LED Resistance

It's something else.

But before we give you an answer you have to explain how you are able to see an IR LED.

Best regards.

Through his digital camera set to black and white...

Try it with your TV remote.

milo

an LED conducts one way when the voltage gets above it's critical voltage. Before that is looks like a very high resistance and microamps or less with flow. Above that voltage it becomes a very good conductor and if the voltage it too high by .1 of a volts, you might fry the diode.

They usually use a currennt limiting resistor of 100 to 200 ohms per volt above 2 volts.

So if you have a 12 volt battery, that is 10 volts above 2 volts. Thus a resistor of 1000 to 2000 ohms is needed.

10 volts will force 10 milliamps through the LED with 1000 ohms, 5 milliamps with 2000 ohms etc.

The IR signal is a complex beast and hooking simple audio up is not likely to provide decodable signals. I suggest you drill here.

http://www.google.ca/search?hl=en&q=IR+%2B%22remote+control%22+%2Bwiki&btnG=Google+Search&meta=

Thanks, but I'm not working with 12 volts (at least I don't think).

I can't get the exact measurements due to a malfunctioning multimeter but an Mp3 player can't put out more than 3 volts (battery size) and that's pretty loud for a headset.

well, it might not put out enough to properly drive the LED. You can record a data stream as audio and drive a LED. Another thing is the impedance of the ear phones. They are typically low, 4-8 ohms, but may well be higher if crystal based.

So more data needed.

How about a circuit diagram? With all values and voltages of course....

This is the closest to a diagram I can make, I wasn't sure how to depict a headphone jack.

Most IR LED's use about 1.9 volts and my multimeter isn't working to get the rest of the values. I knew I would most likely need an amplifier of somekind but i wasn't expecting the results i got.

I know this thread is dead but i felt obligated to reply

Hello Default.

It sounds as if you are overdriving the MP3 player amplifier. This could be caused by to low of an impedance caused by the IR LED. Music players use an audio amplifier. You can try adding a simple external amp to power the LED. There are many designs to be found on the web, and without knowing the exact setup it would be hard to suggest a certain type of amp. Make sure you match the input and output impedance.

Good luck.

My guess is that the output power from your computer (sound plug) is too low, Use a high speed transistor or a Jfet to switch the led on (give the led its own powersource)

mp3 players by their nature will strip out any waveforms that it decides the human ear could not hear, so your recording will not be a faithful copy of the original. secondly the sampling frequency is too low to record the IR signal anyway. you need to sample at twice the highest expected frequency or more in order to get a faithful digital recording. the carrier frequency of most IR remotes are more than double what a human ear could hear so the sampling frequency is too low by half at least.

The overloading of your amplifier with your LED is the least of your problems.

But I am using my computer (for now), and that still doesn't explain why it dims.

I know this thread is dead but I felt obligated to reply

The led gate opens in an avalance effect so its Resistance becomes very low, if you do not insert a resistor to limit the current the led will short your output.

the LED does indeed act like a low resistance short, but only above it's forward conduction voltage. red and green are typically 1.7 to 1.8 volts.

If connected to 12 volts with no limiter then will burn up immediately. If connected to 1.5 volts...not a glimmer.

Inspect this diagram

I gave this a GA for clarity. A picture isn't worth two words, if it isn't clear. But this one is worth every one of its thousand, for clarity alone.

Micah

an audio input is usually around 2.5Volt with the audio signal imposed on it, it think because it is low power the led drains the internal capacitor and the power source cannot keep. (been awhile since i dabbled in these things so correct me if i a wrong)

This sounds right, and is a particular issue where you are using an inherently "DC" supply, like a battery. If it were an AC primary source, feeding a bridge rectifier, thence to a filter capacitor (of necessarily large capacity for the purpose, but not a "super cap"), the filter on the supply would act as a DC "bucking" source, to hold the output amp drive voltage up. With a battery, there is no filter cap, unless it is to eliminate RF hum, which uses only very small capacity devices, and so you have no "buck" source to hold the output voltage up. Hence, a low-on-resistance device like an LED will drag the output down very quickly, and keep it too low, such the the LED is barely able to cycle in and out of the "knee" range in the graph of answer number 15. Certainly the LED will never drive far enough into the avalanche )above the knee) region, long enough, to deliver any useful intelligence, after the possible first on moment. In fact, it saves the output amp that it does not, as if the amp were able to hold its voltage just high enough, it would quickly overheat due to the excessive power load it had to handle.

So, a booster amp, probably even an LM741 OP Amp, would be needed. In fact, since a 741 can handle all of the necessary parameters, including delivering at least a 1/4 watt (no heat sink) or up to a full watt in a power tab (TO-220 or similar) form, it would work perfectly. And there must be a million simple OP Amp audio circuit schematics on the web. So, dig out the iron, or the breadboard, and have at it!

Micah

Thank you for the explanation!

As usual, there are a myriad of approaches to solve an engineering problem. The use of a driving operational amplifier will surely work. Using an operational amplifier will also provide the most flexible collection of circuit choices . But don't forget that an LM741 operational amplifier will also require a bipolar supply (±5V minimum, ±18 maximum) to operate. (This is not one of the amplifiers optimized for a single supply voltage.) Only a unity gain voltage follower circuit configuration does not require additional resistors. Other circuit configurations will require several resistors to configure the circuit. One of the most likely useful configurations would be a voltage summing design that could be adjusted to compensate for the turn on voltage threshold of the LED. In of these configurations a separate series resistor to the LED will still be required.

My earlier suggestion of putting in series a 1.5 volt battery, a 1K resistor, and the LED with the MP3 output may seem too simple to work. But lets not forget the great engineering axiom KISS, Keep It Simple Stupid.

My apologies to Redfred. I didn't read his post well the first time. He is right. His suggestion, simple as it is, will establish a new "zero voltage" rail 1.5 volts up (the single battery), which will provide a "springboard" for the MP3 players output to drive the IR LED into conduction. Simple and elegant. This, is old time radio transmitter circuit design was called "high level keying", and it basically makes any intelligence (the MP3 output) ride on the reference voltage, and thus look like it is 1.5 volts higher than it, in fact, ever is.

Works for me, biasing the circuit artificially. GA, Fred.

Micah