power consumption from 9v battery

Hi guys

Just a quick question. Say im using one of those 9V pp3 batteries with a capacity of say 150mah. If i power something at 5v using a regulator, for example say it uses 150mah at 5v. does drawing through a 5v regulator affect the battery life in anyway different to drawing 150mah @9v?

I know its not a specific question as such, im just trying to understand how it works. In my head, using a regulator would use the same amount of battery, but the excess would be wasted as heat but im not sure if i am correct

hi valrik

150mAh means that the battery can give you 150mA for one hour or 75mA for 2 hours or 50mA for three hours.

The 9V battery does not care where the current is going - it simply does not know. So a 5V voltage regulator will just turn 4V into heat (wasted energy). It will not change the current requirement of your device.

And it is only the current that counts.

If your device needs 15mA your battery will run for ten hours (15mA x 10hours = 150mAh).

Using a 9V battery specified for 150mAh does not make much sense if your device needs 150mA.

May be useful...

What are you planning on doing?

Dropping a 9V source to 5V to power an MCU from a 150mAh battery of that size may not be the best solution.

I think there is no "may" about it. Use a switching Buck regulator for best efficiency because that does work like you hoped the other regulator would. You take the current and the voltage multiply them together and multiply by 0.8 and that is the power it will take from the battery. but the best solution is to use a better battery.

valrik:
Hi guys

Just a quick question. Say im using one of those 9V pp3 batteries with a capacity of say 150mah. If i power something at 5v using a regulator, for example say it uses 150mah at 5v. does drawing through a 5v regulator affect the battery life in anyway different to drawing 150mah @9v?

I know its not a specific question as such, im just trying to understand how it works. In my head, using a regulator would use the same amount of battery, but the excess would be wasted as heat but im not sure if i am correct

The answer is charge is conserved - the same charge goes into the linear regulator as comes out of it (ignoring
the tiny current on its ground lead).

Charge is always conserved, even in a black hole! Switching regulators are not the same as they don't
pass the input charge straight to the output.

Sorry for not yet responding, been away from the computer.

All that is very informative, thank you all for your knowledge.

To clarify, i am going to be powering an arduino pro mini and a few servos. I was thinking of using a 9v pp3 instead of 5 AA's for space and weight reasons only. I dont know much about 'buck' regulators, relativly new to doing my own electronics but i will certainly look into that.

I was thinking of using a 9v pp3 instead of 5 AA's for space and weight reasons only.

Think again. A servo takes a lot of current, that battery will drain very quickly.

See:-
https://www.dimensionengineering.com/info/switching-regulators

Ah i see. well, in one simple post that clears that one up very quickly. thanks.

Consider using 3 AAs without a regulator. Or a single LiPo without regulator.

MarkT:
Charge is always conserved, even in a black hole!

Can you post a link to the research on that? I thought we were clueless about what happened beyond the event horizon?