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Registered Member #4252
Joined: Fri Dec 09 2011, 10:43AM
Location:
Posts: 18
Hello guys. Im new over here and i'd like to know a bit more about coils and how to design its parameters. I am trying to build a circuit for charging a capacitor and then discharge its energy through a coil. The high voltage is 8 kV. What is the minimum section of the wire between the capacitor and the coil and also the coil wire (for them not to brake while discharging)? How could i calculate it? I have searched but, since this is an LCR circuit, when the capacitor is charged and i close the discharged loop, the current is not DC, neither common AC. Its a transient (sinusoidal) with a high peak and this sine wave is attenuated ... This way i dont know how to choose the rated current in order to select the wire...
Registered Member #27
Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058
The rated current for a wire is for a permanent current, it will take a lot higher pulsed current without any problems.
As a crude approximation a single discharge through a simple coil will be extremely lossy. That means that the wire will absorb the energy and convert it to heat. Find the energy contained in your capacitor, then calculate how much copper you need to get a reasonable maximum temperature for your application. Then you know the least amount of copper you can use.
Then figure out what wire to use depending on other requirements, like frequency of oscillation or maximum current.
Registered Member #4252
Joined: Fri Dec 09 2011, 10:43AM
Location:
Posts: 18
Steve: Thank you for the simulator idea. I have used this one in the past. Ill install it again. The formula has the number of dicharges per second but I i just want to discharge manually (like 30 per hour).
Bjorn: The energy i want to discharge is 2 kJ and the peak current it will generate when i close the loop is 35 kA. Which formula do i have for the temperature and consequently the section of copper i need for this application?
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
For a single big discharge, you want to compute I2t (I-squared-t) which has the same relation to RMS current as energy does to power. The energy dumped in a resistor "R" is R*I2t.
Or, Bjorn's assumption that all the capacitor bank energy gets dumped as losses in the coil is a good, conservative one.
Then you need to know the resistivity and specific heat capacity of copper, and so on. Given an energy and a mass of copper, you can calculate the temperature rise, and set it less than what would melt the insulation on the wire.
At 35kA, Lorentz forces may also be an issue. They could tear a flimsy coil apart even if it stays cool.
Registered Member #4252
Joined: Fri Dec 09 2011, 10:43AM
Location:
Posts: 18
Patrick: Yes i think it is relevant. I just didnt know if it was the right formula to use. According to that equation, for t=400us and with Ipeak=35kA the minimum copper section is 2.4022 mm^2
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Coil design
