Trying to calculate the peak current of a Capacitor bank

In summary, the conversation is about building a Tesla Coil and the issues the individual is experiencing with their current setup. They are using a 15,000 volt transformer rated at 30 mA, but they believe it may be too small for their Tesla Coil as it keeps blowing fuses and causing a red light to turn on. The individual also includes specifications for their setup and mentions trying to find a formula to calculate the peak current of the circuit. They also mention having safety equipment, but other members caution them about the dangers of high voltage and recommend additional safety precautions.
  • #1
johnboyman
22
2
Hello I am building a Tesla Coil. I built it well I attached an image of the circuit I used. I bought a 15,000 volt transformer rated at 30 ma. I think it is too small for my tesla coil because it keeps blowing a fuses and a red light turns on. Here are the specs.
-Inductance (L): 17 uH
-Capacitance (C): 0.023 uF
-Resonant Frequency(F): 254.201 KHz

-Diameter (D): 6.5 inches
-Number of Turns(N): 12
-WireDiameter (W): 0.25
-Turn Spacing(S): 0.25

Height: 7.5 inches
Length Of Wire: 282.743
Inductance: 17 uH
.33uF/13 = 0.025 uF

2000 Volts each * 12 = 25,000 VoltsI am trying to find a formula on the internet to calculate the peak current of the entire circuit to find out how much current is going through my transformer. I have found many formulas that are very close but i just can't fully get my head around this. Here are some formulas I found.

This is what someone else wrote:
Q: A 140 micro farad capacitor is connected with 250 volts, 50 hertz. What is the value of the current?
A: The impedance of a capacitor is given by Xc = 1/ (2*pi * f * C)
If there is no other component in circuit, then I = V / Xc = 250 * 2 * pi * 50 * 140*10^–6 = 11 amps.

My ideas on how to finish these formulas for my needs is to:

Vrms = 15000 volts / 13 = 1,153 volts per capacitor
C = .33 uf per cap

calculate the peak current per capacitor with the above formulas, then divide the result by 13 may equal the total peak current in the circuit. That is the direction i am moving in.

This is as far as i can find. I have one string of capacitors in an mmc string. non in parallel just 13 .33 uf capacitors in a string and a 15000 volt transformer. How can i use these equations and factor in my 13 .33 uf Capacitors to find the peak Current of My tesla coil circuit. I could just buy a 60 ma transformer and i think it would work but i need to know the math before i move on because its just driving my nuts.

Thanks a lot guys.
wiring.jpg
 
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  • #2
:welcome:

Can you give us some more information on your background and experience, and on your project. Are you working alone, or with a group, or with a mentor?

Working alone with high voltages can be dangerous. And the fact that fuses blow and you don't know why, indicates that you may not understand the circuits you are building.

Other members may wish to hold back direct answers to the OP until we have more information.
 
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  • #3
I am a student. I am working alone. I know the dangers of high voltages. I have high voltage gloves and glasses. I just recently drew up a schematic. Hope this helps.
 

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  • #5
Welcome to the PF. :smile:
johnboyman said:
Actually I found a post in this forum that has my answer. Thanks for the help.
https://www.physicsforums.com/threads/lc-circuit-and-peak-current.224493/
That's great! What did you find in that thread that answered the questions you were asking in this thread here? Just curious.
johnboyman said:
I am a student. I am working alone. I know the dangers of high voltages. I have high voltage gloves and glasses
Wearing gloves and safety glasses is a good practice in many school lab experiments and at work, especially with chemicals and high pressures, mechanical lab work, etc. I've never seen them recommended for high voltage work. There are other, very important, safety precautions for working with high voltages. Can you say what a few of those might be? :smile:
 
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  • #6
So you have 15 kV /30 mA rated NST( inductive shunt current-limited xfmr). If you didn't overvolt its' primary side and didn't overvolt secondary side, how possibly you could be blowing mains fuses? What's the fuse amperage rating? Did you check if NST worked normally?
 
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  • #7
berkeman said:
I've never seen them recommended for high voltage work. There are other, very important, safety precautions for working with high voltages.
Indeed. I thought I should show the OP a small clip of what a 9 kV arc looks like. I call it "Glove Burner".
glove burn.gif

johnboyman said:
... I am building a Tesla Coil. ... I built it well I attached an image of the circuit I used. I bought a 15,000 volt transformer rated at 30 ma.
From the above clip, now consider what a 15 kV arc could do. I'm sure it would burn through most gloves. Also note that all the Neon sign transformers I've seen generally has a case grounded center tap. Hence an arc can be created from a HV lead to the transformers case.
 
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  • #8
zoki85 said:
So you have 15 kV /30 mA rated NST( inductive shunt current-limited xfmr). If you didn't overvolt its' primary side and didn't overvolt secondary side, how possibly you could be blowing mains fuses? What's the fuse amperage rating? Did you check if NST worked normally?
It works normally. If i hook my cap bank up in series with my spark gap, the transformer will run smoothly and the spark gap will constantly run without overloading my tranformer. Thats not how a tesla coil runs through. So i have to hook up up like the following. My t coil is hooked up like in the schematic. The transformer runs and the spark gap fires for 1 split second really loud and then the tranformer turns off and a red light glows. I checked if anything is shorting it out and I cannot find any stray metal contact or leads. There is only one alternative that the transformer cuts off when the current goes over 30 ma. I am now trying to calculate how much current is running through my LC circuit to find out how big of a transformer I need now. Thanks for the advice.
 

