How to use capacitors in a Solenoid circuit

In summary, the small coil-gun designed by the student has a capacitor that is needed to discharge the bolt within a certain time limit. However, the capacitor loses voltage as it charges, and a resistor is not needed in the circuit.
  • #1
lekh2003
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TL;DR Summary
I need some help with an LRC circuit I have, and charging/discharging capacitors to function as needed.
I've been working on designing an experiment over the past few weeks as part of a school project, under the supervision of a teacher.

I have designed a small low-power coil-gun. I have a coil of roughly 60m 24 AWG copper wire wrapped around a length of 2.5cm of clear PVC pipe. I tested the resistance to be 5.4 ohms. The projectiles are small 5cm iron bolts. The idea is to have a current flowing through the coil for a short period of time to "shoot" the bolt. The speeds of the bolt are under 3 m/s and inside of a long PVC tube, keeping me well within safety. The idea isn't to create an optimal gun, but rather to test out some concepts in physics.

By using a DC power supply at 20V, I did some trial and error (with video analysis) that the optimal time for a current to be flowing through the coil is between 20-30ms. Obviously, I can't use a DC power supply since the trial and error method is too inaccurate. I would have to use an arduino to control the circuit to release current for a short period of time, but I don't have the access or the capability to use an arduino.

Hence, the solution is to use a capacitor. I calculated that a capacitance of 500##\mu##F would be optimal in the circuit to discharge within 20-30ms. My variable for the experiment is to check the impact of adding a resistor to the LRC circuit (which would increase the discharge time and decrease the current).

My issue is the capacitors themselves (electrolytic ones). I initially purchased a couple of 16 V 100##\mu##F capacitors, but I found that the voltage was too low to power the coil. Hence, I bought a couple of 50 V 100##\mu##F capacitors. So far I've only been working with one of them (for testing purposes), but hopefully I can eventually connect them in parallel to achieve 500##\mu##F as required.

My process has been the following:
1. Set up a circuit with the DC power supply at 20 V, connected with a 3k##\Omega## resistor (I read this was needed to keep the current low so as not to damage the capacitor).
2. Charge up the capacitor using this circuit.
3. Connect the capacitor to the coil.
4. Watch the results.

The problem is in the charging up of the capacitor. Firstly, I can charge it up to roughly 19 V. However, as I am looking at the voltage with my multimeter connected to the capacitor it loses voltage in real time. After a minute or so, it seems like its almost fully discharged. Not to mention, when plugged into my coil, it does absolutely nothing, ever.

Am I doing something wrong with my capacitors? Like some sort of special important rule that I'm missing? Are my capacitors just crap? I know they are the problem since everything else works fine, the coil and everything "fires" fine when connected to the power supply.

I can provide any images if necessary, I would greatly appreciate the help.
 
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  • #2
lekh2003 said:
Am I doing something wrong with my capacitors? Like some sort of special important rule that I'm missing? Are my capacitors just crap? I know they are the problem since everything else works fine, the coil and everything "fires" fine when connected to the power supply.
You may have too little capacitance. We can work out how much you need.

The energy stored in the capacitor; E = ½·C·V², will, as the current rises, be transferred into the magnetic field of the coil; E = ½·L·I², and then some energy will return to the capacitor as the current falls back through zero.

You should not have a resistor in that circuit.

When you connect the coil to the capacitor, a current with the wave form of a sinewave, will increase to a maximum and then fall back to zero. I expect the time that takes will need to be about 25 millisec. The capacitance and inductance decide that time. You may need to adjust the capacitance to produce the optimum pulse.

Have you calculated or measured the inductance of the coil, or can you let us know the number of turns and the dimensions of the solenoid so we can work out the approximate inductance.
 
  • #3
Try letting the capacitors sit while charged with the meter disconnected. Then after the minute or so that they would normally discharge, connect the meter and see if they still hold a charge.

This is to find out if the meter is discharging them.

