IR Phototransistor pinout/hookup question

raschemmel:
I've used the circuit described in this article before

While he could do that and use the photo-diode, it's probably better to just buy a 'real' photo-transistor.

I am an old campaigner and I love a good campaign.

So, here, my "Detector" (as D.U.T.) findings --

I have the IR emitter pretty close to the detector, 0.1 inches or so.
I do have a light on, so that I can see what I'm doing, which influences the results a tad.
Uninterrupted, unobstructed, there is 0.8V at Vout and 1.4V when I interrupt, or obstruct, the optical path between.
If I place translucent material between (esd mylar bag, old pink anti-static bag), Vout varies with the material's transmissibility.

[quote author=Runaway Pancake date=1448159056 link=msg=2489337]
I am an old campaigner and I love a good campaign.

So, here, my "Detector" (as D.U.T.) findings --

I have the IR emitter pretty close to the detector, 0.1 inches or so.
I do have a light on, so that I can see what I'm doing, which influences the results a tad.
Uninterrupted, unobstructed, there is 0.8V at Vout and 1.4V when I interrupt, or obstruct, the optical path between.
If I place translucent material between (esd mylar bag, old pink anti-static bag), Vout varies with the material's transmissibility.[/quote]
In the dark, I imagine the collector would go much closer to the 5V rail. (When the beam is interrupted.)
I'm surprised that the output doesn't go closer to 0V than 0.8V with them that close together though.

How hard are you driving the emitter? 100mA?

OldSteve:
In the dark, I imagine the collector would go much closer to the 5V rail. (When the beam is interrupted.)
I'm surprised that the output doesn't go closer to 0V than 0.8V with them that close together though.

How hard are you driving the emitter? 100mA?

I'm only running about 20mA through the emitter. (The "spec card" states 40mA max for this.)
Changing the Vout pullup to 1kohm: obstructed Vout = 3.9V (also dark, under an old 35mm film holder) and unobstructed = 2.3V

[quote author=Runaway Pancake date=1448166092 link=msg=2489408]
I'm only running about 20mA through the emitter. (The "spec card" states 40mA max for this.)
Changing the Vout pullup to 1kohm: obstructed Vout = 3.9V (also dark, under an old 35mm film holder) and unobstructed = 2.3V[/quote]
You'd get better results with a high-power IR LED. Mine are rated for 100mA continuous, 1A peak. (Everlight IR333-A)
I'm running a 25% duty-cycle, (38kHz), in my current project and driving the IR LED at 375mA, (10 ohm series resistor, 5V supply). Modulated at 1kHz. Great range.

OldSteve:
You'd get better results with a high-power IR LED.

Likely so, but perhaps somebody else.
My interest is only bringing clarity to the subject at hand.
Over the years they probably used several devices presenting these pairs.
My pack was pretty old. The "blister" had yellowed as well as the phototransistor encapsulant, but I opened it up for the sake of Science - to shed some badly needed light on this matter.

[quote author=Runaway Pancake link=msg=2489823 date=1448197577]
Likely so, but perhaps somebody else.
My interest is only bringing clarity to the subject at hand.
Over the years they probably used several devices presenting these pairs.
My pack was pretty old. The "blister" had yellowed as well as the phototransistor encapsulant, but I opened it up for the sake of Science - to shed some badly needed light on this matter.[/quote]
I think that light was shed when it was realised that 'interestingfellow' really had a photo-diode, not a photo-transistor.
There wasn't actually a need for you to show the performance of your photo-transistor and low-power IR LED.

Well, that makes me feel better; I was making a mistake, just not the one I thought.

Thanks, agian, y'all!

(and, no, I didn't notice the lack of transistor symbol, but I won't anymore* ;D )

OldSteve:
I think that light was shed when it was realised that 'interestingfellow' really had a photo-diode, not a photo-transistor.
There wasn't actually a need for you to show the performance of your photo-transistor and low-power IR LED.

You like to be a real wiseguy.
"Realized" by whom? You? So what?

The OP has no interest in experimentation, nothing is conforming to his preconceptions. And you want to speculate endlessly, from your conceit of authority, based on nothing more than how you figure things ought to be - this part you like, that part you don't, so you make up the rest.
The RadioShack package that the OP's using has a phototransistor, albeit mismarked, but you can persist contrariwise all you like. The only authority here is the data and if the OP could be pulled away from playing with his toys, his admitted primary concern, and actually participate in his subject - then we might have more data.

