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Boost converter 12V to 36V (25 mA load) using 555

KrisBlueNZ

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That's great! Well done!

The MOSFET is not critical. Assuming you want a compact through-hole part, an excellent and very cost-effective option from Digi-Key is the Fairchild FQU13N10L at USD 0.71. Ratings are 100V, 10A, 0.18Ω, IPAK/TO-251AA package: http://www.digikey.com/product-detail/en/FQU13N10LTU/FQU13N10LTU-ND/1053609

The next better device would be the International Rectifier IRLU3410 at USD 1.05 which has an RDSon of 0.105Ω instead of 0.18Ω which would give only a fraction of a percentage point of improvement in efficiency: http://www.digikey.com/product-detail/en/IRLU3410PBF/IRLU3410PBF-ND/812424

That circuit is not a good basis for building a high-power boost converter. Not at all! I would look at an IC solution from a company like Texas Instruments, Linear Technology, Micrel, Fairchild or STMicroelectronics. Also look for application notes from those companies that will tell you important things about inductor/transformer design, MOSFET selection, and PCB layout.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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For higher power, you normally use a transformer instead of just an inductor.

I would advise trying to find an old PC power supply and grabbing the transfomer out of that. If you get one with more than one 12V output (these are pretty common these days) then you can drive a push pull circuit using these windings and take your output from what was the primary winding.

Alternatively you could get the transformer from a UPS.

If your soldering iron is just a dumb heating element, I'd go for the SMPS transformer and deliver something like 220VDC to the iron. If it's smarter, you may need to generate 50/60Hz AC. That's not as simple.

Either way, the driving is going to be different and the potential danger (both to yourself and the components) will be at least an order of magnitude higher. Note that 220V AC or DC could easily kill you.
 

abuhafss

Aug 3, 2010
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Okay, got it.

Just another query, can this set-up be modified to get an output 300-400VAC @ few mA for motorcycle CDI?
 

KrisBlueNZ

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Possibly, but CDI needs a lot more than a few mA...
 

abuhafss

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Also with the original design, the ON time of Q1 is too long for a 100 µH inductor at 12V; with C1 = 2n2 the ON time is around 16 µs, and the increase in inductor current can be calculated from:

dI = dT V / L
= 16×10-6 × 12 / 100×10-6
= 1.9A

That's far more peak current than you need; for a 25 mA load and a step-up ratio of 3, your peak inductor current should be about 300 mA.

How do we know the peak current for inductor for a given load and output voltage?
Or simply, how did you worked out 300mA?
 

KrisBlueNZ

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The peak inductor current is roughly equal to:

IL(pk) = 2 × IOUT × VOUT / VIN.

Do a worst-case calculation with minimum VIN and maximum IOUT and add at least 50% to the calculated peak current, and use that to specify your inductor.
 

KrisBlueNZ

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As I said in post #81, that circuit is not a good starting point for a high-energy, high-voltage converter. Yes, you should start a new thread. In the thread title, mention boost converter for CDI.
 
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