how to get IR emitter/receiver to work

Other Hardware General Electronics
1

So, I bought an infrared LED and photodiode in a package from Radio Shack a while ago, and now that I've finally found them in my messy room again I can't seem to get them to work. I have a test circuit (no arduino involved yet) of 2 AAs, to a 330 ohm resistor, to a normal LED, to the photodiode, back to the AAs, and the LED lights only slightly dimmer than without the photodiode.

Then on another circuit I have 1 AA, to another 330 ohm, to the IR LED, back to the AA. Opening and closing this circuit pointing the LED to the photodiode doesn't seem to make any difference in the brightness of the normal LED.

Here are the specs on the back of the package word for word:

Detector (photo diode--clear package)
Standard T-1 3/4 package

Maximum Voltage and Currents
Reverse breakdown 60V
Foward voltage 1.2V@l F 1mA
Total power dissipation: 100mW
Peak sensitivity wavelength: 900nm
Spectral bandwidth range: 750 - 980nm
Angle of half sensitivity: ±80°

======================
Emitter (tinted package)
Standard T-1 3/4 package

Maximum Voltage and Currents
Reverse voltage: 5V
Continuous foward current: 100mA
Foward voltage: 1.5V typ., 1.8V max @50 mA
Radiant power output: 75.5mW/sr (typical)
Wavelength at peak emission: 850nm

So yeah am I doing something wrong? I might be mixing up forward/reverse current (by the way they misspelled "forward" as "foward" on the package LOL :P), and yes I have checked and double-checked my wiring, everything is solid and connected + to anode - to cathode.

I CAN tell you that I'm pretty sure the problem lies with the emitter, because I hooked up another normal LED to it in series (running on 2 AAs and a 330 resistor) and they did not light. If I bypass the IR LED with a jumper, the normal LED lights just fine.

I suppose the emitter could be faulty, but could there be another problem? Thanks in advance!

2

Your test of the IR LED was flawed. When you put it in series with the LED, the combined junction drops could easily meet or exceed the power supply voltage, 3V.

Also you need to research photodiode circuits. It doesn't sound like you are using it correctly. The output from it is a very weak current, it usually needs to be amplified.

3

True, but I don't want to fry the IR LED. Is the max I can put through it the forward voltage or reverse? If it's reverse then I'll try it with 5V. As for the photodiode, if it's so weak then why was the LED at nearly full brightness?

4

Squirt_5432:
True, but I don't want to fry the IR LED. Is the max I can put through it the forward voltage or reverse? If it's reverse then I'll try it with 5V. As for the photodiode, if it's so weak then why was the LED at nearly full brightness?

Two things could fry the IR LED. Exceeding either the maximum reverse voltage, or the maximum forward current. But I seriously doubt that the IR emitter is faulty, so you're wasting your time.

The photodiode probably turned on with the LED because the combined forward voltage of the photodiode and the LED was less than 3V.

You really need to follow my advice and research photodiode operation. The two common modes are photoconductive and photovoltaic. Either way will work, but you're not using either.

Find and download some photodiode data sheets. They often have test circuits and example application circuits that can help you get on the right path.

5

aarg:
Find and download some photodiode data sheets.

Would something like this work? If so, would the 47K resistor in the example be a good choice? (sorry, I'm still really new at a lot of electronics things and still learning)

Also, I tried pointing the emitter at a normal photoresistor circuit (I tested it with a different IR LED that I knew worked as well and that one made a difference) and that didn't do anything either. Does that change anything?

image.jpg

6

That phototransistor circuit would work, but you don't have a phototransistor, you have a photodiode. Regarding your second statement, photoresistors don't work well with IR emitters because (if memory serves me) their peak sensitivity is in the green-yellow range. Have you found any datasheets yet?

Also, I just googled "photodiode circuit" and the first hit was a usable circuit diagram. :slight_smile:

7

Hmm.. ok I'll try that when I get home. And actually what I was saying is I found an IR remote and pointed it at the photoresistor and it DID effect it, so I know that one does detect light in the IR range. BUT it occured to me that most cameras display IR light as white so I tried pointing it at my iPod camera and found that it actually does work.

Is this the diagram you found? If so what's the difference between gnd and -5V?

image.jpg

8

Squirt_5432:
Is this the diagram you found? If so what's the difference between gnd and -5V?

One of many. The difference between gnd and -5V is 5V.
Great idea about the camera!

9

I want to post, "Try this"...
but it may all be a bridge too far.
It requires a controlled environment (placement in a box.)

