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I have a PCB which is nominally powered by a custom battery pack. Specifically, this battery pack:

enter image description here

So five cells connected in series, at 3.6V each, which I assume means that the nominal output voltage is 18V.

Speaking generally, would it be possible to do away with the battery pack and replace it with a mains-driven DC power supply? And assuming so, what would happen if a slightly mismatched power-supply is used (for instance, one that provides a 19V output, or one that provides a 15V output)?

Could the PCB generally be expected to function, or does the input voltage from the power supply need to exactly match the 18V provided by the battery pack?

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    \$\begingroup\$ These sort of questions are almost impossible to answer precisely without knowing the exact design. I'd guess 15V wouldn't cause damage because the cells will discharge below that level, and it'll probably work but hard to know for sure without trying it. At 19V it's unlikely to be that touchy on the input voltage, but once again hard to know for sure and going over-voltage is more likely to cause damage. \$\endgroup\$ – PeterJ May 14 '14 at 10:04
  • \$\begingroup\$ @PeterJ - Thanks for the suggestion. I tried it with the lower voltage power source, and appears to work just fine. \$\endgroup\$ – aroth May 14 '14 at 10:39
  • \$\begingroup\$ A higher power source could be dropped with a diode or two in series. Just mind how much current is pulled, and size the diode accordingly. \$\endgroup\$ – Passerby May 14 '14 at 16:22
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A battery doesn't operate as a constant-voltage source at the nominal output voltage, the voltage will be within some usable range. In your case the cells pictured are lithium thionyl-chloride cells made by Saft, which are NOT rechargeable. This simplifies things a great deal as there won't be circuitry on the board to deal with battery charging. The photo shows only two wires going to the battery pack, so there aren't extra wires for e.g. authentication against counterfeit packs. It appears to be simple plus and minus.

The cell data sheet claims an open-circuit voltage (when new at 20C temperature) of 3.67V and nominal voltage of 3.6V. This chemistry has a very flat voltage profile at 20C compared to some other chemistries, so it will spend most of the time around 3.6V and fall steeply as it gets close to full discharge. At other temperatures it may operate at a somewhat different voltage. In the data sheet they discharge to 2.0V for the discharge curves, so it's safe to assume the cell is considered dead below that voltage and should not be discharged further.

It should be a reasonable assumption that your circuit board will run from any applied voltage between 2.0V-3.67V per cell, multiplied by 5 cells gives 10.0V-18.35V. It's possible the board doesn't run as low as 2.0V per cell, since there isn't much energy in the cells below a somewhat higher voltage, unless operating at high or low temperature which changes the voltage profile. The safest guess would be to power the board from 5 x 3.6V = 18.0V, but it would probably work at least over the range down to 5 x 3.0V and possibly down to 5 x 2.0V.

I wouldn't be surprised if the board could operate from an even lower voltage, since the battery may simply run into switching or linear regulator(s), which might only need to have enough headroom above the output voltage to keep running properly. If the board has a regulator creating a 12VDC power rail then the minimum input would be at least 0.5V or even 1.5V above that. If the power rails on the board are all 5.0V or 3.3V then it might power up from an input as low as 6.0V. If you measure across any large bulk capacitors that appear to be on power rails it would tell you what those rail voltages are. Looking up the voltage regulator IC(s) would also tell you something, if you can read them.

For safety of yourself and the board, you should add a fast-blow fuse in line with the V+ wire, so if something goes wrong the power supply doesn't keep frying the board. Measure current into the board to be sure it's about the same as current from a battery. Measure the output of your power adapter before applying it to the board to be sure it's really the voltage you think it is. If using a big wall wart rather than a regulated wall adapter or bench supply, the voltage won't be exactly what it says on the outside except at a certain load current.

More information from the manufacturer's website:
http://www.saftbatteries.com/battery-search/ls-lsh

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Assuming you have a constant voltage source, the circuit will draw as much current as needed.

If you increase the voltage, it will draw more current. But when it comes to 1 volt increase of a 18 V supply, the current wouldn't be enough higher to produce permanent damage. It will increase power dissipation. But that can't be said sure without knowing the actual circuit.

If you decrease the voltage, it may work because it is a circuit designed for batteries. The voltage of batteries decreases over time. So 15 V wouldn't do any harm to try with.

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