LM317T Current Regulator: Why Is Vadj-Vout Only 0.73V and LED Dimmer Than Expected?

60 13

#1 21660774 19 Jan 2012 15:07

Scott Wohler Scott Wohler

Anonymous

Post #1
21660774 19 Jan 2012 15:07

To make the regulator function properly, you actually need two fixed ADJ resistors, referenced to the negative side of the battery (ground). Currently there is no reference, the way you've got it wired.

Also, you should probably have a current limiting resistor in series with the LED.

It's not really clear what your battery pack source voltage is (it appears to be a two-cell AAA pack, but I can't tell.) For this circuit, with only two series AAA's you will on have 3.0V from the battery pack. The linear regulator dropout is probably anywhere from 1.5V - 2.0V. This means that to generate the required 3.5V output you would need an input voltage of at least 5.0V - 5.5V to the regulator. This won't work with the two-cell pack you have there (3.0V input). Moreover, you need to current limit the LED so you actually need an output voltage higher than the required LED forward voltage drop, and then you would limit the current with the series resistor.

I recommend reading the datasheet for the LM317:
http://www.ti.com/lit/ds/symlink/lm317-n.pdf

ADVERTISEMENT
#2 21660775 19 Jan 2012 15:17

Steve Lawson Steve Lawson

Anonymous

Post #2
21660775 19 Jan 2012 15:17

The LM317 has a fairly high "dropout voltage", esp. when using it in the current regular configuration. The voltage regulator dropout voltage is around 1.5V (it could be more, and appears to be if you measured it correctly [2.35V]) for 20ma and 25 degrees C. But there is also the 1.25V reference voltage that is also dropped before current is delivered to the load. If the LED forward voltage settles at around 3V, then you need 3V + 1.25V + 1.5V = 5.75V minimum to run the thing. Adjust those values as appropriate, e.g. 3V + 1.25V + 2.35V = 6.6V or whatever.

Your photo shows a 3V battery pack, which isn't enough voltage to drive this circuit for ANY LED, even an IR LED which runs at the lowest voltage of all (around 1.2V).

The fact that you are measuring .73 volts when you connect up the LED is because there isn't enough overall voltage for the LM317 to produce 1.25 volts.

Also, are you sure that LED can take 125ma? In the photo, it looks like a standard low current device which usually has a max current of 30ma. The LED in the instructables article is a high power LED, which is different than the one in your photo.

For more on driving LEDs check out the following:

http://www.eeweb.com/electronics-forum/really-simple-question-leds-with-series-resistor/

Also, using an LM317 to drive a high powered LED is an inefficient way of doing it, and if you are going to use a battery, then you will get more life out of the battery if you use some kind of switching current regulator, or even a low dropout current regulator with a battery voltage as close to the minimum supply requirement as possible.
#3 21660776 19 Jan 2012 15:26

Steve Lawson Steve Lawson

Anonymous

Post #3
21660776 19 Jan 2012 15:26

Scott,

Actually, only one resistor is required. He is using the LM317 as a current regulator and only one resistor from Adj to Output will set the current: 1.25V/R = I.

In this arrangement, the LED takes the place of the second resistor. Basically, this takes advantage of the functionality of the LM317. It regulates the current to ground in order to keep 1.25V across the Adj/Output pins. To use the LM317 as a voltage regulator, a resistor is connected from Adj to ground. The constant current though the resistor converts the current to a constant voltage. This voltage, plus the 1.25V reference voltage is what is seen at the LM317 Output pin. Thus, the Output voltage is constant.

Replace the ground resistor with anything else, and the LM317 will still keep the current constant, but if it isn't a pure resistance, the voltage will fluctuate to keep the current constant.
ADVERTISEMENT
#4 21660777 19 Jan 2012 15:33

Scott Wohler Scott Wohler

Anonymous

Post #4
21660777 19 Jan 2012 15:33

Steve,

My apologies, I missed that. I feared what I might see if I read the Instructables article...

Then I agree that it's just an issue of dropout. He needs at least another 3V at the input.

-S
ADVERTISEMENT
#5 21660778 19 Jan 2012 15:35

Christopher Telford Christopher Telford

Anonymous

Post #5
21660778 19 Jan 2012 15:35

Just to clear things up the battery pack is a 4 x AAA battery back, unfortunately photographed from dead on to make 2 of them invisible. When i say it was a 5.3 volt supply i mean that is what i measured the output of my mostly charged 4 x AAAs as producing at that time.

Steve, If you are saying i'd need 5.75 volts or possibly 6.6V minimum to make the circuit work properly then that certainly helps me a lot, thanks. I will try it again at a higher voltage and see what happens. Eventually i will use a DC wall adapter as i wish to make an LED table lamp as a simple electronics project so the fact it's not the most efficient way of doing it doesn't matter too much as long as it works. The next project will be a torch and i will likely choose a different circuit for that as you suggest, thanks.

The LED in the picture is a 5 chip 10mm LED rated at 100mA that i decided could probably take 125 for a brief period for testing purposes. I have other high power LEDs that i will be using once i get the circuit working properly. When i have the current limiting bit working as it should i intend to wire it up to a simple 555 timer based PWM circuit for diming via a transistor.

