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# Coilgun, Mosfet problem and joules? Answered

I have two questions:
when a mosfet is damaged, it conducts electricity without putting energy to the gate?
I have the 75N75 mosfet and I want to use it on the rwilsford07 boost converter, but the mosfet didn't work and I don't know why it conducts energy without putting energy to the gate. I read that the mosfet can be damaged with static electricity, but I use always an anti-static wrap.

The formula to get the joules from a coilgun is this?:
0.000005 * C * V * V
C= uf
V= voltage

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You do know all mosfets have have an intrinsic inverse diode ?

how the intrinsic inverse diode works?, what it do?

A diode is a one way valve for electric current flow and always
lets current flow in reverse to the mosfet desired switch direction.

The mosfet's control Gait is glass insulated but can only withstand
about 20 VDC before shorting out.

A

Thanks, I make the mosfet work, and also finish the circuit.
Did you know if the formula to get the joules is correct

Thanks

1/2 C V ^2 yes, it right, if your decimal point is right.

I get the ecimal point from rwilsford07 but I dont know if it is right, how I can know the right one

Here is how to work the math ;     Joules = Energy

Energy = (1/2)×C×V×V = (0.5)×C×V2 = (0.5) × (C / 1000,000) × V2

Joules = 0.000005×C×V2
---------------------------/\-uF

My capacitor is 450v and 1000mfd. It give me 1012.5 joules, it is right?

J = C * V2 / 2
J = (1000 × 10-6 / 2 ) × 450 × 450 = ( 10-3 ) × 202500 / 2 = 101250 × 10-3

J = 101.25 Million microJoules

Storage of 101 Joules is only going at 450 VDC the capacitor failure
voltage so stay around 425 VDC maximum about 90 Joules.

A

Ok, thanks
I will puttwo stgesnon my coilgun. Can you recomend me the length of the coils and the separation between them?. I will use infrared leds.

Never made a staged coil gun, You may have to read or resort to trial and error.

My unit was a ball bearing in a paper tube wound with loose wire about 10"
spread and a single fine 2" wire as a fuse that blows before the bearing
reaches half way and continues on into sheetrock.

Well A. agrees with you.

Engineers would usually say 0.5 E - 6 x C (in uF) x V^2