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Forums
4hv.org :: Forums :: High Voltage
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Capacitor current-limiting MOT
Move Thread LAN_403
HVPaul
Thu Nov 11 2010, 08:44PM Print
HVPaul Registered Member #2321 Joined: Fri Aug 28 2009, 05:13PM
Location: Boston, MA
Posts: 34
 		Has anyone tried capacitor current-limiting a MOT?
How effective would that be?

I calculated that I'd need a 312nF cap in order to reduce the current to
20mA. Or would that not even be enough to drive the transformer?

I'm not an EE major so pardon my ignorance :)
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monokel
Fri Nov 12 2010, 02:04PM
monokel Registered Member #2981 Joined: Thu Jul 08 2010, 01:47PM
Location: Germany
Posts: 35
 		The current depends on the circuit. E.g. with a level shifter circuit which is commonly used in microwave ovens
you could also limit the current if the cap is smaller.
I assume you connect the cap in series with the MOT and connect the load directly to this (no level shifter circuit). Assuming that the max. current is not bigger than the
short-circuit current (this is true e.g. with a resistive load), with 312 nF you would get a max. current of about 290 mA. If you want to limit the current to 20 mA the capacitance must be about 26 nF.
Regards, monokel
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HVPaul
Fri Nov 12 2010, 03:33PM
HVPaul Registered Member #2321 Joined: Fri Aug 28 2009, 05:13PM
Location: Boston, MA
Posts: 34
 		How did you come up with 26nF?

My original calculation was for peak voltage of 170V on a 120v line.
If i rewrite the formula to limit 120V to 0.020A that would require 6kOhm resistance.

C = 1/(2*PI*60*6000) = 442nF

Am I missing something here?

Thanks
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ScotchTapeLord
Fri Nov 12 2010, 05:11PM
ScotchTapeLord Registered Member #1875 Joined: Sun Dec 21 2008, 06:36PM
Location:
Posts: 635
 		A lot of that capacitance goes toward canceling the inductance.

impedance = sqrt((1/(2*pi*C*60))^2 + (2*pi*L*60)^2 + R^2)

where impedance is in ohms, C is your primary series capacitance, L is your primary leakage inductance, and R is your wire's ohmic resistance.
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Carl Pugh
Fri Nov 12 2010, 06:08PM
Carl Pugh Registered Member #1064 Joined: Tue Oct 16 2007, 05:04PM
Location:
Posts: 42
 		Assuming transformer secondary voltage of 3000 volt and neglecting transformer inductance..

Isec will be less than 1 milliamp.
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monokel
Fri Nov 12 2010, 06:42PM
monokel Registered Member #2981 Joined: Thu Jul 08 2010, 01:47PM
Location: Germany
Posts: 35
 		@HVPaul:
I thought you wanted to connect the prim winding of the MOT to the mains and limit
the current of the sec winding by connecting a cap in series.

ScotchTapeLord wrote ...

A lot of that capacitance goes toward canceling the inductance.

impedance = sqrt((1/(2*pi*C*60))^2 + (2*pi*L*60)^2 + R^2)

where impedance is in ohms, C is your primary series capacitance, L is your primary leakage inductance, and R is your wire's ohmic resistance.

That's not right. The impedance is
impedance = sqrt( ( (2*pi*L*60) - 1/(2*pi*C*60) )^2 + R^2)
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ScotchTapeLord
Fri Nov 12 2010, 07:34PM
ScotchTapeLord Registered Member #1875 Joined: Sun Dec 21 2008, 06:36PM
Location:
Posts: 635
monokel wrote ...

@HVPaul:
I thought you wanted to connect the prim winding of the MOT to the mains and limit
the current of the sec winding by connecting a cap in series.

ScotchTapeLord wrote ...

A lot of that capacitance goes toward canceling the inductance.

impedance = sqrt((1/(2*pi*C*60))^2 + (2*pi*L*60)^2 + R^2)

where impedance is in ohms, C is your primary series capacitance, L is your primary leakage inductance, and R is your wire's ohmic resistance.

That's not right. The impedance is
impedance = sqrt( ( (2*pi*L*60) - 1/(2*pi*C*60) )^2 + R^2)


Sorry, I was in a hurry. You're right. The impedances wouldn't cancel out with my equation!
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radiotech
Fri Nov 12 2010, 07:43PM
radiotech Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546
 		At 1250 volts secondary, forget the inductance, Use a 0.04 ufd
capacitor. That will give you ~~ 19 mA at 60 Hz.

Why forget it? because the inductive reactance is about equal to the MOTs capacitor Xc so by Pythagorus, your hugh reactance compared to the transformers small reactance means the numbers wont even show a difference.. As far as you are concerned, its a constant
voltage source.
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