L7805 current draw

crutschow

Joined Mar 14, 2008
32,931
Thanks
How long would 2 x 2200mAh 18650's last with the L7805 sitting idle, no load?
I assume the batteries are in series, so the typical battery life would be 2200mAh / 4.3mA = 511 hours or 21 days.

If that's not sufficient, could you add a circuit to remove the power to the L7805 when there's no output load?
Alternately there are very low quiescent power (microamps), low-dropout 5V regulators that likely would work for you.

What's the maximum 5V load current when the circuit is active?
 
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Thread Starter

liteace

Joined Mar 7, 2012
162
I assume the batteries are in series, so the typical battery life would be 2200mAh / 4.3mA = 511 hours or 21 days.

If that's not sufficient, could you add a circuit to remove the power to the L7805 when there's no output load?
Alternately there are very low quiescent power (microamps), low-dropout 5V regulators that likely would work for you.

What's the maximum 5V load current when the circuit is active?
I think I might have to put the batteries in parallel, I want to use 4 x 18650's that'll = 16 ish volts then Ive got to drop that back down to 5v so large heat sink and lost energy
The max load will be 700ma
 

crutschow

Joined Mar 14, 2008
32,931
I think I might have to put the batteries in parallel, I want to use 4 x 18650's that'll = 16 ish volts
That's if they are in series, not parallel.
You could use two each in series, in parallel with another two in series.
That would give an output of about 7.5V with double the Ah capacitor of one battery.

But the batteries would need to be individually charged to avoid overcharging and battery damage.
 
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