Jim, like I said it is tricky and you have to make the measurements like you are doing. Every case is different and what is safe for one instance may not be for another.
A question that I have is are you running the shunt in parallel with your ematches with a series resistor between the power source and your parallel circuit?
In otherwords do you include a dropping resistor that the shunt works against to drop the voltage across the match?
Actually, I mispoke on the method I use. I don't assume the nominal voltage of the battery - that wouldn't work. Instead, I use published values for the maximum current capability of the battery. I also don't try to measure the resistance of the shunt wire itself - again, this is from published information.
A few years ago, I posted the calculation I developed for the case of a 9 volt battery with a shunt. I've copied it below. I think a review of it will show the process isn't that complicated, although I have always recommended ground testing of the exact circuit just for peace of mind.
More recently, I have gone to LiPo batteries (not having a choice with the EasyMega altimeters). There, I solve the current problem by using small LiPo's with limited current capability, again, combined with ground testing (per my previous post).
Jim
...From February 2015 ...
It's relatively easy to determine if your shunt is sufficient. A simple example.....
I use a 9-volt alkaline battery for my altimeters (I specifically do not use LiPo's in this service). The maximum current a 9-volt battery can produce is around 5 amps. The statistics for the JTEK ematches I use are:
- Maximum no fire - 300 ma
- Minimum all fire - 750 ma
- Maximum recommended test current - 40 ma
- Recommended firing current - 1000 ma
Let's say that the resistance for the ematch part of the circuit is 2 ohms (the match and some wire). If my shunt wiring is 1 foot long and I use 24 gauge wire, the resistance of the shunt could be as low as 0.026 ohms, depending on the switch used. If the altimeter fires and shorts the battery to the ematch, the maximum possible current through the ematch would be 0.026 / 2.026 x 5 = 64 ma. This is above the maximum recommended test current but well below the maximum no fire current. And, since the circuit has some actual resistance, the battery won't actually produce 5 amps.
In the above scenario, the shunt would prevent the ematch from firing if the altimeter fires. However, if you used a battery or cap that could source more current, or a longer piece of wire for the shunt, or finer wire, you could approach the maximum no fire current. I ran this test case with 4 feet of wire (0.104 ohms) (edit - for the shunt wire) and a shorter ematch (1.4 ohms perhaps), which might have given 0.104 / 1.504 x 5 = 346 ma. That fired the match.
Although my example case would be safe, I have gone to adding a resistor to the ematch pathway for sustainer ignitors. For a 9 volt battery, and with a recommended firing current of 1 amp, the ematch circuit could have a resistance of as much as 9 ohms. I use a 3 ohm resistor (plus or minus), giving 5 ohms total, to drop the maximum current to below the recommended maximum test current.
Jim