Voltage question

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Bunnykiller

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Working out amps with a regulated mod like you have is not totally straightforward.

Say you have 35W at 4.7V, Watts=Volts x Amps, so Amps = Watts / Volts. You are drawing 35/4.7 = 7.4 amps from the mod.

However the mod is drawing more than that from the battery. If the battery is fully discharged, it might be supplying only 3.2V, but still 35W. So the amp draw from the battery is 35/3.2 = 11 amps. It will be a little more than that, since the voltage conversion is not 100% efficient, but you don't want to get too close to the rated current anyway.

The thing to remember is that power is conserved. No matter what happens to volts or amps, the power taken from the battery is the same as the power delivered to the coil.

well most of it is... there are some losses in the circuit... capacitive, inductive, resistive losses occur...
 
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