I'm installing a solar powered weather station at a remote Rc field. It reports to the Internet, not available there, via APRS ham radio.

The equipment I use has the ability to turn off the power to the 4watt tx and power it on prior to transmission of collected wx data.

Wx equipment drain 12v @50 ma/hr. 24/7

Transmitter 7.4v @ 75ma receive, 1.4A transmit. TX every 3 Min 24/7 for 10 seconds.

The receiver does not need to be on so the 75ma current can be eliminated.

If I use the power/voltage control function, I can power down the transmitter for 2 min 30/40 seconds, have it power on to stabilize the tx circuits, make the 10 second data transmission and power down again, repeat.

I'm trying to calculate the smallest size solar panel I can use and it's wattage?

During the winter months and night time, the stored voltage will run it. I need to calculate that overnight drain, and the charge current required to recharge it on the shortest days or overcast days. I have a 12v car battery as the power source that I'm guessing is around 80A capacity, not cranking current.

I also have a 10A gel and lipos.

So to break it down to the minimum.

50 Ma/hr for the wx instruments.

75ma every 2 min 30-40 seconds (20-30 seconds)

1.4a for 10 seconds every 3 min (300 seconds)

Maybe 10 hours of sun max winter months here in Los Angeles, maybe shorter with overcast or if we get rain? Nights, dark, 12-13 hours. So 50max13 hrs = 650 ma

Summer, long sun hours!

I also have a higher powered tx that draws 10a on transmit that I could use math help with, same setup.

Thanks for any help on the math.

## Need some battery math help

- ShowMaster
**Posts:**4327**Joined:**Thu Dec 29, 2011 3:44 am**Country:**-**Location:**Los Angeles, CA USA

### Re: Need some battery math help

Just stumbled on this post, (never actually ventured into the Pub before!) and saw there were no replies...

Its been a while since you posted so you may have already sorted all of this by now?

If not, this is how I approached the same problem in the past. I have remote Temp/Hum sensors run off 1S LiFe cells reporting back to my Arduino based weather station - they sleep for 30 seconds consuming < 0.01mA (the lowest I could measure), then power up, read sensors, transmit, then sleep - the current draw while awake is the order of 22mA for about 100 milli-seconds).

so that calculation went:

22 mA for 0.1 second in every 30 = 22 * 0.1 / 30 = 0.073mA + 0.01mA (from the constant sleep power) = 0.083mA average constant current = 0.083mA/h of battery capacity = 1.99mAh / day.

I have used 700mAh LiFe packs, so the assumption I've been working to is they would last about (700 / 1.99 =) 352 days assuming no self discharge. So far they've dropped less than half a volt from their fully charged 3.3V in the first 6 months of operation. The Arduino's are monitoring the battery voltage each time they wake up and transmitting that data along with the weather so I can track it over time

So that's how I did my maths... here's my calcs on your numbers.

Standard disclaimer of any responsibility applies!

I'm assuming everything is running from 12V and your current (mA) values include any voltage regulation losses where lower voltages are required for the components.

50mA/hr = 50mA (easy one)

75mA every 2 min 30 (150 seconds) for 30 seconds (assuming worst case) = 30 / 150 * 75 = average of 15mA

1.4A for 10 seconds every 300 seconds = 10 / 300 * 1400 = average of 46.6 mA lets say 47mA (rounded up)

So total average constant current = 50 + 15 + 47 = 112mA. So for 1 hour, you would need 112mAh of battery capacity.

If you need to run that for 24 hours, you'll need a battery with (112 * 24) 2.7Ah capacity (again, rounded up)

Lets assume you want to replace that entire charge each day and assume an average of 4 hours of useful sunlight per day over the year (I'm not familiar with weather/pollution in LA but that's probably very conservative), then you need to supply roughly (because charging is not 1:1 with discharging) 12V * 2.7A / 4 = 8.1W, so you would need a 12V, 10W solar panel and solar charge controller capable of about 1A (all rounded up)

If you've already put all of this together, I'd love to know how close experimental data comes to the rough maths above

Its been a while since you posted so you may have already sorted all of this by now?