Did you bother to read Dave Evans' post, #18?

And no need to get personal.

OldSteve:
Did you bother to read Dave Evans' post, #18?

Yes, as a matter of fact.
How critically did you read it?
Did you catch the part where he basically didn't know for certain?
"and the detector was a photo-diode (or so the package said - I haven't tested it or used it)! "
In both cases (the OP's and mine) there are mismarkings. So, we fill our boots and do some intelligent experimenting.
Or just bin the lot.

interestingfellow:
Well, that makes me feel better; I was making a mistake, just not the one I thought.

So what could I do with the set I have? Or rather, a typical intended purpose?

Thanks, agian, y'all!

(and, no, I didn't notice the lack of transistor symbol, but I won't anymore* ;D )

According to the label, and Dave Evans in post #18, (Edit: concession to 'Runaway Pancake's last post.) it's most likely a photo-diode and not a photo-transistor. I think that's pretty clear, despite what someone else thinks. That's no doubt why it wouldn't work for you, when connected as a phototransistor.

You could make use of it, but you'd get much better performance from a photo-transistor.
There are some photodiode circuits here, but they require additional components to be really useful:-
Photodiode Circuits

For your optical tacho, it would really be best if you could get a phototransistor. (Unless you do feel like adding a transistor or op-amp to the photodiode for experiment.)
Either way, let us know how you go. :slight_smile:

Edit: You could set up a test circuit, I guess, just to verify exactly what it is. Just copy the photodiode/transistor circuit from the Google image search link I posted. This one or similar:-

I think what I showed in #26 is the way to go (despite "phototransitor" (sic).
Have you tried that, exactly?
[I won't guaranty that "digital-worthy" performance can be had from these components alone.]

This is kind of sad the whole problem here is not knowing what the op has forsure.
But be it a photodiode or a phototransistor they both will work for caching the blinks of a ir led. Even a red led will work.

A phototransistor is "A junction transistor which responds to incident light by generating and amplifying an electric current. "
A photodiode is "A semiconductor diode which generates a potential difference or changes its electrical resistance when illuminated. "

Now here are some more expanded explanations.

A phototransistor is a photosensible semiconductor device comprising three electrodes. Light or ultraviolet light activates this bipolar junction transistor.
Illumination of the base generates carriers which supply the base signal while the base electrode is left floating.The emitter junction constitutes a diode , and transistor action amplifies the incident light inducing the signal current.

A photodiode is a semiconductor diode which gives an important photocurrent under illumination.
One type of photodiode uses a p-n reverse biased junction operated below the breakdown voltage. Excess charge carriers or electron hole pairs are generated by photoconductivity with exposition to electromagnetic radiation.Carriers usually recombine quickly , but the ones produced near or in the depletion layer of the junction can cross the junction and give a photocurrent which is superimposed on the small reverse saturation current.
Photodiodes allow the manufacture of devices with a depletion width convenient for best sensitivity and frequency response.

Using a led is good for some things but heres a little wirte up from makezine How to Use LEDs to Detect Light - Make:

Runaway Pancake post #26 is the best start at getting this to work the only problem is hooking the IR whatever the OP has right.
If it's a diode it conducts only when light hits it and backward you still get nothing
I would set it up like that and shine IR light on it and test output nothing then swap it's pins around one way it's going to work.
If it doesn't throw it in the trash and get new one.
Oh and post 26 works for me with radio shacks bagged set.

I don't have time to try it right now but as soon as I can, I will.
I know I have some 2n7000s with me but I'm not sure about BC547's...

Thank you all again.

I found a guy on ebay who had these (276-0142) for sale. I ordered two packages.
I expected them to be the same. As fortune would have it - they aren't, exactly.
And who's surprised?

From the front, same label, looks like the same components.

But, on the back - Differences!

Anybody want to continue the "conversation"?

I gave up on that dern radiocrack crap. I managed to go home for the holiday, went ahead and pulled out my parts bin, and salvaged some proper printer emiter/phototransistor pairs. I have 2 pair. So far, I have verified that my 16x2 lcd works, and have since moved on to testing the phototransistor.

I'll report back, sirs.

Yup - looks like my first and second packages. My first one had a phototransistor and the detector in my 2nd one really is a photodiode (I checked recently).

You all do know there are two types of photodiodes Photovoltaics and Photoconductors. And for what most people ask to do the photodiode would be better than a phototransistor. They turn on faster nS where as phototransistors take mS to turn on.