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12vdc Solenoid driver off of 9v battery

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Hablomos

New Member
I'm working on a circuit for a paintball gun that will drive a 12v solenoid. I have the timing circuit completed with a 555 timer (for full auto). My problem is that I want to charge up a 2200mf capacitor to 12v from a 9v battery and discharge the the capacitor through a transistor or section of transistors to fire the solenoid. I had planned on using a single NPN transistor but if anyone has any idea's I'd love to hear them.

 

k7elp60

Active Member
What is the resistance of solenoid coil, and how much time do you have to charge the 2200uF capacitor? Your post says 2200mf, but I think you mean microfarad which the correct notation is uF. There are circuits that will boost the 9V to 12V but the charging current of the 2200uF capacitor is important.
 

Hablomos

New Member
yes, it seems to be a 2200 microfarad cap. I don't Exactly have a key for the micro symbol and haven't really ever had to try to express it in text. Anyway, I don't have long to charge the capacitor. The gun cycles the solenoid at 17Hz as is. id like to improve upon that if at all possible, say somewhere around 20-25Hz.
The coil ohms out at .5 ohms. I guess at 12v the coil alone would pull 24amps though I'm sure the inrush current would be much higher.
The odd thing is that when I took a reading of the capacitor on the existing board and got a reading of around 9v, but If i charge a "loose" 2200 microfarad capacitor with a 9v battery then discharge it into the solenoid, it doesn't pull all of the way in.
 

Hero999

Banned
I don't Exactly have a key for the micro symbol and haven't really ever had to try to express it in text.
Use u.

If you click on advanced there's a symbol panel just to the right of the text box which includes the µ symbol as well as others, Ω, Π, √ etc.
 

Sceadwian

Banned
Full auto fires at what 15 shots a second? You're not going to be able to get that much power out of a 9 volt battery, it simply isn't there.
 

duffy

Well-Known Member
Yes, I think that may be asking too much from a 9V. What is the average current draw in full auto?
 

Hablomos

New Member
Oh, wow how blind was I? thanks hero9999, for the tip aboµt the symbols. The current draw is about 400mA at 13 balls per second and a little over 500mA at 17bps. It seems like the existing board seems to only fire the solenoid for just long enough to release the bolt. I'm sure its in the miliseconds.
 

duffy

Well-Known Member
Measure the voltage on a new 9V battery while the circuit is pulling 400-500ma. It will be loaded down pretty heavy, maybe only read 7V or so.

Timing is important. The peak current would be 18A off a 9V, so 162W - it probably never gets that high. If your 555 isn't delivering the same pulse (too narrow) the coil probably won't move enough current to operate.

See if you can find out why the loose 2,200uF doesn't fire it. The valve may be a "pilot valve" and needs air pressure to function correctly (probably not, the resistance seems way too low for a pilot valve). I suspect the cap only provides part of the energy pulse, and the rest is provided by the battery, in parallel.
 

Sceadwian

Banned
Why aren't you using a 9 volt solenoid anyways?
 

Hablomos

New Member
To answer Sceadwian's question. I'm just using the solenoid that the gun was manufactured with. I'm not a hundred percent sure that it is a 12v solenoid I just know that a 9v battery will not put out enough current to completely pull it in.
As far as weather or not the battery would be able to supply the current required to cycle the solenoid at the desired rate. I checked the voltage of the battery against the voltage drop when its being loaded. I dont have a fresh battery handy, the one I used has cycled the solenoid about 500 or 600 times (a three hour game of paintball) It measures 8.78v with no load on it what so ever. When cycling the solenoid full auto the voltage drops to about 8.18 right away then down to about 8.00v over the course of 5 seconds of firing.
I'm actually quite familiar with industrial solenoid valves that need air pilots. This is not actually a solenoid valve but just a solenoid that releases releases a latch to let the bolt cycle.
this is the way the firing system works;

when the gun is put into full auto mode and the trigger held

First: the solenoid fires once only for a few miliseconds to release the bolt.

Second: the bolt spring pushes the bolt forward

Third: the bolt strikes a the air/co2 valve to release a burst of co2 to fire the paintball.

Fourth: the blast of air/co2 pushes the bolt back into the locked position because the solenoid has already released and the bolt catch was able to lock the bolt again.

that is one cycle

then a timer times out and it starts all over again.


I just want to make it clear that in full auto mode the solenoid is NOT pulled in 100% of the time and the bolt allowed to just "runaway"

If I was asked to guess about the duty-cycle of the solenoid I would say it's about 5%

Thanks again guys for being so patient. I'm new to the site and my knowledge base is a bit more toward the electrical end at this point. I'm a first year electrical engineering student and at this point the only things I know about solid state are what I read about on my own time.
 

Sceadwian

Banned
Have you tested it under pressurized conditions to see if the 9V pulse works to trigger the spring, or are you just assuming it won't because it pulls back further with 12 volts?

I've seen a few electric paintball markers like this before and they all used 9 volt batteries. The more recent one's don't even use that much.
 

duffy

Well-Known Member
It's a little risky, but you can get a faster turn-off time out of a solenoid by introducing a resistance in the freewheeling diode circuit. The time constant is L/R, and for a low resistance coil like that, with no appreciable resistance from the diode, the "off" time can be quite large.

Downside is that the peak turn-off spike voltage goes higher. You can mitigate this somewhat by replacing the diode (probably a crappy 1N4004 or something) with a zero-recovery time schottky and choosing the resistor carefully by monitoring the PIV spikes on your driver. A 5Ω resistor will reduce the time constant to a tenth of what it is now, allowing faster solenoid drop-out and maybe faster cycle times.

Of course, if you are getting around that 18A peak current, you will see close to a 90V spike, and you have to consider the breakdown voltage of the driver. I suspect it won't be close to 90V, though. Got a scope?
 
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