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Anyone able to help me out for my invention?

bpall

Dec 21, 2014
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I apologize in advance if this sounds stupid. I'm new to this technology and haven't done experimented with it at all. I am huge into a sport called airsoft, which is similar to paintball but involves shooting plastic bbs instead of paint. One of the top brands is named Polarstar and offers top quality High Pressured Air airsoft markers. This is a newer type of power source and all Polarstars cost $600+. I have designed my own variant which uses the same power source but has cheaper homemade pieces. I built all of these pieces my self and found they work; but need help with the electronic phase. I have a solenoid valve which opens when connected to a battery, and stays open until the power source is disconnected. That is the problem; I need the valve to open and shut immediately even if the power is still connected. I need this to happen every time the battery is connected (this happens when the trigger is pulled). The point of this timing is so that a small burst of air can flow through and launch the round foward.

Again, sorry if this sounds bad I'm new to the electronics.
Please reply if you have any idea on how to help. You can also contact me through my instagram page @airsoftunloaded
Thanks!
 

davenn

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Sep 5, 2009
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Hi
welcome to EP :)

you need to tell us a lot more about the gear you have
what sort of solenoid?
what sort of battery, voltage and current ?
how about some sharp and well lit pic's so we can see what you are talking about :)

cheers
Dave
 

bpall

Dec 21, 2014
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Wow thanks for the fast response. I'm not able to take pics at the moment, but I can tell you that the solenoid is normally closed 12v DC :) I'm not sure which battery would work best yet so if you have an idea on that it might help! Thanks
 

Externet

Aug 24, 2009
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There is many unknown particularities to your solenoid, its characteristics, the supply you are using, the time you want it open in order to suggest a decisive way.

But try a 'large' capacitor in series with the solenoid. There will be a value that should work.
A resistor of an unknown value in parallel to the capacitor will discharge it between trigger pulls.

Bat(+)------------------triggerswitch----------------capacitor--------------solenoidvalve----------------(-)

The formula is time in seconds = capacitor charge in Coulombs divided by solenoid current in Amperes
And, as capacitor charge in Coulombs = capacitor voltage times its capacity in Farads; then,
Time = capacitor voltage times capacitor capacity divided by solenoid current.
 

Merlin3189

Aug 4, 2011
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Try a switch operated by the solenoid. When the solenoid has moved its armature to the full extent, it operates a switch to break the current.
That way you do not have to know the operating time of the solenoid (which I expect would vary with conditions such as temperature and pressure) - it is negative feedback.
If the gun trigger switch were to generate a brief gate pulse to an SCR and the solenoid operated switch simply interrupted the current through the SCR, then the solenoid would remain de-energised until the trigger generated a new pulse.
 

KrisBlueNZ

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But try a 'large' capacitor in series with the solenoid. There will be a value that should work.
You would need some way to discharge the capacitor quickly when the trigger is released. If the trigger is a SPDT type, that might be possible. Otherwise I don't think that's the right approach. Also, using a capacitor won't give cleanly defined ON and OFF transitions to the solenoid; I'm not sure whether that would be a problem or not.
Try a switch operated by the solenoid. When the solenoid has moved its armature to the full extent, it operates a switch to break the current.
That means the solenoid would deactivate as soon as it reaches its fully open limit. I think the solenoid needs to be opened for a settable length of time.

Here's my suggestion. I didn't really want to use a 555 but it's a pretty good fit for the application. I've included Digi-Key catalogue numbers for all the parts.

271916.001.GIF

When the trigger is pulled, CT is initially discharged, so pins 2 and 6 of the 555 are at 0V. Therefore the 555 drives its output (pin 3) high. This forward-biases power MOSFET Q1 and energises the solenoid.

CT then begins charging through RT. The 555's upper voltage threshold at 2/3 of the supply voltage will be reached after a time that can be calculated by multiplying the value of CT (in farads) by the value of RT (in ohms) then multiplying by 1.1, with the result in seconds. This formula is given on the schematic as well.

With the values given, the time period will be 0.11 seconds. Vary CT to vary this period by changing the value of CT, to suit the solenoid and the desired activation time. I've specified a capacitor with 5% tolerance. You can also vary RT to adjust the solenoid activation time, but keep the value between 68k and 150k.

