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Registered Member #3114
Joined: Sat Aug 14 2010, 08:33AM
Location:
Posts: 608
I don’t want to divulge what I’m making (I will share later), but it consists of a flyback core and magnet wire. The wire I am using is getting really hot (current). Can i put a resistor on the output of the hv side?!? Will that help anything or will the current be dissipated as heat in the resistor. If it will work what wattage and resistance would I be looking for (ohms or kohms)? I am not concerned with a specific current just a lesser one.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
I think we may need a bit more information before we can help, ie a 'standard' flyback puts out a few mA at most and I use suppressed automotive spark plug lead as resistors.
What are you driving? what is the output current and voltage?
Registered Member #9252
Joined: Fri Jan 04 2013, 06:27AM
Location: Andromeda
Posts: 253
Well. The resistors should have the voltage rating of that thing your making. And i actually bet that won't work. to decrease current.. well i cannot be sure what i will be talking about as i do not know what your doing. But i bet its winding a home made flyback. I know you made a 3000 turns one before :D so if it is a flyback. To decrease heat. Increase windings. So less current but more voltage. Or just give it less input. or make it oil cooled. :D..
Registered Member #834
Joined: Tue Jun 12 2007, 10:57PM
Location: Brazil
Posts: 644
A resistor in series with the arc will reduce the power dissipation. How much is needed is difficult to determine without more data, but let's suppose that the output voltage is 10 kV and the arc is draining 10 mA. This corresponds to an effective load resistance of 1 Mohm. If you add 1 Mohm of resistance, the current will be reduced to 5 mA, supposing that everything is linear and that the voltage doesn't rise. The resistor will dissipate 5 kV x 5 mA = 25 W. Use a series association of several physically long resistors.
Registered Member #3114
Joined: Sat Aug 14 2010, 08:33AM
Location:
Posts: 608
Antonio wrote ...
A resistor in series with the arc will reduce the power dissipation. How much is needed is difficult to determine without more data, but let's suppose that the output voltage is 10 kV and the arc is draining 10 mA. This corresponds to an effective load resistance of 1 Mohm. If you add 1 Mohm of resistance, the current will be reduced to 5 mA, supposing that everything is linear and that the voltage doesn't rise. The resistor will dissipate 5 kV x 5 mA = 25 W. Use a series association of several physically long resistors.
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