Another thing that occurs is that when electrolytics have been unused for months or years, they get electrically leaky. This is because the dielectric (insulator) in an electrolytic cap is a thin chemical film on the plates. The film is an electro-chemical reaction between the plates and the electrolyte. It is created when voltage is applied and will, over time, slowly dissolve back into the electrolyte. The cure is to keep voltage applied to the caps for a few days so the film can regenerate.

Very old or unused electrolytics can also have the electrolyte dry out. This of course doesn't allow them to maintain the dielectric film so they will have a lower than expected capacitance.

And of course another possibility is, as you suggested, that they really are crap! If there are impurities in either the electrolyte or in the metal foil that make up the plates, they just don't work as expected.

Hope this helps and gives you a few ideas to help track things down.

Fun project. Please keep us updated on your results.

Cheers,
Tom

p.s. "...the coil and everything "fires" fine when connected to the power supply."
That's likely because you are using the filter capacitors in the power supply. If it is a high current supply they may be in the 1,000's of uF. Ask your instructor if a service manual, or schematic, is available. If not, perhaps it can be opened to find out what the capacitor values are.
 
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  • #4
Baluncore said:
You may have too little capacitance. We can work out how much you need.

The energy stored in the capacitor; E = ½·C·V², will, as the current rises, be transferred into the magnetic field of the coil; E = ½·L·I², and then some energy will return to the capacitor as the current falls back through zero.

You should not have a resistor in that circuit.
Alright, so as part of my project, I need to have some variable that I am changing, to see it's impact on the circuit. My idea was to find the capacitance which creates an optimal shot when the circuit has a resistor in it. Currently, the solenoid has a resistance of 5.4 ##\Omega##. With a 4 ##\Omega## resistor, that makes 9.4 ##\Omega##. My idea was to find the optimal capacitance for this artificially higher resistance, and then reduce the resistance by increments of 1. This would decrease the discharge time, but increase the current in the circuit. My aim is to find out which of these effects is more significant to the design. I hope this makes sense.

Hence, I did some calculations, assuming that the capacitor discharges fully at 5RC, at a capacitance of 0.0006 F and a resistance of 9.4 ##\Omega##, discharge will be 28.2 ms. I calculated that as I reduce resistance down to 5.4 ##\Ohm##, the discharge will reduce to 16.2 ms (however at a larger current). I will investigate the changes as I reduce the resistance.

Tom.G said:
p.s. "...the coil and everything "fires" fine when connected to the power supply."
That's likely because you are using the filter capacitors in the power supply. If it is a high current supply they may be in the 1,000's of uF. Ask your instructor if a service manual, or schematic, is available. If not, perhaps it can be opened to find out what the capacitor values are.

One important note I would like to make, referring to the quote above. My power supply was a DC power supply with consistent current. When plugged into my circuit, it was at about 20 V with 3 A. I actually took a slow-motion video of me repeatedly closing the circuit by simply grazing the contacts, then analysed the video footage to find the exact times which cause the projectile to fire. It was a painful experience (metaphorically).

Baluncore said:
Have you calculated or measured the inductance of the coil, or can you let us know the number of turns and the dimensions of the solenoid so we can work out the approximate inductance.

The length of the wire is between 50-60m (I measured by considering the resistivity of the wire and total resistance of the coil). It is wrapped around a tube of diameter 2.5cm. Here's a schematic. I changed out the nail at instruction of my teacher :P
1606313243572.png


Anyways. I'm going to take Tom.J's advice about checking whether my capacitors are discharging on their own or whether it's due to the multimeter. Just figure out if they're working as intended or not. I'm pretty sure I blew out two of them earlier today by running too much current through them. I'll check the rest.

On another note, what's a good way to conect several capacitors in parallel? I can't have several wires connecting 6 capacitors in parallel. Should I use some bare wire and tape them all into parallel? I was considering a breaboard but saw that they can only withstand up to 1 amp. However, would 3-5 amps through them be fine if it were for under 30 ms?

Once again, I strongly appreciate all the help, I'll keep you guys updated. The support from here is encouraging.
 
