Changing Capacitor Discharge Solenoid Parameters

[bobledoux]

I have a capacitor discharge circuit. A 12 volt input boost converter charges a 100 uf capacitor to 160 volts. This capacitor discharges into a solenoid.

I want to reduce the discharge voltage so the FET that controls the discharge operates in a more efficient operating area. (Most FETS seem to reach current peak handling in the 40 to 60 volt range.) I want to reduce the operating voltage by increasing the coil turns and/or current to achieve the same impact.

I want to reduce the charge voltage to one quarter or, 40 volts. How should I change the coil and capacitor to provide the same solenoid force?

First thought: "Energy in joules equals capacitance in farads times voltage squared."

Am I trying to keep the energy constant? If so, this suggests I need sixteen times the capacitance at one quarter the voltage.

The present coil is 276 turns and has a 1 millisecond peak current of 11.9 amps for a rating of 3300 ampere turns.

How should I approach resolving these questions?

 

[ronsimpson]

You do know the solenoid will operate slower.
I think your solenoid current will be about 3 amps for 4 times longer.

 

[bobledoux]

The firing pulse is 2 milliseconds, about once a minute.

 

[ronsimpson]

I believe your new pulse will be about 8 milliseconds.

 

[bobledoux]

The discharge period is controlled by a microcontroller that shuts the pulse down after 2 milliseconds. By then the solenoid armature has completed its motion and any additional discharge is not needed.

I've attached the actual discharge curves for voltage and amperage in the present, high voltage configuration.

My intent is to generate the same solenoid "punch" with lower voltage.

 

[ronsimpson]

The discharge period is controlled by a microcontroller that shuts the pulse down after 2 milliseconds. By then the solenoid armature has completed its motion and any additional discharge is not needed.

I see the voltage is down to 10 volts in 2nS and the current is down to 50%. The capacitor shuts down the solenoid in 2 ms.
If you increase the capacitance and decrease the voltage the time will increase. It will take longer to get the same punch in the solenoid.

 

[bobledoux]

Does the following work?

I need to keep the short firing pulse period.

I calculated the power delivered by multiplying the voltage by the current. Taking the product of the area under the curves I get 1100 watt milliseconds or about 1.1 watt seconds of energy.

In order to deliver this amount of energy with a quarter of the voltage, the current must be multiplied by 4. Keeping the ampere turns constant means the number of coil turns can be reduced by a factor of 4.

The coil is currently wound using gauge 29 AWG wire. Subtracting a gauge by 3 doubles the current carrying capacity. So 29-6 gives me 23 gauge for the new coil with four times the current capacity. I will increase the winding volume (the weight of the wire) slightly because larger wire has more open space between the turns.

The old coil coil contained 276 turns of number 29 wire, passing 12 amps for 2 milliseconds. While the fusing capacity of this wire is rated at 12 to 14 amps, the voltage rise in my application was very small. But according to a wire table, as the wire size increases the fusing amperage increases more slowly. A size 23 wire is suppose to be only good for 32 amps, and I need 48. But that standard doesn't appear to consider the pulse length.

The old coil had 276 turns; the new coil has 69.
The old coil had 4.6 ohms resistance; the new coil has 0.29.
The old coil had 55 feet in 5 layers; the new coil has 14 feet in three layers.
The mass of copper in each coil is about 9.5 grams.

 

[ronsimpson]

I did not see you can make your own Solenoid coils.

V=1/4 C=1/16

If you want to keep the resonant frequency the same (time) and the cap is 16x then the L needs to be x/16. (1/16) To get L=1/16 then the turns =1/4.

I think we think the turns =1/4. You did it from amp*turns and I did it with time. C*L=C*L

 

[MrAl]

Hi,


You have 16 times the capacitance, so you need 1/16 times the resistance.
You've quartered the length of the wire, so that quarters the resistance, and you've doubled the wires working area so that quarters the resistance again which means the resistance is then 1/16 of the original.

The 2ms fusing current of #29 copper wire is over 400 amps. The 2ms fusing current of #23 copper wire is over 1600 amps. The thing is, if the copper got up to 1082 degrees C and didnt melt yet, the insulation would be in trouble. So i dont think the fusing current is going to matter that much, but could help comparatively. Since we do see about 4 times the fusing current for a 2ms pulse, it should be ok.

Then we have the heat buildup to consider. The fusing current (and the wire insulation degradation) is based on an ambient temperature of 30 degrees C, but if the coil heats up during operation then that looks like an ambient temperature rise which brings the fusing current down. It would help then to estimate the coils temperature rise.

If the max current is 48 amps and the max voltage is 40 volts that's a peak power of nearly 2000 watts, but it's at a duty cycle of only one third of one percent (0.0033 percent) which brings us down to about 6 watts average power dissipation, at worst case. The full power wont be present for the full 2ms so we'll estimate at one half of the peak, which brings us down to about 3 watts average power dissipation (we could double check this).

The remaining question then is what size is the coil. We need to know the diameter and the total length of the coil turns end to end. We can then figure out if the coil will get too hot or not.
The coil sounds like it is about 0.8 inches diameter by about 0.6 inches long but that's just a guess.

