Power drain, coin cell, while we're on the subject

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

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

The one second per hour was a quicky guesstimate on my part.
10 characters @ 2400 baud once per hour is more like the real situation. Add that to some RF data-slicer set up time and the real amount would probably be more like 62ms (just a tad bit different from the one second I originally planned!).
But the math...does it sound right, feel right? Theoretically of course...
I don't really expect a coin cell to run for 2 years with my building ability...2-3-4 months would be just fine with me.
JDG