Should I reduce the voltage or find a different inverter?

[HR19]

First of all, I hope this is the right place.

OK so briefly, we have pretty regular power outages where I live, so I set out to build a couple inverters (a low-power one basically and a high-power one). I got an inverter that's really nice and pretty well priced, but it only has a voltage input range of 40-60V. Apparently I didn't realize how strict it is, because at 61.2V it'll beep saying the voltage is too high and it won't work. My power source is (planned to be) three Milwaukee M18 batteries in series. If they're fully charged they're slightly over 20V, about 20.8-21. Is there an easy, cheap, efficient way to reduce this voltage? My first thought was just to buy a whole bunch of cheap diodes and do them in combination series-parallel to get 30A rated 3V drop over my input but my recollection is that diodes are not terribly efficient. Is there a good, efficient way to use diodes? Is there a better way to do it? Is there a better way altogether?

 

[ChrisP58]

Using diodes to drop 3 volts at 30 Amps will produce 90 watts of heat, so yes, they're not very efficient for that purpose. A better way of dropping the voltage is with a buck converter. But it's unlikely that you'll find on off the shelf that has the rating that you'd need, so you'll probably need to design and build one. Not really a job for a beginner in electronic circuits.

That said, your battery voltage will drop fairly quickly as soon as you start drawing current. If you make up a relay circuit that bypasses the diodes as soon as the battery pack voltage drops below 60V, then their inefficiency goes away.

 

[HR19]

I'm familiar with a lot of the basic concepts of micro electronics as a hobbyist and journeyman electrician. That said, how would I build such a circuit? Specifically, how would I monitor the voltage so the circuit could shut off automatically? (Alternatively, since it has an input voltage readout, I could just do a SPDT manual switch and flip it when the battery voltage drops; I would hardly run it at full capacity, most likely just a few amps, less than 10, but I don't want to risk forgetting and frying it when I don't have materials nearby to fix it.) So I guess that's the real question, is it worth the hassle of building such a circuit, or since it would only be a few minutes usually, or should I just use a manual switch?

Sidenote: what sort of diodes would be recommended? Any old diode off the shelf, and just use enough to get the right voltage and amperage?

 

[rjenkinsgb]

what sort of diodes would be recommended?

Look at large bridge rectifiers.

50A ones are generally rather cheap from many suppliers, eg.

GBPC5010 T0 | Taiwan Semiconductor Bridge Rectifier, 50A, 1000V, 4-Pin | RS

uk.rs-online.com

Ue just the + and - terminals and you have two high current diode voltage drops. For a single diode drop, connect the two AC terminals together and use those with + or -.

 

[HR19]

Wow, that's an awesome idea! So to use it, which direction would I connect it? Should I just use all 4 terminals, like my battery + and - to the "AC" lines and use the load + and - ? Would that be the best way? If like you said just using the + and -, would I connect the battery - to rectifier + and rectifier + to load -, or batt - to rect - and rect + to load -?

Or alternatively, thinking about a rectifier diagram, would I get the best results using one of the +/-, and then one of the A/C terminals? That would let me capture 3/4 of the diodes, would it not? If I only use +/-, that would mean only half of the diodes being used, wouldn't it?

 

[rjenkinsgb]

Just + and -.
That gives two diodes series, with two pairs in parallel.

Battery pos to rec negative.

 

[HR19]

Thank you so much. Drew out the diagram and I was picturing it wrong, you're 100% on the money.

 

[HR19]

Just + and -.
That gives two diodes series, with two pairs in parallel.

Battery pos to rec negative.

I did it. I'll start a new post with pictures and stuff soon to show it off.