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4hv.org :: Forums :: High Voltage
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FERRITE CORE COIL CALCULATION

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Newton Brawn
Fri Dec 16 2011, 01:27AM
Newton Brawn Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311
Hi Patric !

The magic number is the corrected rod permissivity u_rod'. (something like the golden number)
Regards

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Patrick
Fri Dec 16 2011, 01:57AM
Patrick Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639
Newton Brawn wrote ...

Hi Patric !

The magic number is the corrected rod permissivity u_rod'. (something like the golden number)
Regards


Ok thank you i was really racking my brain trying to figure that out! so its what ever "u rod" comes out too, or in your case "5"!?
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Tony Matt
Mon Dec 19 2011, 01:11AM
Tony Matt Registered Member #3700 Joined: Sat Feb 19 2011, 12:59PM
Location:
Posts: 107
Hi Newton:

You wrote;

""The next step is calculate the maximum induction in the ferrite core when a 0.001 uF capacitor discharge 5000V on the 2 turns primary. I understand the core shall be not saturated during the discharge.
if you can help me it will be nice. ""

Let me try...

The hollow ferrite core cross section could be:
Ac = (0.017^2 - 0.009^2) X (pi / 4)
Ac = 0.000163 m2


The characteristic impedance: Zo = (L / C)^(1/2)

Zo= ((0.5 x 0.000001)/(0.001 x 0.000001))^0.5

Zo =22.36 ohms

Assuming the resistive component of inductor and ckt <<Zo,

Peak current Ipk = Ec / Zo

Ipk =5000 / 22.36

Ipk = 223.6A

Discharging time t = (C x L)^0.5

t = ((0.001 x 0.000001)x(0.5 x 0.000001))^0.5

t = 0.0224 x 0.000001 s

t= 0.0224 us

Fo = 1/(2 x pi x t)

Fo = 1/(2 x pi x 0.0224) = 7.11Mhz

Maximun induc tion in the core = Bmax = (Epk x t)/(n x Acx)

Bmax = (5000 x 0.0224 x 0.000001)/(2 x 0.000163)

Bmax = 0.345 T

I gest that your ferrite core material is good up to 0.35~0.4T before saturation.

So, if this is true the core induction does not saturate during the capacitor discharge.

By the way, what is the ferrite core material ??

Anyone have any comment?

Tony


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