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Author Topic: Toroid  (Read 507 times)
KE6DUB
Member

Posts: 3




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« on: June 01, 2005, 07:19:51 PM »

Hello,

I have this toroid choke with has the following
dimensions.

OD  42MM
ID  20MM
W   18MM
Black Core, Powdered Iron
Number of Turns 106. 1MM wire diameter.

The windings are interwoven. The first with a gap
between the first winding which is filled in with
the second winding. In other words, one winding of 53
turns and the second of 53 turns.
I was told this is a common mode choke which has the
first winding in series with the second increasing the
inductance.

Please can someone compute the inductance?
What is the formula for same?

Thanks,

Larry, KE6DUB
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KE6PKJ
Member

Posts: 256




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« Reply #1 on: June 02, 2005, 03:34:54 PM »

Without knowing the A/L value of the ferrite this would be next to impossible. Here's a quick trick.

Toss a capacitor (try .01 ufd) onto the toroid's leads then add a couple of turns of spare wire through the toroid. Couple these few turns into a GDO and find the resonant frequency. Take the KNOWN frequency and multiply it by itself (Freq. squared). Multiply this value by your KNOWN capacitance....now multiply all this by 39.5. Take this number and divide it into 1.

This will give you the inductance of the toroid in uH.

Kirk
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KE6DUB
Member

Posts: 3




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« Reply #2 on: June 02, 2005, 05:02:45 PM »

Kirk,

I forgot to include the fact this toroid is used
in a DC to AC inverter. The OUTPUT of the H bridge
produces the AC. The gates of the mosfets are oscillated at 60Hz with 50KHz PWM. These are the devices on the positive HVDC side of the bridge. The negative side of the bridge has no modulation, except the 60Hz on the gates, to allow the devices to switch properly.
A 5 mfd, 250 VDC non electrolytic is at the output of the choke to the neutral side of the AC line. I believe this choke is part of the filtering of the
high freqs from the PWM of the square wave, thus
producing a sine wave of 60Hz. I presume the choke has a rating of at least 8 amps of current. 600W inverter.
I'm sorry I don't have the equipment to run the test for the A/L.

Larry, KE6DUB
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KE6DUB
Member

Posts: 3




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« Reply #3 on: June 02, 2005, 05:17:51 PM »

Kirk,
Woops! It is 15.625Khz not 50Khz.
Larry, KE6DUB
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KE6PKJ
Member

Posts: 256




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« Reply #4 on: June 02, 2005, 05:49:46 PM »

Well I hate to take an (educated?) guess but here goes...
I believe it MIGHT be close to an Amidon 150-57 or 150A-57
toroid with an A/L value of 1250-1300. From this number you can calculate the inductance in milliHenry's.

http://www.amidoncorp.com/aai_compositetoroids.htm

Good luck,
Kirk
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