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Recently my electric paintball gun broke so i decided to come up with my own circuit. I tried to reverse engineer the one my gun had but it was covered in black glue...

I have been trying to design a circuit that will power a solenoid to fire my paintball gun. My problem is that the capacitor does not charge fast enough after being discharged to fire the gun. The capacitor the gun uses is a 6800uF electrolytic cap and the battery is a 9.6 volt alkaline battery specifically for my gun. The solenoid has a resistance of about 1.2-1.5 Ohms. I did the calculations and you need .313 Joules of energy in order to fire the solenoid. Normally i would like to use a regular 9V alkaline battery instead of the 9.6 one. Either way, I'm having trouble keeping the capacitor charged at 9 volts so that it fires the solenoid repeatedly without dropping the voltage too low.I know its possible because i used the circuit from my friends gun to fire my solenoid while measuring the output of the capacitor. The voltage would never drop below 8.8 volts if i used the 9 volt battery. Any idea on what i can do?

[ I included a picture of the original circuit my gun had. It looks like they have 1 IC, 2 transistors, a few caps and resistors and 1 or 2 diodes.1[The other side just has a button a capacitor and an led [2]] enter image description here

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  • \$\begingroup\$ What kind of transistor did you use to charge the capacitor with? A bjt? a mosfet? What is it called? Or did you salvage it from the gun? \$\endgroup\$ – Harry Svensson Jul 12 '17 at 6:34
  • \$\begingroup\$ Instead of pulling teeth I'd love to see that circuit you were trying to design. - You'll get much more help that way. \$\endgroup\$ – Harry Svensson Jul 12 '17 at 7:02
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    \$\begingroup\$ Ok first time using the software from here, but heres what i have tried. I have also tried it without connecting the P channel mosfet and just having the battery connected to the capacitor. \$\endgroup\$ – IvanR10 Jul 12 '17 at 7:37
  • \$\begingroup\$ The trigger signal, what kind of signal is that? Is it a simple physical switch or is it some IC sending pulses? And V3 looks.. fishy. Try out this if you want to play around with circuits in the future btw. I strongly recommend this: tinyurl.com/y84stbey \$\endgroup\$ – Harry Svensson Jul 12 '17 at 7:56
  • \$\begingroup\$ Yes an IC will control that and will control V3. well i have a p-channel for V3, would that not work as a high side switch? what if i just take that off? \$\endgroup\$ – IvanR10 Jul 12 '17 at 8:08
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I am not sure how you came up with your Joules requirement but your capacitor has about 0.275 joules of energy at 9 volts (W=0.5*C*V2). And that assumes you take the cap down to 0 volts which will not be the case. If you care to elaborate on your Joule calculation, perhaps this could be refined.

The ESR (Effective Series Resistance) of the cap will also play a role as it effectively converts some of the capacitor's energy to heat during the discharge cycle. It also slows the charge time of the cap. And it slows the fire time of the solenoid. Select a capacitor with the lowest possible ESR to minimize these effects.

The RDSon of the FETs are important as they too will work in the same way as the ESR of the cap. Choose FETs with low RDSon.

The timing sequence should be to have both FETs off initially. Then turn on the left FET to charge the cap. When you want to fire, turn off the left FET and then turn on the right FET. If you have both FETs on to fire, you will not get the desired high energy effect from the cap.

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  • \$\begingroup\$ I got .313 joules because i used 9.6 volts instead of 9v. When you cock back the gun, it wont fire if the cap is charged at 9 volts. It needs at least 9.3-9.6 volts. This is also another problem im dealing with because i dont know where to get these 9.6v batteries. i want to use a standard 9volt alkaline, so what would be the best way to step up the 9 volt battery to like say 10v?. I would need a low ESR but im trying to use the cap that was originally on my gun since i know its possible because they used it to begin with. Thank you, i will try to look for a FET with low Rdson. \$\endgroup\$ – IvanR10 Jul 12 '17 at 22:08

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