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Please check my simple circuit for errors

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ceefna

New Member
Hi everybody, do you think this circuir will work as variable current power source. If you spot any errors or if you just think its rubbish please let me know what mods I need to make.
Thanks

Ceefna
 

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Mr RB

Well-Known Member
Hmmm... The LM117 with 120 ohm adjust resistor is a 100mA constant current drain. It will draw 100mA throught the 150 ohm trimpot, giving an adjustment of 0v to 1.5v. You need an adjustment of 0v to 1.2v to give full range adjust of the LM338s.

Looks workable, and pretty clever. Is that circuit from a design note or is it yours?

I think you should always add a 10uF caps on the adjust pin from the top of trimpot to wiper of the trimpot to stabilise the adjustment voltage.

And don't forget some real large heatsinks...
 

ceefna

New Member
Hi Mr RB, the circuit is from the 338 data sheet that I just added more 338's in parallel to give more current. I was the one with the anodising power supply thread but that doesn't seem to be getting anywhere so i thought i would just stick my plan on to get some feedback. Wish I knew what i was doing with electronics. I understand the very basic stuff but as for working out what the correct resistance on the LM117 should be??????
Can you help.

Ceefna
 

Mr RB

Well-Known Member
Whoops that's 10mA through the LM117, I said 100mA sorry.

Your resistor values don't seem too bad, the LM338 datasheet says it only needs about 50uA from the adjust pin, which typically changes less than 1uA so using 10mA for the biasing circuit seems fine.

Maybe someone else can see a problem? It's real late here and I need some sleep. :)
 

bountyhunter

Well-Known Member
Hi Mr RB, the circuit is from the 338 data sheet that I just added more 338's in parallel to give more current.
I have never tried it, it's possible it may work. It's been my experience in general that linear regs get unhappy when you parallel them becaue they try to "buck" against each other. maybe this will work because it's trying to be a curret source and not a voltage source (?)
 

ceefna

New Member
Hi, think I will build it and see what happens. Like in "Field of Dreams,build it and they will come" the amps I hope!! :D

Ceefna
 

Mr RB

Well-Known Member
I have never tried it, it's possible it may work. It's been my experience in general that linear regs get unhappy when you parallel them becaue they try to "buck" against each other. maybe this will work because it's trying to be a curret source and not a voltage source (?)

Since each regulator has its own current sense resistor they should act like emitter resistors and balance the current nicely. If you tried to do it with just 1 current sense resistor it could get ugly.
 

ceefna

New Member
Hi guys, do you think these heat sinks will be ok? Rated at 8.6 DEG C/W. I have got one for each LM338, with a 120cm fan for extra flow.
 

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Mr RB

Well-Known Member
Well even if you had a 120cm fan those heatsinks would be many times too small... (I'm guessing you meant 120mm ????)

If you're running 5A per device and they might be dropping 20v each that's 100w actual dissipation per unit x3 for the total so up to 300W in the cabinet.

When I said "real large heatsinks" that wasn't a joke. 1 deg C/W is almost in the "large" category. Think 3 heatsinks, and each is BIGGER than that. Or stay with 1 deg C/W and use some serious fan forced airflow...

As for your teensy weensy heatsink, that's suitable for about 3W dissipation, not 100W....
 

ceefna

New Member
Hi Mr RB, told you I am clueless, didn't know there would be that much heat. Will it be ok to mount the 338's off the pcb so I can get some huge cooling. Does it matter that there will be distance from resistors etc? Am thinking about mounting them to 3 cpu coolers as they are relativley cheap for 100w cooling performance. Think I will make my wife some earings for Christmas with the heatsinks I was going to use!!!!! :) Will the LM117 need such big cooling?
Thanks again for your input, without it I would have made the worlds most short lived garage heater :D

Ceefna
 

fernando_g

New Member
To emphasize what Mr RB is saying the very first step in selecting a heatsink is to calculate how much power you will be dissipating. You need four pieces of information: input voltage, output voltage and current, and ambient temperature. Worst case for all values!

Let's say, in an hypothetical example that:
Vin = 12, Vout = 5, Iout = 15 and Tamb = 35C

Then power = (Vin-Vout)*Iout or (12 - 5)*15 = 105 watt.
Since you have 3 devices paralleled, each will dissipate 1/3 of the power or 35 watt.

