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4hv.org :: Forums :: High Voltage
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type of core and frequency

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Andre
Wed Oct 03 2012, 03:49PM Print
Andre Registered Member #6921 Joined: Wed Sept 26 2012, 07:47PM
Location:
Posts: 109
I'm building a transformer with a ratio of 1:1666(24 to 40kV) is very low output current 0.3mA , how can I calculate the frequency that I should use, and what type of core should I use? round or square?)
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Dr. ISOTOP
Wed Oct 03 2012, 11:42PM
Dr. ISOTOP Registered Member #2919 Joined: Fri Jun 11 2010, 06:30PM
Location: Cambridge, MA
Posts: 652
Andre wrote ...

I'm building a transformer with a ratio of 1:1666(24 to 40kV) is very low output current 0.3mA , how can I calculate the frequency that I should use, and what type of core should I use? round or square?)

It will be very hard to build such a transformer; even with 1 turn on the primary (which is doable - the large cores can do 30-50V/turn at 100KHz) you'll need to insulate a 1600 turn secondary for 40KV.
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Andre
Thu Oct 04 2012, 01:12AM
Andre Registered Member #6921 Joined: Wed Sept 26 2012, 07:47PM
Location:
Posts: 109
Dr. ISOTOP I see your post but is not showing the formula to calculate the freq and the type of core I should use, can you post again plz.
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Patrick
Thu Oct 04 2012, 05:52AM
Patrick Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639
Andre wrote ...

Dr. ISOTOP I see your post but is not showing the formula to calculate the freq and the type of core I should use, can you post again plz.
the core area depends on freq, so if you want a low turn primary youll need large area with high freq.
and volume largely depends on turns needed and power.

transformers for our purposes can be exceptional difficult to build.
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Dr. Slack
Thu Oct 04 2012, 07:05AM
Dr. Slack Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Andre,

If you are asking whether you should use a round or square core, then you already have the only formula that's relevant to your apparent state of knowledge, 1:1666.

Before round or square comes core area, design field strength, magnetic length. Frequency is in the mix there along with self resonant frequency - which depends on number of turns, winding configuration and insulation levels.

Before those comes topology - forward or flyback?

For 40kV output, I'd buy an auto ignition coil, and then experiment with driving it. It's a transformer that's wound for 300v to 30kV when used forward, but will easily do 24v to 40kV as a flyback. Not only that, it's ready-wound, ready-insulated, and not too expensive, or free if you're lucky.
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Andre
Thu Oct 04 2012, 02:02PM
Andre Registered Member #6921 Joined: Wed Sept 26 2012, 07:47PM
Location:
Posts: 109
Thanks guys but I'm not looking to do trial and error, I been reading and I know the core size would matter, I was looking for advice between the round core and the square core, and how to calculate the frequency for this type of voltage. that way I can understand the physics of it if something does not work.
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Dr. Dark Current
Thu Oct 04 2012, 02:33PM
Dr. Dark Current Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
The induced voltage V on a winding with N number of turns is V = N * dphi/dt, where phi = B*A is magnetic flux (A = core area, B = magnetic flux density in Tesla).
For linear signals V = N * delta phi / delta t. From this you can already calculate the required number of turns, if you know the peak flux density, frequency and core area.
If you give me your core params I can calculate this for you.

P.S. If you input a 24V square wave into a HV transformer with a ratio of 1:1666, you can theoretically get up to 120kV of ringing/overshoot on the output, more if you drive the transformer near its resonant frequency or on its subharmonics.
Also, the resonant frequency of such transformer can be as low as few kHz, not more than few tens of kHz and you must drive a transformer below its resonant frequency.
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Andre
Thu Oct 04 2012, 04:19PM
Andre Registered Member #6921 Joined: Wed Sept 26 2012, 07:47PM
Location:
Posts: 109
THANK YOU Dr. Dark Current!!! you seem to know this inside out,
I am missing some information, I will use the round ferrite core since it seems easy to assemble, I'm trying to make it as compact as possible, I did some research about finding the flux density and I got the formula for the frequency t= (V * B)/(N * Ae)
Link2
I'm getting 4 cores from there "64-40-24" "69-24-14" "81-13-16" "81-33-16"
I been reading and the flux density in Tesla should be a maximum of 0.25T right?
so t= (V * B)/(N * Ae)
t = (24V * .25T) / (20 * 0.032)
t= 9.375hz
That's a very low number. so I'm thinking I have the wrong formula. or most likely I'm putting the wrong numbers.
20 is the number or turns I was thinking to use in the primary and .032m2 is the Ae from 64-40-24
eventually I will like to use those cores to output +40kv and -40Kv on each core.
what are you thoughts on using those core for +40kv and -40kv and equation for this transformer.
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Dr. Dark Current
Thu Oct 04 2012, 04:38PM
Dr. Dark Current Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
OK, as an example I'll take the "64-40-24" core, which has an effective area of 3.20 cm2 = 0.00032 m2.
The magnetic flux density of 0.25 T is fine, for most cores you can use 0.3 T so I will go with that. Note that above a few tens of kHz, core heating becomes the limiting factor over saturation.
If you solve the equation I posted for N you get
N = V * delta t / delta phi.
V is clear, phi = B*Ae. As for t: For 50% duty cycle square wave signals you use t = T/4 where T is the cycle length (why T/4? If you plot the waveform of flux density in the core this becomes clear, if you want I will explain).
T = 1/f so after substituting you get

N = V / (4*f*B*Ae)

Now you need to select the switching frequency. For such low power you would usually go with something like 50-100 kHz, but with low power high voltage transformers the resonance is a serious problem and might really make you use a frequency as low as several kHz here.

So just as an example solving for f=20 kHz:

N = 24 / (4*20000*0.3*0.00032) = ~3 turns, the secondary has 3*1666 = ~5000 turns.
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Andre
Thu Oct 04 2012, 05:12PM
Andre Registered Member #6921 Joined: Wed Sept 26 2012, 07:47PM
Location:
Posts: 109
thank you that explain a lot!!, I was confused about the Ae calculation, and it was cause the conversion from cm2 to m2 I did was wrong.
so I will stay in the khz frequency deep everything in oil
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