  • #5
lekh2003 said:
I will investigate the changes as I reduce the resistance.
When you connect the capacitor to the coil you will have a resonant circuit.
The resistance will cause the oscillation to be damped. It will not change the period of the first magnetic pulse. That is determined by L and C.
You must measure the inductance of the coil to understand the response.

If you wound the coil as say 4 sections, you could connect all 4 in parallel, two in series in parallel, or all 4 in series. That will give you three different resistances, numbers of turns and currents. The different combinations of winding inductance and amp*turns gives a wider and more interesting range of tests than simple resistance variation.
 
  • #6
Baluncore said:
When you connect the capacitor to the coil you will have a resonant circuit.
The resistance will cause the oscillation to be damped. It will not change the period of the first magnetic pulse. That is determined by L and C.
Wouldn't resistance change the period anyways? The capacitor in a circuit discharges in 5RC, which is proportional to resistance, right?
 
  • #7
lekh2003 said:
Wouldn't resistance change the period anyways?
An RC circuit will discharge towards, but will never reach zero.
An LC circuit will ring like a bell. Resistance in the LC circuit, RLC will not change the period of oscillation, but it will change the rate of decay of the oscillation.

You want to propel the bolt in the first half cycle of the oscillation, over approx 25 msec.
LC oscillation frequency in Hz; f = 1 / ( 2·π √(L·C) ); ∴ period; Tsec = 2·π √(L·C)
You want about; 25 msec = π √(L·C) ; which is not a function of R.

Measure the inductance of your coil, or wind it with 4 parallel windings so it can be changed.
 
  • #8
As @Baluncore noted above (while I was typing this response), the coil inductance is important. If you do not have the equipment to measure the inductance it can be approximated by calculation; read on.

The image you supplied above in post#4 seems to be missing some information. If the given dimensions are correct and it was carefully and neatly wound, the coil has about 600 turns in 14 to 15 layers. The coil inductance is affected by its physical dimensions and the number of turns.

If you can supply the outer dimension of the winding and the size and material of the projectile, we can show you how to approximate the coil inductance. Since the inductance, resistance, and capacitance all affect the peak current and the time to discharge, it will be easier to understand the whole problem.

Cheers,
Tom
 
  • #9
π * Sqrt( 1200 uF * 50 mH ) = 24.335 ms
To simulate a capacitor, charged to 20 volts, then connected to the coil, the capacitor is driven with a 20 volt step. Notice that, as predicted with these values, the first half cycle of current (red) lasts for about 25 msec. Notice also the decaying sine and cosine waves of current (greenred) and voltage (redgreen) in the circuit.
LCR.png


In the second circuit I have added a diode that blocks the second half cycle, so it produces a single magnetic pulse peaking at 2.25 amp, lasting 25 msec. ( It returns the unused energy to the capacitor).
WithDiode.png
 
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  • #10
Thanks for the help so far.

Using this new information, I will probably have to readjust some of my methodology. This happens to be a practical which contributes largely to my final grade in high school. Everything needs to be documented and changes to my methodology involve large changes to the write up.

I will attempt to measure and capture the inductance. However, the technicalities and complexity of the experiment at this stage makes it disadvantageous towards getting a higher grade (i know, it annoys me as well).

I will probably attempt to get the capacitors working. Or, I will hook up an arduino to a relay module and run the coil-gun like that. Then, I can attempt to adjust some other factor in my experiment, possibly the length of the projectile.

As a result of COVID, I need to ensure that my experiment is quick and simple, so I can actually get it done in the very limited lab time available. I apologise, but that really limits me.

Baluncore said:
To simulate a capacitor
Do you mind sharing what software you used for the simulation? I have been unable to run any theoretical simulations, I'm quite inexperienced with them.
 
  • #11

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  • #12
Baluncore said:
Thanks for this resource.

In the time since this post, I attempted to use a relay module to control the circuit rather than any capacitors at all. However, I think I am realising that the relay module can't actually stay on for 20ms, by design. It can only turn on and off within a second.

