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I am using a tracer battery BP2603(12V,3.5A) to turn on my camera. The camera needs 12V and it consumes max 3A. Maximum voltage the camera can accept is 13.5V. But I've found that when the battery is fully charged it has a voltage of approx 13.7v and it is not able to turn on the camera. So I need to drop approx 0.2V-0.3V to turn on the camera. I had a thought to use shunt resistor but the power dissipation is nearly 1W. The second option I thought about using a voltage converter. Is there any other way I can drop 0.2V-0.3V? Please let me know if there is any other way.

Thanks in advance

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  • \$\begingroup\$ Use a DC to DC converter, easily avaliable on ebay etc But you don’t want to drain the battery as per title... \$\endgroup\$ – Solar Mike Mar 21 '18 at 9:51
  • \$\begingroup\$ Or use 4 resistors in parallel to split the 1W into 250 mW (which most resistors can handle) as a temporary solution. \$\endgroup\$ – Michel Keijzers Mar 21 '18 at 9:55
  • \$\begingroup\$ @SolarMike Thanks. yes I've found one. Probably I think the drop would be better than drain? \$\endgroup\$ – Dhinesh Mar 21 '18 at 10:24
  • \$\begingroup\$ @MichelKeijzers Thanks. I can try this but I am not sure this is going to be a permanent solution \$\endgroup\$ – Dhinesh Mar 21 '18 at 10:25
  • \$\begingroup\$ The solution by MCG is better (I will upvote), but this is just an alternative in case you don't have a (schottky) diode at hand. \$\endgroup\$ – Michel Keijzers Mar 21 '18 at 10:27
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If you want to drop a small voltage like that, you can simply use a diode. That will have a (almost) consistent 0.7V(ish) drop. If you want a little less of a drop, use a schottky diode.

Checking the datasheets for the Vf vs current graphs should enable you to find one suitable for what you want.

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