I was asked to make a 24V battery cart with eight 12V batteries to boost some of our equipment on site. No big deal. Now they want to be able to flip a switch and get 12 volts out of the cart without a DC to DC converter. I don't want to pull 12 volts from just one or two batteries. So I was wondering if anyone could tell me how to wire up these batteries so I can get 12 volts from all of them with a flip of a switch. They also want just one cable reel for both 24V and 12V.
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\$\begingroup\$ Not possible without breaking some connections and making others, or let relays do the job for you. Are you ok with that? \$\endgroup\$– winnyJun 27, 2020 at 16:52
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8\$\begingroup\$ Only a matter of time until someone accidentally connects 24V to a load that wants 12V and fries it. Are they sure they want this? \$\endgroup\$– user57037Jun 27, 2020 at 17:42
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4\$\begingroup\$ 12V @ what current? For relatively low power, just add a buck convertor. For starting engines ... no. \$\endgroup\$– user_1818839Jun 27, 2020 at 21:33
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\$\begingroup\$ For high power you should be able to get away with a multi-phase converter and a decently sized bank of output capacitors. But we do need to know the intended load current. \$\endgroup\$– Drunken Code MonkeyJun 28, 2020 at 1:42
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\$\begingroup\$ I considered something like this, to have three 12V SLA batteries seried giving 36V to a bike, but reconfigure to parallel to charge all three at once from a 12V car-charger to care for the batteries. It turned out too complex, and the possibility of error would have been small but potentially catastrophic charging problems. I dropped the idea. \$\endgroup\$– CriggieJun 28, 2020 at 13:13
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2 Answers
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It's possible, using a heavy-duty DPDT switch.
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4\$\begingroup\$ This would need to be a high current contactor or relay, driven by a lower current switch. It may have to be rated for hundreds of amps. \$\endgroup\$– user105652Jun 28, 2020 at 3:33
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How about the following?
simulate this circuit – Schematic created using CircuitLab
Close SW1 to charge, or deliver, 24 V.
Open SW1 to deliver 12 V, less a diode drop. For Schottky diodes, this need not be more than 400 mV or so.
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\$\begingroup\$ I think D2 is backwards for 12 volt operation, and if corrected would short V2 when the switch is closed for 24 volts. Also, I suspect this cart is used to start heavy equipment, so he'd need >1000 Amp diodes if it could be made to work. \$\endgroup\$ Jun 27, 2020 at 19:33
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\$\begingroup\$ @PeterBennett Conventional current flows in the direction of the arrow. The diodes conduct during discharge to the load, not during charging. This configuration essentially copies the rather nice alternative rectifier capacitor I found when I tore down a LED light. It was a bridge from the mains, into a reservoir like this, with SW1 replaced with a downward conducting diode, driving a buck to the LED. So the caps are charged in series, but allowed the voltage to drop to half, good for power factor, and then take over in parallel at V/2. \$\endgroup\$– Neil_UKJun 27, 2020 at 19:53
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1\$\begingroup\$ Looking harder, I'll agree that the diode is the right way around, and your circuit will work. \$\endgroup\$ Jun 27, 2020 at 20:20
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1\$\begingroup\$ This is the best solution, assuming the DC converter request to be misguided... If all they want is a 12/24v, this is the cheapest and most robust solution. That diode will need to be pretty beefy though, assuming they will use the cart to crank motors... \$\endgroup\$ Jun 28, 2020 at 1:46
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\$\begingroup\$ The voltage drop across the diodes takes too much of the available 12V, I think. \$\endgroup\$ Jun 28, 2020 at 13:13