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  • #9
dlgoff said:
Indeed. I thought I should show the OP a small clip of what a 9 kV arc looks like. I call it "Glove Burner".
View attachment 254820

From the above clip, now consider what a 15 kV arc could do. I'm sure it would burn through most gloves. Also note that all the Neon sign transformers I've seen generally has a case grounded center tap. Hence an arc can be created from a HV lead to the transformers case.
Surely, that 9 kV xfmr can kill easily, but 30 mA NST is relatively forgiving and can deliver just quite a jolt.However a decent HV capacitor connected to 15 kV source can send you thru the roof
 
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  • #10
Yes I am aware of the danger. I am no where near this thing when it turns on plus i have bleed resistors on my capacitor bank for my safety aswell.
 
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  • #11
johnboyman said:
It works normally. If i hook my cap bank up in series with my spark gap, the transformer will run smoothly and the spark gap will constantly run without overloading my tranformer. Thats not how a tesla coil runs through. So i have to hook up up like the following. My t coil is hooked up like in the schematic. The transformer runs and the spark gap fires for 1 split second really loud and then the tranformer turns off and a red light glows. I checked if anything is shorting it out and I cannot find any stray metal contact or leads. There is only one alternative that the transformer cuts off when the current goes over 30 ma. I am now trying to calculate how much current is running through my LC circuit to find out how big of a transformer I need now. Thanks for the advice.
If you open the spark gap too wide and the capacitor is near resonant than not only current can go over 30 mA but peak voltage over 30 kV. That can kill NST. RF can also do a damage to the winding so better use some filter circuit to protect it.
 
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  • #12
Ok thanks for the advice. I am going to make a filter circuit for it. I don't think that the spark gap is too far apart but i will work on that too. Do you think that the transformer could be to small.
 
  • #13
johnboyman said:
Do you think that the transformer could be to small.
No
 
  • #14
zoki85 said:
Surely, that 9 kV xfmr can kill easily, but 30 mA NST is relatively forgiving and can deliver just quite a jolt.However a decent HV capacitor connected to 15 kV source can send you thru the roof
Agreed. But a fire danger still exist.
 
  • #15
According to your schematic, the spark gap is directly across the transformer secondary.

Realize that the impedance of an arc is extremely small, putting a direct short across the transformer secondary. In fact an arc has negative resistance! This means that as the arc current increases the voltage drop across the arc declines.

The usual solution is to put some impedance (resistance) between the transformer and the arc.

An alternative solution to try would be to widen the spark gap enough that it just sparks. That should yield a sparks only near the peaks of the power line voltage, reducing the average load on the transformer. A convenient way to do that is to disconnect the capacitors and set the gap just wide enough to support a continuous arc. Then reconnect the capacitors. You may have to slightly reduce the gap with everything reconnected.

You have not told us the the current rating of the fuse that blows. Realize that the transformer is rated for 450 Watts continuous. Too start, I would use a fuse rated at least 2x the rated current... maybe even a Slo-Blo (time delay) fuse often used for motor loads. To pick a fuse rating, divide the 450 Watts by your line voltage, then double it. if the line voltage is 120V that would be:

450/120 = 3.75 . . . 3.75x2 = 7.5Amp or 10Amp Slo-Blo fuse.

Cheers,
Tom
 
  • #16
Thread closed temporarily for another safety review...

EDIT/ADD -- After further review, this thread will remain closed for safety reasons.
 
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Related to Trying to calculate the peak current of a Capacitor bank

1. How do I calculate the peak current of a capacitor bank?

To calculate the peak current of a capacitor bank, you can use the formula I = C * dV/dt, where I is the peak current, C is the capacitance of the bank, and dV/dt is the rate of change of voltage. This formula assumes that the capacitor bank is being charged or discharged in a linear manner.

2. What is the capacitance of a capacitor bank?

The capacitance of a capacitor bank is the measure of its ability to store electric charge. It is typically measured in farads (F). To calculate the capacitance of a capacitor bank, you can add the individual capacitances of each capacitor in the bank.

3. How does the voltage affect the peak current of a capacitor bank?

The voltage of a capacitor bank directly affects the peak current. As the voltage increases, the peak current also increases. This is because the rate of change of voltage (dV/dt) in the formula for peak current is directly proportional to the voltage.

4. What factors can affect the peak current of a capacitor bank?

Aside from the voltage and capacitance, other factors that can affect the peak current of a capacitor bank include the type of dielectric material used in the capacitors, the temperature, and the frequency of the AC source.

5. How can I reduce the peak current of a capacitor bank?

There are a few ways to reduce the peak current of a capacitor bank. One way is to increase the capacitance of the bank, as this will decrease the rate of change of voltage. Another way is to use capacitors with a lower dielectric constant, as this will also decrease the rate of change of voltage. Additionally, using a higher frequency AC source can also reduce the peak current.

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