I will have to use a capacitor, but this is now slowly getting ridiculously difficult.
 
  • #13
lekh2003 said:
I will have to use a capacitor, but this is now slowly getting ridiculously difficult.
Am I correct in thinking you want to press a button, that dumps the charge in a capacitor through the coil for 20 ms only. Or do you have some other trigger ?
 
  • #14
lekh2003 said:
I have designed a small low-power coil-gun. I have a coil of roughly 60m 24 AWG copper wire wrapped around a length of 2.5cm of clear PVC pipe. I tested the resistance to be 5.4 ohms. The projectiles are small 5cm iron bolts. The idea is to have a current flowing through the coil for a short period of time to "shoot" the bolt. The speeds of the bolt are under 3 m/s and inside of a long PVC tube, keeping me well within safety. The idea isn't to create an optimal gun, but rather to test out some concepts in physics.

hi, I'm trying to grasp your aim behind the experiment you are designing. so far, I'm understanding that you want to be able to demonstrate how changes in one variable affect changes in another?

basically, i'd like to understand which "concepts in physics" you are trying to test with this railgun (edit: my mind wanders to railguns with materials like these) coilgun setup, and see if there's any way we can reduce the complexity to something that would be doable in your time frame.

it would be helpful if you could distill it to the smallest possible variables, or simplest possible variables (e.g. projectile range vs current for a constant circuit ON-time, projectile range vs current for a variable circuit ON-time determined by constant bolt position (let's say current switches off when bolt's centre of gravity reaches coil center)) so we can find out which complex stuff we could bypass.

cheers!
 
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  • #15
Update time! I have a functioning coil gun, that is the good news. But it was a bit different from what I initially wished to do. I used an arduino and a relay module combination to turn on a circuit for however long I wished. I have it attached to a DC power supply, outputs about 19V when connected, and the optimal time so far is roughly 34ms.

I'm not sure I want to go down the capacitor path much further. I did see some examples of a coil gun which had a combination of a relay module and a capacitor rather than a DC power supply, but I don't really see the point of doing this. If I had an oscilloscope, maybe I could try and analyse the different types of current flow and their impact on the strength of the coil gun, but I don't.

Luckily, I have all the equipment set up in my garage now, as opposed to at school. I can test as much as I like and work on this. I'm going to try and find the simplest possible variables, as maxwells_demon suggested. I have a couple ideas to try out, so I'll get to it!
 
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  • #16
I have encountered a somewhat minute concern with my setup.

Here's the current circuit for anyone who's curious:
circuit.png

Here's the code I'm running on the arduino:

C++:
int relay2 = 8;
int relay1 = 7;

// Timer Variables
int trigger = 1;
int ontime = 38;

void setup() {
  // Pin for relay module set as output
  Serial.begin(9600);
  pinMode(relay1, OUTPUT);
  pinMode(relay2, OUTPUT);
  digitalWrite(relay1, HIGH);
  digitalWrite(relay2, HIGH);
}

void loop() {
  if (trigger == 1) {
    digitalWrite(relay2, LOW);
    delay(ontime);
    Serial.println("now");
    digitalWrite(relay1, LOW);
    trigger = 0;
  }
}

I had to do something like this because the relay modules had too much of a delay in switching.

Anyways, this has been working well, and I'm using a DC power supply. However, my power supply maxes out at 19V, which is problematic for actually testing much. I would like to increase the voltage to 30-40V. I tried adding two 9V batteries to the circuit alongside the power supply only to see it absolutely fail.

I started checking my batteries, and overall, they seem to be pretty pathetic. Together they have about 15V of power. I tried to replace the DC power supply in the above circuit with the batteries, and even that didn't work (I thought that the two types of power sources were "interfering"). What's happening? Am I missing something about how batteries work in contrast to a power supply?
 
  • #17
Do you have flyback diodes across the relay coils ?

