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LM317 Variable power supply

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Nice attempt at math, audioguru.

(and before you say it, NO, I do not disagree with the manufacturer's data sheet)

This is what I refer to in my post about crappy design techniques.
I can get a half amp out of this regulator easy without a heat sink. (and using the manufacturer's design parameters).

MP

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MP,
Certainly you can get 0.5A from an LM317 in a TO-220 package with an ambient temperature of 25 degrees C, without a heatsink.
Just use a regulated input voltage that is 3.0V (no less) to 3.99V (no more) than its output voltage!

Since it will be operating at its thermal limit, it probably won't last long.
That is a crappy design technique.

Just use a reasonable input voltage and put a little heatsink on the thing.

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audioguru, you seem to be agitated. I design products for the real world, not as a hobbyist, so I can tell you that proven application backs up what I have told you. I see this attitude a lot in the hobbyist world. A guy wants to make sure he will get all the current he needs for a project, so he buys a BIG transformer just to be sure. But he is not getting more current by doing this. In fact, he is just dissipating the excess voltage as heat. That regulator is not turning that extra voltage into current. The extra is turned into heat. Nothing else. No other benefit.
If the regulator spec sheet says you need 3 volts more input than output, there is no reason to give the regulator more.
You save money with a smaller transformer and you save money with not having to purchase a heat sink.

I am sure you will not be convinced of anything here, but since others will also read this and can learn, let's start with the basics:

The regulator gets hot because it regulates the energy delivered to its load. A linear regulator drops an input voltage to a lower, regulated output, while conducting a current nearly equal to the load current. It must dissipate the excess energy, which is:
Pd = (Vin - Vout)*Iout (Pd is in Watts)
We see from the spec sheet the regulator needs 3 volts. We calculate Pd=(3volts) * 1/2 amp, which is 1.5 Watts.
This energy is dissipated in the form of heat - the regulator die heats up above the ambient temperature with a temperature rise that's proportional to the power dissipated, and to the thermal resistance from inside the die to the ambient environment. This formula is:
Trise = Pd*ThetaJA (ThetaJA is in degC/Watt). For this regulator, ThetaJA is 50 degrees. Thus, Trise = 1.5watts * 50, which equals 75 deg. C. This regulator has a MAX of 125 deg C. in which it shuts down to protect itself from heat over load. Our spec is a little over half of the manufacturer's Max spec. I am not worried about problems with this temperature. It will be hot to the touch and will burn your finger, but it is well within spec.
(by the way, 1 full Amp only brings you a little above the spec. Only a small heat sink needed)

I hope that others can take note of what I am saying here for their design work. You save money not having to buy a heat sink. You save money not having to buy a bigger transformer. And you are not generating a lot of heat that might affect other things in your project. You also benefit from having a smaller package in your design with no bulky heat sinks added. Good design work.
Now you have the formula. Do the math and determine if you really need the heat sink. Think twice about that big transformer.

MP

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Sure MP,
But you forgot that the IC's die heats 50 degtees C above the ambient temp of a nice and comfy 25 degrees, so the die's temp is 100 degrees (not 75) which is pretty close to its limit. Don't enclose it!

Are you going to make a custom transformer to give exactly 3.0VDC above the output voltage? Hobbyists just pick one off the shelf.
A stock transformer doesn't cost more to have a few volts more output voltage, and certainly isn't any bigger.
Transformers have a tolerance and so does the mains voltage. Will your LM317 shut-down when they are both on the high side?

Are you going to use a huge input capacitor so that additional input voltage is not needed to allow for ripple? Hobbyists just use a reasonable cap, and let the regulator reject the ripple, the way it is designed to do.

Hobbyists just use a heatsink when required, to keep down the costs of those other things and improve reliability. Maybe you should do the same. ;D

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Well, like I said....I did not think that you would get much out of it..... and I think your original comment was that it could not be done at all? You seem to be changing as we go on.
More voltage is not needed to allow for ripple rejection. As I said earlier, the added voltage only goes to heat transfer. You get no benefit from it at all.
I really spent the additional time to explain this for the others who were interested.
BTW, not special transformers to tweak the input voltage to the reg. There are other cheap ways to do this also if it is needed. It is all part of the good design process.

