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Portable transformer help! 11v to 800v

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leodahsan

Sat Jan 22 2011, 12:06AM

Registered Member #3638 Joined: Fri Jan 21 2011, 11:56PM
Location:
Posts: 10

Hello all, this is my first post so..
Hi, my name is Leonardo and I'm from Brazil. I'm designing a coilgun and I need a 800v source for charging the caps. I'm working with Zero (at LaserPointerForums). My cap bank is 247,6J (1st stage) and 147,2J (2nd stage). I need 400W power to charge both banks (in 1 second or less), I assume.
I want 1 second maximum charge time. Less would be great but, I don't know capacitors limitations (0.25 sec > 800v 2A? :P)

That isn't easy to get in a portable, I know. But I'll use a 11v 4Ah 35C battery (does more than 80A! 200A possible :P), I assume I'll need ~38amps (depending on converter efficiency).

I think I need a oscillator circuit (DC to AC) and a handwinded transformer. I don't know what circuit to use to get AC from 11v and how to handwind the transformer (either to use a old transfomer, ferrite bar, ferrite toroid, HOW MANY WINDS, wire thickness etc).

Can anybody help me? And sorry for making my first post a question. :(

Thanks!

edit: I can change the source to 18650 batteries or such (for higher voltage) if needed.

Patrick

Sat Jan 22 2011, 12:34AM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

I dont know if what you ask is possible.

when you say 247,6 J does this mean 247.6?

247.6 J at 800V must be about 700uF.
147.2 J at 800V must be about 480uF.

well 700 + 480 = 1180uf, so if R is 10ohms, and V is 800, while total C is 1180uF, then your RC is 0.012, you will need 166ohms or less total impedance for your charging circuit,( if R=166, C=1180uF, then 5RC is 0.06 seconds, so 1/0.06rc = 166ohms for a one second charge rate)

Your initial charge currrent at 4.9 amps, and decaying (I = (Vb/R)e^(rt)/(RC)) to 0 amp across 5RC (or 1sec)

the red bold text is bogus math: so you will need a input of about 2kW continuous input and about 4kw peak (for 0.3 sec) so from 10 volts (a 12 volt battery with a drop) 4kW = 10V @ 400A, and 2kW = 10v @ 200A.

your voltage should rise from 0 to 800 V across 5RC (RC = 0.19 sec)
others should check my math. im half asleep.

leodahsan

Sat Jan 22 2011, 12:38AM

Registered Member #3638 Joined: Fri Jan 21 2011, 11:56PM
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Posts: 10

Why?
I've just made 160v with a rubbish handmade inductor (transformer) by 4v input. Oscilations were made by hand (on/off cycle, 10HZ max :P)...
150 turns (thin wire) 8 turns (thick wire).

hboy007

Sat Jan 22 2011, 12:59AM

Registered Member #1667 Joined: Sat Aug 30 2008, 09:57PM
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Posts: 374

Designing a power supply that suits your needs an easy thing to do *cough*. Take two transformers from ATX Power supplies, add two unipolar mosfet drivers, pump 60-80A through the windings and wire the secondaries in series, then build a nice 1.5kV rated rectifier (add a choke to slug the charge pulses behind the rectifier).

Enjoy the current-mode converter design for this 600W+ design plus all the regenerative snubber and clamping circuitry needed to avoid pieces of hot silicon flying across the room

ps. forget about the thick wire. Use litz wire / a bunch of thin wires in a multifilar winding instead if you insist on building your own magnetics.

leodahsan

Sat Jan 22 2011, 01:05AM

Registered Member #3638 Joined: Fri Jan 21 2011, 11:56PM
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Posts: 10

@hboy007
Thanks! but.. 60-80A at 11v thats 660-880W input! I only need 400W output and do not want extra 260-480W extra as heat (only in winter lol). 2 Transformers are a little heavy arent they? Thanks again.

@Patrick
Yes, 247.6 (replace the commas with dots :P)
well you say R is 10 ohms? The ESR of the caps? They are all parallel, didn't tested the R but it is lower I think..

Patrick

Sat Jan 22 2011, 01:16AM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Hold on let me see about the math.

leodahsan wrote ...

@Patrick
Yes, 247.6 (replace the commas with dots :P)
well you say R is 10 ohms? The ESR of the caps? They are all parallel, didn't tested the R but it is lower I think..

the charging circuit you will build will have resistence, this resistence will be in series between the Vsource, and Cbank, thus you will have an RC circuit, and a time of charge will be decided by that RC number. (IE.. V will go from 0% to 99% in 5RC's.)

oops! i did the integration wrong, [(Q=CV[1-e^(-t/RC)] it looks like youll need 500W for 0.2 sec, then 240W for the 2nd RC and so on, so you can get 500-700 watts out of ferrite. however you will still need to be certain your total charging impedence is 166 ohms or less,
(if Vb=800, R=166, I=about 0.8 amps into Cb=1180uF then your caps will charge in 1 sec or less.)

leodahsan

Sat Jan 22 2011, 01:23AM

Registered Member #3638 Joined: Fri Jan 21 2011, 11:56PM
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Posts: 10

Sorry I'm half asleep too. You say that the time needed for charging depends on the caps resistance too?

Patrick

Sat Jan 22 2011, 01:52AM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

If you dont understand linear math/passive components, and the the non-linear math related to C's and L's and RCL's then you need to look at this website...

leodahsan

Sat Jan 22 2011, 01:57AM

Registered Member #3638 Joined: Fri Jan 21 2011, 11:56PM
Location:
Posts: 10

Thanks! that helped a lot.
But I still need someone that help me draw a inversor circuit.. if possible of course.

Dr. ISOTOP

Sat Jan 22 2011, 02:11AM

Registered Member #2919 Joined: Fri Jun 11 2010, 06:30PM
Location: Cambridge, MA
Posts: 652

H-bridge inverter, maybe?
If you use rectified high frequency AC, you can use a lightweight ferrite transformer.

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