Calculating power consumption

Thread Starter

Lprkon

Joined Mar 8, 2021
4
When calculating power usage of a device using rechargables am I to use nominal voltage, charged voltage, or device voltage rating
 

MrChips

Joined Oct 2, 2009
23,069
None of the above.
Power is calculated from:

P = I x V
P = I x I x R
P = V x V / R

i.e. operating V and I.

For approximation, you can use any of your three suggested voltages if it is not critical.
 

wayneh

Joined Sep 9, 2010
17,027
I'd say the charged (peak) voltage is the least useful. Once placed under a modest load, the battery voltage will sag. The peak voltage is only briefly "real".

The nominal battery pack voltage and the device voltage rating are usually the same.

Personally, I'd use the information provided with the device to calculate its power consumption. But the devil is in the details. Tell us more if you need a better answer.
 

Thread Starter

Lprkon

Joined Mar 8, 2021
4
None of the above.
Power is calculated from:

P = I x V
P = I x I x R
P = V x V / R

i.e. operating V and I.

For approximation, you can use any of your three suggested voltages if it is not critical.
I think operating voltage is what I meant by voltage rating, thanks for the reply
 

Thread Starter

Lprkon

Joined Mar 8, 2021
4
I'd say the charged (peak) voltage is the least useful. Once placed under a modest load, the battery voltage will sag. The peak voltage is only briefly "real".

The nominal battery pack voltage and the device voltage rating are usually the same.

Personally, I'd use the information provided with the device to calculate its power consumption. But the devil is in the details. Tell us more if you need a better answer.
It's a home built light powered by an 18650 regulated to 3.3v.
Thank you for the reply
 

wayneh

Joined Sep 9, 2010
17,027
It's a home built light powered by an 18650 regulated to 3.3v.
Thank you for the reply
The most accurate estimate would be the current draw actually measured at the battery, along with the battery voltage under load. This would include all losses in the regulator and any other circuitry.

Otherwise a back-of-the-napkin estimate is the power used by the load, plus about 20% for all losses. That's probably a good-enough estimate if all you're doing is comparing to battery capacity, because it's hard to know actual battery capacity with accuracy.
 

MrAl

Joined Jun 17, 2014
8,063
When calculating power usage of a device using rechargables am I to use nominal voltage, charged voltage, or device voltage rating
The most accurate calculation can not use any constant value of anything because everything changes over time and these changes must be accounted for one way or another.

The things that change are voltage and current. Under load, the voltage falls and that means the current falls except in the case of a constant current load, but then the voltage still changes because it still falls as the battery becomes more and more depleted. The real challenge comes in because these changes are usually not linear but their rate also changes with time such that the average power is not a simple linear calculation but involves a sum of individual calculations.

The instantaneous power is easy to calculation:
p=v*i

where both v and i are allowed to change with time, and to calculate 'p' we need to measure both of these as frequently as possible. If they do not change much over say 1 second, then a decent estimate is to measure them both once per second. We can then calculate the average power although the energy is usually more important.

If we had consistent measurements over one hour intervals then if we calculated say p=10 over the first hour then the energy is 10 watt hours. Since we had 10 watts the whole time, the average power is 10 watts.
If we had 10 watts the first hour and 30 watts the second hour, then the total energy is 40 watt hours and the average power is (10+30)/2=20 watts.
If we had 30 watts the first hour and 10 watts for the next three hours, then the total energy is
30*1+10*3=60 watt hours
and the average power is:
(30+10+10+10)/4=60/4=15 watts.
Note that above we measured over a total time of 4 hours and so we must get the same energy when we multiply the result by the number of hours, and this means that 15*4 (watts) is the same as 60 watts.
Making the time intervals the same always makes the calculation simpler. The result is simply:
energy=sum(p) {energy in watt hours}
where p is the power in watts measured every hour and N is the number of hours.
and:
average power=energy/N {average power in watts}

The exact expression is an integral, but we more often take individual measurements so that wont be of much use. The summation is useful because it allows individual measurements to be summed.

I have a USB measuring device that measures watt hours and watts and voltage and amps and ampere hours. Ampere hours is a good measurement because that allows us to estimate device run times when we get a new device or a new battery or something.
The ampere hours calculation goes the same way as the energy we just add up all the individual readings and express the result in the units we wish to use. Usually ampere hours are used but also milliampere hours are used and you might even find ampere seconds to be useful.
 

Thread Starter

Lprkon

Joined Mar 8, 2021
4
The most accurate calculation can not use any constant value of anything because everything changes over time and these changes must be accounted for one way or another.

The things that change are voltage and current. Under load, the voltage falls and that means the current falls except in the case of a constant current load, but then the voltage still changes because it still falls as the battery becomes more and more depleted. The real challenge comes in because these changes are usually not linear but their rate also changes with time such that the average power is not a simple linear calculation but involves a sum of individual calculations.

The instantaneous power is easy to calculation:
p=v*i

where both v and i are allowed to change with time, and to calculate 'p' we need to measure both of these as frequently as possible. If they do not change much over say 1 second, then a decent estimate is to measure them both once per second. We can then calculate the average power although the energy is usually more important.

If we had consistent measurements over one hour intervals then if we calculated say p=10 over the first hour then the energy is 10 watt hours. Since we had 10 watts the whole time, the average power is 10 watts.
If we had 10 watts the first hour and 30 watts the second hour, then the total energy is 40 watt hours and the average power is (10+30)/2=20 watts.
If we had 30 watts the first hour and 10 watts for the next three hours, then the total energy is
30*1+10*3=60 watt hours
and the average power is:
(30+10+10+10)/4=60/4=15 watts.
Note that above we measured over a total time of 4 hours and so we must get the same energy when we multiply the result by the number of hours, and this means that 15*4 (watts) is the same as 60 watts.
Making the time intervals the same always makes the calculation simpler. The result is simply:
energy=sum(p) {energy in watt hours}
where p is the power in watts measured every hour and N is the number of hours.
and:
average power=energy/N {average power in watts}

The exact expression is an integral, but we more often take individual measurements so that wont be of much use. The summation is useful because it allows individual measurements to be summed.

I have a USB measuring device that measures watt hours and watts and voltage and amps and ampere hours. Ampere hours is a good measurement because that allows us to estimate device run times when we get a new device or a new battery or something.
The ampere hours calculation goes the same way as the energy we just add up all the individual readings and express the result in the units we wish to use. Usually ampere hours are used but also milliampere hours are used and you might even find ampere seconds to be useful.
So nominal voltage as it would give me an average , but a regulated circuit say 3.3v I would use the regulated voltage
 
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