LM317 question.

[sureshot]

Merry Christmas to the Community, hope every one is enjoying the holidays.

Just a quick general question on the popular LM317 voltage regulator.

For better precision, can you use 0.25 watt +/- 1% metal film resistors for R1 and R2 or do they have to be 0.5 watts, which I can only get in +/- 5% Thanks for reading, any help appreciated.

 

[AnalogKid]

1/4 watt should be fine. You know the resistance value and the voltage across each resistor, so you can calculate the actual power dissipations. The components should be rated for at least twice those values.

ak

 

[GPG]

Recommended minimum load is 10mA. This is the current through the divider chain. So the top resistor is 120Ω The dissipation of this resistor will be ~ 120 x I² = 12μW. Repeat for the bottom resistor. so ¼W will be fine

 

[AnalogKid]

Dude, I was prodding him to do the math himself...

 

[dorke]

You can do better by using a multi-trim pot. in the place of R2.
Actually split R2 to a fixed part ,and 10-20% or so pot. to allow for the output voltage setting.
In this case you can use 5% fixed resistors.

 

[GPG]

Dude, I was prodding him to do the math himself...

While I was composing, you posted. Lousy typist
But I left OP with one to do

 

[sureshot]

Thank you for the help, maths is not my strongest point lol, but I understand what you've laid out in a formulae. I'm looking at this simple circuit, but struggling with the transistors base. Most have a direct base resistor for series pass transistors, this one has me stumped to be honest.

 

[sureshot]

Recommended minimum load is 10mA. This is the current through the divider chain. So the top resistor is 120Ω The dissipation of this resistor will be ~ 120 x I² = 12μW. Repeat for the bottom resistor. so ¼W will be fine

So my lower fixed resistor is 2200k so power dissipated 22uW .

 

[hevans1944]

Recommended minimum load is 10mA. This is the current through the divider chain. So the top resistor is 120Ω The dissipation of this resistor will be ~ 120 x I² = 12μW. Repeat for the bottom resistor. so ¼W will be fine

This time I remembered to use the calculator! 10 mA = 0.010 A. Square that and you get 0.0001. Multiply by 120 Ω and you get 0.012 watts or 12 mW (not 12 μW!). Still, that's considerably less than 250 mW or 1/4 W.

On resistor power dissipation: you can connect two resistors (of the same power rating) in series or in parallel to double the power rating. If in series, each one will be half the original value. If in parallel, each one will be twice the original value. Just make sure you leave enough room around each resistor to dissipate the power. I've done this for power resistors in the 10 watt and higher ranges, but free-air convection paths become important at higher power levels. Not so much a problem for 1/10 watt up to 1/2 watt sizes: space the resistors about a body-diameter apart.

So my lower fixed resistor is 2200k so power dissipated 22uW .

This doesn't sound right. 2200 kΩ is 2.2 Meg-ohms. Ten milliamperes through that much resistance would require an output voltage in excess of 22 kV! I think someone has misplaced a decimal point somewhere.

 

[sureshot]

This time I remembered to use the calculator! 10 mA = 0.010 A. Square that and you get 0.0001. Multiply by 120 Ω and you get 0.012 watts or 12 mW (not 12 μW!). Still, that's considerably less than 250 mW or 1/4 W.

On resistor power dissipation: you can connect two resistors (of the same power rating) in series or in parallel to double the power rating. If in series, each one will be half the original value. If in parallel, each one will be twice the original value. Just make sure you leave enough room around each resistor to dissipate the power. I've done this for power resistors in the 10 watt and higher ranges, but free-air convection paths become important at higher power levels. Not so much a problem for 1/10 watt up to 1/2 watt sizes: space the resistors about a body-diameter apart.


This doesn't sound right. 2200 kΩ is 2.2 Meg-ohms. Ten milliamperes through that much resistance would require an output voltage in excess of 22 kV! I think someone has misplaced a decimal point somewhere.

Thank you for the explanation, yes sorry my mistake typing 2k2 not 2200k yes my mistake.

