LM317T, using it properly

Thread Starter

PeteHL

Joined Dec 17, 2014
267
In my circuit including a LM317T configured as a current regulator, the input and output voltages to the IC are respectively 43V and 40V. Is that high of voltages permissible? In the data sheets for example from ON Semiconductor, it is stated that the input-output voltage differential must not exceed 40V, but that both voltages can be any number above ground. Is that actually the case would you say with this regulator?

Also, output current is 0.5A. Would 40V times 0.5A equals 20W be more output power than this regulator can accommodate?

My problem is with overheating of LM317T even with a substantial heat sink attached to the regulator. This might be related to the entire circuit including the regulator, but I'll leave showing the entire circuit to a later date, perhaps.

Thanks in advance for your feedback,

Pete
 

Evil Lurker

Joined Aug 25, 2011
116
Well if memory serves me correctly the LM317 is rated at 36 volts tops unless it is of the HV variety, however, since you are using it as a current regulator I'm thinking that it only "sees" about 3 volts difference between the input and output voltages. It should definitely be getting warm but nowhere near it's maximum rated thermal dissipation limit. What kind of heatsink are you using?
 

MikeML

Joined Oct 2, 2009
5,444
Normally, the voltage rating refers to the difference between the voltage at the input pin and the voltage at the output pin. Power dissipation is figured by taking the same difference voltage and multiplying by the current flowing through the regulator.
 

crutschow

Joined Mar 14, 2008
23,824
That will work as long as there's no possibility that the input-output voltage difference will never exceed the LM317's maximum rating during startup or other transient conditions.
For example, if you accidentally short the output to ground you could zap it.
 

Thread Starter

PeteHL

Joined Dec 17, 2014
267
Normally, the voltage rating refers to the difference between the voltage at the input pin and the voltage at the output pin. Power dissipation is figured by taking the same difference voltage and multiplying by the current flowing through the regulator.
Thanks, I stand corrected about power dissipation relative to the regulator. -Pete
 

Thread Starter

PeteHL

Joined Dec 17, 2014
267
What I've built is a step-down rectifier ultimately to drive a string of twelve 3W LEDs. Instead of using a transformer, there is a capacitor on the AC side of the bridge rectifier of the circuit to divide the 125 VAC at input to the circuit between the capacitor and the LM317T configured as a current regulator supplying current to the load of 12 LEDs in-series.

Here is my hunch as to what is happening. As the LM317T IC heats up, impedance of the IC increases. This results in an increased input- output voltage of the IC, which in turn causes an increased temperature of the IC. The increased temperature of the IC causes a further increase of the IC's differential voltage, and a greater increase of temperature. .... and so on, until the heat sink attached to the IC is no longer adequate to dissipate the heat generated by the IC, and the IC's thermal protection is activated.

-Hope that this is clear. I know, a schematic would help. If necessary, I will try to post one. Any thoughts about my hunch?

Thanks if anyone can weigh in,
Pete
 

studiot

Joined Nov 9, 2007
4,998
Forgive me but how can you say the output voltage is 40 volts and it is a current regulator?

A current regulator varies its output voltage to hold the current constant.

So the output voltage will be whatever it needs to be to achieve the set current.
 

Thread Starter

PeteHL

Joined Dec 17, 2014
267
Forgive me but how can you say the output voltage is 40 volts and it is a current regulator?

A current regulator varies its output voltage to hold the current constant.

So the output voltage will be whatever it needs to be to achieve the set current.
LM317T IC doesn't vary its output voltage to hold current constant. In the current regulating configuration of the IC, the voltage between the adjust and output terminals of the IC is 1.25V regardless of voltage across the load. So for example, connecting a 2.5 Ohm resistor between adjust and output pins of the IC produces a constant 500 mA output current.

When the string of 12 LEDs in-series are given enough current, then the voltage drop across each LED equals 3.3 V. So 12 times 3.3V equals 39.6V. Specified operating current for the 3W LEDs is 700 mA, and I'm planning to have 500 mA through the string of LEDs.

For testing, I have a 80 Ohm resistor as load substituting for the string of LEDs. Again, the LM317T is set to regulate current at 0.5A, and 80 Ohm times 0.5A equals 40V. Actually the output voltage of LM317T is 41.25V including the 1.25V drop across the current setting resistor connected from adjust to output terminals of the IC.
 

ronv

Joined Nov 12, 2008
3,770
You need to post your circuit. Please include an isolation transformer or the thread will be closed because the circuit will be "hot".
Anyway I think you may be missing a few things.
We need to know the filtering after the bridge (if any) and the size of the cap in series with the regulator.
 

Thread Starter

PeteHL

Joined Dec 17, 2014
267
Then how do you explain the green plot in this simulation?
I'm sorry, but I don't completely understand your graph. It looks as if the X axis is resistance of R2. But then Y is the voltage drop across R2, is that correct? If so, then Y (the green plot) should increase with increasing resistance of R2. Also, X is the constant input voltage of V1 or 35v?
 

MikeML

Joined Oct 2, 2009
5,444
I'm sorry, but I don't completely understand your graph. It looks as if the X axis is resistance of R2. But then Y is the voltage drop across R2, is that correct? If so, then Y (the green plot) should increase with increasing resistance of R2. Also, X is the constant input voltage of V1 or 35v?
No, the green plot is the voltage drop across the LM317, from X to Y. This is to show that the voltage drop across the LM317 is effected by the load resistance, pointing out that your statement: "LM317T IC doesn't vary its output voltage to hold current constant" is false.
 

bertus

Joined Apr 5, 2008
20,189
Hello,

If you keep using the capacitor to lower the voltage in stead of a transformer I will have to close the thread, as it is not allowed to use the "watt less voltage dropper" capacitor.

For safety reasons, why do you not look for a led driver?
http://www.newark.com/recom-power/racd30-500/led-driver-ac-dc-cc-0-5a-56v/dp/62X9593

I have selected this as your input voltage is 120 V and your wanted current 500 mA.
Knowing the specs of your leds a better advice can be given.

Bertus
 

Thread Starter

PeteHL

Joined Dec 17, 2014
267
Here is the schematic of the step-down rectifier, including an isolation transformer. Please forgive me if it doesn't show up properly, given that this will be the first time that I'm attempting to post graphics. I'm showing the circuit with a 80 Ohm dummy load.

MikeML, thanks, I understand your graph now. The disagreement might be just semantics, but I do not understand the inner workings of LM317. However, given that the designers of that IC show how to configure it as what they call a source of constant current, not variable voltage, it makes more sense to me to say that Vout increases with increasing resistance of the load simply because of Ohm's law and the fact that the IC is configured to put out a constant current regardless of load. That assumes of course that there is sufficient dropout voltage. STEP-D~1.png
 
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