Power Regulator with LM317

Discussion in 'The Projects Forum' started by carlosgoac, Nov 16, 2010.

  1. carlosgoac

    Thread Starter New Member

    Nov 16, 2010
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    Hi everyone,

    I'm trying to make a voltage regulated power source to test some other circuits i have to make, first as my instinct said I built one and then used a simple voltage divider with a potentiometer to regulate the output voltage, it worked fine until I connected a load and it just died and also burned a few potentiometers in the process, then found some literature on the LM317 to build a simple power source just as I want so I bought one and downloaded the data sheet. Found in that data sheet a voltage regulator, it's not from 0 - 30V like I want but it was from 1.2 - 30V and I could definetly live with that. Built it with the same components as the data sheet and it still died when I connected the load, thought about the capacitors so I connected at the output of the rectifier a 3300uF to have a heck of a filter. The circuit works fine without load but when I connect it it doesn't die but it's really diminished, have been using 6V to make some tests with LEDs and when I connect them the output voltage goes down to 3.2V.

    Any suggestions?

    Cheers!
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    The venerated LM317 has been around for many years, but that doesn't mean it's the best solution for a voltage regulator.

    The problem is power dissipation. If the unregulated voltage supply is within a few volts of the output, the power dissipation isn't too bad. However, if you have a 35v input and are trying to output 5v at high current, most of the power gets dissipated in the regulator as heat.

    This is usually pretty difficult for "electronic noobs" to understand.

    Let's just say that you have an unregulated input of around 35v that will stay at that voltage no matter what.

    Let's also say that you have a load that requires 1A @ 5v, just to keep things simple.

    An LM317 would have to dissipate (35v-5v)*1A power, or 30 Watts, while your load would receive only 5v*1A = 5 Watts power. Your LM317 would go up in smoke unless you could immerse it in liquid nitrogen; even then it would be hard-pressed to dissipate all that power.

    I have an ancient Lambda linear supply that I modified to go from 0.25v to 54v @ 2.3A. It's inefficient as the dickens. It has a HUGE heat sink. It was made to military specifications. I love it for it's clean power. But, what you need is a switching supply.

    You're not going to get there with an LM317.
     
  3. windoze killa

    AAC Fanatic!

    Feb 23, 2006
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    I dissagree Sarge. We don't know what his requirements are. An LM317 maybe perfect for what he is doing. he may just be hooking it up wrong.

    OP. How about posting your input and output requirements and the type of load you are putting on it. You maybe doing something very simple that is wrong. It actually sounds like your input current may not be high enough. Can you post your schematic?
     
  4. Markd77

    Senior Member

    Sep 7, 2009
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    Have you got it bolted to a nice big heatsink?
     
  5. carlosgoac

    Thread Starter New Member

    Nov 16, 2010
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    Thanks for your replies, it's a power source in which I should be able to control the output voltage, according to the LM317T specs I can get up to 1.5A but my circuits are very simple and don't really need much current because they are made to use just the right amount of power for I'll be using batteries but as charging and discharging them takes too long to make tests I need this power source to speed up my tests.

    Right now I'm using a load based on LEDs and resistances that consumes about 0.5A, also an OpAmp and some other resistances and passive circuitry, the whole load doesn't even get over 0.5A.

    I want to paste the image of the circuit but it asks me for a link, any help on that too?
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Below the reply box, you'll see a button that reads "Go Advanced". Click on it.
    On the next screen, down near the bottom, you'll see a "Manage Attachments" button. Click on it. (if you have a popup blocker installed, have it always allow popups from this site.)
    Then click on one of the the "Choose file" buttons, and navigate to where the image is stored on your local system, and click on the file to attach it. Then click the "Upload" button.

    .png files are the best for schematics and other line-drawing type images; they are compact and require no extra software to view. .jpg is good for photos, but poor for schematics, as it is a "lossy" format. Avoid uploading .bmp images, as they are large and require the use of external software. .pdf files are OK, but also require external software.
     
  7. wayneh

    Expert

    Sep 9, 2010
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    As Sarge has detailed, dropping a high voltage across the LM317 will limit its current handling due to power dissipation. Without seeing the details of your power supply, it doesn't surprise me at all that you're having trouble squeezing even 0.5A thru. The LM317 may be cutting out on heat, to protect itself, or your supply may simply not have enough oomph to do what you want. When you're seeing your voltage drop below the regulated setting, have you checked the supply to the LM317 under those conditions? I'll bet it's lower than you think.
     
  8. Audioguru

    New Member

    Dec 20, 2007
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    The datasheet for the LM317 shows that its output current is 1.5A minimum (2.2A typical) only when the voltage from its input to its output is 15V or less. When its input to output voltage is 40V then it reduces its output current to as low as 150mA minimum or 400mA typical. The LM350 and LM358 also reduce their output current when they have plenty of voltage across them.