In "darkness" there will be leakage current, but it should be a greater impedance than the resistor (maybe my suggested values are a little high, but don't go to less than 1K) - effectively an open circuit between "sig out" and +5.
The signal out should be low (no) volts till the photodiode conducts.

irattempt.JPG

10

Not really too far out... it's worth a try. But the photodiode is shown forward biased. I think it should be reverse biased. A higher resistor value might be needed in the photodiode circuit. If you add a pot in series, you could adjust the sensitivity.

11

aarg:
I think it should be reverse biased.

Why?
"Reverse bias" results wee changes, there's typically a "very" high series resistance limiting reverse current necessarily, and is for sensitivity-based applications.
As I've shown, or attempted to show - the photodiode, not unlike a photocell, in darkness presents a "great" resistance compared to the series resistor and illuminated presents not much at all in comparison. It's important to not get carried away with the current allowed through it - and with the values I've suggested everything should stay reasonable.
The proximity of the IR LED to the IRphotodiode plays a part too (vis-à-visworking vs not working, not "fried" vs OK.)

All I can find in my stash of stuff is phototransistors.

12

In photodiodes (indeed, all diodes are light sensitive), the light induced current flows in the opposite direction of the diode "arrow".

Here is a definitive guide to using photodiodes: http://ecee.colorado.edu/~ecen4827/hw/hw1/PhotodiodeAmplifers.pdf

13

WELL... now I feel like an idiot; I didn't know what I was dealing with. I looked up the difference between photodiodes, photoresistors, and phototransistors (which I probably should have done a while ago), and found that I was thinking the photodiode had the same properties as a photoresistor! In that case, I should probably go buy some amplifiers. Thanks for your help, guys!

14

...On that topic, would anyone happen to have a good link that gives basic knowledge of how to use an 8-pin amplifier? :stuck_out_tongue: I'm finding all sorts of different setups on google images...

15

See figure 39 of a LM339 PDF data sheet for a photo diode cct that may meet your needs.

Edit: http://bitsavers.trailing-edge.com/pdf/national/_appNotes/AN-0074.pdf

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16

Well, the interesting thing about the paper tape reader circuit, is that the actual diode bias and output part of the circuit really consists of nothing more than the diode and a 100k resistor (because there is no negative feedback). The op amp just sets a threshold at VCC/2 and adds some hysteresis. So you could probably skip the op amp and connect the same circuit directly to a digital input.

In the era of paper tape, all the logic was TTL, or even worse, RTL. TTL doesn't have a high impedance input.

17

This is not an OP amp it is a high speed voltage comparator ! !
Response Time ~1.3uS.
"there is no negative feedback"
Consider looking up how a voltage comparator works.

So you could probably skip the op amp and connect the same circuit directly to a digital input.

No you cannot.

Look at the data sheet for the LM339/139.
The o/p is open collector so you can do whatever you want with it.

18

LarryD:
This is not an OP amp it is a high speed voltage comparator ! !
Response Time ~1.3uS.
"there is no negative feedback"
Consider looking up how a voltage comparator works.No you cannot.

Look at the data sheet for the LM339/139.
The o/p is open collector so you can do whatever you want with it.

I'm not a noob. I am intimately familiar with the differences between op amps and comparators. Give me a blackboard and I could design one for you out of transistors and resistors, off the top of my head without consulting any books. Doubtless you are already aware of the functional similarities. They are both infinite feedback differential amplifiers. I have used the 339 and op amps in hundreds of designs, so I don't need to look at it today, unless I were starting a new design. I apologize for the error, however in the context of my reply what I said still holds true for a comparator.

Perhaps you should review the circuit and identify how there can be negative feedback, since there is absolutely no signal path between the output and the inverting input.

Of course you should use a comparator, for well, comparing, but you don't have to. An op amp will perform the same function when inserted in a similar circuit. A comparator can be configured as an op amp as well, but isn't designed for closed loop feedback stability so it's a mess.

Whether the output is open collector or not, has no bearing on the issue at hand, which is interfacing a photodiode to the Arduino. The 339 was probably also selected for the original application versus an op amp because the output voltage range of an op amp would not have been sufficient to drive a TTL input.

Regards my suggestion. compare an Arduino input with the comparator input. The comparator is a high impedance input. The comparator has a threshold near VCC/2. The comparator has a digital output. The Arduino input satisfies all these criteria. The only thing missing is hysteresis. If it's debounced that won't be an issue.

Also, I'm trying to help the OP. How does your post do that?

19

accidental post

20

The cct will help the OP get his future projects to work with an IR receiver (which he rediscovered on his desk).
The cct does indeed work well and I urge th OP to try it.