Is a LM335 a direct replacement for a LM317T? - diyAudio
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Old 17th April 2010, 08:18 PM   #1
phrarod is offline phrarod  United States
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Default Is a LM335 a direct replacement for a LM317T?

I changed my filament supply from a 12V to a 6V and the 317T in there is overheating and can't regulate the voltage. I know the 335 is higher wattage.

Is it a direct replacement?
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Old 17th April 2010, 08:23 PM   #2
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You mean LM338
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Old 17th April 2010, 08:30 PM   #3
Bill_P is offline Bill_P  United States
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If the LM317 and LM338 are in the same package (TO-220), the heat dissipation capability is the same for both. To improve the power handling, add a heatsink or put a power resistor in series with the input to take up some of the wasted power. How much current do you need?
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Old 17th April 2010, 08:42 PM   #4
phrarod is offline phrarod  United States
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It has a large heat sink. I was able to adjust it down after changing one resistor off the LM317T divider network. Problem is as it heats up the voltage drops and I can't adjust it anymore.

I'm running separate supplies for each tube so the current should be minimal. One 9 pin 6922 style for each fil supply.

So even thought the 317T is a 1.5A and the 338 is a 5A that won't make a difference?

Last edited by phrarod; 17th April 2010 at 08:45 PM.
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Old 17th April 2010, 09:08 PM   #5
Bill_P is offline Bill_P  United States
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The 6922 only needs 0.3 Amp which is within the current rating of the LM317T. The symptoms suggest that your heatsink is completely ineffective. If that is the case, the LM338 will not work any better as it needs an effective heatsink too.
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Old 17th April 2010, 09:12 PM   #6
phrarod is offline phrarod  United States
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I guess I need to figure out what size power resistor to install on the input of the supply to take the heat off the 317T.
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Old 17th April 2010, 11:23 PM   #7
Matt BH is offline Matt BH  United Kingdom
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Whats the input voltage to your 317? A 5A reg in the same case wont help it still has to dissipate the same amount. A resistor in parallel accross the reg will help but it needs to be calculated carefuly. Is there no way you can drop the input voltage?
Cheers Matt.
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Old 18th April 2010, 12:12 AM   #8
tomchr is offline tomchr  United States
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Quote:
Originally Posted by phrarod View Post
I guess I need to figure out what size power resistor to install on the input of the supply to take the heat off the 317T.
It sounds like you have a wiring error somewhere. First off, the output voltage should not drop as the LM317 heats up. Not until it hits the thermal limit at Tj=150 C anyway. If the resistors are heating up, you're doing something wrong.

I suggest the following: Set the resistor from the output to the adjust pin to 120 ohm. Then set the resistor from adjust to ground according to the equation in the data sheet (Vout = 1.25(1+Rx/120)). You'll end up at 485 ohm for the second resistor. 470 ohm (standard value) is close enough. Vout = 1.25(1+470/120) = 6.15 V.

There will be 1.25 V across the 120 ohm resistor and negligible current flowing in the adjust pin of the LM317. Hence, approx 10 mA will flow in the two resistors. The 120 ohm resistor will dissipate about 12.5 mW (1.25 * 10E-3) and the 470 ohm resistor will dissipate 47 mW (10E-3^2 * 470). The normal 0.25 W resistor types will be just fine.

The LM317 will dissipate some amount of power. You can calculate how much if you know the current draw and the input voltage. Pdiss = (Vin-Vout)*Iout. If you dissipate more than, say, 10 W in the regulator, I'd suggest lowering the input voltage if you have that option. Otherwise, you'll need a big heat sink. Big as in several cubic inches of aluminum. A different regulator will not help you there as it'll dissipate the same amount of power and heat up to the same temperature under the same conditions.

I suggest going with the lowest input voltage you can as this dissipates the least amount of power in the LM317. However, you must have a drop of about 2.5~3 V across the regulator in order for it to work correctly. So with 6.15 V out, you need at least 9 V in.

If I interpret your first post correctly, you're applying 12 V on the input of the LM317. So the dissipated power will be on the order of 1.8 W (0.3 * (12-6.15)). You can dissipate that in one of these here heat sinks. At room temperature (~25 C), the temperature of the heat sink will be approx 40 deg C. That's slightly higher than body temperature (37 C). You should not have issues with thermal shutdown.

Hope this helps...

~Tom

Last edited by tomchr; 18th April 2010 at 12:20 AM.
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Old 18th April 2010, 12:22 AM   #9
phrarod is offline phrarod  United States
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I think everyone here is right. I have to drop the voltage at the input.

I measured and the PT is putting in 20VAC into the fil not the 13V that the board has imprinted. I measured .9A current draw which is bizarre give the draw of the tube is ~.3A Nothing is going to ground.

Right now I have a pair 7 ohm 10W on one fil supply and it seems stable and the 317 doesn't go much above 110F while the other side without the dropping resistors read 20F higher and starts losing voltage as it heats up.

Once I finish this circuit I need a more elegant solution so I don't have giant sandbox resistors hanging on the input - of course that probably means a new PT.

Thanks everyone.
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Old 18th April 2010, 12:24 AM   #10
Matt BH is offline Matt BH  United Kingdom
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Absolutely, completely agree Tom.
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