Parallel Circuit - Fireworks Firing System

[bluemarshall]

Hi,

Just a heads up, I'm a noob when it comes to circuits so I could be asking a very complicated question and not know it, but I am technically savvy so hopefully if anyone does offer any help, I should understand an explanation easily enough.
So here is my problem:
I am building a push button firing system for a fireworks display. I have (let's say) 10 NO mom push switches wired in parallel from the +/- posts of a 12V 1700mA power supply. I attach one of the 10 parallel loops to an electric match, and when the loop is closed ( I push the button) the match lights instantly. If i push 2 buttons at the same time, the 2 matches do not go off. I am familiar with V=IR and P=IV and understand why this happens.
My question is:
How do you build a circuit that will supply each loop with 12V 1700mA regardless of how many buttons are pushed (loops are closed)? I realize that this will likely require a custom transformer, and capacitors for Potential storage, but other than that I figure I should ask some professionals.



Sorry for straight outta Compton nature of my diagram I'm at work and I threw it together in photoShop

 

[CocaCola]

The first thing you must understand is that the 1700mAh is a not how much current the battery can supply at any given time, it's an overall capacity vs time rating... In short of it is that it can supply 1700mA for 1 hour... That isn't to say it can't supply 3400mA for ½ an hour or 850mA for 2 hours, or 50mA for 34 hours... BTW that is all perfect math, not real world drain times

The internal resistance of the battery and the resistance of the circuit will dictate how much instantaneous current will flow...

In short get a battery that will provide more current in an instant to solve your problem... Something like a car battery...

 

[bluemarshall]

Thanks for the reply CocaCola. I'll give that a try tonight. I'll let you know what happens.

 

[KrisBlueNZ]

CocaCola, he isn't using a battery and his figures are mA not mAh.
The simple answer is that you need more current. With two (or more) matches in parallel, the combined resistance is too low and the power supply voltage drops.
If you need to be able to fire all ten matches at once, you need a power supply that can deliver 17 amps, which is quite a lot of current!
My first suggestion would be to use more than one power supply. Obviously if your power supply can only light one match, you would need another nine similar power supplies. But if you can increase the current available from each supply, you could get away with two or three power supplies. If the matches are normally activated in sequence, you can interleave the power supplies to reduce the chance that many matches will be powered simultaneously from each - for example, power supply 1 would power buttons 1, 4, 7 and 10; supply 2 would power buttons 2, 5 and 8, and supply 3 would power buttons 3, 6 and 9.
Are you using a DC power supply? An AC power supply (just a transformer) would be simpler, and you might be able to pick up suitable transformers second hand or in a bulk lot. Also, DC power supplies often have "foldback" current limiting - as soon as the preset current limit kicks in, the output voltage drops steeply. With a simple transformer, there is no intentional current limiting of this kind.
A car battery as CocaCola suggests would definitely be able to fire all ten matches simultaneously... they are typically rated for over 300 amps CCA (cold cranking amps)! Watch out though, a car battery can easily deliver enough current to turn normal wires into heating elements, then into a smoking mess! I strongly recommend insulating the posts reliably and using fuses, as close to the battery as possible, to minimise the chance of heavy current flow in your circuit. Also, remove any rings and other jewellery while you work with a car battery. You don't want to be the unwilling recipient of an immediate finger amputation!

 

[bluemarshall]

Hey Kris,

Thanks for the advice. I guess that the biggest issue in this project is the variability in the volts/current requirements. If there was always only one button, then you build for that. If it would always be 13 buttons simultaneously, then build for that. Based on what I'm reading from you guys it's the current that is most important, not the volts as the matches are being lit using nichrome resistance wire. If we were to go the car battery way, what gauge wire should be the minimum? Also, even though there are 300 cca's available, won't the circuit only pull what it needs? So if the e-match needs 1 amp, then that's will flow through that loop? And if i press all ten buttons then 10 amps is pulled from the battery but across each parallel loop i'm still only seeing 1 amp? is this correct?

thanks

 

[CocaCola]

CocaCola, he isn't using a battery and his figures are mA not mAh.

Missed that.. As you said there simply isn't enough current using you current power source, best to use one that has excess...

it's the current that is most important, not the volts as the matches are being lit using nichrome resistance wire.

Both are important to you in this application as a sum

what gauge wire should be the minimum?

If the matches are rated for 1A @ 12 volts, I would say stick to 28 gauge or heavier...

won't the circuit only pull what it needs? So if the e-match needs 1 amp, then that's will flow through that loop? And if i press all ten buttons then 10 amps is pulled from the battery but across each parallel loop i'm still only seeing 1 amp? is this correct?

That is the short of it...

 

[gorgon]

I don't know about these electric matches, but I know that explosive actuators has a tendency to short when fired. If you press more than one button at once, you may end up with one short circuit and the rest not getting fired, before you release the shorting switch.

Depending on how safe you want your acctivation to be you need to think about this. I would think the best would be separate feeds, or some sort of pulsed drivers.

