How to calculate runtime when...

kosPap

Flashlight Enthusiast
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...you have the batt mAh, and the working voltage and the flashlight mA draw?????

Case 1. Li-Ion cells...
I am making a guess...a protected batt will work from 3.6-4.2V. that means a drop of voltage which when multiplied with the current draw will give us watts...compare that with the batt amphours and you have runtime hours.


Case 2 Li primary cells.
Now I am troubled, cos from a voltage value and below that the driver will drop out of regulation. But if you know what voltage is this (both nominal and working batt voltage under load), then you can measure the regulated runtime with the case 1 method...

am I correct????
 
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I've been fairly accurate by measuring battery current and voltage simultaneously to get watts consumed from the battery.

Runtime ~= (nominal battery voltage * rated mA*h) / (measured current * voltage under load)

Battery mA*h is rated until a voltage cutoff, so you don't need to worry too much about termination voltage.
 
I am making a guess...a protected batt will work from 3.6-4.2V. that means a drop of voltage which when multiplied with the current draw will give us watts...compare that with the batt amphours and you have runtime hours.
Power calculations do not work this way actually. The decline in voltage when the battery runs down cannot be multiplied by the current to work out how much power has been supplied. To get power, you have to look at the voltage difference between two terminals, and the current flowing from one terminal to the other.

Further to this, you cannot compare watts with amp hours, as they are different things. It is like comparing apples with oranges. Either you can compare amp hours contained in the battery with amp hours consumed by the light (fairly accurate if the voltage does not change much over time), or you can compare watt hours contained in the battery with watt hours consumed by the light (accurate all the time).

Comparing watt hours is sort of what EngrPaul is doing above.
 
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marduke you are a savior...I had all the data in my mind and I got the runtime (I used actual battery working voltage), and fot the xpected runtime...I got 2.18 hours with it...also the calc got me a 597mA draw which is exactly what i measured,(0.60A)

engrpaul.
I will study your formula at home. I got 2.457 hours with it.

MrHappy...
I do need a good tutorial...wonder when I am gonna tax our ham club's electronics man on this....By and large I got what you mean since amphours x voltage is watthours...

all the calculations are made with ohm's law (am I not correct?) but I got to get the feeling of doing it...

Thanks for your patience, kostas
 
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