For reliable operation, a semiconductor junction should not be operated continuously higher then 85C.
Subtract the ambient value and you get 125 - 35 = 90C maximum temperature rise due to self heating.
Now divide that temp rise by your power and you get 90/35 ≈ 2.57 C/watt as your minimum thermal resistance requirement.

Please note: this 2.57 C/watt is the TOTAL thermal resistance...by total meaning the sum of the heatsink, thermal pad and regulator's junction-to-case resistances, In other words θt = θhs+θpad+θjc

Of course, you need to put YOUR ACTUAL VALUES in the calculations above.
 

bountyhunter

Well-Known Member
For reliable operation, a semiconductor junction should not be operated continuously higher then 85C.
Not true at all. max ratings on junction temps for most ICs is 125C and discrete transistors typically allow 150C as maximum ratings. While it is true semiconductor life reduces as temp increases, a well designed semiconductor operated at 125C will outlive most of the other electronic components in the assembly.
85C is WAY too conservative for maximum junction temp.


Then power = (Vin-Vout)*Iout or (12 - 5)*15 = 105 watt.
Since you have 3 devices paralleled, each will dissipate 1/3 of the power or 35 watt.

For reliable operation, a semiconductor junction should not be operated continuously higher then 85C.
Subtract the ambient value and you get 125 - 35 = 90C maximum temperature rise due to self heating.
Now divide that temp rise by your power and you get 90/35 ≈ 2.57 C/watt as your minimum thermal resistance requirement.


This is the wrong way to do it. You have to first know the thermal resistance from junction to sink for the transistor and calculate the rise of the junction temp above the heatsink: with a TO-3 device, it might be 2C/W total including the mica insulator. That is in series with the haetsink's thermal resistance. With 35W in each transistor, the junction temp is going to be about 60 - 70C hotter than the heatsink. If the heatsink is handling 105W total, and (for example) you only wanted the temp of the heatsink to rise 30C above ambient (which means the transistor junction would be 100C above ambient since it is 70C above heatsink), the thermal resistance of the sink would have to be 30/105 or about 0.3C/W. That is a MONSTER heatsink.

A design with 35W in each transistor is not feasible in most cases.

I will attach the text of a book I wrote for a class I taught on linear and switching regulators. Appendix B has a very good explanation of power dissipation and how to select a heatsink.

NOTE: I will attach that course document if they ever fix the stupid server here. All of that is dead, my screen has nothing but garbage on it. I posted it in another thread here a couple of weeks ago.

Let's try the attachment again:
 

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fernando_g

New Member
85C was a typo error.
If you read my calculations, it has 125 in them.

And with respect to heatsink calculations, I wanted to show the original poster what is involved, if you read my post carefully you'll note that I mentioned TOTAL thermal resistance, which is the sum of all resistances, junction to case included.
 

bountyhunter

Well-Known Member
85C was a typo error.
If you read my calculations, it has 125 in them.

And with respect to heatsink calculations, I wanted to show the original poster what is involved, if you read my post carefully you'll note that I mentioned TOTAL thermal resistance, which is the sum of all resistances, junction to case included.
I read it, but the error is that you have to address the temp rise for the single transistor (or IC) first using the power it dissipates to get the allowable temp rise of the heatsink, and then calculate the R value of the heatsink based on the total power it dissipates and the allowable temp rise for it alone. If you have multiple transistors or ICs (as this design does), you can't calculate the total resistance without doing that first. The three IC's thermal resistances are effectively in parallel in getting to the surface of the heatsink, but the heatsink thermal resistance is a single value to get to the ambient from that point.

Your value:

Now divide that temp rise by your power and you get 90/35 ≈ 2.57 C/watt as your minimum thermal resistance requirement.

Does not specify how much of the toal R value is for the transistors theta to the heatsink and how much is the heatsink to ambient. That is critical because there is 3X as much power flowing through the latter. The two values have to be specified separately.
 
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Mr RB

Well-Known Member
Meanwhile the poor OP has probably blown a gasket trying to do heatsink calcs... ;)

Ceefna, since you already do metalwork and stuff, why not go down to the metal store or scrap metal yard and just look through their offcuts of aluminium. Something in 5/8" plate about a foot square should be plenty and if it's an off cut maybe scratched etc you can get it cheap, you might get it for less than $20 which is a LOT less than 3 large "proper" heatsinks with or without fans.

