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#1 |
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diyAudio Member
Join Date: Sep 2007
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Hello!
I've had such a converter in mind for a while, but I've had problems with the stability. Now it looks better, and the converter starts taking final shape. The specifications: -Input voltage : 12 V - 14 V (car battery) -Output voltage : 25 V (maximum) -Output current : 10 A (maximum) -Output power : 250 W (maximum) -Average input current @ Pmax : ~22 A I built previously the 95-W laptop boost converter shown in Elektor magazine, but I wanted more available power. Why? Well, take a look at the new laptops - more and more power will be demanded. So this converter could be used to feed a laptop from a 12-V car battery. Or, as might be in my case, from a 12-V solar battery bank. I dare not say this converter is completed, but more closer to it. At this point I would like to have some comments on this design. Maybe there is something crucial I've missed. I tried to take many aspects into account, for example by implementing current-mode control with slope compensation. How do you feel about this?
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-Juha |
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#2 |
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diyAudio Member
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Hi
It looks very good to me, don't know if there are any errors, but you can simulate it, so it must work in real life too, you've done good |
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#3 |
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diyAudio Member
Join Date: Sep 2007
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Thanks Luka, that pdf file doesn't show the RCD turn-off snubber which I'm currently adding. LTSpice revealed that the MOSFET switching losses make more than 20 W total when averaged over one switching period! Turn-off losses make the most, approx. 16 W. I got these values with the maximum output power.
If I roughly calculate the efficiency at various output power levels, I get something like ~82 % @ 250 W, rising up to ~92 % @ 5 W. That isn't too good, I'm trying to find ways to improve it at high power levels. One trick is reducing the switching losses, then the diode losses etc. The reason for 2 switches is to reduce the on-state losses - for the same reason I'm using 2 diodes in parallel. I intend to buy some really over-rated parts from eBay to make the losses as small as possible. The goal is to have ~10 mOhm on-state FETs and 0.4 V forward voltage diodes @ Imax. Lot of work to be done!
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-Juha |
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#4 |
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diyAudio Member
Join Date: Oct 2005
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How about a synchronous design? That would lower the losses over your diodes by replacing them with a MOSFet. Or possibly using a loss less current sensing design? Use the MOSFet's Rds(on) as the sense element?
I don't know a hole lot about boost converters but those should lower the losses you are getting. |
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#5 |
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diyAudio Member
Join Date: Sep 2007
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That's correct! I've been thinking about that as well, but it would need a special circuit to drive all the FETs correctly. I happen to have almost all necessary components for a basic boost converter - that's why I'm designing this kind of converter.
I might consider using synchronous converter if I find a suitable controller for it. Thanks for comment!
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-Juha |
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#6 |
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diyAudio Member
Join Date: Jan 2003
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Have a look at the ripple current through C11... :P
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#7 |
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diyAudio Member
Join Date: Feb 2002
Location: Toronto
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If you want to greatly increase your efficiency, you will need to change the architecture. You could go with a synchronous controller replacing the output diodes with FETs. I would also replace your bipolar transistor FET driver with a high speed off the shelf FET driver.
I would also look at paralleling up some higher speed caps on the input and make sure your cap layout is more "star" like to ensure you are pulling equally from each cap. What is the switching frequency? I would assume you have minimized the resistance of your inductors? You may even want to consider winding your own to get lowest DC resistance. If you are running this plugged into a car that is on, I would suggest putting some protection against load dump on the input, some good MOVs/zener clamp. Alvaius |
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#8 | ||||
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diyAudio Member
Join Date: Sep 2007
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Quote:
Quote:
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-Juha |
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#9 |
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diyAudio Member
Join Date: Feb 2002
Location: Toronto
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Power = I * I * R. I think you will find when you look at the current on the boost inductor, it is quite high and when coupled 80 mohms will be huge...on the order of tens of watts.
Alvaius |
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#10 |
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diyAudio Member
Join Date: Sep 2007
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Correct. My DMM doesn't show resistances below 0.5 ohm correctly, so that 80 mohm was just a guess. Then again, this inductor has only a few turns, and as it has dozens of hair-thin wires wound in parallel, the total resistance must be low. I have to measure it with proper equipment.
Using 50 mohm yields about 20 W copper loss within the boost inductor @ Pmax. That is indeed way too much, and it seems to be the major source of dissipated power. Fine, still more simulation to be done
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-Juha |
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