TOK

 

[bluemarshall]

@ gorgon
The electric match is basically a wooden match with a positive and negative 16 gauge wire on either side of the match head. The wires are bridged with 44 gauge nichrome ignition wire designed to heat rapidly and ingnite the match head. Once the match lights the ignition (bridge) wire is toast and the circuit is open even if the button is still down. Each e-match is wired in parallel off the main +ve and -ve posts from the power supply. Since the e-matches all share a common ground, their negative side each have a diode so that to voltage or current can move along another e-matches -ve lead.
@Kris & CocaCola
We are testing things out with a lawn tractor battery and so far it's looking good. We first tried a marine battery with 750 CCA's and it fried the nichrome bridge wire so fast it didn't light the match head of the e-match. Then we added a 10 ohm resistor to the loop (using V=IR we figured 12V supply, 10 ohms resistor + 2 ohms nichrom wire would give us 1A which should be more than enough to light the e-match) and the resistor fried. Oddly enough, the resistor fried and the e-match bridge wire did not light the match head in this instance. Should the bridge wire have heated the match head (or fried) before the 10 ohm resistor in this case? Why would a 10 ohm resistor burn out before the super thin 44 gauge bridge wire?
Thanks for all the input to everyone, it is all very helpful and informative

 

[CocaCola]

The resistor was likely rated too low, you are likely dealing multiple Watts here to fire the matches, did you use a beefy power resistor our did you try a ¼ Watt one?

 

[bluemarshall]

The resistor was 10R 1/2W 5%. The highest we could find was 2W. Which based on P=IV still wouldn't be enough. So let me get this straight; V=IR, so 12V/(10ohm+2ohm) = 1A.
Therefor 1A x 12V = 12W ? so we would need a 12 watt rated resistor? Am I using these formulas correctly?

 

[CocaCola]

I'm not 100% familiar with commercial matches (I have make my own) what is the recommended voltage? May I suggest that instead of a resistor (you can get huge used ones on Ebay if you must) you just lower the voltage with a series diode, or two or three or four... You will get about a .7 volt drop with each diode, with some experimentation you can probably get a reliable ignition... This will have the same effect on slowing the heat up and burn of the wire...

And yes you are in theory going to need large wattage resistors, or a huge one on the main supply...

 

[KrisBlueNZ]

I guess that the biggest issue in this project is the variability in the volts/current requirements. If there was always only one button, then you build for that. If it would always be 13 buttons simultaneously, then build for that.

Yes, your supply has to be able to provide its specified output voltage when driving the maximum load (current) that you will be drawing from it. If each e-match draws 1A at 12V and you want to be able to power ten at once, it must be able to supply 10A.
If you have mains available, one or more transformers would be a good way to get that kind of grunt.
The e-match shouldn't care whether you use AC or DC, but you mentioned that it has a diode in series with it. I don't see any reason to do that; if you're using DC and the match goes short, it will still short out the supply and prevent other matches from firing (while the shorted match's button is held down), so this suggestion that if the match goes short, the diode will allow other matches to fire, is not true. If you use AC though, you should remove the diodes. If you're worried about the e-match going short (and you say you aren't), you could add a polyswitch rated at, say, 150% of the e-match's rating in series with each loop. These are resettable fuses. They take a little while to open, though, depending on the short-circuit current as a proportion of the rated current. Several seconds is typical. So they might not be enough. Lucky that the e-matches don't go short circuit!

Based on what I'm reading from you guys it's the current that is most important, not the volts as the matches are being lit using nichrome resistance wire.

Both are important. The power source produces a certain voltage, and will maintain that voltage provided that you don't draw too much current from it. If you draw too much current, the voltage will drop. For a simple transformer, it will drop fairly steadily. For a DC power supply with foldback limiting, exceeding the current rating, even briefly, can cause the output voltage to drop way down. For a lead-acid battery, exceeding the current rating will make the wiring melt

If we were to go the car battery way, what gauge wire should be the minimum? Also, even though there are 300 cca's available, won't the circuit only pull what it needs? So if the e-match needs 1 amp, then that's will flow through that loop? And if i press all ten buttons then 10 amps is pulled from the battery but across each parallel loop i'm still only seeing 1 amp? is this correct?

Yes, that's exactly right.

The resistor was 10R 1/2W 5%. The highest we could find was 2W. Which based on P=IV still wouldn't be enough. So let me get this straight; V=IR, so 12V/(10ohm+2ohm) = 1A.
Therefor 1A x 12V = 12W ? so we would need a 12 watt rated resistor? Am I using these formulas correctly?

Not exactly; it depends on the resistance of the e-match. When you close the button, the 12V will be split between the resistor and the e-match according to their relative resistances. If the e-match is rated for 12V and 1A then its resistance is about 12 ohms (Ohm's law: resistance equals voltage in volts divided by current in amps). With a 10 ohm resistor, the total resistance is 22 ohms. 10/22 of the total voltage will be dropped across the resistor and 12/22 of the total voltage will be dropped across the e-match.
If the e-match blows too quickly I would use a lower voltage instead of series resistors, if possible. Otherwise, yes a series resistor for each loop would be OK. Another possibility would be a PTC (positive temperature coefficient) thermistor, which is like a polyswitch fuse I mentioned earlier but has a significant resistance. It will heat up due to the current flow, and when it gets hot enough, it will go open circuit. This would be a convenient way to prevent problems if an e-match goes short. You would need some pretty big thermistors though.
If you have mains available I really would recommend using AC straight from a transformer rather than a battery that can deliver 600 amps! They really are dangerous.
Food for thought...

 

[gorgon]

The resistor was 10R 1/2W 5%. The highest we could find was 2W. Which based on P=IV still wouldn't be enough. So let me get this straight; V=IR, so 12V/(10ohm+2ohm) = 1A.
Therefor 1A x 12V = 12W ? so we would need a 12 watt rated resistor? Am I using these formulas correctly?

I would have used a 5W wirewound power resistor. The few seconds the power is applied will average out, even if it is higher than 5W.

TOK