I've used those combo packs from RadioShack and IIRC the tinted one is the LED and the clear one is the sensor, which is actually a photodiode, not a phototransistor.
Some of the reviews on the RS site said that too.
I tried the diode test on both components.
1. tinted component - hooked leads up to DMM - display showed '1' or open. shined light - no change. switched leads and shined light - no change
2. clear component - hooked leads up to DMM 0 display showed 617. Shined light - display changed to 670. reversed biased the leads and the display read '1' or open. Shined light and the display read -300.
Based on this, does this prove the tint component is the LED and the clear one is the photodiode?
Maybe
The way to be sure is to watch the LED using a cellphone camera, and trying both components, both ways round, with a 220 ohm limiting resistor. Only the LED will emit light, and it will emit light only when connected the right way round.
If this is a photodiode, would this not produce the results I was looking for in my initial post? Do I need to get an actual phototransistor in order to get those results?? Again, the results I'm looking for is a voltage of 800 mV when the sensor path is clear and 4-5 volts when the sensor path is blocked (i.e. golf ball has passed sensor)
You need a phototransistor. If you have a photodiode, you can add an external transistor. Connect the photodiode's cathode to the transistor's collector, and the photodiode's anode to the transistor's base. This uses the photodiode in photoconductive mode, which is the way normal phototransistors work. kpatz's earlier suggestion uses photovoltaic mode and I'm not sure if it will work very well.
I connected the photodiode as you suggested - reverse biased and +5V supply and the resistor. It measured 5volts. I shined my flashlight directly onto the photodiode and the voltage went practically to 0
That sounds like it's a phototransistor.
I hooked up the LED and moved the photodiode very close - there was no change in voltage. I did this to both of the 'tinted' components. I then tried a normal LED and when I moved the photodiode close, the voltage dropped by about 2 mV
I would expect the voltage to drop pretty low with a normal LED shining on it. Is your flashlight a lot brighter than a normal LED?
I just found another camera and when I plugged in both 'tinted' LEDs, I could see a little purple light on each so I guess they are emitting.
You should see a good clear bright light from the IR LED on your camera display.
That would indicate that it is a phototransistor and was connected the way you want it.
Yes, I agree.
Also, the emitter is rated as 1.3V and 150mA. To get that current with a 5V supply, the resistor would be 25 Ohms. With a 220 resistor you may not be getting a lot of light out of it. Try lowering the resistor.
I don't think that's a good idea - I don't trust Radio Shack's specifications, and in any case, 20 mA should be plenty of current to get a good visible indication on a cellphone camera display.
The action of the phototransistor is the opposite of what I'm actually looking for. I'm looking for the following: when +5V is applied and the path is clear, I want about 800 mV and when something is blocking the sensor, I want 4-5V.
That's what you're getting. With no light falling on it, the output voltage is high, and with light falling on it, it drops low. That's what I would expect with a phototransistor, and it matches the description of what you need, AFAIK.