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If I've got two batteries of the same capacity, one is 12V and the other is, say, 18V with a step down buck converter, will I essentially be able to power my load at 12V for longer? I know that a battery's voltage will decrease as it discharges, so it seems to me that a better way to power a load with a more stable voltage (and maybe for a longer period of time at this voltage) would be to power it with a higher voltage battery and step that voltage down. Is this true?

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The other guys are right in theory or abstract. And certainly, whichever battery has higher energy storage will likely last longer.

But from a product design perspective it is a good idea to use a battery pack that is a bit higher in voltage than what you need and buck down to your desired voltage. So I think you are very much on the right track in your thinking and approach. A simple buck will be easier to find and cheaper and more efficient than a buck/boost.

Let us consider 12V. If you use a 3S lithium ion battery pack, your span will be something like 12.6-10V. If you need 12V, then you will have to buck/boost. But if you go 4S, then your span will be 16.8-13.2V. So you can use a buck. That is going to be easier and more efficient.

In low-current, low cost applications, a linear regulator may also be considered instead of a buck. Typically linear regulators can be designed with lower quiescent currents than a buck. So the overall system battery life may not be any worse when the low quiescent current is factored in. I doubt that is the case for you, though. Usually 12V applications are not low-power.

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  • \$\begingroup\$ Will the higher voltage battery setup be more of an all or nothing type of deal, in contrast to say a lead acid 12V? If the 12v battery gives it's power over a long period of time while steadily lowering in voltage, it seems to me that the downside of the 4S bucked lithium setup would be that while it does keep you at a solid supply of the 12v you wanted to push your load to the max, after it had discharged beyond it's capacity to provide 12v, it would be useless (not speaking literally as to the max discharge of a lithium cell, just theoretically speaking about batteries.) Is this correct? \$\endgroup\$ Mar 16, 2020 at 2:07
  • \$\begingroup\$ Well, a dead battery is a dead battery. The system should ideally be designed so once the battery is dead, it does not attempt to discharge the battery anymore until the battery has been recharged. There may be some lead acid battery systems out there that don't have any under-voltage shutdown feature. Maybe there is a motor, and it just spins slower and slower once the battery is dead. But I would always prefer to design it so that the battery cuts out completely below some threshold. If possible. \$\endgroup\$
    – mkeith
    Mar 16, 2020 at 3:46
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What matters in not voltage, it's energy.

The energy in a battery is quoted in Wh (watt-hours). Many still don't specify the watt-hour capacity but do quote the Ah (amp-hour) capacity. With this and the battery voltage you can calculate the Wh capacity from:

$$ Wh = V \times Ah $$.

Compare the two.

If you use a step-down converter you need to factor the efficiency of that into the calculations.

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  • \$\begingroup\$ If I wanted to drive, say a light bulb at maximum current, wouldn't using a higher voltage source, bucked and regulated, ensure that my bulb was always being driven at the max, because it's always being supplied with it's specified max voltage? In this example my 12V battery after not much discharge would start dropping below 12. So now I'm not able to drive my bulb to it's maximum. \$\endgroup\$ Mar 16, 2020 at 2:00
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    \$\begingroup\$ That is correct. \$\endgroup\$
    – Transistor
    Mar 16, 2020 at 2:20
  • \$\begingroup\$ Okay, thanks transistor! \$\endgroup\$ Mar 16, 2020 at 3:10
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Assuming we are comparing apple with apples, ie. both batteries have the same energy capacity and power, there is no fundamental difference between them. So you are looking at secondary effects such as converter efficiency and the load's tolerance for voltage variation.

The primary advantage of a higher voltage stepped down is that the regulator can maintain constant output voltage throughout the discharge. If your load requires at least 12V for proper operation then this could extend run time significantly (if the '12V' battery drops below 12V toward the end of its discharge).

On the other hand the step-down regulator itself wastes some power, so the higher voltage setup may actually perform worse. Achieving high converter efficiency at both high and low currents is difficult. A load that needs high power for short periods (eg. DC motor) and/or spends long periods in standby (eg. telemetry transmitter) might be better powered directly from the battery.

Finally you should consider the extra bulk and weight of the regulator. To be truly equivalent the two systems should be the same size and weight, which means the higher voltage battery may have to be smaller to accommodate the regulator. A good example is a drone, where any weight increase results in lower run time. A higher voltage stepped down would suffer from extra loss in the buck regulator, plus the extra weight of the 'oversized' regulator required to handle peak motor currents.

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  • \$\begingroup\$ Doesn't the maintaining of a bucked voltage from the higher voltage battery cause your load to be able to pull it's max wattage through the battery's life? And if it's supplied by the regular battery (in this case, 12v battery for a 12v load) as it discharges and the voltage drops, the load won't be running at maximum power anymore, since it isn't being supplied the full 12v? \$\endgroup\$ Mar 16, 2020 at 2:13
  • \$\begingroup\$ It depends on the load. A simple light bulb or heater would draw a bit less power as the battery discharges. A heater with thermostat would just run a higher duty cycle, and a motor speed controller could just increase the throttle to maintain rpm. A device with internal regulator would not even notice the reduced voltage. Generally, devices designed to run from a 12V battery can handle the voltage variation, while those designed to run from a regulated 12V supply might not - which is where the "primary advantage of a higher voltage stepped down" comes in. \$\endgroup\$ Mar 16, 2020 at 3:23
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You are forgetting that there are converters that can both step up or step down the voltage as appropriate. So you don't actually need a higher battery voltage than your load voltage. But you do need the battery to have enough energy left in it, whatever its actual voltage is relative to the desired load voltage.

A dead battery is a dead battery, no matter how high or low its voltage is. But just because a battery has a lower voltage than you need for you load doesn't mean it has no energy left inside it that can't be used to step up the voltage.

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Only something to add:

Higher voltage battery has more cells in series. The total voltage with a regulator probably stands usable when the weakest of the cells is exhausted. That weak cell starts to get charged reversely when the rest of the cells still output current. The weakest cell can get serious punishment during that process. Advanced battery packs have individual cell monitoring circuits to prevent self-destructive usage like this.

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  • \$\begingroup\$ Which is why most 'power banks' use a single cell with a step up converter. Maintaining a single li-ion cell is much easier - don't need to have matched cells, no balancing required, simpler discharge monitoring, can charge from 5V without having to boost the charge voltage etc. \$\endgroup\$ Mar 16, 2020 at 3:31

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