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Registered Member #1497
Joined: Thu May 22 2008, 05:24AM
Location: Toronto, Ontario, Canada
Posts: 801
Well I recently got my caps for my coilgun, max 450v total 4800uF, and I did a quick simulation of the circuit.
The end result (in some detail): Coil: 0.277mH, 0.273 ohm with 18AWG wire LCR simulation: 950-1000A peak, zero crossing after 4ms, reverse peak at 100A at 5ms.
My question to the other coilgunners out there, what are the major selection criteria for the quenching diode+resistor combination? Speed, peak current? How should I choose my resistor also, since P=IV for resistor wattage doesn't make sense (ie: 4500W resistor??), and what would a good resistance be.
Registered Member #95
Joined: Thu Feb 09 2006, 04:57PM
Location: Norway
Posts: 1308
You'll be fine without a resistor, the current will be limited regardless. If you do use one it only needs to stand the energy it will absorb. W = J/s, so calculate how many joules will be dissipated in a given time span and you have the power rating of the resistor. The simulation gives you a ballpark figure of the reverse current, so rate your diodes well over that number. Since the current rises in a nice sine wave over the course of a few ms pretty much any diode will work.
Registered Member #1497
Joined: Thu May 22 2008, 05:24AM
Location: Toronto, Ontario, Canada
Posts: 801
'antiparallel', that is in parallel with the capacitor, but in reverse so that prior to discharge, zero current flows through the diode. Once the caps are emptied and the coil voltage peaks and begins to drop down and pass through zero, the collapsing magnetic field will cause reverse current. If the diode is placed right next to the coil (more or less) then the reverse current will just be dissipated as heat in the diode and coil, instead of charging the electrolytics in reverse.
Registered Member #7669
Joined: Mon Oct 29 2012, 02:55PM
Location:
Posts: 19
Sorry to resurrect this thread, but you said 'pretty much any diode will work', Is that actually true? Surely the surge from the coil back into the capacitor needs to be dissipated and a little PCB diode will just get fried?
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