Understanding DC-DC boost converters

Question

I’m currently working on a solar powered boat with a modest sized team of people. For our competition we are only allowed to have 3 12v batteries on board but we are allowed to have up to 52v at any point in our setup. We are hoping to kick up the speed of our outboard motor for the sprint portion of the completion and are looking into using a boost converter for this.

After doing a little research I’m a bit confused about how they fundamentally work. From what I can tell they take in a voltage and current and put out a higher voltage with a lower current with some level of efficiency. But it seems to be independent from the load( in this case the motor) which doesn’t make much sense to me because I thought the current would be determined by whatever my load is. How do loads actually interact with boost converters?( The simpler the better.)

For reference we are using the Saietta 119r motor system and I’m considering using the MPS MPQ3425 boost converter.

Answer 1

Basically, the way electrical loads work is that you give them a voltage and they pull a current. The current pulled is based on the state of the load and its environment, so things can get complicated, but in general the higher the voltage, the higher the current.

So if you make a boost converter that puts out 52V, then your outboard motor will draw whatever current it draws it that voltage and boat speed.

A typical boost converter has a closed-loop controller in it that sets the voltage. So if you set the converter to output 52V, then under normal operation, 52V is what you'll get. (Note that it's usually poor system design to do this when there's a DC motor in the mix -- it's absurdly easy to source a motor with a different winding, or even rewind a motor, to get more current at lower voltage. But, rules.)

Answer 2

Tim's answer is good. Considering where you are on the learning curve it is probably appropriate. But for the sake of being thorough, there are a lot of cases in the real world where the load and power source are not easily approximated this way.

The real answer is this: when a two-terminal power source is connected to a two terminal load (assuming both are time-invariant) then they settle on a voltage and current that satisfies both of them, unless the system is unstable in which case a fire or crash or blown fuse will put a stop to it.

If the load is a resistor, then it will be satisfied when V=I*R (Ohm's law). If the load is an LED then the load will be satisfied by a more complicated equation. But it is not just the load that can be complicated. If the source is a solar panel, then the voltage output will not be fixed, and you once again have a complicated situation (even for a resistor as a lod).

Usually you solve this in practical applications by some form of successive approximations until it is close enough. There is also a concept called the "load line". This is where you draw the V vs I characteristic for the source and load on the same graph, and find the point of intersection. The intersection is where the circuit will operate.