A higher voltage will draw a higher current from the 19 V supply and batteries.
Measure your battery voltages.
Unfortunately we need to know the inductance of the coil to model the situation.
 
  • #18
The coil's inner diameter is 2.5cm and the outer diameter is 4.5cm. It is 60m of 24AWG copper wire. The length of the coil is 6cm. Is this enough to estimate inductance?

I also do not have flyback diodes. The circuit just disconnects after the discharge time.

Edit: I tried calculating the inductance myself. I assumed the coil radius was the average of the outer and inner radii. Based on that, I assumed the turns were roughly 550. I used the length of 6cm. I found the inductance to be 0.0061. Does that seem correct?
 
  • #19
lekh2003 said:
I also do not have flyback diodes. The circuit just disconnects after the discharge time.
I am not talking about the gun coil. You really need a power diode across the solenoid coil. When the output D7 or D8 turns off the current to the relay coil, there will be a voltage spike produced by the relay coil that will damage the D7 or D8 output circuit.
https://en.wikipedia.org/wiki/Flyback_diode
 
  • #20
Baluncore said:
I am not talking about the gun coil. You really need a power diode across the solenoid coil. When the output D7 or D8 turns off the current to the relay coil, there will be a voltage spike produced by the relay coil that will damage the D7 or D8 output circuit.
https://en.wikipedia.org/wiki/Flyback_diode
I haven't considered that. More importantly, I don't think I have the time to consider that either. I've seen plenty of similar projects which do not use a flyback diode on the relay. I'm more interested in understanding why the battery doesn't work correctly.

I have one other observation, for some reason the gun coil suddenly loses efficiency for voltages under 15V, it's quite bizarre. Obviously there's a decrease in distance that the gun shoots the projectile, but it's not just a normal decrease, it's just absolutely nothing. I calculated the energy efficiency, and it's roughly 30% transfer from energy in the coil to the object (based on the distance traveled). But for under 15V it drops to 20% efficiency. Any possible reason? I'll do some more trials to make sure.
 
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  • #21
lekh2003 said:
The coil's inner diameter is 2.5cm and the outer diameter is 4.5cm. It is 60m of 24AWG copper wire. The length of the coil is 6cm. Is this enough to estimate inductance?
...
Edit: I tried calculating the inductance myself. I assumed the coil radius was the average of the outer and inner radii. Based on that, I assumed the turns were roughly 550. I used the length of 6cm. I found the inductance to be 0.0061. Does that seem correct?
Maybe.
Start by getting some estimates of coil parameters.
24 AWG = 0.54 mm diam with insulation.
(OD-ID)/2 = thickness = ( 45 - 25 ) / 2 = 10 mm thick => 10 / 0.54 = 18.5 layers.
60 mm long => 60/0.54 = 111. turns per layer.
18.5 * 111 = 2053.5 turns

Then go to this calculator; http://www.pronine.ca/multind.htm
Search for the inductance by matching OD and turns.
Inductance; L = 65 mH.
Coil ID = 25 mm.
Coil length = 60 mm.
Wire gauge = 24 AWG.
Number turns = 2051.
Turns per layer = 110.9
Number layers = 18.49
Coil OD = 45.56 mm.
Wire diameter = 0.54 mm.
Wire length = 222.07 metre.
DC resistance = 18.7 ohms.

But:
The number of turns is not 550.
The inductance is not 6.1 mH.
The wire length is not 60 metre.
The coil resistance is not 5.7 ohms.

What gives ?
 
  • #22
Your code contains...
Code:
digitalWrite(relay2, LOW);
delay(ontime);
Serial.println("now");
digitalWrite(relay1, LOW)
How long do you expect the Serial.println("now") will delay digitalWrite(relay1. LOW) ?
Should that Serial statement not be outside the timing loop ?
 
  • #23
lekh2003 said:
I'm more interested in understanding why the battery doesn't work correctly.

