24VDC to 12VDC using LM317T questions

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by qsiguy, Sep 10, 2007.

1. qsiguyMember

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I need to reduce 24VDC to 12VDC. The load draws <.6 amps @ 12VDC. The LM 317T is supposed to handle up to 1.5A. I made the simple circuit with a couple resistors and have the desired voltage in/out voltage but the LM317 is getting too hot still even with a heat sink on it. I reduced the load to about .319 amps and it still gets hot. Is this because I am dropping the voltage so much? Is there any way I can keep the temperature down by reconfiguring the circuit? Possibly different value resistors in different spots? Below is a diagram of what I put together, R2 is 1.2K ohms, and R1 is 135 ohms. If I put a higher resistance on R1 and drop R2 would that make any difference? I know I can achieve the desired output voltage with various configurations but would like some input.

Any tricks to parallel two or more LM317's together to get more current capability? Also, if anyone has another way to drop 24v to 12v reliably feel free to post it. I'd like to keep it simple and inexpensive. What I'm doing is adding a small neon transformer/lights made for PC's to an industrial control panel with a build in 24V power supply. I use them in some other panels I make that can use 12v throughout but I have some where the primary voltage requirement is 24v and I don't want to have to add another power supply.

2. audioguruWell-Known MemberMost Helpful Member

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The LM317 has 12V across it and up to 0.5A through it.
Simple arithmatic calculates 6W of heat. Not much.

A black aluminum finned heatsink isn't huge if thermal grease is used and the heatsink is in free air or has a fan.

But if the "heatsink" is just a piece of tin can and is inside a box then it won't cool the regulator.

3. qsiguyMember

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I'll have to check the actual current on the output. The current tests were done on a 12volt supply without the regulator being used to determine what was required. When I run the Neon light with only one bulb (can use 2) the amperage was only .319 straight from a 12v supply. When I have it running through the regulator with a 24v supply, dropped to 12volts into the neon running one bulb the LM317 gets too hot for me to keep my fingers on it. For a temp heat sink I have 4 - 1" diameter washers bolted to it and they all get quite hot. I'm sure a finned heat sink will be better but it gets "too hot to handle" in less than 1 min. of operation so I am doubtful that a different heatsink will make that much difference with the exception of a really large one.

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5. audioguruWell-Known MemberMost Helpful Member

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A heatsink is aluminum with fins to transfer the heat to the air.
You don't have a heatsink, you just have a bunch of washers

Get a real heatsink to cool the regulator..

6. qsiguyMember

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lol, ok. Did some actual output readings. 12V/3.77W/.318A as measured with a digital power analyzer in series with the output. Temp got up to 147*F after about 2 min.......I'll pick up a proper heat sink before my next bench test. I'm sure I have a few laying around, just have to find one.

Thanks for the input. BTW, is this the best/simplest way to accomplish my goal? Seems like it but want to make sure there isn't a better way. Assuming the proper heat sink does the job I doubt I could make it work for less \$\$ but if there is a more reliable way to do it a cheap price isn't the main determining factor for this. I do have some leeway on the budget. Any comments?

7. audioguruWell-Known MemberMost Helpful Member

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The 12V regulator needs an input voltage of at least 14.5V. Your input voltage is too high and causes too much heat in the regulator.

You could add a power resistor in series with the regulator to drop 9V (28.3 ohms, use 27 ohms) so the resistor will dissipate 2.9W and the regulator will dissipate 1W without a heatsink. If it is in a TO-220 case then its max allowed dissipation without a heatsink is 2W.

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10. Hero999Banned

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Did you check the degree C/W rating of your heatsink? It should be adequate to stop the die from exceeding the maximum rated temperature and shutting down.

11. qsiguyMember

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So my original theory of too high input voltage creating the heat is the issue. I did get a proper heat sink and it still gets way too hot. kchriste, that part you suggested is actually a really good option. Off the shelf and ready to go. I actually did some checking and there is a replacement version of the that Mouser stocks. Here is the datasheet

http://www.electro-tech-online.com/custompdfs/2007/09/pt78st212.pdf

At \$24.45 from Mouser it's a great turnkey option for me.

