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Voltage Dropper question

Hi there,

This is a basic question about electrics I'm hoping someone here might
be able to help me with. I have a 24V 20AH LiPo battery I am planning
to use for a device that runs at 12V. I am planning on using a
voltage dropper that takes 24V down to 12V and gives two cigarette
lighter outlets at 4A x 2 or 8A x 1. The battery has a rated
discharging current of 20 Amps. If the battery gives out 20A and I am
only pulling max 8A is there a dangerous heat build-up / discharge
likely to blow the voltage dropper? Does 20A flow from the battery
even though there is nothing to consume it or is it on 'tap' and just
taken when required by a device using the battery? Hopefully the
question makes sense.

Thanks.
 
R

Rich Webb

Jan 1, 1970
0
Hi there,

This is a basic question about electrics I'm hoping someone here might
be able to help me with. I have a 24V 20AH LiPo battery I am planning
to use for a device that runs at 12V. I am planning on using a
voltage dropper that takes 24V down to 12V and gives two cigarette
lighter outlets at 4A x 2 or 8A x 1. The battery has a rated
discharging current of 20 Amps.

If the battery gives out 20A and I am
only pulling max 8A is there a dangerous heat build-up / discharge
likely to blow the voltage dropper?

The 8A rating on the "dropper" should represent a current that could be
continuously drawn through the device under its rated ambient
environmental conditions without hazard.
Does 20A flow from the battery
even though there is nothing to consume it or is it on 'tap' and just
taken when required by a device using the battery? Hopefully the
question makes sense.

The battery only supplies what's called for by the loads connected to
it. If your dropper is a linear regulator then, to a first
approximation, current in from the battery equals current out to the
load. If it's a switching regulator, it will be somewhat more efficient,
by taking less current (at a higher voltage) to supply more current (at
a lower voltage).
 
J

Jim

Jan 1, 1970
0
The 8A rating on the "dropper" should represent a current that could be
continuously drawn through the device under its rated ambient
environmental conditions without hazard.


The battery only supplies what's called for by the loads connected to
it. If your dropper is a linear regulator then, to a first
approximation, current in from the battery equals current out to the
load. If it's a switching regulator, it will be somewhat more efficient,
by taking less current (at a higher voltage) to supply more current (at
a lower voltage).

Thanks for the reply Rich - that clarifies things.

Regards,
Jim.
 
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