    They do not burn up, they simply shut down if they get too hot. When they cool enough then they begin working again. On and off many times like this will probably cause failure due to thermal and mechanical fatigue.
     
  9. carlosgoac

    Thread Starter New Member

    Nov 16, 2010
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    Thanks, here's the schematic.
     
  10. carlosgoac

    Thread Starter New Member

    Nov 16, 2010
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    Actually that was the first thing I looked up, but when the load is connected at the input is just the same as if it wasn't, about the supply it shouldn't have any trouble because I'm connecting it directly to the power outlet to a 127/30VAC transformer, from there the circuit I uploaded begins.
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    R1 is really too large. In order to provide guaranteed output voltage regulation, R1 should be 120 Ohms, as it will then have at least a 10mA current flow.

    If your transformers' output is 30VAC, then the voltage on C1 will be about 40v DC with no load. At about 400mA load, C1 will have about 1v of AC ripple on it.

    If the LED being tested has a Vf of 3.2v and no resistor in series with it, with 400mA current flowing, then the input-output differential of the regulator will be roughly 39.5v - 3.2v = 36.3v, and the regulator will be dissipating 36.3 * .4A = 14.52 Watts of power, while your LED will dissipate 3.2v *.4A = 1.28 Watts.

    The LM317 will very quickly overheat and shut down, even if you have provided a rather large heat sink for cooling.

    I've attached a schematic of a switching buck-type regulator I worked up awhile back. It has an output range of 1.5v to 30v at 2.5A, and is quite efficient. You need something like this instead of a linear regulator.
     
  12. Audioguru

    New Member

    Dec 20, 2007
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    1) Your 30VAC transformer will make rectified and filtered 41VDC(!) input to the LM317, not 30VDC.
    2) The LM317 needs the resistor from its output to its ADJ pin to be 120 ohms or less (not 1k) so its output voltage does not rise without a load.
     
  13. Audioguru

    New Member

    Dec 20, 2007
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    I bought some mains-input Name-Brand 5V/2A tiny switching power supplies at a surplus electronics parts store for $1.25 each. The label says they have a voltage adjust pot inside that has a wide range but I didn't cut one open yet.
     
  14. SgtWookie

    Expert

    Jul 17, 2007
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    I bought three PWD-80D Nemic-Lambda +/-15v and +5v switching supplies for $1.50/ea at a local electronics surplus store. The power cords cost more than the supplies, let alone the enclosures and binding posts.

    If our OP wants something other than a room heater, they'll have to go to a switching supply.
     
  15. windoze killa

    AAC Fanatic!

    Feb 23, 2006
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    OR a lower voltage transformer.
     
  16. hobby16

    Active Member

    Aug 30, 2010
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    I remember when I used my first LM317 (years ago, lol). Had the same type of problems. After many trials and false diagnostics, realised I had the pins connexions pitifully wrong, maybe because I'm so used to the LM78xx regulators pinouts.
    So I suggest you check, recheck, and re-recheck the wires ;)
     
  17. carlosgoac

    Thread Starter New Member

    Nov 16, 2010
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    Thanks!

    In fact when I measure C1 it's 38VDC, I'm using a range between 5-7V to polarize the LEDs with a resistor of 220 Ohms in series with each LED.

    At the end what I really want is the power source I can control with Rvar from 0 - 30V to make tests with serveral circuits I'm designing.

    So I have several questions because I really don't understand why the rectifier bridge increases the voltage when it makes the conversion from AC to DC and how can you calculate it because for what I know due to the diodes the voltage should be 30 ideally but a bit less on the field because they need the 0.7 or whichever voltage (according to the semiconductor) and it should be decreased from the actual value not increased.

    Next, I just started with the LM317, this is the first time I work with it and I don't quite get how to calculate the value of the resistors for the Adjustment terminal, I saw there's a formula in the data sheet to obtain the Vout which is Vout = 1.25(1 + R2/R1) + I_Adj(R2), is that the one I should use? Because right now I'm making it with one fixed resistor of 220 Ohms and then a potentiometer, when I get the Vout I want I just measure the value of the potentiometer and then look for a commercial value resistor near to that value I got.
     
  18. carlosgoac

    Thread Starter New Member

    Nov 16, 2010
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    And what about that 1VAC ripple you mentioned?
     
  19. Wendy

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  20. #12

    Expert

    Nov 30, 2010
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    AC voltage is usually labeled as its RMS value. The peak value is RMS times the square root of 2, or 1.414. 30 radical 2 = 42.426 volts peak. From that you subtract the diode loss and you will have the voltage that will accumulate on your first capacitor if there is no load.
     
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