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Gabriel Caffese
Joined: 09 Oct 2003 Posts: 39 Location: LA PLATA, ARGENTINA
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Calculating power consumption and battery duration |
Posted: Sun Jun 13, 2004 8:47 am |
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Hello everyone,
I´m involved in a project, in wich there is a 16LF88, and two I2C perpheralls. Timer 1 is being used along with a 32,768Khz crystal oscilator, and Pic micro configured to count and wake up every 15 seconds. There´s also an infrared module, for sending data to a PC at 57,6kb, and reconfigure everything.
I want to measure power consumption, and estimate battery duration (after choosing the right battery).
This is what I measured for the complete circuit, @ 3,6V:
During 15 seconds, current is 5uA
During .1 seconds, current is 3ma
Twice a week, during 90 seconds, current is 4ma.
The cycle repeats for ever, and once started, will NEVER be turned off.
Now, how do I calculate "final" power consumption, and select the apropiate battery ?
Hope someone can help. Thanks in advance.
Gabriel.- |
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Ttelmah Guest
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Re: Calculating power consumption and battery duration |
Posted: Sun Jun 13, 2004 2:55 pm |
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Gabriel Caffese wrote: | Hello everyone,
I´m involved in a project, in wich there is a 16LF88, and two I2C perpheralls. Timer 1 is being used along with a 32,768Khz crystal oscilator, and Pic micro configured to count and wake up every 15 seconds. There´s also an infrared module, for sending data to a PC at 57,6kb, and reconfigure everything.
I want to measure power consumption, and estimate battery duration (after choosing the right battery).
This is what I measured for the complete circuit, @ 3,6V:
During 15 seconds, current is 5uA
During .1 seconds, current is 3ma
Twice a week, during 90 seconds, current is 4ma.
The cycle repeats for ever, and once started, will NEVER be turned off.
Now, how do I calculate "final" power consumption, and select the apropiate battery ?
Hope someone can help. Thanks in advance.
Gabriel.- |
Basically, you need to sum the current over a 'standard' time period (normally 1 hour). So in the case of the 'twice a week' event, only 1/84th of this occurs in the average sampling period (given that there are 168 hours in a week), and the mAHr figure for this is:
(4*90)/(3600*84) = 0.0012
Then for the 0.1 second 'event', this occupies 0.66% of the time (0.1/15.1), giving:
3*0.00663 = 0.020
Then the 15 second event, occupies the remainder of the time (15/15.1) giving:
0.005*0.994 = 0.00497
Round these up, and add them, giving an average Ma rate at:
0.02617 => 0.03mA
So to run this for a year, you need:
365.25*24*0.03 MaHr useable capacity in the battery = 263mAHr.
Remember then that battery capacities are quoted at particular discharge rates/enviroment conditions, that may well not be met (in your case, the battery self discharge will reduce the available capacity, and depending on the enviroment, the full capacity may not b available - if cold, most batteries give a lot less than their quoted 'room temperature' rating).
Best Wishes |
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Gabriel Caffese
Joined: 09 Oct 2003 Posts: 39 Location: LA PLATA, ARGENTINA
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Thanks a lot.... |
Posted: Mon Jun 14, 2004 9:27 am |
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Ttelmah,
Thank you very much for this explanation.
I bought ER14505 batteries, wich are 3.6v, 2400mah and would have expected at least 2 years of duration.
For what you explain, I think there is a quite a bit more duration |
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Ttelmah Guest
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Re: Thanks a lot.... |
Posted: Mon Jun 14, 2004 10:28 am |
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Gabriel Caffese wrote: | Ttelmah,
Thank you very much for this explanation.
I bought ER14505 batteries, wich are 3.6v, 2400mah and would have expected at least 2 years of duration.
For what you explain, I think there is a quite a bit more duration |
Yes. The capacity on these drops to about 1.6Ah, at 0C, and the self discharge will perhaps halve the available capacity, but you should get four or five years relatively easily. :-)
Best Wishes |
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