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View Full Version : Power drain, coin cell, while we're on the subject

- 29th April 2006, 08:20
How does my math look here...
I've got a circuit going for a keyfob type project, using a PIC16F688, internal clocks (both 32khz and 4mhz), RF transmitter from Rentron, an LED for display, 2032 type coin cell for power.
PIC draws about 200nA sleeping using ULPWU module...
PIC draws about 20uA total while check buttons once per second at 32khz internal...
Circuit draws about 40mA (PIC plus transmitter plus an indicator LED) total while transmitting (figure one second per hour)...

So,
3,595,000 ms @ 200nA = 719,000,000 nAmS (nano-amp milli-seconds)
3600ms @ 20,000 nA = 72,000,000 nAmS
1000ms @ 40,000,000 nA = 40,000,000,000 nAmS
Total = 40,791,000,000 nAmS (call it 41,000,000,000)
divide by 3,600,000 (ms in an hour) = 11,330.83 nAh

A good coin cell capacity is 200mAh = 200,000 uAh = 200,000,000 nAh

200,000,000 nAh / 11330.83 nAh = 17,651 hours = 735 days = 2 years

IF my circuit ran according to spec's, it should last 2 years on a cell, right?
JDG

Ron Marcus
- 29th April 2006, 18:17
Not exactly. The current drain of 40 mils will kill the battery much faster than the rated capacity. A 200 mil battery has a maximum drain rating of 15 mA/H. Over this, and the battery projected run time( and chemistry) is severly affected. A one second burn is actually a pretty long time for these critters. Using a sizeable cap will help, but size will probably be at a premium. Reducing on time is the best way to stay within specs.Look at a couple of months for a reliable estimate.

Ron