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I am currently working on a circuit in order to ignite E-Matches such as these in order to light up the engine of an amateur rocket. In order to have control over when to actually ignite the E-Matches I would like to connect the circuit to a microcontroller and control the ignition from there.

I have seen this question that is very similar but for me the connection of the microcontroller is still unclear. E.g. why do we need a separate switch connected to the microcontroller if we can use the transistor as a "switch".

That's why I tried to come up with my own circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The idea is that if the I/O pin of the microcontroller is set to high (\$3.3V\$) a small current flows through the base of the transistor and therefore we have a closed switch. Because now the capacitor has a connection to ground the capacitor begins to charge up and at the beginning the charging current would be $$I_{c_{max}} = \frac{9V}{10\Omega} = 0.9A$$ as at the beginning of the charging process the capacitor acts as a bypass and therefore all the current flows through the capacitor. Because the E-Matches need a minimum fire current of \$0.705A\$ the \$I_{c_{max}}\$ current should be enough. The value of the capacitor was chosen in order to have a slower charging time so the time the current through the capacitor is higher than \$0.705A\$ could be expanded in order to guarantee an ignition. The resistor \$R_2\$ is needed in order to discharge the capacitor when there is no connection to ground through the transistor.

Does this circuit actually work as I just described or am I thinking completely wrong about this problem? Is there some additional protection/safety needed especially for the microcontroller?

Any help is greatly appreciated.

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  • \$\begingroup\$ Did you consider using something like a capacitive discharge ignition system? \$\endgroup\$
    – SteveSh
    Dec 5, 2021 at 22:00
  • \$\begingroup\$ I really don't like the specifications given at that website. A proper all-fire specification will state the number of Joules required and the time within which they must be delivered. (Action integrals are what's important.) And time is VERY important when talking about rocketry and safety, not to mention multi-engine coordination (if that's something that applies.) The minimum all-fire current alone isn't a spec, my opinion. Looks like one. But isn't. \$\endgroup\$
    – jonk
    Dec 5, 2021 at 23:22
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    \$\begingroup\$ @Nick Only that you make a phone call and talk with the product engineer and ask for the details. That's what I did with Estes. I had a long and very useful discussion. Very productive. The engineer also sent me thousands of ignitors to use in my testing -- no charge. I recommend you do the same. Call and ask \$\endgroup\$
    – jonk
    Dec 6, 2021 at 7:02
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    \$\begingroup\$ Related: if you require ignition at altitude be VERY aware that ignition requirements can be substantially more demanding at altitude with lower pressure. Even though the e-match has its own oxygen supply (Chlorate or whatever) pressure often has a major effect. Numerous recovery failures have occurred when always reliable igniters failed to ignite at altitude. \$\endgroup\$
    – ATCSVOL
    Dec 6, 2021 at 8:43
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    \$\begingroup\$ @Nick I do have another thought. There are safety concerns which relate to a rocket located some distance away from the controls. Having a circuit capable of igniting rockets located at the rockets (whether one or more than one) means that a malfunction (false activation due to local unexpected events) could be a problem. Consider ways that are fundamentally safe based on physical laws of nature and not solely based upon engineered safety. You want nature on your side and a violation of fundamental laws of physics to be required for an "accident." That makes it "unlikely" to happen. \$\endgroup\$
    – jonk
    Dec 6, 2021 at 17:13

2 Answers 2

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You need to include the resistance of the e-match, the ESR of the capacitor, and the saturation voltage of the transistor in your circuit model. These will cause a significant reduction in the current.

You should also describe the characteristics of your 9V source. If you are thinking about using a common alkaline battery then the ESR of the battery needs to be included as well.

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  • \$\begingroup\$ Thanks for the advice. I looked at some specifications of the required components and added all the ESR needed (battery (\$8\Omega\$, capacitor \$0.08\Omega\$, E-Match \$1\Omega\$). Also I added a saturation voltage for the transistor of about \$0.9V\$. With all these resistances by my calculations a \$9V\$ battery is barely capable of providing the necessary current even when leaving the top resistor completely out. So it is save to assume that I need a battery with a higher voltage? \$\endgroup\$
    – Nick
    Dec 6, 2021 at 7:15
  • \$\begingroup\$ That would be a safe assumption. \$\endgroup\$ Dec 6, 2021 at 11:40
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R2 should be connected in parallel with C1 in order to discharge the capacitor. As you have drawn it, R2 will not discharge C1.

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  • \$\begingroup\$ Ah I see the problem with my circuit there. Thanks for the info. \$\endgroup\$
    – Nick
    Dec 5, 2021 at 21:44

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