Igniter test circuit design

[TX-Redleg]

I found a great paper by Robert Briody titled Electric Current Requirements of Model Rocket Igniters dated 2000. I plan on reproducing his test setup, but I'm not an Electrical Engineer and want to understand the design of the "Current Sense Resistor" in the bottom right of the attached diagram. I plan on testing both large igniters (up to QuickBurst Fat Boy) requiring 12V batteries down to MJG e-matches using 9V batteries or small LiPo (either 1S or 2S) batteries used in Dual-Deploy. How is the resistance value of the "Current Sense Resistor" calculated? How do I calculate the proper wattage for the resistor? Should it be 1 watt, 10 watt, 20 watt, etc... Or, is there a better test circuit design in general for this test? I realize this design is based on a launch controller circuit and could be simplified. Any guidance and advice will be greatly appreciated.

 

[Grog6]

The "Current Sense Resistor" will develop a voltage suitable for the o scope, 6A would give you a 1V signal, and the resistor would need to be a 10W resistor.

 

[OverTheTop]

A current sense resistor just turns a voltage into a current that is suitable for measuring. That is how most multimeters measure current, and an easy way for a scope to display current since they typically display voltage.

 

[waltr]

The current sense is based on Ohm's Law. E = I*R
Which is the Voltage across the resistor = the resistance times the current through the resistor.
Just rearrange to get current from Voltage measurement: I = E/R

 

[UhClem]

This is what I used to test igniter current requirements: http://davesrocketworks.com/rockets/rnd/igniter/current/adjsink.html

 

[Handeman]

The most important thing about the Current Sense Resistor is that you know exactly what the resistance is. As @waltr said, it uses ohms law to determine the current and the more accurate you know the resistance, the more accurate your reading can be.

The value of the Current Sense Resistor is a compromise. Since it is in series with the igniter under test, it adds to the resistance of the circuit, so smaller values are better. At the same time. The smaller the value, the more accurate your measurement instrument has to be when reading the voltage drop across it to determine the current flow.
That will be especially true with ematches that may have less then 1 ohm of resistance. A Current Sense Resistor of 0.15 Ohm would be over 15% of the resistance of the circuit and could have a significant affect on the total current developed when using low voltage batteries like a 1S.

 

[Kelly]

You asked about the wattage of the resistor. To calculate the required wattage, you multiply the square of the max current by the resistance: P = I²R.
The current can be hard to know, since that's what you're trying to find with this circuit, and worst case the igniter could be (theoretically) a dead short, meaning your total resistance in the loop would be just the current sense resistor itself, 0.15 ohm.
If you know the total resistance ahead of time, the formula for power is V²/R, so if your ignitor were a dead short the power requirement would be on the order of 12²/0.15, or 960 watts. But, that's a big-a$$ resistor, so we need to be more reasonable.
Your igniter is probably on the order of 1ohm, so the current might be 12/1 = 12 amps. Now we need a sensing resistor of 12²*0.15, or 21.6 watts. That's more reasonable, but probably still oversize. You can probably de-rate this a bit, since a 20-watt (say) resistor is meant to disperse 20 watts continuously, and yours only need disperse that power for the amount of time it takes to burn through the igniter, or the length of time you keep your finger on the 'fire' switch - so, milliseconds to a second. Worst case, you burn out the resistor and need to buy a new one (and smaller-wattage resistors cost less than big-wattage resistors).

If it were me, with a 12v battery, I'd use something like a 1/2 to one ohm resistor, and size it at about 10 watts. This should be a good balance between getting sufficient current through the igniter, and not needing a giant resistor.