As far as making my own circuit work, what if I stepped down the voltage gradually using 3 LM317's in series? Drop the voltage by 6 volts each step. 24->18->12. More parts but in theory that would also work, correct?

Jospfh, I've seen the diagram you link to. I don't quite understand how the regulator keeps the output voltage where it needs to be with a bypass transistor going around it. Can someone clarify this circuit.

12. Hero999Banned

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Have you measured both the input and output voltage?

If you're getting 12V on the output then I wouldn't worry about it.

What are you using it to power?

Have you considered upping the output voltage slightly?

If your device is designed to work from a lead acid battery then it should work from 13.8V with no trouble.

13. ericgibbsWell-Known MemberMost Helpful Member

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hi

>> Jospfh, I've seen the diagram you link to. I don't quite understand how the regulator keeps the output voltage where it needs to be with a bypass transistor going around it. Can someone clarify this circuit.

Does this help?

Last edited: Jul 7, 2008
14. qsiguyMember

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Power is from an IDEC 24VDC control panel power supply which is what will be used in the panel this regulator would be used in. I did try to take the voltage up to about 14 and it still got too hot. Somewhere I read to keep the voltage differential below 7 volts which would make my output 17 volts and I'm pretty sure that would fry my neon transformer.

Thanks for that explanation Eric, makes more sense now. So I assume that the capacitor on the output is much more critical in that circuit than it would be in a basic regulator circuit since there will be some oscillation?

What about the "daisy chain" option to get the voltage down or running two or three in parallel? Can the LM317 be ran in parallel?

15. kchristeNew MemberForum Supporter

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The problem is that the total heat will still be the same. ie: you'll need 3 heaksinks the same or bigger than the one you are already using so you are not getting anywhere. You are just spreading the heat around.

That'll work great and run a lot cooler than any linear regulator such as the LM317s.

16. ericgibbsWell-Known MemberMost Helpful Member

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hi,
You could, if there is no other more efficient way, use a dropper resistor in series with the +24Vdc and the LM317 input.

A rough calculation.
Drop from +24V to say, +15V [ leave 3V across the LM317]
To drop 9V at 0.6A, [9/0.6], requires a 15R resistor, watts = 9V * 0.6A = 5.4W, [use a 10Watt] resistor.

So dissipation across the LM317 3V * 0.6A = 1.8Watts.
A small heatsink would keep the LM317 happy.

I would choose to take some turns off the transformer secondary winding, if thats possible..
so you get about +16V from the FWB/smoothing that drives the LM317

Last edited: Sep 12, 2007
17. mvs sarmaWell-Known Member

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Better have a pre-regulator to bring 24V to 15 or even 14.5V as suggested by AudioGURU, perhaps by using another LM371, therafter the final LM317 will take care of the 12V at 1.5Amps.

However, you need to mount both with suitable Heatsinks.

the best is DC-DC converter. with a series pass transistor. You may try with a rather old chip like MC34063-- the datasheet gives applications and suitable layouts also.

Last edited: Sep 12, 2007
18. Hero999Banned

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I agree, a pre-regulator is probably the most ideal solution, it doesn't have to be fancy, the Black regulator previously posted would be ideal. When a linear regulator is added to a switching regulator it forms a hybrid regulator which is often used in large high-quality switching power supplies.

19. mvs sarmaWell-Known Member

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I meant either a set of LM317s or a dirct DC-DC converter considering the current demand of around 0.6 amps.

black regulators appears OK but for component count. please see the datasheet of MC34063A. perhaps the OP could complete the work with reduced components.

Last edited: Sep 12, 2007
20. ericgibbsWell-Known MemberMost Helpful Member

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hi Sarma,
If its a series voltage regulator, series transistor, bypass transistor or 10Watt resistor the 'excess' power is always going to be dissipated as heat.

The 10W resistor is the cheapest and simplest option.

Regards

21. mvs sarmaWell-Known Member

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in the case of series resistor, the input voltage available for LM317 will change as per the load current.

In view of power economy, i feel the switch based solutions are efficient and perhaps may not be so costly. with an additional 5 or 6 components the job is done.
Datasheet of MC34063 attached.pehaps there are cheaper Switches available off late.

as the power requirement is more than the capacity of this IC , an additional pass transistor may be required.

Last edited: Jul 7, 2008