Just a quick question.
I'm building a firing unit and need to know the volt/ampage needed to trigger Estes Ignitors.
Thanks y'all.
Edited by Helmetfire, 05 October 2004 - 05:46 PM.
Posted 05 October 2004 - 05:45 PM
Edited by Helmetfire, 05 October 2004 - 05:46 PM.
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Posted 10 December 2004 - 05:37 PM
Normally a single AA battery (AA cell) gives 1.5V and 1 Amps. If a multiple AA batteries are connected in series then the voltage will inclease multiply by the number of the battries and the Amps will stay the same (1Amps), but if you connect a multiple AA batteries in parallel then you will get the amps multiples and the voltage will stay the same (1.5V).
Examples :
Connecting 4 AA cells in series will give you 6V / 1Amp.
Connecting 4 AA cells in parallel will give you 1.5V / 4Amps.
Hope this will help you to calculate what you need.
Posted 11 December 2004 - 01:25 PM
Good reply!These are all maximums thought.
If you have 4 1.5Vv cells in parallel, yes the potential developed is still 1.5V, but the current drawn by a similar load will be exactly the same as well. The benifit of this configuration is that each cell will only have to provide 1/4 of the current. If an igniter will not blow on a single cell, it will still not blow on 1000 in parallel.
All currents are produced by potential differences. Just because the battery can produce an Amp, dosen't mean it will. V = IR is essential in all cases.
Take the igniter mentioned above. It will definately blow on half an Amp. Say you have 100m of bell wire at 88ohms/km, that amounts to a resistance of 17.6ohms (8.8ohms in both direction), and the igniter is 1.4ohms, total being 19ohms. To ensure the igniter will blow you need to drive a current of 500mA down the line, this will require a voltage of 9.5V, or 6 and a third 1.5V cells in Series, say seven for the poor battery's sake. With 7 AA batteries you will drive a current 553mA (not counting the internal resistances of the batteries) and the igniter will definately blow. All the batteries will have a current of 553mA flowing through them while the igniter blows. If you had 14 batteries in a 7 X 2 configuration the voltage p.d. will be 10.5V and the current through the igniter will still be 553mA. However all the batteries will only have half of the 553mA flowing through them.
Hope this helps
Andrew
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