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With a typical adjustable DC power supply, I can set the current (typically a mode called I-Set) to provide a fixed current by controlling the voltage.
(within the limits of the supply)

When testing a battery, can I do this in reverse?
In other words, hook up the battery to the power supply backwards, and set the supply to draw a fixed current from the battery, instead of supplying current to the battery?

(again, within the limits of the supply and the battery being tested)

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    \$\begingroup\$ In general - no. A power supply is not designed to do that. You could even damage the supply in the attempt. \$\endgroup\$ – Kevin White Feb 17 at 19:55
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    \$\begingroup\$ sure, i do this all the time. not backwards of course, but you hook a dummy load to the PSU's output, then adjust the PSU output current/voltage to draw what you want from the battery, measured from the battery, not the PSU output (which would only be about 90% of the total load on batt). \$\endgroup\$ – dandavis Feb 17 at 20:42
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No you generally cannot. Otherwise there would be no need for "Electronic Loads". enter image description here

These devices can be set to constant current, constant resistance or constant power draw to emulate whatever your real load is.

They are available from a number of sources, generally cost will increase with increasing power capability (and with the accuracy and pedigree of the maker). Usually you can use them with suitable interfaces to plot battery discharge curves, examine response to pulsed loads and that sort of thing.

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Normally laboratory power supplies do NOT tolerate reverse current, they would get damaged. As Sphero mentioned, ideally you would use a electronic load if you have it, but this can be a relatively expensive equipment.

A cheaper solution is to build a current sink circuit, using one opamp and a power transistor as explained here, the supply in this case would be your battery.

But you must be very careful with the amount of power dissipated by the transistor, otherwise it will explode. The power leaving the battery must go somewhere so, as mentioned by user263983, you will likely need a heat sink, which as to be properly sized.

Also you can use a power resistor as the load for the current sink, so the power is shared between the resistor and the power transistor, but be careful not to saturate the transistor or the opamp. Simulating first is a good idea to check for these things and power dissipation.

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That depends by what you mean power supply. DC-DC converter in constant current mode with load in output and battery as input can serve your purpose. But quiscient current of converter should be taken in account. LM317 comes in mind. Heatsink may be needed.

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