For added batteries, a 6 Volt Lantern battery would be a better choice, it can supply much higher current. If the batteries you tried are the small rectangular ones with snap terminals used in transistor radios and smoke alarms, they are rated for up to 0.2amps. They also have a relatively high internal resistance (around 2 Ohms) which limits their peak current into a short circuit to around 4 Amps; with of course an extremely short life.

The current rating of the power supply you are using is not clear. According to the link you supplied above, the headline says 3A but the text says 1A. Your circuit can draw about 2.5A steady state, somewhat less with short pulses. So the power supply will be the limiting factor for the current regardless of the voltage you add with batteries.What do the Ammeter and Voltmeter on the supply read with the output terminals shorted and the voltage and current turned all the way up?

What do the Ammeter and Voltmeter read when powering the circuit (both relays set to power the solenoid)? Make this reading quickly, the coil may rapidly get too hot to survive.

Cheers,
Tom

p.s. There seems to be conflicting information about the coil. The early posts show its length as 2.5cm but your post #18 states 6cm.

p.p.s. What insulation is used on the wire in the coil; varnish, cotton, plastic, other?
Also, is it wound neatly with the turns next to each other on each layer, or is it "scramble wound" with the turns crossing over each other and no real separate layers?
 
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  • #24
Tom.G said:
For added batteries, a 6 Volt Lantern battery would be a better choice, it can supply much higher current. If the batteries you tried are the small rectangular ones with snap terminals used in transistor radios and smoke alarms, they are rated for up to 0.2amps. They also have a relatively high internal resistance (around 2 Ohms) which limits their peak current into a short circuit to around 4 Amps; with of course an extremely short life.
Ah, thanks, I didn't know that.
Tom.G said:
The current rating of the power supply you are using is not clear. According to the link you supplied above, the headline says 3A but the text says 1A. Your circuit can draw about 2.5A steady state, somewhat less with short pulses. So the power supply will be the limiting factor for the current regardless of the voltage you add with batteries.
I've measured that it draws around 2.49 amps when the circuit is connected at maximum voltage.

Voltage MaxCurrent Max
192.49
192.46
182.33
182.34
172.19
162.07
151.94

These are the currents I measured through the circuit for different voltages adjusted on the power supply. I did it quickly as well, within a few seconds, kept that coil safe.

Tom.G said:
p.s. There seems to be conflicting information about the coil. The early posts show its length as 2.5cm but your post #18 states 6cm.
Yes, I had to change the coil length because there was no way to pack together the coil more tightly.
Tom.G said:
p.p.s. What insulation is used on the wire in the coil; varnish, cotton, plastic, other?
Also, is it wound neatly with the turns next to each other on each layer, or is it "scramble wound" with the turns crossing over each other and no real separate layers?
It's an enamel coating. It is a scramble wound, doing it tightly was much too difficult and I didn't have very good stoppers. I'm not averse to making another one if needed, but when I tried it was practically impossible, the entire process was hellish.
 
  • #25
Baluncore said:
Your code contains...
Code:
digitalWrite(relay2, LOW);
delay(ontime);
Serial.println("now");
digitalWrite(relay1, LOW)
How long do you expect the Serial.println("now") will delay digitalWrite(relay1. LOW) ?
Should that Serial statement not be outside the timing loop ?
That's a very good point, thank you, I will adjust that and check.
Baluncore said:
Maybe.
Start by getting some estimates of coil parameters.
24 AWG = 0.54 mm diam with insulation.
(OD-ID)/2 = thickness = ( 45 - 25 ) / 2 = 10 mm thick => 10 / 0.54 = 18.5 layers.
60 mm long => 60/0.54 = 111. turns per layer.
18.5 * 111 = 2053.5 turns

Then go to this calculator; http://www.pronine.ca/multind.htm
Search for the inductance by matching OD and turns.
Inductance; L = 65 mH.
Coil ID = 25 mm.
Coil length = 60 mm.
Wire gauge = 24 AWG.
Number turns = 2051.
Turns per layer = 110.9
Number layers = 18.49
Coil OD = 45.56 mm.
Wire diameter = 0.54 mm.
Wire length = 222.07 metre.
DC resistance = 18.7 ohms.

