# Battery with a step down converter - what is the final Ah?

I guess it is a simple electronics question, but I couldn't find anything really clear out there.

I have a 12V 70Ah car battery connected to a step-down-converter, which brings the voltage down to 5V to power an Arduino.

When calculating the battery lifetime, should I consider to multiply the time for 12V/5V = 2,4 - since the battery reported value is intended for 12V, or not?

Thank you

Depends on a number of factors.
How much current is the Arduino drawing from the battery?
What temperature is the battery stored at?
Lead Acid batteries self discharge at around 5% a month, but the rate goes up if the battery temperature goes up.

I actually donâ€™t need to have an exact estimation. I only need to know if the value of the AH reported on the battery should be multiply for 12/5 since we are operating at 5V instead of 12V.

Assuming your step down converter is 100% efficient then that would be right. Obviously it won't be and there will be other losses. But for guesstimation it should work.

You can measure this. With the load attached measure the current into the
converter and the current out of the converter - the ratio of currents is what
you want.

You do mean a switchmode converter, don't you?

For a "linear" regulator, the current at 5V is the same as the current going in. More or - in fact - less.

Thank you so much for the answers.

AH of the battery remains the same since that is the product of amps from the battery times the duration of the load (usually based on a discharge rate of C/20 ie discharged over a 20 hour period) So a 20AH battery should supply 1 amp for 20 hours, by which time it'll be totally flat !

Fitting a step down unit (switched mode) to provide 5 volts doesn't alter the battery chemistry so the AH capacity of the battery doesn't change

However, what you really should be talking about is the WH (watt hours) when you are considering power usage.

Say the battery is 12 v and it's AH capacity is 20AH, then battery energy capacity is 12 x 20 = 240WH

Now say your switch mode PSU is rated at 5v 6A and assume it's 80% efficient then the energy requirement is 5 X 6 x 100/80 = 37.5W so in 1 hour it'll consume 37.5WH

Now the original battery was 240WH so your system will run for 240/37.5 = 6.4hours.

But this is well above the C/20 discharge rate and in general you lose 20% capacity for each time the C/20 is halved (load is doubled). 20AH at C/20 equates to 16AH at C/10 and about 12AH at C/5

So your above system will actually only run for about 6.4 X 12/20 = 3.8 hours

Conversely, if your discharge rate is less than C/20 you get a 20% lift in capacity for each halving of the load.

Your battery is 70AH and your load is obviously less than 6 amps so I leave you to work out the duration

Yes, it's all a bit mathematical but you have to get the sums right to avoid disappointment.

@jackrae: I think he was just speaking of considering a modified battery capacity just for commodity.

Since the step-down regulator would supply more current than the battery actually does, seen from its circuit, it's LIKE he had a battery with a higher capacity but a lower voltage. That's just a model, a "view of mind" (I don't know the equivalent expression in English sorry) do simplify reasoning, since in many cases, when you design a circuit, you're actually looking at the current drawn rather than the actual power.

So a 20AH battery should supply 1 amp for 20 hours, by which time it'll be totally flat !

No, it won't be totally flat... I believe the standard is 80% of the rated voltage. And, the 5V DC-DC converter (or regulator) will probably work down to 6 or7 V, so you should be able to get more than the rated Amp-Hours.

So a 20AH battery should supply 1 amp for 20 hours, by which time it'll be totally flat !

Or 0.1 A for 200 hrs etc...

DVDdoug:
No, it won't be totally flat... I believe the standard is 80% of the rated voltage. And, the 5V DC-DC converter (or regulator) will probably work down to 6 or7 V, so you should be able to get more than the rated Amp-Hours.

Probably but if you pull a 12volt lead acid battery down to 6 volts it'll not only be flat but also on its way to the big battery store in the sky

By totally flat I meant that its capacity has been used - there is still life in it but pulling more out will be detrimental to the battery

If you only pull 100mA from the battery (C/200) you'll probably get around 400hours out of a 20AH unit