If not, this is how I approached the same problem in the past. I have remote Temp/Hum sensors run off 1S LiFe cells reporting back to my Arduino based weather station - they sleep for 30 seconds consuming < 0.01mA (the lowest I could measure), then power up, read sensors, transmit, then sleep - the current draw while awake is the order of 22mA for about 100 milli-seconds).

so that calculation went:

22 mA for 0.1 second in every 30 = 22 * 0.1 / 30 = 0.073mA + 0.01mA (from the constant sleep power) = 0.083mA average constant current = 0.083mA/h of battery capacity = 1.99mAh / day.

I have used 700mAh LiFe packs, so the assumption I've been working to is they would last about (700 / 1.99 =) 352 days assuming no self discharge. So far they've dropped less than half a volt from their fully charged 3.3V in the first 6 months of operation. The Arduino's are monitoring the battery voltage each time they wake up and transmitting that data along with the weather so I can track it over time

So that's how I did my maths... here's my calcs on your numbers.

Standard disclaimer of any responsibility applies!

I'm assuming everything is running from 12V and your current (mA) values include any voltage regulation losses where lower voltages are required for the components.

We need to work out the average constant current, the rest flows from that.ShowMaster wrote:So to break it down to the minimum.

50 Ma/hr for the wx instruments.

75ma every 2 min 30-40 seconds (20-30 seconds)

1.4a for 10 seconds every 3 min (300 seconds)

Maybe 10 hours of sun max winter months here in Los Angeles, maybe shorter with overcast or if we get rain? Nights, dark, 12-13 hours. So 50max13 hrs = 650 ma

50mA/hr = 50mA (easy one)

75mA every 2 min 30 (150 seconds) for 30 seconds (assuming worst case) = 30 / 150 * 75 = average of 15mA

1.4A for 10 seconds every 300 seconds = 10 / 300 * 1400 = average of 46.6 mA lets say 47mA (rounded up)

So total average constant current = 50 + 15 + 47 = 112mA. So for 1 hour, you would need 112mAh of battery capacity.

If you need to run that for 24 hours, you'll need a battery with (112 * 24) 2.7Ah capacity (again, rounded up)

Lets assume you want to replace that entire charge each day and assume an average of 4 hours of useful sunlight per day over the year (I'm not familiar with weather/pollution in LA but that's probably very conservative), then you need to supply roughly (because charging is not 1:1 with discharging) 12V * 2.7A / 4 = 8.1W, so you would need a 12V, 10W solar panel and solar charge controller capable of about 1A (all rounded up)

If you've already put all of this together, I'd love to know how close experimental data comes to the rough maths above

### Re: Need some battery math help

That is actually a lot, at 2.8V an LiFe is basically empty. The voltage difference between full and empty is much smaller than on Li-Pos.Greebo wrote:So far they've dropped less than half a volt from their fully charged 3.3V in the first 6 months of operation.

Absolute lowest voltage for Li-Fe is 2V, recommended 2.5, I like to use 2.8 as when you've got there you've basically only got about 5% left anyway, it starts dropping like a cliff at this point.

### Re: Need some battery math help

You're right of course... I was thinking in my head they've not yet dropped halfway between 3.3 and 3.2 and typed half a volt... so I SHOULD have said dropped less than half of 100 mV! My Weather station actually marks the battery as "flat" at 2.8V.Kilrah wrote:That is actually a lot, at 2.8V an LiFe is basically empty. The voltage difference between full and empty is much smaller than on Li-Pos.Greebo wrote:So far they've dropped less than half a volt from their fully charged 3.3V in the first 6 months of operation.

Absolute lowest voltage for Li-Fe is 2V, recommended 2.5, I like to use 2.8 as when you've got there you've basically only got about 5% left anyway, it starts dropping like a cliff at this point.

I did check the rest of the maths though!