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17th March 2009, 06:08 PM  #1 
diyAudio Member
Join Date: Jul 2008
Location: Sofia

CFL inverter
Hi,
This is my CFL inverter and he works fine, except¡K.. (of course nothing work fine without exceptions ƒº) The temperature of the transformer is 75¡¦C. Is that normal? I tried with large wire diameter. Nothing. His dB swing is 400mT (for Bmax= 200mT) Np = (3.14/2)*(0.707*Vin*3.14)*(20uS /2) / (Ac*dB) Ac = 0.83cm2 DB = 0.4T IM I wrong or missing something ? 
17th March 2009, 10:27 PM  #2 
diyAudio Member
Join Date: Apr 2006
Location: Connecticut

FWIW: I am not a CFL expert. Off the top of my head, I would guess you have a flux walking problem that is leading to saturation. You are using a pushPull circuit to drive your transformer. If there is small difference in inductance or coupling between the two halfs of the primary it can cause the transformer to walk in one direction leading to saturation. You should be able to determine which half is walking into saturation by comparing the current levels between each half (or perhaps by simply measuring which transistor is hotter).
You can correct this by either adding some deadtime, changing topology (halfbridge or Fullbridge), or using a PushPull controller that can correct fluxwalking problems. 
18th March 2009, 04:44 PM  #3 
diyAudio Member
Join Date: Jul 2008
Location: Sofia

This is a current fed converter and they normaly do not have flux walking. I wound a quite unbalanced traf  the temp is 75C! I tryed with free runing caps from 150nF to 2nF  same thing. I wound a traf on p36x22 (Ae = 2cm2) with dB = 200mT (Bmax = 0.1T)! Now the temperature is 65C. I thing that this is becouse he have bigger surface!

26th June 2009, 05:39 PM  #4 
diyAudio Member
Join Date: Jun 2008
Location: Dhaka

Hi,
Honestly this is a poor design. You could improve the circuit by adding a controller IC. For the CFL Inverters that I made, I used SG3525 with MOSFETs in pushpull topology(no feedback/voltage regulation needed) switching a ferrite trafo at around 70kHz with the output rectified to DC using ultrafast rectifiers and filter caps and inductors. Hope this helps.
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1st July 2009, 08:11 PM  #5 
diyAudio Member
Join Date: Sep 2007

The reason why the transformer gets hot is the huge flux density. Check the core data sheet  there are graphs that help in defining the core losses at given frequency and flux density. Basically you can reduce the flux by adding more turns, but this brings other concerns.
Then again, if the transformer stays at these temperature levels, I wouldn't worry too much  it is still at relatively safe level. When the core heats up, the loss density within the core reduces  less core losses with the same flux density.
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Juha 
4th July 2009, 06:17 AM  #6 
diyAudio Member
Join Date: Jul 2008
Location: Sofia

The wire is hot, not the core.
I make a couple of another designs. This is a voltage feed case with IR2156 and variations with E25 and E32 cores. In second case the core temp is less, but the wire is still hot. Secondary wire dia = 0.36mm Primary wire dia = 0.6mm Wound: First is Secondary in 3 layers, insulation, then Primary in 2 layers. 
4th July 2009, 06:20 AM  #7 
diyAudio Member
Join Date: Jul 2008
Location: Sofia

Picture on www.efob.hit.bg
It is in bulgarian. 
4th July 2009, 06:26 PM  #8 
diyAudio Member
Join Date: Sep 2007

Then you have to choose a wire with greater diameter. Usually a value 3 A / mm^2 is used  but you should still calculate the wire resistance (first calculate the dc resistance, then find out the ac resistance, check wikipedias article on Proximity effect). You should also be aware of the coil current rms values.
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Juha 
4th July 2009, 08:43 PM  #9 
diyAudio Member
Join Date: Jul 2008
Location: Sofia

Hi Juha,
There are my quotations: J = I/S S =I/J = 0.4A/3 = 0.13 D = sqrt(4*S/pi) = 0.41mm D(skin) = 65/sqrt(f) = 65/sqrt(45kHz) = 0.3mm Last formula is from IRs AN1038 So I pick up some wire between these values – 0.36. 
5th July 2009, 05:34 PM  #10 
diyAudio Member
Join Date: Sep 2007

Yes, these calculations seem to be ok. Only that commonly the area is denoted with 'A' instead of 'S'. 'S' means Siemens, that is unit of conductance (inverse of resistance, ohm).
It seems that your current will be 0.4 A. Now find out the mean length per turn for your transformer  then you can calculate the dc resistance: Rdc = (rho*N*MLT)/Ae, where rho = resistivity of copper (@ 100 degrees celsius), N = number of turns, MLT = mean length per turn, Ae = wire area.
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