i have been trying to create a simple circuit to make an LED sign, i have two resistors, a 100 Ohm and a 10 Ohm i will use them in series to make 110 Ohm, radioshack didnt have any 110’s… I am trying to use a transformer out of a broken digital clock, and Blue LED’s with a max forward voltage of 3.7, i am reading 6.5 AC volts on my multimeter, however no DC voltage… which isnt a big deal becuase that just means that the LED’s will blink super fast an no ones eye will notice… but here comes the wierd part, i looked up how to measure voltage with the multimeter in AC, it said i was supposed to break the circuit and complete the circuit with my multimeter, but i read the same voltage (6.5) with, or without the resistors, and the 6.5 voltage is burning my LED’s out after about 18 hours…

I replaced my in-door lights of my old car with some blue LEDs a few years back. To test the setup, I used a variable AC-to-DC transformer I got at Walmart for $15. It was one of those “universal power plugs” that had a switch to go anywhere from 1.5V to 12V at 500mA. I cut off the universal plugs and put on some 1A-rated insulated gator clips. To help with troubleshooting, you may start with something like this (you’ll at least stop burning out LEDs right now), but just be careful that the clips DO NOT TOUCH when it’s plugged in.

I didn’t understand the results of my testing – since the door light’s source was 9VDC I needed to be able to use 9V. I accomplished this with a small-value resistor (100-300ohms, trial and error to find the right one) and no less than 6 LEDs in series. Any less and I’d get a nice POP POP POP out of the first LED within seconds. At most I could use 8 LEDs before all of the LEDs became too dim to see, and at no point could I used LEDs in parallel with each other. I had my 6 LED strip setup in my car for 2 years before I sold it, and none ever burned out.

I’m not sure about your AC voltmeter question. I hope this helps a bit. It was fun to turn out the lights and watch a single LED blow up when I plugged in the power at 12V

Just a note of caution about AC voltages… an AC voltage refers to the “root mean square”, or RMS voltage. The peak voltage of an AC waveform is commonly about 40% higher than the measured RMS voltage, so you are giving your LEDs a tiny little blast of 10V sixty times each second.

Your voltage readings are correct. With the resistor in place the circuit should draw approximately 65mA. That should not be enough to sufficiently drop the voltage coming from the power supply, so you should not expect to see a difference in voltage whether you use the 100 ohm resistor or an open circuit (infinite resistance, no current.)

To prolong the life of your LED’s you need to drop the current flowing through them. I would try using a 200 ohm resistor in series with each LED. Alternatively, finding a lower voltage power supply should do the trick, too.

In the meanwhile, take some satisfaction in knowing that you have been successful in creating the very advanced device known to my students as the DED, or “Darkness Emitting Diode”.

Sorry for posting twice in a row, but I should have addressed this point as well.

LEDs are non-ohmic devices. In other words, they don’t follow Ohm’s law. In fact you will find that most LEDs appear to have an extremely high resistance when they see less than about 1.5 volts. This means that if you have 8 LEDs in series, you probably won’t have any significant current flow with anything less than 12 volts.

When that Voltage is reached, current flow tends to increase exponentially… thus a small fluctuation in voltage can cause a huge change in current. This increase in current can quickly burn out an LED. That is why it is important to have a resistor in series with the LED… it provides an ohmic (linear) increase in current with voltage and keeps LEDs from becoming DEDs. (See above)

Wikipedia has some interesting information on LEDs that explains this in much more detail, and also mentions that many LEDs have a maximum reverse voltage of about 5v… meaning that the application suggested by the original poster where the diodes are hooked up to a 6.5VAC supply may be over stressing the LEDs in the reverse direction. This is not something I have observed, but then again, I tend to use my LEDs in DC circuits and thus would not expect to have observed it.

