circuit-designled
I have this IR LED
IR LEDs emit when forward biased right? And we all know that the longer leg is positive.
When I forward-bias this LED it acts as an open circuit. And it emits when reverse biased!
Would you please help me sorting this out? If you want further explanation please let me know.
You're getting the expected result. What you see is the normal behavior of diodes in series and it's completely normal to have one resistor and a string of LEDs connected after it.
What's basically happening is this: When they told you that the forward voltage is 2 V, they lied. It actually depends on the current going through the LED and you can consider the 2 V some sort of nominal value, but the exact drop should be read in the datasheet (if it's available).
In general case when you want to connect diodes in series, you use this formula for resistor:
$$ R= \frac {V_{supply}-NV_{f}}{I_{f}}$$ where the N is number of diodes you have.
This way it turns into simple Ohm's law. But in your case, you're approaching the border at which the above formula will not hold. You basically have a circuit with one branch only and the current going through that branch isn't going to much change with the number of LEDs if the voltage of the supply is high enough to be higher than LED forward voltage.
Take a look at this diagram from Wikipedia:
Notice the point marked \$ V_d\$. For this diode, once the voltage at the diode terminals reaches that point, the current will start quickly increasing with only a small change in voltage. That is why adding more LEDs doesn't immediately affect current. The voltage is high enough that all LEDs will conduct. Should you for example put 10 LEDs in series, the voltage will be too low and they will either show barely noticeable light levels or stay off.
Next, let's take a look at the different voltages you got at the LEDs. Again take a look at the curve for the diode from the Wikipedia. The \$V_d\$ point for each diode made is different and there are some tolerances here. So some diodes of same model number will at same current have a bit larger voltage drop and others will have a bit smaller voltage drop.
Next about LEDs in series. There is nothing wrong with that, but you're still not doing it right. Using the formula I provided, you should set the resistor so that the LEDs will be within their rated current. If you fulfill that condition, there's absolutely nothing wrong with having multiple LEDs connected in series, should you have voltage to spare.
I can make a couple of guesses about what might be going on.
You aren't actually measuring the LED current, but instead you are assuming \$I_{LED} = (V_{in} - V_F)/220\$ and assuming \$V_F\$ is a constant. You are operating in a regime where \$V_F\$ will vary significantly (100's of mV), and neglecting this would explain (qualitatively, anyway) the shape of your graph.
The LED has some parasitic conductive path through it that does not cause light emission. For low currents, this path is taking more of the current and resulting in reduced light emissions until some threshold is reached and the proper path begins to dominate.
At very low currents, the optical emission pattern of the LED is changing, causing less of the light to reach the detector and more to be lost into the package.
The way you are hooking up the detector is causing funny behavior. Hooking up the base but not the emitter is not a common way to use a phototransistor (AFAIK), and is not what the device designers would be designing for. Typical phototransistor circuits are shown in an App Note from Sharp starting on page 13.
Best Answer
Quite simply it is backwards from how standard LEDs are configured.
Usually the long leg is the anode and the leg that connects to the bigger metal object inside the plastic case is the cathode, but that is not always the case.
For example:
You'll have to trust me that the leg with the resistor soldered to it was shorter than the other one, but the LED only illuminates when the longer lead is connected to battery negative.