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Old 11th September 2008, 05:42 PM   #1
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Default 250 W boost converter for 12 V car battery

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?
Attached Files
File Type: pdf circuit.pdf (33.9 KB, 374 views)
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Old 12th September 2008, 03:48 PM   #2
luka is offline luka  Slovenia
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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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Old 12th September 2008, 04:29 PM   #3
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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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Old 13th September 2008, 04:23 AM   #4
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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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Old 14th September 2008, 05:47 AM   #5
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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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Old 14th September 2008, 02:01 PM   #6
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Have a look at the ripple current through C11... :P
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Old 14th September 2008, 02:40 PM   #7
alvaius is offline alvaius  Canada
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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.

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Old 14th September 2008, 04:01 PM   #8
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Quote:
Originally posted by megajocke
Have a look at the ripple current through C11... :P
Yes, that's quite large. I intend to place several smaller (~100 uF) caps in parallel.

Quote:
Originally posted by alvaius
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 hoped I could manage with this topology, but it seems I have to consider others as well. I remember IR had some driver circuits, I'll check them. And Linear has this LT1339 which may be exactly what I'm looking for.

Quote:
Originally posted by alvaius
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.
Switching frequency will be something like 50-70 kHz. At first I thought about 100 kHz or a bit more, but then the switching losses will be too high. Yes, I have several salvaged inductors suitable for this converter. The resistances are about 80 mohms, so they are not a major source of power loss.

Quote:
Originally posted by alvaius
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
Good point, will add that.
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Old 14th September 2008, 10:28 PM   #9
alvaius is offline alvaius  Canada
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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.

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Old 15th September 2008, 10:35 AM   #10
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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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