But:
The number of turns is not 550.
The inductance is not 6.1 mH.
The wire length is not 60 metre.
The coil resistance is not 5.7 ohms.

What gives ?
I reckon it is that way because the coil isn't tightly wound and layered?
 
  • #26
Although I am not sure if its actually wise to provide you with something approaching a coilgun schematic...

Here it is anyway:

[Schematic deleted by the Mentors]

This circuit provides a 20+A burst in a very short amount of time in a 5V environment.

This makes for a good starting point in understanding or changing the circuit behaviour. Also read up on some core component datasheets to understand what they are all about. You need some power-resistors, one to limit the powersupply in operation when feeding the SuperCap and the other to limit and tune the current going through the coil. You can swap out the relay with logic type components which can be driven directly from the modulated voltage, although they need to be able to take the current and possible voltage spikes. The Zener diode is there to prevent overvoltage on the SuperCap foremost. The smaller capacitor takes in the voltagespike at the start of recharging the circuit. The regular diode is there to limit the direction of flow.

There is no need for two relays and you need the switch to be on the low side of the coil.

Too make it even easier, here is the link, notice and match the configuration for each component: [link deleted by the Mentors]

What kind of teacher do you have anyway, did he suggest the coilgun? Cant you start designing an electric pottery wheel or something? There is a market for the latter. And you know what Winston Churchill said: the whole of society floats on the cork of manufacturing (or something like that anyway).

I don't accept any responsibility for the application of this circuit by you or anyone else.

They actually seem to limit the sale of large SuperCaps to prevent these types of projects from going too far.

Possibly the admin just needs to take down this topic. Honestly.

[Mentor Note -- While the circuit posted was not inherently dangerous, it contained some errors that may confuse folks viewing this thread in the future. Hence, the circuit information has been deleted.]
 
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  • #27
Sender123 said:
What kind of teacher do you have anyway, did he suggest the coilgun?
You can do more damage with a rubber band than with a single stage coil gun.
Making a simple coil gun teaches more electrical power engineering principles.
Try to avoid powder operated nail guns and chainsaws, they are far more dangerous.
 
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  • #28
Sender123 said:
Too make it even easier, here is the link, notice and match the configuration for each component: [link deleted by the Mentors]
Honestly, this is beyond me. I don't think I can attempt to replicate this and doubt I have the resources either. I'm not trying to make an actual working coilgun. I'm testing some concepts, that's all.

Additionally, I initiated the experiment, and designed everything myself. I was only under supervision, but got no advice or directions.

Furthermore, I've tested the maximum capabilities of this coil gun, it can shoot at maximum a projectile at a speed of 3m/s, which gets stopped almost immediately by friction in the "barrell"I'm using. Theres no dangers, I've made sure of that. A large component of my grade in the project is ensuring that I've been safe as well.
 
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  • #29
Sorry for taking so much of everyone's time, but I think I have a final concern/question.

For my solenoid, it turns on because I close the circuit using a relay. But then, I open the circuit to turn it off. Does this still mean I have pullback? I'm aware that a solenoid in a capacitor circuit will have a current flowing in the opposite direction to oppose the magnetic flux when it turns off. But is this still possible in a circuit that's just open?

My circuit diagram is still above. Maybe I'm missing some technicality. But surely the solenoid isn't perfect right? What are some energy losses or inefficiencies? Maybe it is the pullback?
 
