Thanks guys i moved everything to the correct pins and it works great now. The problem would have never happen if i had the pinouts for the '317 which i didnt, i just found it at the bottom of a big electronics junk draw..next time i wont let this happen :) thanks again mIkE On Mon, 8 Nov 1999 23:04:02 -0800 Bryan Minugh wrote: >To work as a current limiter, the load must be connected to the adjustment >terminal, not (as it is in the incorrect schmatic) to the output terminal. >This URL is one of many I found showing how to connect the 317 *correctly* >as a voltage requlator: >http://esaki.ee.boun.edu.tr/~ogrenci/EE206/project.html > > -Bryan Minugh > >----- Original Message ----- >From: Kipton Moravec >To: Mike M >Cc: >Sent: Monday, November 08, 1999 10:03 PM >Subject: Re: SRS: Adjustable power supply > > >> It is quite simple actually, the circuit drawing is WRONG! >> >> The 10 Ohm resistors should be hooked to the OTHER SIDE of the 390 Ohm >> Resistor. >> >> The way you have it hooked, is the configuration for a current limiter. >> Set at 1.2 / 390 ~= 3 mA. However it does not pull enough current to >> regulate correctly. They recommend a resistor from .8 to 120 Ohms. >> >> Given that, I do not know why you are smoking your 10 Ohm resistors. If >> the regulator is not working as a current limit because the 390 Ohm >> resistor is too high, then you may be trying to dissipate an (18 - 1.25 >> = 16.75V) drop across the two 10 Ohm resistors. >> >> V = I * R >> 16.75V = I * 20 >> .8375A = I >> >> W = V * I >> W = 18 * .8375 >> W = 15+ Watts (between two 1/4 W resistors.) = SMOKE! >> >> So assuming you fix the circuit by putting the 10 Ohm connection to the >> other side of the 390 Ohm resistor, Then your equation is correct and >> you are dissipating 1.31 V across (390 + 20 =) 410 Ohms >> V = I * R >> 1.31V = I * 410 >> .0032A = I >> >> So >> V = .0032 * 390 = 1.246 V is dropped across the 390 Ohm Resistor >> and .0032 * 10 = .032 V is dropped across each 10 Ohm resistor >> >> So the 390 dissipates 1.246 * .0032 = .004 Watts = 1/250 Watts >> and the 10 ohm dissipates .032 * .0032 = .0001 Watt -- so a 1/4 Watt >> resistor will have no problem. >> >> Kip >> >> Mike M wrote: >> > >> > Hey guys, i built a real (cheep) quick adjustable power supply using an >LM317...i basically followed the schematic at: >> > >> > http://expert.cc.purdue.edu/~bymaster/power_supply.gif >> > >> > anyway, when i check the output voltage on the 317, no matter what >values of resistance i put the adjust pin and the output pin always read >16.8v. There is a 18v input. The 5v 7805 regulator works fine... Now i >know i probly blew out the chip because while testing i turned the 5k pot to >0, there should be 20ohm in series wit the 390ohm resistor and when u do the >math >> > >> > 1.25(1+ 20/390) = 1.31v >> > >> > In my circuit..the two 10ohm resistors started to smoke and turned >black. Why?? They are 1/4 wat resistors (probably why but all i had on >hand)..??? >> > >> > mikE >> > >> > Send someone a cool Dynamitemail flashcard greeting!! And get rewarded. >> > GO AHEAD! http://cards.dynamitemail.com/index.php3?rid=fc-41 >> > To unsubscribe, send email to majordomo@seattlerobotics.org >> > In the message body, put unsubscribe srs >> > For more information, www.seattlerobotics.org/contact.html >> To unsubscribe, send email to majordomo@seattlerobotics.org >> In the message body, put unsubscribe srs >> For more information, www.seattlerobotics.org/contact.html >> > > Send someone a cool Dynamitemail flashcard greeting!! And get rewarded. GO AHEAD! http://cards.dynamitemail.com/index.php3?rid=fc-41