MP

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hello...i have read all the comments adn i have some questions:

1- i want to design a pSU with 1.2-30V output at the maximun load possible, so what can i do to improve the ripple rejection?, more filter caps?

2- if the soulutio is to use more filer caps, i must use a bigger bridge, do not I?

3- if i use a 15 + 15 seconadary transformer, i can use only one side of it if i need a voltage in the ouput between 1.2 - 15V, and the two sides of the transformer if i need more than 15V, i think that it can improove the current output, but how can i do a voltage detector or selector or something like that?....

OK Thanks!!!!

:)

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Emiliano,

1.Yes bigger filter capacitors will improve the ripple rejection but so will a series inductor (choke).
2.The inrush current will increase with bigger caps so a bigger bridge is necessary.
3 It is possible to use one half of the transformer for a

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Hi everyone! I saw there are several different projects for power supplies and I'm not sure what the real differences between them are. In particular, what is the difference between using a high current regulator (say lm317 or lm350) and using a low current one (lm723) with external transistors to boost the output current? Which design is cheaper?
Also, what's the easiest way of getting +/- output voltage with the same current requirements? Do I need to double the trafo voltage or can I just flip the input to the regulator with a switch?

Thanks in advance!
ffrige

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You cannot just flip the input to get a negative voltage if you are using a regulator. You will need a positive regulator for the positive voltage and a negative regulator for the negative voltage. I hope I did not misunderstand what you meant. If you mean a switch to use the same AC voltage off the secondary side of the transformer to go to either of the two regulators, then yes this could be done.
The power supplies which use a transistor to boost current will give you more current output but at a little more expense obviously because you need more parts.

MP

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MP, thanks for the answer and sorry if I was confusing. I guess I didn't want to use two different regulators since I don't need positive and negative voltage at the same time. But I need a switch to be able to flip the output somehow. If I can't do that with the ac voltage using only one regulator, can I maybe do that directly on the output? What kind of relay would I need to do this automatically?

Thanks,
ffrige

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The higher the power dissipated by the regulator, more the ripples at the output.

It is what I learn from real product design.

So my experience is the input voltage usually 20% more than the specification required in order to give that 20% safety margin. It will not cause much heat generated.

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The 317 can also current limit.

This feature could be added to this simple yet elegant circuit.

http://www.electronics-lab.com/projects/power/010/index.html

The math is R=E/I

"The Key feature of the LM317, and its family, is that the regulation voltage between the

post-6155-14279142041107_thumb.jpg

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So... I finally managed to finish my variable power supply, but unfortunately it didn't last long. I used it to drive a DC motor and, as you guys suggested, I simply reversed the output wires to change direction. After just a few switchings the LM317 croaked. I imagine some inductive spikes from the motor were too strong for the poor regulator. (although I thought it had an internal protection...)
What can I do to prevent this from happening again? The thing is that the output from the LM317 does not go to 0V ever. Which means that even reducing to minimum the voltage applied to the motor, this will always go nuts with a sudden input reversion. Do I need to disconnect the motor from the regulator before switching the polarity?

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i agree,you should use diodes regardless of the application. use a diode with a peak inverse voltage (PIV) of at least twice the max output voltage. 1N4002 or hgher, if you plan on alot of transients and spikes, a schotty diode (SB180) may be a better choice. also the lm317 series and most adjustable regulators do not drop to 0V

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Hi Enigma,

The regulator should also be backwards bypassed by a diode to play it safe. If the voltage is set to 5volts and you connect a cap that is charged to 15volts for example you can blow the regulator. Another scenario could be while running a small motor at high rpm suddenly turn down the output voltage the emf from the motor can kill the regulator. But I am sure you knew this! ;D

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I certainly wasn't thinking on that one. :)

In my mind's eye, I saw the diode being implemented at the motor in the same way as a clamping diode, instead of in the usual manner for regulator input shorting protection.

More coffee is required.

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Hey, i'm building the variable power supply using an LM317T as well. I only need the output to be 1.5 - 12V. I've got a couple of questions.

1. Using the 2200uF capacitor in the circuit, would the inrush current blow the 2A rectifier?

2. If i used a bigger capactior, say 4700uF, what would i have to change the rectifier because of a bigger inrush current?

3. What secondary voltage should i get for my transformer, since my output is not going to go above 12v.

4. What size of fuse would i need for the primary side of the transformer?

Cheers people.

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