Inline Capacitor for Ignition Circuit? - Page 1
Forum section: electronics / projects

author: eerocketman
author_rank: Contributor
timestamp: February 03, 2018, 02:33:27 am
content: Hi Everyone,I am currently working on an E-Match ignition circuit (see schematic below). In order to avoid pulling too much current from the main power source in case the ematch shorts, I would like to push the current through a capacitor so that once DC is reached the circuit will be essentially open. What sort material capacitor can I use for this circuit? And are there any blatant issues with this circuit that I am overlooking?Edit: I have also attached the current output of the circuit through the ematch (R1).Thanks!Ali

author: jbb
author_rank: Super Contributor
timestamp: February 03, 2018, 10:50:07 pm
content: There are a lot of safety worries with things that go ‘woosh’ or ‘bang.’ Think carefully about short circuits, broken wires, programming mistakes, static discharges, sparks from metal tools, impact from dropping tools etc.And storage. Oh my yes. One match going off is a problem. A spark falling into a jar of matches could be a disaster.I suggest you have an electrolytic cap store all of the firing energy for the circuit. It should normally be discharged (use a high side power switch) by a high value pull-down resistor. Before firing, the capacitor is charged through a large charge resistor (bonus points: monitor the cap voltage with a micro to confirm it’s healthy). The match is then fired by a firing transistor using only the energy in the capacitor to avoid bouncing the main rail.If the charge resistor value is high enough, it could limit input current to such an extent that the match doesn’t fire even if the firing transistor is faulty (shorted closed).

author: eerocketman
author_rank: Contributor
timestamp: February 27, 2018, 02:18:55 am
content: Hi JBB thank you for your response. I was hoping to use a large capacitor INLINE with the E-Match in order to restrict the current trough the component to several milliseconds. Do you forsee an issue with this?

author: CopperCone
author_rank: Super Contributor
timestamp: February 27, 2018, 06:01:46 pm
content: I think it would be a better idea to use a heavy relay switching network to charge a cap or inductor (more reliable?) then isolate it and then discharge it.Unless you require massive parallelism i would not recommend using solid state components for this job unless you are trying to make it small for a missile or whatever. The circuit should run at a low impedance and be heavily magnetically shielded to prevent accidental activation.If your firing control system is connected to another device (obviously using a heavy mechanical switch is best) then you should put heavy emi and rfi filtering on it. I would however highly recommend running on batteries and using a fiber link, however eventually you will run into some kind of weakpoint in your system. If its a fiber link i would recommend a digital control to prevent detonation due to a cut fiberoptic that is exposed to light.You also want to shield your blasting wire and put some kind of protection right near the blasting cap or rocket motor to prevent spark ignition from transients.

author: floobydust
author_rank: Super Contributor
timestamp: February 27, 2018, 07:00:49 pm
content: I think you don't want a super fast high-current pulse due to the thermal mass of the e-match. It needs time to heat up and ignite the pryrogen, instead.Vishay 0805 pyrotechnic resistorswould just fuse open before transferring the heat? I would say a longer pulse is better, 100's msec.The Spice simulation should include field wiring, NiCr element, and the interconnects resistance. It's not 1 milliohm even shorted out. The total is several ohms.

author: CopperCone
author_rank: Super Contributor
timestamp: February 27, 2018, 07:41:48 pm
content: A series inductor with the load could help you form a pulse.is the main reason for this that a wire can be vaporized without heating the element causing a thermal release in the wrong part of the circuit?

author: mikerj
author_rank: Super Contributor
timestamp: February 28, 2018, 11:47:21 am
content: As drawn I suspect the circuit is incorrect. With the capacitor on the source side, the gate-source voltage will decrease as the cap charges, which will quickly move the transistor into a linear region (high power dissipation), rapidly reducing current through the load as a function of time.It sounds as though you actualy want the capacitor on the drain side, so the transistor operates as a saturated switch for the entire pulse duration.