Chief,
Jason is close. The voltage you are reading is an approximation of RMS and for 60 Hz sine wave it is pretty good. To convert to the peak you multiply the RMS reading by 1.414 to get the peak and then by 2 to get the peak to peak or 18.32 volts. This is the full voltage swing available across the output terminals of the transformer. Calculating for 40 ma of current through the diodes, 18.32-3.7(diode forward drop)=14.682/.040 amps=367 ohms. The next closest standard value is either 390 or 270. I would go with the 390.
Without a full wave rectifier, the diode is only going to conduct for half the cycle of the sine wave giving you half the average current at a 60 Hz repetition rate. You can try the 270 and see if it makes it any brighter. You didn’t give any current specs on the diode you were using so I guessed it to be 40 ma. If your diode is different than recalculate the series resistance. At your current resistance the diode is being forced to sink about 150 ma or more than 0.5 watts.

To measure current with a multimeter, you have to break the circuit and put the meter in series with the load. Maybe that’s what the instructions that you read were talking about.

As Al suggested, you need to figure out what the recommended current is for the LED, then use Ohm’s law to calculate how much resistance you need to limit the current to that amount.

Instead of a full wave rectifier, I think you could simply place another LED anti-parallel to your first one. That is to say, put it in the same place as LED1, but backwards.

You should check its reverse voltage rating. It is possible that you are exceeding it for the other half of your cycle. Remember, since the resistors are not conducting, they are not dropping voltage. Therefore, you see the full voltage in reverse. Adding the anti-parallel LED would fix this, as it would clamp the reverse voltage seen by LED1 to the forward voltage of LED2 (3.7Volts).

For reference, I’ve found a 3.7Volt LED with a reverse rating of 5Volts. If you were using this part, it would probably fail in a similar manner to what you are experiencing, as you hit it with 9.2ish volts reverse.

Please post the part number or the data sheet for the LED you are playing with.

Lastly, forgive my intrusion Al, but I’m not convinced on your second doubling of the voltage. I think our “disagreement” stems from semantics, and is highly dependent on what “Chief Pride” actually has in front of him. Since his multimeter says 6.5VRMS, I believe his Resistor/Diode circuit will only see 6.5V*sqrt(2) instantaneous peak voltage, so long as he is connecting at the same points the multimeter is. I do agree that the full range swing is 2x that, but I don’t think it will ever “feel” it.

My reasoning goes as follows: If we pretend to ground one of the two outputs of the transformer, the other output will swing from +9.2ish Volts to -9.2ish Volts. The peak to peak is 18.4, but there is no single point in time that has that. However, I have been wrong at least twice in the past, so please correct me.

In addition to what was said before, some other considerations:

Be sure to consider the power dropped by the resistor. As an example, because P=I*V, 40 mA at 18 volts = 0.72 watts - use a 1 watt or larger resistor or it’ll overheat quickly.

For LED dropping resistors, close enough is fine. If your calculations call for 310 Ohms, anything from 270 (a bit broght) to 390) a bit dimmer) is fine.

You must always have a resistor in series with LEDs, or they will go into hard conduction and overcurrent very rapidly, letting out a tiny pink glow and a small pop as they fail forever. Start too large and move smaller.

LEDs are diodes, too, so running two in opposite polarity, but in series, will mean no current flows.

The short answer to what you did wrong is, your 110 Ohm resistor is too small, Al gives good values.

Erik,
By definition, RMS of a pure sine wave is .707 * the peak AC voltage. Since peak is one half peak to peak then the correct method of working back from RMS to P-P is Volts(peak to peak)=1.414 * Volts (RMS) * 2. Remember that RMS is intended to be an approixmation for the “heating” equivalent of an AC wave compared to a DC voltage that would produce the same heat in a load. By using the RMS method, this takes into account the varying AC voltage of a sine wave. Now, some people will say that the average voltage of a sine wave is zero since the positive portion of the cycle is exactly the same magnitude but opposite phase as the negative portion of the cycle. However, anyone can observe that a light bulb does heat to incandescence on an AC voltage. So by observation we know that AC voltage does have some ability to do work.
A quick search gave me this link…http://www.bcae1.com/voltages.htm. It show the relationship of P-P, Peak and RMS values superimposed over a sine wave.
One thing to note as well, the 120 volt AC house power supply is also measured in RMS. So that gives 165 volts peak and 330 volts P-P. Hence the need to have high voltage insulation in house wiring. As an aside, line voltage operated TV sets use a full wave rectifier and a simple capacitor filter right off the power line, making most TVs primary power supply 165 volts DC.

