I only use Lipos nowadays.

If weight is not an issue, I recommend using these:

- For 1S 3.7V batteries - https://www.amazon.com/dp/B01B4EUP6Y/?tag=skimlinks_replacement-20

This is true with the Stratologger, though not with other Altimeters. The Eggtimers don't have this limitation and have functioned fine on those. I use the 1200mah 2S lipos because the eggtimer units use 80ma from the WIFI. They are overkill for the stratologger though. The OP did not mention a specific altimeter, so I gave a general recommendation.

For the Stratologger in particular, 10C 500mah batteries are hard to come by. A 250mah 20C battery like

https://www.amazon.com/dp/B00W9F6VAC/?tag=skimlinks_replacement-20 will do well. Max current is 5A and it has 2.5 times the current capacity of a normal 9V. Given the stratologgers minimal consumption of 3ma it would run for days.

Nate

A 10C 500 mah cell or a 20C 250 mah battery will deliver way more than 5 amps. In reality the short circuit is likely to be 25 to 50 amps!

What the C rating means for these batteries is that you can draw 5 amps continuously from the battery without significant loss in capacity due to internal resistive heating of the cell due to internal IR drop. Alternate interpretation is the higher internal conductance of the cell, the higher the C rating, and the lower the I^2 R heating.

Ohm's law of series resistance always applies. I = V/R where R = R battery + R wires + R FET + R e-match.

The R battery is 1.8 ohms for a 9 volt alkaline battery, but below 0.1 ohms for many LiPo batteries. R wires is typically less than 0.1 ohms, R FET is usually below 0.1 ohms, and the resistance of the e-match is ~ 1.5 ohms if good, less than 0.1 ohm if shorted, and infinite if open unless there is a plasma arc and then the resistance of the plasma will be less than 0.1 ohms.

When you use an 9 volt alkaline battery, I < 9/(1.8+1.5) = 2.7 amps under normal conditions and I < 9/1.8 = 5 amps there is a dead short.

If you use a single 4.2 LiPo cell, then I < 4.2/(1.5 + 0.1) = 2.6 amps but if the e-match shorts then the current could be as high as I ~ 4.2/0.1 = 42 amps if the battery and FET resistance is really low! The only simple remedy is to put a resistor in series with the e-match return to the -side of the battery. A wire would ~1.6 ohm resistor with 2S battery pack should work: I < 8.4 /(1.5 + 1.6 + 0.1 ) = 8.4/3.2 = 2.6 amps under normal conditions and I = 8.4/(1.6+0.1)=5 amps with a shorted e-match.