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PIC Burner


napierzaza
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Does no one have any idea why the LM317 would be getting so hot? Or why the voltage woul be so low. I'm getting about 12 volts off of the two batteries in series and I've moved around the dial (P2) to see if the voltage changes but there is nothing significant happening.

I've checked all the pinning and traces. Has anyone built this? Should I used a PSU instead of the batteries?

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It sounds to me like you may have your regulator and diode wired backward.

Double check your wiring against the schematic .

Also measure the battery voltage with the batteries out of circuit. First alone (9v each) and also in series (18v total) if you get significantly less than this, you may have a problem.

I would recommend using a PSU for this EVENTUALLY, but I would fault-find using batteries first (as they are less likely to burn your house down  ;)

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I had checked them already, I did it again and they're fine. Even though the LM317 is the same part I still consulted the pinout and there are no reversals. The voltage from the batteries are not 18v, they are less. So why should the LM317 get so hot though? If it's not exactly the right voltage I don't see why it would get so hot.

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I'm not sure where you are measuring the 3v "out of" the 7805, But according to the spec sheet, the 7805 will drop out if it sees less than 7.2 volts. Also, carefully check the adjustment resistors R1 and P1 (near LM341), as these will determine your output voltage directly. Ensure they are in the correct orientation (ie that R1 is conn. to "out" and "adj" and that P1 is conn. to "adj" and ground) if these are in reversed positions, you might well get a lower output than desired. If the P1 is not wired correctly, you may well get a voltage as low as 1.2V! (This would be a short between ADJ and GND). Also, Does the "power" LED turn on? I would think not, as it is derived from the 7805, which should be dropping out...

Also, your LM317 will get hot as a result of dissipating excess voltage as heat. This is why it is vital that the adjustment components are correctly oriented. If your battery voltage is (close to) 18v, and your regulator is adjusted to 3v, the regulator has to get rid of 15 volts, which is not impossible, but is a challenge ;). If you can still touch the reg without burning yourself, it is no real cause for concern.

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I made very certain there are no shorts in the LM317 and that it's the right way around.

Also, your LM317 will get hot as a result of dissipating excess voltage as heat. This is why it is vital that the adjustment components are correctly oriented. If your battery voltage is (close to) 18v, and your regulator is adjusted to 3v, the regulator has to get rid of 15 volts, which is not impossible, but is a challenge ;). If you can still touch the reg without burning yourself, it is no real cause for concern.

It is way too hot to touch, not sure if that means it's zapped or not, I don't think it was left like that for very long.

Can I just use a 12V psu? Or do I need that extra half volt?

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I'm sorry, yes I have indeed been changing the values on the Pot. It's a 2k pot and I've moved it extreme left and extreme right. It goes from 3 volts to 1 volts. Should I put a largish resistor between the pot and the LM317 so that it (permanantly) adjusts to a higher voltage setting?

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Perhaps. I would double check the resistance of R1. It should be around 220 ohms.

If you have the means, I'd even suggest measuring R1 and P1 with a meter to be sure. (Those bands can be tricky - Red-Red-Brown-other, or Red-Red-Black-Black-other while were on the subject... ;) ) If the voltage changes when you adjust P1, then the reg is (sort of) working, but there seems to be a problem in the resistor network.

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Right, I understand this, so you agree it might be advisable to add a more powerfull resistor? Also I can change P1 into a much higher value pot, not sure I have any trims, but I can use a regular 100k pot or something.

Thanks again for the help.

I checked to make sure the trim pot is 2k, and the resistor 220 ohm (not 220k or anything). The two legs closest to the LM317 are connected to each other.

So would it help to replace the trim pot with a larger one or not? I can put it left and right and the voltage changes, but never all the way up to anything above 3v.

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Okay,

So with new batteries and after checking the orientation of all ICs and any shorts but it still is broken.

With the dial all the way left (measured no resistance) there's no heat coming off the IC and the voltage is ~2v. Even after being left for some time there is no significant heat.

With the dial half-way right (957 Ohm) the LM317 starts becoming extremely hot.

It seems like this works with the balance of the 220Ohm Resistor and the pot. Is this correct? If so which should I be increasing (pot side or resistor side) to possibly stop this from happening?

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Okay, I removed the very LM317 from the board (this thing is starting to look pretty roughed up) and I assembled a test circuit.

http://www.electronics-lab.com/articles/LM317/

The only difference I can see in these circuits is that there is that diode, and the Pic Burner uses all 3 legs of the pot (which I think makes no difference). Does anyone know what the diode is supposed to do?

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Found out what it was, there was something wrong with the 78L05. I removed it and the LM317 worked fine. I then tested it on a breadboard and no orientation made it work (tried 4-5 different possibilities). So something was wrong with it. It's now replaced with a regular sized one and the whole circuit works, didn't know that could have affected the circuit!

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Dunno,

looks like it was, but somehow it was interfering with the circuit. Like I said I couldn't find any pinning that make it work. Strange that it disabled everything else, would have thought it would only affect things after it and not things (voltages) before.

Thanks again for the help BTW.

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No, it would absolutely affect the whole circuit.

If the 7805 were faulty or wired incorrectly, (say, it causes a short or near short) the Lm317 would be desperately "trying" to feed the full regulated voltage to a shorted 7805. In this case, to compensate for extreme voltage drop, more current is allowed through. This soon becomes an extreme example of ye old Ohms Law:

I=V/R

if R (resistance) is very low, I is very high.

if R is zero (or a real world version of zero ;) ), then I runs off in the general direction of infinity.

of course, R won't reach zero with a correctly working 317, hence your voltage pulled down to 2-3v, not 0v.

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