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Registered Member #90
Joined: Thu Feb 09 2006, 02:44PM
Location: Seattle, Washington
Posts: 301
3760uF and 350v is about 230 joules. Nice.
Inductance is millihenrys (or microhenrys) not amps.
Whether or not you need a freewheeling diode depends on the total series resistance of the discharge path. With just a little resistance, the oscillation may be damped enough that the capacitor is not reverse-charged.
To find out, fire it at low energy and no diode. Measure capacitor voltage before and after. If it's more than a dozen volts negative, then you need a protective diode. Edit: If the remaining negative voltage is more than 10% of the positive starting voltage, then you need the diode.
Better yet watch the capacitor with an oscilloscope. This tells you both the total series resistance, and the discharge time (or resonant frequency). Scopes can really help you tune the inductance for a desired discharge time.
Let's call it a "freewheeling diode" since it is the ringing (not back emf) that damages capacitors.
I strongly suggest minimizing net resistance. It makes a big difference. Use short fat wires and soldered connections. Overbuild the copper discharge path. You only need a freewheeling diode if you don't have a spare capacitor.
Cheers, Barry I'm trying to take one day at a time, but sometimes several days attack me at once.
Registered Member #1787
Joined: Thu Oct 30 2008, 12:33AM
Location:
Posts: 13
if i can get my hands on an oscilloscope, what kinds of numbers should i be looking for? "fire it at low energy and no diode" what would be low energy in your books? like how much voltage?
Registered Member #1792
Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527
The lower the better: you can always increase energy but once you blow something up you can't go back. I would guess something on the order of 12-20V is a good low power starting point, using E=0.5*C*V^2 your energy at those voltages will be under a joule. You're looking to see what portion of the starting voltage ends up negatively charging the capacitors, so the higher this voltage is the more you might stress them.
Registered Member #1787
Joined: Thu Oct 30 2008, 12:33AM
Location:
Posts: 13
Barry wrote ...
To find out, fire it at low energy and no diode. Measure capacitor voltage before and after. If it's more than a dozen volts negative, then you need a protective diode. Edit: If the remaining negative voltage is more than 10% of the positive starting voltage, then you need the diode.
so at 90v, i'm getting about 5v bak into my caps. I should be safe? I gotta take quite a few turns out of my coil too.
the projectile flies about 5 feet at 90v, but doesn't leave the barrel with 140v. thats with the protector diodes
Registered Member #968
Joined: Fri Aug 24 2007, 04:54PM
Location: Norman, Oklahoma
Posts: 104
What voltage do you get back after 140v? The protection diodes help prevent the negative capacitor charge so that makes the test Barry offered difficult.
You will probably have to move the projectile starting position after such a large change in bank energy(90v->140v). This may be one reason why your projectile doesn't shoot at higher voltages.
Registered Member #90
Joined: Thu Feb 09 2006, 02:44PM
Location: Seattle, Washington
Posts: 301
casey wrote ...
if i can get my hands on an oscilloscope, what kinds of numbers should i be looking for? "fire it at low energy and no diode" what would be low energy in your books? like how much voltage?
I use ten-percent as a general rule of engineering.
Charging your capacitors to 35v can be considered "negligible" compared to their 350v rating. You can safely dry-fire and test to your heart's content without fear of damage.
The deal here is that the flyback diode will greatly extend the time for the magnetic field to die away. It is a drastic change of timing. Think of it this way ...
the coil will be energized in approximately a sinusoidal waveform according to the LC frequency
But then the diode kicks in at maximum current ...
Now the coil is essentially shorted, and it doesn't see the capacitor ...
the current dies away at a roughly linear rate according to copper's resistive losses and the 1.4v forward-biased diode drop
You found an oscilloscope? Awesome! It will really help show you what's happening in there. Cheers, Barry Today is a day for firm decisions! Or is it?
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