I want a top speed of 6 MPH or 528 ft/min, and each rotation of the wheel is 3.14*16 = 50.2 inches = 4.18 feet. So I need a shaft speed of 126 RPM.
I'll use a 3450 RPM motor since that is similar to the ICE speed, (and its what I already have), so I need a reduction of 27:1. The rated torque of a the motor is 2HP*5252/3450 = 3.04 lb-ft. So with the reduction drive I will have about 82 lb-ft of torque. I needed 67 so it seems more than adequate. Of course there are losses due to friction, but 81% efficiency seems reasonable. But this is for the rated motor torque, and it should have short term overload capacity of 2x to 3x. So I'm good!
Now I'll calculate the amount of energy I will need. For a tractor, Rolling Resistance
and grade are the overwhelming factors. I can estimate the Rolling Resistance
on the expected rough surface by simply pushing or pulling it with a spring scale. I need to do this, but for now I'll estimate that it takes 20 pounds to push it. For a round trip, the average grade will be zero, but since regeneration is only good for about 20% and I might not even implement it, I'll assume a 5% grade over half the distance, or effectively 2.5%. This requires a force of 12.5 pounds plus the 20 for Rolling Resistance
, or 32.5 pounds. This is a torque of 21.6 lb-ft and at top speed of 126 RPM that is about 1/2 HP or 389 watts. So if I want to run at this speed for 1 hour I need 389 W-Hr. If I use a single 12V 105 A-Hr deep cycle battery, that is 1260 W-Hr and 389 watts will draw 32 amps. This is 1/3C so I can probably get at least 50 A-Hr and maybe 1.5 hours of use. So far, so good. Even allowing for 67% overall efficiency I meet my 1 hour goal.
Enough for now. I want to break this up into reasonable chunks, so I'll get into the design considerations for the DC-DC converter in another post in this thread. BTW, as a reality check, I have already taken this vehicle on a couple of test runs and it drew about 15 amps at 24 volts, or 360 watts.