Or grab an old car engine cast iron cylinder head and bolt your LM338's to that. You wouldn't be the first person to use brute force heatsinking for 300W of semiconductors. :)

Anyway you have a large messy anodising tank full of acid and other washing/dyeing tanks etc so it's not as though you need to have a tiny pretty high tech fan-forced heatsink...
 

bountyhunter

Well-Known Member
Meanwhile the poor OP has probably blown a gasket trying to do heatsink calcs... ;).
I don't know how anybody could do power calculations since he has yet to specify what the input voltage range will be.

Ceefna, since you already do metalwork and stuff, why not go down to the metal store or scrap metal yard and just look through their offcuts of aluminium. Something in 5/8" plate about a foot square should be plenty and if it's an off cut maybe scratched etc you can get it cheap, you might get it for less than $20 which is a LOT less than 3 large "proper" heatsinks with or without fans.
Depends. If the power dissipation total is in the ballpark of 20W per regulator (guess since there are no specs), that would not provide anywhere near adequate heatsinking. There's a reason they put fins on heatsinks, you have to have a lot of surface area to pass the heat to the air.
 
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Mr RB

Well-Known Member
I don't know how anybody could do power calculations since he has yet to specify what the input voltage range will be.

Because some of us have messed with anodising in our past lives and may have also read his other thread. ;) You need quite a bit of voltage overhead especially if trying to get a deep anodised layer ie "hard anodising". A 20v supply is about a minimum, 24v is probably better.

...
Depends. If the power dissipation total is in the ballpark of 20W per regulator (guess since there are no specs), that would not provide anywhere near adequate heatsinking. There's a reason they put fins on heatsinks, you have to have a lot of surface area to pass the heat to the air.

Now your sounding like a professional power supply designer and not a guy that's done some anodising. At the start of the process the load voltage is low, almost a short circuit and worst case the regulators will be dissipating maybe 20v at 5A each (that's why I said 100W each).

BUT as the anodised layer forms it acts like a resistor as anodised aluminium doesn't conduct, so the load voltage rises pretty rapidly and the dissipation in the regulators goes down a lot. After 20 mins or so its getting about done and the regulators will be dropping very little voltage at all. That's basically how you know when the anodising is done, you can't maintain the current. The system needs a pretty big heatsink but lends itself nicely to a thermal mass type (which is why I suggested a big slab of metal) since by the time the mass gets hot after 5 to 10 mins the heatsink wattage has dropped to very low value at that part of the cycle.

And the size of the slab I suggested was based on experience with a big hotplate I use here that runs at 160W and takes about 10 mins to get to 100'C, so with a thicker material like the 5/8" alloy AND the nature of the diminishing heatsink dissipation curve during the anodising cycle the size I suggested should do the trick even given the 300W starting heat. ;)
 

bountyhunter

Well-Known Member
Now your sounding like a professional power supply designer and not a guy that's done some anodising. At the start of the process the load voltage is low, almost a short circuit and worst case the regulators will be dissipating maybe 20v at 5A each (that's why I said 100W each).
I am a professional power supply designer, and I can tell you without fear of contradiction that if any of those IC's dissipate 100W for more than a few seconds, they will go into thermal shutdown regardless of heatsink (even attached to the legandary "infinite" heatsink). And when they go into shutdown, the current is cut off or drastically reduced.

Even assuming the LM338 devices are in the large TO-3 package (which is the best for heat transfer) the thermal resistance from IC to the heatsink will be about 1.5C/w. If the device did not have thermal shutdown, 100W of dissipation would spike the die temp to over 175C even if the case was held at 25C using an infinite heatsink. Thermal shutdown will kick in at 160C, which will cut off or reduce output current very quickly.
 
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ceefna

New Member
Hi guys, have a look at the heat sink I have got. That was the biggest I could afford. Going to also have a couple of fans. I have got a 24v 25A transformer, and an old ATX power supply to run fans and ammeter etc. As for how it looks you are right Mr RB, I would bolt it to a turd if it kept it cool!!!!!

Having trouble finding a 150 Ω pot can I replace with 100 Ω

Ceefna
 

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