Scott, i don't understand what you mean about needing 2 resisters. I've got it wired up like this;

http://www.instructables.com/image/FNVCWO9FDYPTCX6/Super-simple-high-power-LED-driver.jpg

and it's being used to limit the current not the voltage. The voltage regulating circuits i've seen made using the LM317 had 2 resistors in them.
ADVERTISEMENT
#6 21660779 19 Jan 2012 15:49

Scott Wohler Scott Wohler

Anonymous

Post #6
21660779 19 Jan 2012 15:49

Chris,

Yes I thought you were using it as a voltage regulator instead. Ignore the comment about using two ADJ resistors.
#7 21660780 19 Jan 2012 15:49

Christopher Telford Christopher Telford

Anonymous

Post #7
21660780 19 Jan 2012 15:49

Ok i just tried it with a 6 volt 4 x AAA non rechargable pack;

I measure 3.2 volts across the LED and 2.76 across the LM317T regulator. The LED was definitely brighter as well. I measured the current as 110mA with the LED attached and 120mA short circuited so i'm guessing my previous battery pack was either low on voltage, charge or possibly both?

I'm only surprised that the dropout voltage across the LM317T is so high. 2.76 seems to be a lot of voltage dropped. Is this normal?

Thanks for all the advice - i'll be coming back to this tomorrow with new batteries now i think.
#8 21660781 19 Jan 2012 16:07

Christopher Telford Christopher Telford

Anonymous

Post #8
21660781 19 Jan 2012 16:07

Hm - just tried another battery pack that was giving out 8.something volts and still got 3.3 volts across the LED with the other 5.2 dropped across the LM317T chip. Bit mystified by what's going on with this one. I've not been running the circuit long enough for heat build up to be an issue.

Just tried the circuit again with 2 LEDs - got 3.13 across each LED and 2.27 across the LM317. Is the drop out voltage supposed to vary depending on the load in order to keep the current constant?
#9 21660782 19 Jan 2012 17:05

R A R A

Anonymous

Post #9
21660782 19 Jan 2012 17:05

Entonces trata de usar LM7805, es un regulador de voltaje de 5 Vcc, con una corriente maxima de 1A, pero para ello debes polarizar bien para no malograr el CI, como dato del led tienes: Vled e Iled, aplicas Ley de Ohm,
R = (Vcc-Vled)/(I led). Vcc= 5.
Puedes usar un circuito :
- Con Transistor darlington
- Thief joule
- Rc a 220Vca 0 110 Vca directo, *tener cuidado con este circuito *puede causar una descarga electrica en tu persona, este circuito requiere de mas analisis (formulas y por ende calculos)..............
Puedes usar un lenguaje simulador para probar tus circuitos.......como es proteus, orcad, ............
Disculpa por escribir en castellano, es mi idioma oficial...........................
#10 21660783 19 Jan 2012 17:09

R A R A

Anonymous

Post #10
21660783 19 Jan 2012 17:09

Then try using LM7805 is a voltage regulator 5 Vdc, with a maximum current of 1A, but this must not fail to polarize well for the CI, as the data have led: Vled and ILED, apply Ohm's Law,
R = (Vcc-Vled) / (I led). Vcc = 5.
You can use a circuit:

- With transistor darlington
- Joule Thief
- CR 0 110 Vac 220Vac direct, * be careful with this circuit * may cause an electric shock on your person, this circuit requires more analysis (and hence load formulas )............ ..
You can use a simulator to test your language ....... as proteus circuits, OrCAD, ............
Sorry for writing in Castilian, is my official language ...........................
#11 21660773 19 Jan 2012 18:28

Christopher Telford Christopher Telford

Anonymous
Post #11
21660773 19 Jan 2012 18:28

I made a simple circuit using the LM317T to regulate the current going through an LED. The voltage drop between ADJ and Vout is supposed to be 1.25V and the current is limited based on a resistor placed between the adj and output terminals where current out = 1.25/Resistance like this one;
http://www.instructables.com/id/Super-simple-high-power-LED-driver/
But when i measure the voltage across the output pins of the LM317 I get 0.73 between the ADJ and Vout and 2.35 volts dropped across the LM317 in total meaning my LED isn't getting enough volts to light up as brightly as it should. The supply was measured at 5.3 volts and the LED is only seeing 2.8 volts across it when i measure that using my multimeter, the rest being dropped across the LM317T. The LED is specified for 3.5 volts and should be a lot brighter than it is and i don't understand why so much voltage is being dropped across the LM317T. Also when i measure the current out of the LM317 to ground i get 125mA as i should based on the fomula with a 10 ohm resistor, until i connect up the LED then it drops to 50mA.
Can anyone offer any suggestions as to what i'm doing wrong? I really need help understanding this i'm pretty new to electronics, Thanks.
Edit - My supply voltage was too low. I wasn't allowing for the reference voltage of 1.25 volts across the sensing resistor + the dropout voltage of 1.5 volt (ish). My LM317T wasn't getting enough voltage to function properly and i was getting strange results because of this.

