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Topic: LM317 voltage regulator R2 is getting very hot? why? (Read 11482 times) previous topic - next topic

calvingloster

I have set up a LM317 to output 13.7v. From the output to adjust I have used a 5ohm resistor and then my R2 which is from the R1 to ground I have used a 50ohm resistor. When I put a 16v power supply on the 50ohm resistor gets smoking hot!!! Like I can even smell it! Why is this?

KeithRB

Use ohms law to calculate the dissipation. The current through them is 16.7/55 = 300 mA. The power in the 50 ohm resistor is .3*.3*50 = 4.6 W.

You can safely increase the resistance by 10 or even 100 times to minimize the power dissipation.

calvingloster


Use ohms law to calculate the dissipation. The current through them is 16.7/55 = 300 mA. The power in the 50 ohm resistor is .3*.3*50 = 4.6 W.

You can safely increase the resistance by 10 or even 100 times to minimize the power dissipation.


If I increase the resistance the power dissipation barely decreases. If I use a 100ohm for R1 and a 1000ohm for R2 I get 12W of heat dissipation off R2!!!!!!!!

I can't understand this? Why do they say u can use the LM317 as a voltage regulator. It's ridiculous

fungus

So... you've got 55 Ohms between 13.7v and GND.

a) Can you calculate the current that's going to flow through the resistors?

b) Using the answer from (a), how much power will each resistor dissipate? (volts * amps)

c) Do the answers from (b) seem reasonable?


No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

calvingloster


So... you've got 55 Ohms between 13.7v and GND.

a) Can you calculate the current that's going to flow through the resistors?

b) Using the answer from (a), how much power will each resistor dissipate? (volts * amps)

c) Do the answers from (b) seem reasonable?





A) yes I can 250mA
B) R1 will dissipate 1.25W. R2 will dissipate 12.5W
C) No 12.5W does not seam reasonable for a 1/8W resistor

However if I increase R1 to 100ohms and R2 to 1000ohms then:

A) 0.0125A
B)R1 will dissipate 1.25W and R2 will dissipate 12.5W
C) oh look it's exactly the same BUT the R2 doesn't get hot at all unlike the first case where they burn up

123Splat


calvingloster


E/R= I
I*I*R=P, not I*R*R



Can you please talk in words and not letters? I don't know what you trying to say?

Heat dissipation is Amps multiplied by resistance.

JChristensen


I can't understand this? Why do they say u can use the LM317 as a voltage regulator. It's ridiculous


Because it is a voltage regulator :D

Have a glance at the datasheet. 240? is the usual value for R1. (What would ever possess a person to use 5?.) The LM317 develops a 1.25V reference voltage across R1, resulting in 65mW power dissipation. Use 2400? for R2, this will set the output voltage to 13.87V, and 66mW dissipated by R2.  Where is the problem.

Use 1% resistors and change R2 to 2370?, this will come closer to the target voltage, 13.71?.

calvingloster



I can't understand this? Why do they say u can use the LM317 as a voltage regulator. It's ridiculous


Because it is a voltage regulator :D

Have a glance at the datasheet. 240? is the usual value for R1. (What would ever possess a person to use 5?.) The LM317 develops a 1.25V reference voltage across R1, resulting in 65mW power dissipation. Use 2400? for R2, this will set the output voltage to 13.87V, and 66mW dissipated by R2.  Where is the problem.

Use 1% resistors and change R2 to 2370?, this will come closer to the target voltage, 13.71?.


Ok so let me ask you this, if I use a 240 ohm resister as R1 then I can supply a voltage whatever depending on R2 but my maximum current I can provide is 1.25/240 is 0.005 Amps. Now what if I need 250mA?

JChristensen

#9
May 27, 2014, 07:54 pm Last Edit: May 27, 2014, 07:56 pm by Jack Christensen Reason: 1


E/R= I
I*I*R=P, not I*R*R


Can you please talk in words and not letters? I don't know what you trying to say?


A person needs to know a little algebra and common variable names to be successful in this hobby. Do the substitution yourself, E = voltage (or you can use V), I = current, R = resistance, P = power.

Quote

Heat dissipation is Amps multiplied by resistance.


No, power (heat) is volts times amps. Or: P = V * I

With a little algebra (you passed algebra?), we also have  P = I2 * R  and P = E2 / R

Very useful and essential formulas, should be memorized.  Used properly, they are guaranteed to prevent that burning smell ;)

JChristensen


Ok so let me ask you this, if I use a 240 ohm resister as R1 then I can supply a voltage whatever depending on R2 but my maximum current I can provide is 1.25/240 is 0.005 Amps. Now what if I need 250mA?


Incorrect.  Where in the world did you get such an idea?

calvingloster



Ok so let me ask you this, if I use a 240 ohm resister as R1 then I can supply a voltage whatever depending on R2 but my maximum current I can provide is 1.25/240 is 0.005 Amps. Now what if I need 250mA?


Incorrect.  Where in the world did you get such an idea?



Sorry I am abit slow I really do wish I was as clever as many of you people, please forgive me.

I see now that I made the mistake of multiplying amps and current to get power dissipation instead of squaring amps and then multiplying it by current.

I got that idea from if you put a 240ohm resistor between output and adjust pin then u get a constant current source of 0.005A. So there is no way you can get more amps from using a 240ohm resistor. I am however confused why the LM317 does not act like a current limiter if you wire it up as a voltage regulator.

123Splat

AMPs am CURRENT,,, It's Voltage , Resistance, and Current (in AMPs).
The 1.25 V and 50uV (not 5mV) are what is passed by the Adjust pin of the regulator. You can read the datasheet to find out how it works. But that current path is totally different from the reference loop setup by R1 and R2.  That sets the Highend voltage the regulator will pass.  with proper heat sinking, the 317 will pass a max of something like 1 AMP, as long as you source it with Vin=Vout+1.25V, or more and at least 1 AMP.

JChristensen

#13
May 27, 2014, 08:25 pm Last Edit: May 27, 2014, 08:27 pm by Jack Christensen Reason: 1

Sorry I am abit slow I really do wish I was as clever as many of you people, please forgive me.


No worries, just slow down and try to understand the fundamentals like Ohm's Law. Not difficult at all and as ever, GIYF.  It's also very important to read the datasheets of the devices in the circuit.

Quote

I got that idea from if you put a 240ohm resistor between output and adjust pin then u get a constant current source of 0.005A. So there is no way you can get more amps from using a 240ohm resistor. I am however confused why the LM317 does not act like a current limiter if you wire it up as a voltage regulator.


It doesn't act as a current limiter (well, except to protect itself when current exceeds its limits, or it gets too hot) because it's a voltage regulator.  Which do you want?  If you want both, then something other than a simple LM317 will be needed.

The main current to the load flows through the LM317 from its input terminal to its output terminal.  The Adjust terminal works in conjunction with the resistors to set the desired output voltage.  The currents involved in doing this are relatively small and have nothing really to do with the output current to the load. Does that help?

123Splat

The 317 can also be configured as a current regulator (for relatively Constant Currents BELOW 1A). And,, you canconfigure one 317 as a voltage regulator (for desired Vout+1.25V) and follow it with a second 317 configured as a current limiter.  It works. It kills batteries. It wastes parts.  I've done it many times and with regret, will probablly continue to do so...

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