Changing Capacitor Discharge Solenoid Parameters

Hi, RonV;

The measured inductance for the 4.6 ohm coil is 3.66mh.

Would you display the schematic you are using so I can try it in LTSpice? It was cut off on the bottom of your attachment.

In my system a boost converter charges the capacitor up to desired discharge voltage. This takes about 7 seconds. At discharge the primary energy source is the capacitor.

Please explain your sweet spot. I've laid out three scenarios that I would like to try. Each is based on a 2 millisecond discharge period. Scenario #1 is the actual layout I have built:

#1 is existing scenario with a 100 uf capacitor charged to 160 volts. The coil is 276 turns of #29 wire with a measured resistance of 4.6 ohms. Max current draw is about 12 amps. The discharge characteristics are in the graph in prior posts.

#2 is a new scenario with a 560 uf capacitor charged to 80 volts. The coil is 139 turns of #26 wire with a calculated resistance of 1.14 ohms. Max current draw is about 24 amps.

#3 is a new scenario with a 2200 uf capacitor charged to 40 volts. The coil is 69 turns of #23 wire with a calculated resistance of 0.21 ohms. Max current draw is about 48 amps.

Notice that I wind a new coil for each scenario. As stated earlier the duty cycle is less than 1% so heat buildup is not a limiting factor. I haven't calculate the inductance for coils in scenarios #2 and #3. Because my coils have a ferrous core the standard formula doesn't apply.

As I change the coil inductance and voltage at full charge, I would expect the voltage and current discharge curves to change. I would like to simulate those curves.

Thanks for the input.

Solenoid

That was an interesting picture - sorry.

Here is another one.

Here is what I was saying about the sweet spot.

If you want to get the same current thru the coil at a lower voltage you need to lower the inductance of the coil and increase the size of the capacitor. The capacitor is no problem, but to lower the inductance you need to remove turns from the coil. When you remove turns the amp turns goes down so you have less pull in power.
I'm not sure what function the capacitor serves if the micro turns it off after 2 ms. You might consider taking it out. The curve will look different but you may get the results you want.
Did you measure the inductance with the plunger in the coil? It may also make a difference (Oh good, another variable. )

bobledoux said:

MrAl: Where do I find the fusing current for wire as a function of gauge and pulse length? I took my values from a standard AWG wire table.

Click to expand...

Hi again,


You find this data in an Electrical Engineering handbook. I could probably post a graph or table if you want to take a look.

I agree that there is some resonance here. The new coil would be about 70uH and with 1600uf that would resonate at abut 475.6Hz, and the half cycle time for this is very close to 1ms which is half the total pulse time. However, this is the bare coil with nothing inside. Once the solenoid shaft is placed inside the coil the inductance will go up quite a bit, leading to a much longer half cycle period of resonance. I would not be surprised if the inductance went up by 10 times, which would increase the cycle time to over 3ms which is then beyond the pulse width time. And that's still not the last word. With the shaft outside the body or partly outside (before it is pulled in) the inductance will be lower, and as the shaft moves through the body of the coil the inductance will increase more and more until the shaft totally fills the coil. This means the construction will basically be a non linear inductance. So the cycle wave might look pretty strange, but one thing is for sure, the current will not rise as a linear function of the resistance of the coil. That doesnt mean the resistance should be ignored though, but i think you got that covered.

So the best approach here is probably the empirical one that you had been more or less using anyway. With the average coil inductance being down by 16 times and the capacitor bank being up by 16 times we maintain the same average resonant frequency.

The other consideration is the coil heating. Outside surface area is about 1.7 square inches, and power heating due to the resonant activity is probably going to be quite low, so it is doubtful that this new coil will overheat. However, the empirical approach is probably the best bet again due to the large number of variables, thus it would be very good to know how hot the original coil got under long term operating conditions (like several days). It would also be good to know if this setup is to be run 24 hours a day 7 days a week or just 8 hours a day or something like that.

Ronv:

In the three scenarios in my previous post the ampere turns of each coil is the same, about 3300. So I am expecting the same force from each coil.

The capacitor provides the power for the discharge. This is most visible in scenario #1 where a 12 volt battery charges the 100uf capacitor to 160 volts through a boost SMPS converter. The discharge of the capacitor provides 1.28 joules of energy. The battery cannot provide this amount of energy for a 2 millisecond period, so the capacitor stores the energy for rapid discharge.

I measured the inductance with the plunger in the coil, because that is the real application.

My expectation, and it may be false, is as the wire becomes bigger and the turns become fewer, the inductance becomes less so the reactance is less so there is less resistance to the capacitor discharge, thus keeping the discharge period about the same.

MrAI:

The coil fires no more than once a minute for a total of 80 discharges over a two hour period.

The discharge time is based on the time it takes the plunger to move. Once it has reached the end of its stroke there is no further need for energy discharge.

Okay, I think I fianlly get it.

If you don't have other restrictions I would just make the cap as big as reasonable (say 4500 Ufd. 3X 1500 Ufd.) and use your second coil at 40 volts.

Thanks, folks.

If there are no other comments, I'll order some parts and test it out.

My real measures will be the actual voltage and current curves over the discharge time.

Hi,


I had asked about the temperature rise of the original coil so we could compare that to the new ones projected temperature rise. It doesnt matter how low the duty cycle is, it depends on the average power. You can have a low duty cycle but still have significant average power.

On page 1 of this thread is a graph showing voltage and current over the 2 millisecond discharge period. The coil delivers one watt second of energy. It reaches a peak of about 1KW, but this lasts for about a millisecond.

In its application the coil is buried. But for a typical 80 discharges during a two hour period there is undetectable heating of the coil surround.