When the rising threshold of the 555 is reached, its output goes low and the MOSFET turns OFF, deactivating the solenoid.

When the current flow into the solenoid is interrupted, the solenoid generates a voltage due to "inductive kickback", which brings Q1's drain positive. This voltage is clamped at about 48V by D1 and D2. Traditionally, a single diode (D2) is used to clamp inductive kickback, but that approach slows down the deactivation of the solenoid noticeably. Adding a zener (D3) speeds up the deactivation while still protecting Q1 against overvoltage.

When the trigger is released and power is removed from the circuit, R1 ensures that the circuit's power rail falls to zero quickly. This is needed to ensure that CT is rapidly discharged, through D1. Without these components, CT would remain partly charged between consecutive trigger activations and this would shorten the solenoid activation time.

C1 is required to decouple the 555's power supply and ensure reliable operation. C2 decouples the sense voltage of the 555 and is also recommended for reliable operation.

I've included Digi-Key part numbers for all components except the battery, the trigger button, and the solenoid. I've included three options for Q1. The FQU20L06LTU is a fairly cheap device with good performance in a small through-hole IPak package. The HUFA76429D3 has a lower ON-resistance (i.e. better performance) in an IPak package for about 50% higher price. The IRF510 has poorer performance and is in a larger TO-220 package, but it's more widely available.

The circuit can be constructed on a small piece of stripboad. The parts of the circuit shown with thick lines carry heavy currents and should be constructed with thick wire to minimise resistance and impedance.
 

chopnhack

Apr 28, 2014
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Nice solution Kris!
What is the purpose of the 47v zener diode?
I had assumed that one diode would be the snubber to take care of the negative draw when the magnetic field collapses in the coil, but with two anode to anode, there will be no flow of current through that at all, correct?
 

KrisBlueNZ

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No, zener diodes pass current in the reverse direction when operating in the zener region. They conduct in the forward direction just like a normal diode.

So when the drain of Q1 goes more positive, due to the inductive kickback, the normal diode is forward-biased and the zener is reverse-biased into its zener region. So the combination clamps the voltage to about 48V.

The reason for allowing 48V across the inductor, instead of the usual 0.7V, is that it allows the magnetic field to collapse much more quickly.

The formula for current change in an inductor is dI/dT = V/L. This is a direct analogy of the capacitor formula, dV/dT = I/C, with current and voltage exchanged. If you want a rapid change of current, you need to minimise L, and maximise V. With a standard diode inductive kickback suppression diode, V is only 0.7V and this really slows down the collapse of the magnetic field. For more information Google relay coil zener diode.

I just realised! The MOSFET sees the supply voltage (12V) as well as the clamping voltage. So the total peak voltage across the MOSFET is 60V which is its maximum rated voltage! I should have specified a 39V zener instead of a 47V zener. Here's a corrected schematic.

271916.002.GIF
 

(*steve*)

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The purpose of the 47V zener is essentially to convert the energy stored in the solenoid to heat as rapidly as possible consistent with maintaining the voltage across the mosfet below its maximum.

Normally you see a diode here. This essentially allows the current in the inductor to continue to flow until the energy is dissipated. The primary means of dissipating the current are the resistance in the inductor and the voltage drop across the diode. Increasing the voltage drop across the diode will dramatically reduce the current in the inductor by bleeding energy off faster.

The zener needs to be placed in series with a regular diode as the zener is forward biased in normal operation of the solenoid.
 

(*steve*)

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I just realised! The MOSFET sees the supply voltage (12V) as well as the clamping voltage. So the total peak voltage across the MOSFET is 60V which is its maximum rated voltage! I should have specified a 39V zener instead of a 47V zener. Here's a corrected schematic.

I have done some tests on a single mosfet (so this is anecdote, not evidence) to determine how much margin there was above the rated Vds(max), In my case the answer was almost zero.

Some mosfets can be safely operated in this region (they are "Avalanche rated"). In this case, the total energy needs to be managed. I guess this actually makes it possible to use the mosfet to switch an inductive load without protection in some cases, but I don't know whether I would design something that way...
 
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