Attachments:

forums-lm317-1326912250.jpg Download (54.59 KB)
#12 21660784 19 Jan 2012 20:00

Steve Lawson Steve Lawson

Anonymous

Post #12
21660784 19 Jan 2012 20:00

Yes, the large voltages across the LM317 are normal. The "Dropout Voltage" I was speaking of is a Minimum Voltage required for the chip to regulate. The important thing is if you are getting the 1.25V across the Adj & Out terminals. Also, the current through the LED should also remain the same as you increase the supply voltage (up to the maximum voltage that the LM317 can take--or the maximum heat it can tolerate, which ever comes first).

And, yes, the minimum dropout voltage for an LM317 does tend to be high when used as a current regulator.
#13 21660785 19 Jan 2012 20:32

Steve Lawson Steve Lawson

Anonymous

Post #13
21660785 19 Jan 2012 20:32

You could also try these:

* LMS8117A -- A "low" dropout version of the LM317. It still suffers from the additional 1.25V when used in current regulation mode. But the gain may be only about a half a volt--but that could make all the difference if at the margin
* LM2941 Even lower dropout voltage but at the cost of extra parts (capacitors). Also, there is no example circuit for using it as a current regulator, but if you replaced "R2" (as shown on the datasheet) with the LED, and if that arrangement is stable, it should work.

But, then, these are only viable for a battery solution and only if the battery voltage is very close to the minimum operational voltage (i.e. Dropout Voltage + Reference Voltage + LED voltage).
#14 21660786 19 Jan 2012 20:50

Steve Lawson Steve Lawson

Anonymous

Post #14
21660786 19 Jan 2012 20:50

To elaborate, the large dropout voltages you are seeing are merely the voltage that is being dropped by the regulator in order the achieve the target voltage. In this case (because of the current regulation arrangement), the target voltage is the Reference Voltage plus the LED voltage. All voltage regulators do this. They are basically a dynamic resistor that adjusts it's "resistance" as needed to maintain the target voltage. If the regulator is a resistor in series with the load (also a resistance whether dynamic or constant), then according to Kirchoff's voltage law, the sum of the voltages across these two resistors will equal the supply voltage.

For instance, a 7805's target voltage is 5 volts. If 12 volts is applied to the input of the 7805, the voltage across the input and output terminals will be 7 volts (12 - 5 = 7).

I think the minimum dropout voltage for a 7805 is around 2.2V (too lazy to look it up So, if 2.2V is the minimum dropout for the 7805 to behave as a regulator and maintain 5 volts on it's output, then, 7.2 volts is the minimum voltage that can be applied to the input of a 7805 and still have 5 volts on the output. The voltage across the input and the output will be 2.2V in that case, but if the input voltage rises, then the voltage across the in and out will increase by the same amount. Thus if it increases to 8V, then the voltage across the in and out will increase to 3V, etc.
Create an account, log in here. You will receive points by participating in discussions.
Join this discussion.

Install Elektroda application

Didn't find an answer? Ask Artificial Intelligence

*I agree to send the question to OpenAI, Anthropic PBC, Perplexity AI, Inc., Kagi Inc., Google LLC - owners of language models in order to prepare the best response. The companies may monitor and log information entered into the form.

*I agree to publicly display my question and answer. The question and answer will be publicly available to everyone. The process may take a few minutes. Upon completion, you will be redirected to the page with the answer.

Wait...(2min)

Topic summary

The discussion addresses the issue of an LM317T used as a current regulator for an LED dimmer than expected, with a measured Vadj-Vout voltage of only 0.73V. The primary cause is the insufficient input voltage from the battery pack relative to the LM317T's dropout voltage and reference voltage requirements. The LM317T requires a minimum input voltage approximately equal to the sum of the LED forward voltage (~3V), the 1.25V reference voltage, and the dropout voltage (1.5V to over 2.3V), totaling around 5.75V to 6.6V minimum. Using a 3V or low-voltage battery pack results in inadequate voltage for proper regulation and current delivery. The circuit configuration as a current regulator only requires one resistor between Adj and Output pins, with the LED acting as the load, maintaining a constant current by regulating voltage across the resistor. Voltage drop across the LM317T can be large and varies with load to maintain constant current, which is normal behavior. Testing with higher voltage battery packs (6V and 8+V) showed improved LED brightness and current near expected values, confirming the voltage supply as the limiting factor. Alternative low dropout regulators such as LMS8117A or LM2941 were suggested for battery-powered applications requiring lower dropout voltages. Additional advice includes using a current limiting resistor in series with the LED and considering other circuit designs for efficiency. Simulation tools like Proteus or OrCAD were recommended for circuit validation. The LM7805 voltage regulator was also mentioned as an alternative for fixed 5V output with appropriate current limiting resistor calculations.
Summary generated by the language model.

LM317T Current Regulator: Why Is Vadj-Vout Only 0.73V and LED Dimmer Than Expected?

Didn't find an answer? Ask Artificial Intelligence

Topic summary