  • #30
lekh2003 said:
I'm aware that a solenoid in a capacitor circuit will have a current flowing in the opposite direction to oppose the magnetic flux when it turns off. But is this still possible in a circuit that's just open?
The current in an inductor continues to flow in the same direction when you disconnect it. It is the voltage that reverses, causing the “flyback” voltage spike that can damage semiconductors.
With metal contacts in a relay, a high reverse voltage arc will form as the contacts separate. That negative voltage makes di/dt = V / L very high so the current flow rapidly reduces.
The energy that was stored in the inductor's magnetic field appears as heat in the arc. E = ½·L·i²
 
  • #31
Baluncore said:
The current in an inductor continues to flow in the same direction when you disconnect it. It is the voltage that reverses, causing the “flyback” voltage spike that can damage semiconductors.
With metal contacts in a relay, a high reverse voltage arc will form as the contacts separate. That negative voltage makes di/dt = V / L very high so the current flow rapidly reduces.
The energy that was stored in the inductor's magnetic field appears as heat in the arc. E = ½·L·i²
Ah, so if I'm understanding correctly, the solenoid still has a pullback affect on the projectile, but the energy from that pullback is transferred to heat energy in the relay module?

If that were the case, it explains why the relay modules heat up during the experiment.
 
  • #32
lekh2003 said:
If that were the case, it explains why the relay modules heat up during the experiment.
The two relays heat due to the resistance of the coil. One relay is also heated by the arc between the opening contacts. If you take the cover of that relay and operate it in the dark, you should see that arc.

A "flyback" or "freewheeling" diode across the gun coil would drop a low reverse voltage, (≈1 volt), so the current continues to flow for a longer time, and the projectile will be pulled back.

The arc at the opening relay contacts can be timed to rapidly kill the magnetic field as the projectile enters the centre of the coil. That greatly reduces pull back of the projectile.

A capacitor discharge through a diode to the gun coil produces a single half sine wave pulse, so it does not cause pull back of the projectile, and it does not produce a flyback pulse that might destroy a semiconductor switch.
 
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  • #33
Baluncore said:
The two relays heat due to the resistance of the coil. One relay is also heated by the arc between the opening contacts. If you take the cover of that relay and operate it in the dark, you should see that arc.

A "flyback" or "freewheeling" diode across the gun coil would drop a low reverse voltage, (≈1 volt), so the current continues to flow for a longer time, and the projectile will be pulled back.

The arc at the opening relay contacts can be timed to rapidly kill the magnetic field as the projectile enters the centre of the coil. That greatly reduces pull back of the projectile.

A capacitor discharge through a diode to the gun coil produces a single half sine wave pulse, so it does not cause pull back of the projectile, and it does not produce a flyback pulse that might destroy a semiconductor switch.
Thank you very much for the explanation. It makes sense.

The fact that a capacitor would have been a better choice is also useful information. I can point out that it exists as a weakness of my experimental setup.
 

Related to How to use capacitors in a Solenoid circuit

1. What is the purpose of using capacitors in a Solenoid circuit?

The main purpose of using capacitors in a Solenoid circuit is to reduce the amount of electrical noise and interference in the circuit. Capacitors act as a filter, smoothing out the current and reducing any spikes or fluctuations that could affect the performance of the solenoid.

2. How do I choose the right capacitor for my Solenoid circuit?

The capacitance value of the capacitor is the most important factor to consider when choosing one for your Solenoid circuit. It should be large enough to effectively filter out any noise, but not too large that it slows down the response time of the solenoid. It is also important to choose a capacitor with a high enough voltage rating to handle the voltage in the circuit.

3. Can I use multiple capacitors in a Solenoid circuit?

Yes, you can use multiple capacitors in a Solenoid circuit. In fact, using multiple capacitors in parallel can provide better noise filtering and stability compared to using a single capacitor. However, it is important to make sure the total capacitance and voltage ratings of the capacitors are appropriate for the circuit.

4. Where should I place the capacitor in my Solenoid circuit?

The capacitor should be placed as close to the solenoid as possible, in parallel with the solenoid coil. This allows the capacitor to filter out any noise before it reaches the solenoid, ensuring optimal performance.

5. What happens if I use a capacitor with the wrong polarity in my Solenoid circuit?

If a capacitor with the wrong polarity is used in a Solenoid circuit, it can cause damage to the capacitor and potentially other components in the circuit. It is important to always check the polarity of the capacitor before connecting it to the circuit and to make sure it matches the polarity of the solenoid.

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