 

[sureshot]

C

Thank you for the help, maths is not my strongest point lol, but I understand what you've laid out in a formulae. I'm looking at this simple circuit, but struggling with the transistors base. Most have a direct base resistor for series pass transistors, this one has me stumped to be honest.

Does this circuit look feasable to the more experienced ? I was thinking its capable of about 3 Amps at 12 volts.

 

[hevans1944]

An emitter follower is not the most efficient way to boost current. Try this circuit:

This was copied from edaboard.com forum and I have not built or tested it. It appears to have been copied from the application notes on the National Semiconductor datasheet, attached. You should read the datasheet.

 

[sureshot]

Many thanks for the schematic, I will try both as a challenge. Thanks again to you all for the help.

 

[sureshot]

Save starting another thread, could I ask opinions on this circuit. It appears to be from a good source, I have a few value 78xx voltage regulators laying about. Its the output tolerance I was wandering about, and how much it might drop voltage under load. Thanks for any input on this question.

http://www.pcs-electronics.com/guid...d1fbc&osCsid=af49776c46cb786d6bafe6a714a5a0be

 

[hevans1944]

Minus the diode connected to the base of the 2N3055, and the use of the fixed-output 7812 instead of the adjustable LM317, this circuit is the same as the first circuit you uploaded. As I stated earlier, I am not a big fan of using emitter-followers without feedback to "boost" current output. The regulation of the output truly sucks, but if you don't mind that, try it.

 

[sureshot]

Ok thanks, i think the regulation is going to be less than perfect, but i've never put one of these simple series pass transistors together using the 2N3055, its a taste it and see thing i think to be honest. the diode at the ground pin on the 7812 is a 0.7 volts boost i think to replace what's lost through the transistors volts drop. by all accounts its probably poor. But i will try one of these using the 2N3055. In the first circuit i posted, i've no idea what that diode is doing, other than dropping another 0.7 volts or there about's.

 

[hevans1944]

I think you are right about the purpose of the diode in the 7812 circuit. Problem is, the forward bias is current-dependent for both the diode and the base-emitter junction of the 2N3055. The forward bias voltage of both junctions is also temperature dependent, so unless the two junctions are thermally coupled, it is not likely the external diode will do anything vis a vis voltage regulation. But as you said, "taste it and see" or, here in Dayton OH, a local pizza chain says, "La prova é nel gusto" or "the proof is in the taste."

If you need really tight regulation of the output voltage under variable load, there is no alternative I know of than comparing the output against a stable reference voltage and using the difference to control the conduction of the pass transistor(s). This is all done for you in both the 7812 and the LM317, so you could consider their outputs to be your "reference" and build on from there.

 

[sureshot]

Yes there's no real short cuts for a proper circuit, it will pay long term to get it right with a decent build. But just to see what kind of poor regulation and for the sake of a handful of components I've got to build it and see, as odd as that might sound lol. Thanks for your help Hevans 1944 (-;

 

[hevans1944]

I am not trying to blow their horn, but while visiting my son here in Atlanta GA he produced a dozen or so back issues of Elektor magazine for me to peruse. One of these (June 2013) has an article that may offer a possible solution to your power supply build: re-purpose a PC ATX power supply by adding an external circuit board with banana jacks and assorted other small components to turn the supply on, provide some LED indicators, and provide short-circuit or overload protection.

I think I am gonna sign up for an Elektor Gold membership ($101.00) when I get back to Dayton OH. Lots of interesting, professional looking, DIY articles there. Of course my wife has already warned me not to sign up for any more magazines, so maybe I will just continue using my son's subscription until I can figure out a way to sneak one more past her... maybe I could get another post-office box and have the magazine sent there. <sigh>

 

[sureshot]

Happy new year. Sorry I've been absent, thank you Hevans 1944. I have converted some atx power supply's, but none are variable at the moment. But they are good for high current uses, allowing some ripple above a linear on the output, and pwm fan control electrical noise, but handy high current cheap source.