E-Match Circuits
jgary
Posts: 4
Hi,
I'm trying to design an E-match circuit that is able to discharge at least 6 amps of current for about 0.1 seconds. The circuit would need a 9v battery, capacitor(s), resistor(s), and transistor(s). The the discharge would probably depend on the status of a transistor acting like a switch. If anyone has a knowledge base on how to design an E-match circuit, I would greatly appreciate your help!
-Jonathan
I'm trying to design an E-match circuit that is able to discharge at least 6 amps of current for about 0.1 seconds. The circuit would need a 9v battery, capacitor(s), resistor(s), and transistor(s). The the discharge would probably depend on the status of a transistor acting like a switch. If anyone has a knowledge base on how to design an E-match circuit, I would greatly appreciate your help!
-Jonathan
Comments
And welcome to our forums.
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Paul Baker
Propeller Applications Engineer
Parallax, Inc.
This is a common thing in rocketry for ejection charges. What is your application? Are you planning on using this in a rocket? You may want to post your question at the RocketryForum at www.rocketryforum.com.
You could use a relay or a power transistor as your switch. The 9-volt battery may be able to fire your ematch. You should test it several times to verify whether this works.
If you use a capacitor you will need a large one to deliver 6 amps for 0.1 seconds. I would suggest testing with various sizes to determine the size you need. I checked Digi-Key, and a 0.1 Farad, 20-volt capacitor costs over $50. I would hope you could get by with a smaller capacitor.
Dave
Dave mentions that having a .1 F or less capacitor may work to discharge 6 Amps of current over .1 seconds. I'm aware that is a lot of current, but I believe it can work if i have the correct size capacitor. What size capacitor do you think is appropriate?
What type of transistor should I use? Dave mentions a power transistor and a relay transistor, however I do not know what either one of these are.
How much resistance do i need (total) if i'm using 9V and 6A of current? Using Ohm's law, it would seem that i only need 1.5 ohms total (V = I*R), but it doesn't seem right.
Jonathan…
I am just curious? Why 6 amps, most E-mach only require 1 amp and some less than that.
If you plan on doing multiple cues at different times a diode for bleed back on one of the connectors lines is a good idea.· This will help eliminate the possibility of the next cues going of prematurely.
There are a few discussions if you do a search in the Parallax Forums.· Think safe, I blow thing up for a living working in film, pyrotechnics and fireworks and I have heard of some bad experiences with home builds·and·depending where you live·legalities,·liabilities.
As SideStreet said, most E-matches require 1 amp or less.· These is some information on commercial E-matches at http://www.info-central.org/recovery_ematches.shtml·.· These will fire off of a 9-volt battery, or even lower voltage.· A capacitor is not needed.· You need a Low Explosive Use Permit (LEUP) to purchase these E-matches.· You also need a LEUP to use black powder for rocket ejection.
The circuit that fires the E-match shouldn't have any series resistance inline with the E-match.· You want all of the energy to go into the E-match.· The switch should also have low resistance.· As I mentioned before, it should be a relay or a power transistor.· Once the E-match blows it should provide an open circuit, just like a fuse.
I suggest that you join your local rocket club if you haven't already.· There should be people there who have already done what you are trying to do.· They should be able to provide some assitance.
Dave
I have a few comments about your circuit.· If you are going to use a capacitor to fire the igniter, then you should use a resistor between the battery and the capacitor.· The size of the resistor depends on how long you want to allow for the capacitor to charge up.· The time constant is R*C, so for a .1 Farad capacitor you would need a 100 ohm resistor to give a 10-second charge up time.· A 10 ohm resistor would give you a 1-second charge up time.· You should allow for 2 times the time constant for the capacitor to be almost fully charged.
I assume you will control the transistor by a logic-level signal.· It would be better to connect the emitter to ground, and connect the collector to the igniter.· If the base current is 10 milli-amps the transistor would need a gain of 600 to drive 6 amps.· A normal transistor doesn't have this high of a gain.· You would need a Darlington transistor pair to achieve this type of gain.
A .1 Farad capacitor is fairly large.· A 16 Volt·capacitor at Digi-Key is 35mm by 80mm, which is·1.4 inches in·diameter and about 3 inches long.· It's not clear if .1 Farad is large enough for your purpose.· You might need a larger capacitor, or you might be able to get by with a smaller one.· I think it is better to not use a capacitor and drive the igniter directly from the battery.
If you want to calculate how large the capacitor should be you must know the resistance of the igniter.· Also, the current is going to go down as the capacitor drains.· As an example, if the resistance is 1 ohm, and the capacitor is charged up to 9 volts it will deliver 9 amps initially.· When the capacitor discharges down to 6 volts it will be delivering 6 amps of current.· The equation for capacitive discharge is
V = V0 * e ** (-t/RC), where V0 is the initial voltage, R is the resistance and C is the capacitance.
The capacitance can be determined by taking the natural log of each side and solving for C.· This gives the following equation
C = t / (R * ln(V0/V))
For·t = 0.1, R = 1.0, V0 = 9 and V = 6 the capacitance would be 0.25 Farad.· I've made some guesses on the resistance of your igniter and the current requirement, so I can't say for sure what size capacitor you actually need.
Dave
Q = C*V
where Q is the total charge on the capacitor in coulombs,
C = capacitor value in farads, and
V = voltage on the capacitor
1 amp = 1 coulomb/second
Q = I*t
where I =current in amps and t = time the current flows
Combining these two equations gives:
C*V = I*t or solving for C:
C = I*t/V
Plugging in the known values of 6 amps, 0.1 seconds and 9 volts gives:
C = 6*0.1/9 = 0.67 farads
This assumes that the resistance in the output side is such that
9 volts on the capacitor yields an initial output current of 6 amps.
As Dave pointed out, the load current will decrease over time,
so you won't have a 6 amp current for the entire time.
You could select an output resistor such that the average current
was 6 amps over a 0.1 second interval.
Analyzing the circuit based on the charge transfered will get
you in the ballpark in determining the size of the capacitor needed.
Looks like you need a significantly larger capacitor than 0.1 farads.
Consider 0.7 farads.
I agree with Dave regarding his comments about a current limiting
resistor in series with the 9 V supply and I also go along with his
recommendation to change the transistor leads so that the emitter
is grounded.
with
phil
resistance will eat up a lot of your charge from the capacitor, leaving less for the e-match.
Hope this helps!
1) What are the the negative aspects of charging a capacitor without a resistor? I don't see any problems with charging up a capacitor quicker than usual. Does heat become a factor?
2) As the capacitor is being charged, what happens to the voltage of the battery? Does it approach zero?
As the capacitor charges its voltage will approach the voltage of the battery.· The voltage difference between the battery and the capacitor reduces exponentially.
I am still puzzled why you want to use such a large capacitor.· If you are using this in a model rocket the capacitor will weigh more than the rocket and motor together.· A 1 Farad capacitor would probably be bigger in diameter than your· rocket, and would cost more than the rocket itself.
You haven't provided much information about your application.· I am guessing that you plan on using an Estes igniter.· You should be able to fire this with a 9-volt battery without using a capacitor.· You could use a capacitor to give it a little extra energy, but I wouldn't use the huge capacitors we have been talking about.· Most controllers use capacitors in the range of 1,000 uF to 9,600 uF.· Check out this thread on the Rocketry Forum --·http://www.rocketryforum.com/showthread.php?t=37750&highlight=ematch·.
I'm interested in your project, and I'd like to get more information on it.· If you don't want to post it here please send me a PM.
Dave