Estes Ignitors

Hello all.

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.

I've got a copy of the Estes bumph but the no fire / all fire amps aren't shown. There are two launch panels, one states that 4 AA cells are required, the other requires one or two ( ) 7.2v NiCad battery packs.

I use to use the tiger tail igniters for my model rocket and they ignited fine off a 12 v rechargable battery. However it must be at top power so if I were you I would try linking 2 nine volts together.

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.

Cool thats quite interesting.

LeMaitre stage pyro igniters are about 1.4 ohms, 20mA no fire 500mA all fire The shrike applies 400v across a load of no more than 400ohms so that allows 1amp under worst conditions. It seasy to get the HV ganerator by cannibalising a used flash disposable camera then it all works off a single cell 1.5v aaa to d sizes. Biggest worry is allowing for the resistance if the command wire, 50m of bell wire will reduce the power available to the igniter, and I guess 50m isnt the largest safety distance you've used!

Thanks all, that give me something to work on.
I've been trying to get hold of a Shrike, but I've only seen them at very silly prices.

I thought it would be expensive and inconvenient to use a 50 metre long roll of wire to ignite rockets / pyrotechnic devices. I have some experience in electronics and circuit board work so I designed my own variable timer circuit. I can set the delay frome between 5 and 70 seconds and it is very accurate. Press the start button with the required time set and it gives you time to retire to watch the launch. If you have a stop watch you can time how long it will be until it will fire. The timer runs off a 12V lead acid battery but a smaller one would work. If anyone would like the circuit diagram then I could post it on here if you could tell me how I would do that?

Chris

LeMaitre ignitors = Bickfords. Very reliable, but ??????

A 50m roll of bell wire from B & Q should be ?5ish. How much you need is governed by the shrapnel range in the worst case of the propellant exploding. Having a long timer lets you retire to a safer distance but also lets others walk into the danger area without you being able to abort the firing. An explosive with a timer would likely attract the wrong kind of attention from the law.

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. 

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

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

Good reply!