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| Electronic Projects Design/Ideas/Reviews Are you building an electronic project or want to? Maybe you need some assistance? Come and submit your electronic questions here and let our experienced members find a solution. |
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Experienced Member
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Sorry the regulator is available from mouser.com. Are you going to cut a PCB?
At the hobbyist level a laser printer and thermal transfer is workable. At 30A I would bulk up the traces with solder since it is unlikely that you would be getting 2+ ounce copper clad. As to CAD packages I am only familiar with the professional ones that are not affordable to the common man |
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Experienced Member
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Sorry that is available from mouser. I am not familiar with the affordable PCB packages.
Are you going to make a PCB? Laser toner transfer works well enough but you should flood the 30A traces with solder since you will not be getting over 1 ounce copper clad. |
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Experienced Member
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Either a PBC made as u mention, tho i do have a friend who could do it on his CNC mill thats built for doing just this!!
Last option would be the prototype boards, with bulked up tracks as u suggest. It should be reasonably easy to get all the connections in the right place, correct?? Also what other size caps do you suggest for the filtering? |
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Experienced Member
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you lucky dog! I have often wished I had one handy to do prototypes of boards and transformers, but i have yet to put together sufficient data to convince my boss to spring for one.
The circuit is CCM and with 33uH will probably have reasonably low ripple. A 300uF cap should be more than sufficient and a 30uF would probably work. At that point it gets to keep the high frequency circulating currents contained. one across the input and one across the output, to keep the loops as small as possible. The big input cap holds up the supply during the low voltage periods of the line cycle and the regulator will compensate for the ripple on the input cap. It functions as a current regulator and following the app note I attached earlier you can find a off time that will keep the ripple current to 5% using the inductor you chose. That is the nice thing about a variable frequency switcher - it compensates so you do not have to worry about it so much. Last edited by Ubergeek63; 2nd May 2008 at 05:28 PM. |
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Experienced Member
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I found an easy to use one!, designWorks express.
Starting to draw it up now. Dont laugh, but i dont even know how some components should be fitted hehe. inductors in series with the + ? diode the same or across + and - .... . Once i get more done i will post for a good belly laugh! My friend built his little 3 axis CNC from scratch, 10 years work, including the drivers for the stepper motors. awsome little machine. he has done some very intricate work with it. |
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Experienced Member
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Heres my first attempt at designing a circuit! Most likely theres a heap of problems that will let it only make smoke as it is
With the experienced help from here im sure we can get it to work. Just let me know what to change a little at a time. |
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Experienced Member
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the AC source is not connected and C1 is shorted on the schematic, but since you are not making a PCB from the netlist I am sure you will be able to work around that.
the LD pin needs to be a voltage input from 0.5-2V for linear dimming, it actually sets the peak current. You are showing it as a simple variable resistor to GND. R1 should be terminated to the gate pin of the IC and not to GND. this changes the regulation mode of this particular chip to peak current constant off time instead of constant frequency. this allows good regulation at duty cycles over 50%. Sorry I thought I had posted the link to the app note but it does not look like I did: http://www.supertex.com/pdf/app_notes/AN-H50.pdf Early in the discussion it tells about the disadvantage of current mode regulation and explaining that constant off time takes care of it and gives figure 4. the rest of the document goes into the design formulas. Tie U1 Vin to 18V instead of 360V. The chip is capable of running that way, but it is unnecessary dissipation when you have a lower voltage rail available. The internal linear regulator is there for running LEDs off line with out extra parts. You do not say how you are obtaining 18V. A 12V transformer can easily reach 25V under no load conditions. 20% for regulation (small transformer spec), 20% for line variation under normal (not fault) conditions, and 40% from RMS to line peak. That is sufficient to blow U2 and the FET. That is all I am seeing off hand. |
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Experienced Member
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Wow, i did much better than i realised
Ive added in the supply for 18V, would i be wise to keep ground isolated from the rest of the circuit? Whats the best way to supply 0.5-2v for the LD pin? use a pot with voltage supplied from a 7805? |
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oh dear, the meter location confused me. the low side of C1 is circuit ground as well. the sense resistor goes between the FET source and GND. the output is between L3 and D2. C5, if needed (I was assuming so for noise), would be across the load.
See figure four of the app note. That would work, but you would want at least one resistor on the high side of the pot. One on the high side and one on the low side would let you set the high and low end of the pot, realizing that the part never quite goes to zero current... The data sheet lists a minimum on time of 450nS. I had forgotten this detail. Your off time would need to be calculated from your maximum rail level, your minimum current, and minimum drive voltage (giving you the duty cycle and hence the off time). The back through to make sure the inductor value is suitable. Last edited by Ubergeek63; 5th May 2008 at 11:28 AM. |
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Sorry I got it confused with something I was working on...the dimming pin is 0-0.25V I am working with a backlight inverter chip that is 0.5-2.5V dimming.
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Experienced Member
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oh dear, the meter location confused me. the low side of C1 is circuit ground as well. the sense resistor goes between the FET source and GND. the output is between L3 and D2. C5, if needed (I was assuming so for noise), would be across the load.
Im a little unsure of this, i tried to draw it up as you have said to here, but something just doesnt look right. Where does the other end of L3 go to? |
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Experienced Member
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Revision 1
Ive added the 12v and 5v supplys., changed the 18V to 12V so i can run a fan from it if needed. Moved the meter to the neg rail, as the resistance of the meter can be utilised as part of the resistance needed for current sensing, so R2 may not be needed. Still unsure of the location of the Inductors.... |
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Experienced Member
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Would you PLEASE look at the app note!!!! The output is ACROSS C5 and R2 is in SERIES with Q1 source
Last edited by Ubergeek63; 6th May 2008 at 02:16 PM. |
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DoH!!
I will get this right!! |
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Revision 2
Unsure how to calculate values of R1 and R2. Theres some real heavy (complex) looking formulas in the data sheets!! Should the -output be tied into the gnd of the rest of the circuit for a feedback loop to the HV990 ? I really appreciate all this help!!!!! |
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