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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.

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    \$\begingroup\$ Leaving this in the comments because it's information in your "for reference" section -- you may want to ask a separate question about "how do I size a boost converter to match a motor". The motor that you call out has a maximum continuous current rating of 200A, and a peak current (which you will use, if the "sprint" is a drag race) of 400A. The switching converter chip that you call out is rated at 3.5A -- do you see a disconnect here? \$\endgroup\$
    – TimWescott
    Commented Jun 16, 2019 at 18:40
  • \$\begingroup\$ Indeed the load determines the current taken from a boost converter. Obviously the boost converter must be able to supply the amount of current demanded by the load. The motor you mention requires a lot more current that the MPQ3425 can provide so this will not work with that motor. You do not mention what power (or current) you want the motor to take, I see 90 A or more in the table, that's expert territory. Unless you know DCDC converters inside out you will be unable to build a proper boost converter for 52 V 90 A (or more). For that you will need experience. \$\endgroup\$
    – Bimpelrekkie
    Commented Jun 16, 2019 at 18:42
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    \$\begingroup\$ Also, check which speed the propeller is optimized for. The faster a propeller spins, the more power is going into the wake vortexes which don't add to propulsion. \$\endgroup\$
    – Janka
    Commented Jun 16, 2019 at 19:28

2 Answers 2

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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.)

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  • \$\begingroup\$ Ok thanks for the clarification. I have a lot to learn. \$\endgroup\$
    – CMSC
    Commented Jun 16, 2019 at 19:20
  • \$\begingroup\$ Nice answer. Another important concept involved is energy. The input power to the boost converter is the input voltage times the input current, while the output power is the power delivery to the load, i. e., the output voltage times the output current. ln a boost converter the output power is equal to the input power less some internal power loss. This explains why the input current in a boost converter is always higher than the output current. If you want to understand how the boost converter works internally then you'll need some basic knowledge of circuits, inductors and capacitors. \$\endgroup\$
    – joribama
    Commented Jun 17, 2019 at 8:29
  • \$\begingroup\$ Actually, in any energy conversion device the output power plus losses is equal to the input power -- it just has to be, thanks to conservation of energy. So motors, light bulbs, boost converters -- they all have that same constraint. \$\endgroup\$
    – TimWescott
    Commented Jun 17, 2019 at 14:48
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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.

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  • \$\begingroup\$ Thank you for your help. Do you or anyone else for that matter, know of any other ways to satisfy the conditions of the competition and possibly boost my voltage from the batteries? It doesn’t have to be efficient the race is only 300m so any ideas are welcome. \$\endgroup\$
    – CMSC
    Commented Jun 17, 2019 at 0:14
  • \$\begingroup\$ Well, if you put your three batteries in series, you will get 36 V. Not sure if that was your intention, but that could be a starting point. Then you could design a boost regulator for the motor, but that is not such an easy thing to do. Also, you will need some type of speed controller for the motor, unless your plan is to just apply 36V from the get-go and let it rip. I am also confused thought, because I don't get if it is a battery race or a solar power race. It seems like for a 300m race, it would be best to just drain the batteries as fast as possible and cross the finish line first. \$\endgroup\$
    – user57037
    Commented Jun 17, 2019 at 0:31
  • \$\begingroup\$ The endurance race is 2 hours so that’s the one that cares about the solar panels. The sprint race is the race I’m concerned about for the Boost converters \$\endgroup\$
    – CMSC
    Commented Jun 30, 2019 at 15:17

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