Help calculate power usage

[john2k]

Can anyone kindly help me calculate power consumption based on the following:

If I have a 5v 3950mAh battery powering a Tablet, and over a 24hr period i lose 4% battery.

I am going to be replacing this battery and run directly from a 12V car battery using a 12v to 5V converter. What would the power drain equivalent be over a 24hr period based on the above 4% loss over 24hrs on a 5v 3950mAh battery?

Thanks

 

[Gryd3]

Can anyone kindly help me calculate power consumption based on the following:

If I have a 5v 3950mAh battery powering a Tablet, and over a 24hr period i lose 4% battery.

I am going to be replacing this battery and run directly from a 12V car battery using a 12v to 5V converter. What would the power drain equivalent be over a 24hr period based on the above 4% loss over 24hrs on a 5v 3950mAh battery?

Thanks

3950mAh means that it can put out almost 4Amps for an hour.
It also means that it could put out almost 1Amp for 4 hours... See? mAh is the product of time multiplied by current.
So.. if after 24hr, you loose 4%, then you have used 4% of 3950mAh which is 158mAh.
How do you get current out of that? simply divide by time. 158mAh / 24h = 6.6mA

I should warn you though... that things like tablets and phones are not a constant current type device...
It just so happens to average out to be 6.6mA . In reality, the current will vary, sometimes in huge quantities depending on what the tablet / phone actually does.

I would suggest doing a different test:
Charge the tablet and use it. Use it hard. Play games, surf the internet, copy files... try to kill the battery as fast as possible. This will tell you the 'peak' power draw from the tablet which you should size for.

 

[Harald Kapp]

If I have a 5v 3950mAh battery powering a Tablet, and over a 24hr period i lose 4% battery.

I'm not aware of any true 5V batteries. Are you talking about a power pack with 5V output? Then the battery inside is most probably anything but 5V plus a 5V regulator to stabilize the output voltage.
When replacing the "battery" by a power supply you should keep the internal battery of the tablet active. The tablet's electronics is normally matched to the internal battery's characteristics. The external 5V supply is only used to re-charge the internal battery.

 

[john2k]

I'm not aware of any true 5V batteries. Are you talking about a power pack with 5V output? Then the battery inside is most probably anything but 5V plus a 5V regulator to stabilize the output voltage.
When replacing the "battery" by a power supply you should keep the internal battery of the tablet active. The tablet's electronics is normally matched to the internal battery's characteristics. The external 5V supply is only used to re-charge the internal battery.

It's not exactly 5V it's just under 5v it's around 4.3-4.5v.

 

[john2k]

3950mAh means that it can put out almost 4Amps for an hour.
It also means that it could put out almost 1Amp for 4 hours... See? mAh is the product of time multiplied by current.
So.. if after 24hr, you loose 4%, then you have used 4% of 3950mAh which is 158mAh.
How do you get current out of that? simply divide by time. 158mAh / 24h = 6.6mA

I should warn you though... that things like tablets and phones are not a constant current type device...
It just so happens to average out to be 6.6mA . In reality, the current will vary, sometimes in huge quantities depending on what the tablet / phone actually does.

I would suggest doing a different test:
Charge the tablet and use it. Use it hard. Play games, surf the internet, copy files... try to kill the battery as fast as possible. This will tell you the 'peak' power draw from the tablet which you should size for.

Thanks for that great write-up. The drain that I was trying to calculate is during the time when the tablet is in deep sleep. So over a span of few days I've been monitoring it and the highest consumption was 4% over 24hrs. While the tablet is on, this will be much higher depending on use but during that time the car will be on with engine running.

So 4% is approx 6.6mA in 24hrs so if we round that to 10mA, is that a good amount for a device to be using from car battery while it is sleeping and car is off?

 

[Gryd3]

Thanks for that great write-up. The drain that I was trying to calculate is during the time when the tablet is in deep sleep. So over a span of few days I've been monitoring it and the highest consumption was 4% over 24hrs. While the tablet is on, this will be much higher depending on use but during that time the car will be on with engine running.

So 4% is approx 6.6mA in 24hrs so if we round that to 10mA, is that a good amount for a device to be using from car battery while it is sleeping and car is off?

If I were you, I'd run the tablet off the accessory line or other switches power line in the car so that it does not continually pull from the car battery.
But... assuming the values are correct, that would be a safe assumption.
I have a hard time believing that it will last 25 days in standby on a single charge though... based on the 24-hour test you did.
Regardless, keep poking around and getting details.
Harald pointed out a big point that could ruin our math and your measurements.
The battery voltage will not be 5V. They are most often 3.7V Li-Ion used in tablets or phones, but could also be a multiple of 1.5V or 1.2V depending on chemistry.
The energy used from the battery will also not equal the energy used to replenish the battery, or to power the device from an external source. Voltage regulators cause additional loss, and they should always be present in the charging adaptor and usually inside the device.

 

[john2k]

I have a hard time believing that it will last 25 days in standby on a single charge though... based on the 24-hour test you did.

It does, because I have the tablet with a custom ROM designed to put it to deep sleep and the main purpose of the ROM is to preserve battery and it does exactly that. It's claim is maximum 4% draw over 24hrs of sleep.

The battery voltage will not be 5V. They are most often 3.7V Li-Ion used in tablets or phones, but could also be a multiple of 1.5V or 1.2V depending on chemistry.

The actual battery says 3.8V but when i measure the voltage draw from the battery it's typically between 4V and 4.5V. The lower the charge state the lower the voltage gets. As I am replacing the battery completely with direct power source, if I provide the tablet less than 4.5V it detects battery charge status 0% but anything over 4.5V it ignores checking battery status and thinks battery is at 100%

 

[Gryd3]

It does, because I have the tablet with a custom ROM designed to put it to deep sleep and the main purpose of the ROM is to preserve battery and it does exactly that. It's claim is maximum 4% draw over 24hrs of sleep.



The actual battery says 3.8V but when i measure the voltage draw from the battery it's typically between 4V and 4.5V. The lower the charge state the lower the voltage gets. As I am replacing the battery completely with direct power source, if I provide the tablet less than 4.5V it detects battery charge status 0% but anything over 4.5V it ignores checking battery status and thinks battery is at 100%

If that's the case, then fly at 'er.
We don't have the device in question, so we suggest steps and advise based on general knowledge and experience with similar devices.
Sounds as though you have a very analytical head on your shoulders though. You'll be fine.
Just ask the questions as they come and we will see what we can do.

 

[john2k]

If that's the case, then fly at 'er.
We don't have the device in question, so we suggest steps and advise based on general knowledge and experience with similar devices.
Sounds as though you have a very analytical head on your shoulders though. You'll be fine.
Just ask the questions as they come and we will see what we can do.

Many Thanks

 

[john2k]

So.. if after 24hr, you loose 4%, then you have used 4% of 3950mAh which is 158mAh.
How do you get current out of that? simply divide by time. 158mAh / 24h = 6.6mA

Sorry, does that mean i'm using 6.6mA per hour? or 6.6mA in 24hrs?

 

[Gryd3]

Sorry, does that mean i'm using 6.6mA per hour? or 6.6mA in 24hrs?

That means you are drawing 6.6mA
If you ran it all day, you would still only be drawing 6.6mA... BUT you would have consumed 158mAh.

 

[john2k]

That means you are drawing 6.6mA
If you ran it all day, you would still only be drawing 6.6mA... BUT you would have consumed 158mAh.

Thanks.