5
\$\begingroup\$

I'm working on my first small electronics project; building a small LM317 based power supply. I've built the circuit as shown in the data sheet (black part of the circuit diagram below) on a perfboard. Now I wanted to add an LED to indicate if the switch is in on (bottom) or off (left) position. I added it to the circuit as shown below (red). However, the LED doesn't turn on if I switch to on position (the power supply itself works fine).

My question is: can this even work if the LED were connected as indicated in the diagram? If not, what would be the easiest solution to get such an on/off indicator LED?

Image link: http://i.stack.imgur.com/X1Jv7.png

\$\endgroup\$
  • 1
    \$\begingroup\$ Are you sure you have the diode oriented the right way? Putting it in backwards is a VERY common mistake. \$\endgroup\$ – W5VO May 30 '12 at 22:57
  • 1
    \$\begingroup\$ Possible scenario: You originally installed the LED backwards, which violated its reverse voltage spec grossly so now it's broken and when you flip it around it still doesn't work. Also, I'd plan for about 10 mA thru the LED. Try a 1.6 kOhm resistor instead of the 4.7 kOhm. \$\endgroup\$ – Olin Lathrop May 30 '12 at 23:19
9
\$\begingroup\$

Since you're new to electronics the power supply is a good way to start. You'll need one for everything you want to do.

AndrejaKo and the others already told you what's probably wrong: you installed it the wrong way, which killed it, because while most diodes can stand the 18V reverse, LEDs can't. What did you do? Just try it one way and reverse it if you saw it was wrong? OK, now you know that you can't afford this with LEDs.

Try to make it a habit to use components you've got information on. No design engineer would use a part if he can't get the datasheet for it. Here's the drawing from the datasheet for a random LED:

enter image description here

There are a few differences between the pins. One pin is longer than the other, the LED's rim isn't complete on one side, and if you look into the LED you may see a thin wire going from the chip to one of the pins. The wire is usually the anode. Usually! Don't try it if you're not sure, remember you don't get a second chance. Use whatever the manufacturer indicates as the anode. In this case that's the longer pin. Only rely on that. That's your anode, which goes to the plus.

If you wouldn't have a datasheet there's still a safe way to try it. Place two of your LEDs in anti-parallel: the anode of one LED to the other's cathode, and vice versa. Place a resistor in series and connect to a 9V battery (or your new power supply!). One LED will be correctly polarized and that will light, the other one will be reverse polarized, but the first one's forward voltage (around 2V for a red LED, 3.5V for a white one) will limit the reverse voltage to a safe level.

Another thing mentioned is your resistor. Nine out of ten it's too high, though you should still see the LED light faintly. Again there's the datasheet. You want to know the LED's nominal current, which is often around 20mA for an indicator LED, but also its voltage drop. The same datasheet says 40mcd at 20mA, and the voltage is 2V typical. Then you use Ohm's Law to calculate the series resistor:

\$ R = \dfrac{V}{I} = \dfrac{V_+ - V_{LED}}{I_{LED}} = \dfrac{18V - 2V}{20mA} = 800\Omega\$

if it's a 20mA LED. The closest value you'll find will probably be 820\$\Omega\$. Note that, due to the high input voltage, the power dissipated will be a bit more than 300mW, so a 1/4W resistor won't do. Pick at least a 1/2W resistor.

There's still another possible explanation why it doesn't work. LEDs are sensitive to ESD, ElectroStatic Discharge. Touching the LED when you're statically charged to a few kilovolt will also kill the LED. It's possible that many of the components you'll use in the beginning won't be so sensitive to this, like resistors or capacitors. CMOS is also ESD sensitive. If you don't have ESD protection (probably not yet) then touch a large metal object before you handle the LED.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Neat way to test LED polarity safely. \$\endgroup\$ – kevlar1818 May 31 '12 at 14:47
  • \$\begingroup\$ Thank you very much for your detailed explanation. It must indeed have been that I'd built in the first one the wrong way around because it worked out with a new one. Unfortunately I got a whole "set" of LED's and for some reason there is no data sheet included so I can't calculate the required resistor. I chose a 2.2K because I only h ave 1/4W ones but it's bright enough. \$\endgroup\$ – simon Jun 7 '12 at 21:09
0
\$\begingroup\$

Well I see two potential problematic points:

First, if we assume say 2 V drop on the LED (and it's probably more), we have just around 3 mA going through the LED with the chosen resistor value. That may or may not be enough to turn on the LED. Test it in a separate circuit if you can. Anyway, I'd start with a lower resistor value. You didn't give us the specifics of the LED, but usual maximum current for say 5 mm LEDs is 20 mA. Try picking a resistor that will give you at least 10 mA, just to be safe. Something like \$2.2 \mbox{ } k \Omega\$ resistor would be a good start. Tweak it from there if you need to lower the current consumption of the LED or if it's too bright.

The other point is to check if everything is installed correctly. Maybe the LED is connected in the wrong direction? In which case it's now dead, LEDs can only stand a couple of volts reverse.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ I don't think the low current explains the problem. Most LEDs should give some light, even at that 3mA. \$\endgroup\$ – stevenvh May 31 '12 at 8:48
  • \$\begingroup\$ Inserting the LED with the correct polarity also lit it up at 4.7K. \$\endgroup\$ – simon Jun 7 '12 at 21:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.