Shouldn’t power calculations be done with RMS voltage and current?

P=IV=V^2/R

For 270 ohms:
P = 6.5^2/270 = .156 W

For 390 ohms:
P = 6.5^2/390 = .108 W

1/4 W resistors should be fine as far as I can tell. I am still learning this stuff though (Sophomore EE major) so please feel free to correct anything here not quite right.

Emitted Colour : OCEAN BLUE
Size (mm) : 5mm T1 3/4
Lens Colour : Water Clear
Peak Wave Length (nm) : 465 ~ 470
Forward Voltage (V) : 3.2 ~ 3.8
Reverse Current (uA) : <=30
Luminous Intensity Typ Iv (mcd) : Average in 6000
Life Rating : 100,000 Hours
Viewing Angle : ±10°
Absolute Maximum Ratings (Ta=25°C)
Max Power Dissipation : 80mw
Max Continuous Forward Current : 30mA
Max Peak Forward Current : 75mA
Reverse Voltage : 5~6V
Lead Soldering Temperature : 240°C (<5Sec)
Operating Temperature Range : -25°C ~ +85°C
Preservative Temperature Range : -30°C ~ +100°C

the data i entered into that calculator is as follows:

Supply Voltage: 6.5
Voltage Drop Across LED: 3.2
Desired LED Current: 30

this is where i recieved my 110 ohm resistor… if i understand AL, he is saying that i calculated the voltage as if it were DC, and didnt take into account the range of AC, so the real current is 18.32, and from a couple people i apparently misunderstood my LED data, and the actual drop across the led is 3.7, the actual resistor needed is a 487ohm and 0.4386 watts…

now, i still have a few questions
how do i know a resistors wattage?
and, why wouldent i be able to use LED’s in parallel? according to this: http://led.linear1.org/led.wiz i should be able to…

James,
The value I calaculated was based on 40 ma, just a guess on my part as max current through your LED. The actual recommended is the 30 ma which is the value you used. The closest standard value resistor is either 470 or 560. The 470 should be just fine. Power=I^^2 * R or V^^2/R or V * I. Since you have already calculated the voltage drop for the resistor, 14.62 and the current is 30 ma, then the power in the resistor is 14.62 * .03=0.4386 watts. A half watt resistor should then be fine unless you severely limit the airflow around the resistor. A 1 watt resistor would work fine as well. If you are unable to find 1 watt resistors, you can use two 910 ohm, 1/2 watt resistors in parallel, to form a 1 watt, 455 ohm resistor. Resistor wattage is more often known by it’s size or the package markings. Less frequently it might be printed on the resistor body. I believe Radio Shack has a few 1 watt resistor choices, but many half and quarter watt choices.
To not confuse terms, the 18.32 volts is the Peak voltage developed by the transformer which is the highest voltage available for the LED during the peak of the sine wave. Real is a term used in complex AC circuits in which there exists reactance (capacitors and inductors) that affect calculations using vector analysis.

Yes, that is correct, because the heat generated is not constant but pulsed, and the important factor is the average power not the peak.

RMS is a mathematical construct used with sinusoidal signals to account for the ‘real world’ effects of such non-constant voltages. Using RMS in this calculation is good engineering practice.

On the other hand, I doubt the meter is really measuring RMS, more likely just ‘close enough’. And that’s OK.

Also, in the LED specs, it states the max reverse voltage is 5~6 volts? I’d pick the lower (5) and make sure you manage that. Put a regular 1N400x diode in series to half-wave rectify it (turning it into pulsed DC) and don’t subject the LED to any reverse voltage.