So then, what are the dimensions of the coil?
Also, what was the temperature rise of the old coil under the said operating conditions?

 

[ronsimpson]

MrAl,

There is more that voltage and resistance happening here.
I see a LC resonance happening here AND the VR. In the old coil the current takes 1mS to reach peak. We need to keep that time constant. I believe the current is not set (much) by the resistance.

 

[bobledoux]

Interesting discussion.

The coil winding area has a width of 0.675 inches with 27 turns per layer. The coil is 3 layers deep with the outer layer of 17 turns. The outside diameter is 0.790 inches. The inside diameter is 0.742 inches. I could complete the outside layer, but I reduced the turns count to keep the coil mass at 9.5 grams.

 

[ronsimpson]

What is the resistance of the coil?

 

[bobledoux]

Resistance is calculated as 0.29 ohms for a wire length of 14 feet.

 

[bobledoux]

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.

 

[ronsimpson]

Resistance is calculated as 0.29 ohms for a wire length of 14 feet.

If the wire resistance is 0.29 ohms then the resistance has almost nothing to do with the current flow.
160V/0.29 ohms = 551 amps
It is simply energy in a capacitor moving to an inductor in a time that is related to pi sqrt(LC).

This is related to why a 110VAC relay will work at 12VDC.

 

[bobledoux]

I agree reactance is involved. The 0.29 ohms is the DC resistance. Looking at the graph above, the 2 millisecond pulse corresponds to a quarter of a sine wave at 125 HZ. So I think we need to consider the discharge curve as AC, with resistance and reactance in the calculation. This means current flow should be much less than the DC calculation of 551 amps.

For example: In the graph above we have a current flow that peaks at about 12 amps. Initial voltage is 150 volts and the coil DC resistance is 4.6 ohms. Because of the inductance, the current doesn't peak until halfway through the discharge pulse, when the voltage is fallen by 50%. Even then the real current doesn't approach the calculated current of 80 volts/4.6 ohms for 17 amps.

I think I need to calculate the inductance of a multilayer coil; build one and test for actual value.

You raise the specter, that, because of decreased inductance I have to consider the possibility that peak current may be greater than four times the current 12 amps when I reduce voltage by a factor of four--keeping power constant.

While the area under the curve of current across time may not exceed four times the present level, the curve may have a different shape, perhaps with a sharper peak so a more robust switching FET may be required.

Thanks for the feedback.

 

[ronsimpson]

Bobledoux,

I think we agree no how this works.

Here is an example of how to think about the coil.

Pretend the old coil has 4 layers of wire. You want 1/4 the turns and 1/16 the inductance.
Take the 4 layers and stop using them in series but use them in parallel. This gives you the 1/4 turns.
The current is 4x and the wire can handle 4x more current.

Same problem again: Use a wire that is 2x larger. In two layers you will fill the winding area. The turns will be 1/4 the old coil and the current ability will be 4x.

In both examples you have the same pounds (Kg) of wire, the same turns, same inductance, same resistance. (close)

I design many transformers each year. If you fill the winding area with small or large wire you have the same power caring capability.

 

[bobledoux]

It took me quite a while to realize the facts you present. They came clear when I read this:

https://www.qsl.net/ki7cx/Coilrewind.htm

 

[bobledoux]

Summary of Discharge Characteristics

When I started with this question of changing capacitor discharge parameters I tended to believe that simple calculations would suffice.

I discovered I was wrong. Actual measurement demonstrated that the dynamics were far more complex. In simple terms I could not treat a capacitor discharging into a coil to be a DC event, even if it was a single pulse event.

The attached graphs make this clear. In this graph I have two events. For both events I have charged a 100 microfarad capacitor up to 148 volts. For both events I am displaying the first two milliseconds of capacitor discharge.

In one event I have the classic discharge curve as this capacitor discharges into a 4.6 ohm resistor.

In the second event I have measured values as this capacitor discharges into a specific coil with a DC resistance of 4.6 ohms.

The black solid line is the classic capacitor discharge voltage curve. The black dotted line is the current curve. Note the line starts off the top of the chart at 32 amps.

The red solid line is the measured discharge line from my coil. The red dotted line is the measured current line from my coil.

By comparing the two voltage and two current lines we can see the dramatic, and classical impact the coil pays in diminishing the peak current flow.

The current over time graph is especially important in selecting a FET to perform the discharge switching. The length of time the FET has to pass significant amounts of current makes FET selection important.

In the course of this discussion I have intended to reduce the capacitor charged voltage down to 40 volts and increase the current. I don't know how the shape of the current discharge curve will change. If, for example, the quality factor of the coil should improve I might have a higher current peak for a shorter period of time.

 

[ronv]

Pulsed solenoid

I don't think you can get there from here.
When I simulate your coil I can pretty well match your curve with an inductance of 6 Mh & 4.6 ohms.
The problem being that to get the inductance down you need to reduce the number of turns and to keep the same power you need to have more and more current. Once you go past the sweet spot your dead. If it was DC and you didn't care about the pulse timing it would be easy but in this case the design probably got to 160 volts to get the time period needed.