LEDs and Photodetetor emitters ???

Thread Starter

Mathematics!

Joined Jul 21, 2008
1,036
Ok , I am trying to understand some of the characteristic's stuff.
This package of radio shack part 276-142 Infrared Emitter / Detector.

It has Detector__________________________________________
Max voltage and current
Collector to emitter 70 V
Emitter to collector 5V
Collector current 50mA
Total power dissipation 150mW
peak sensitivity wavelength 850nm
spectral bandwidth range: 620 - 980nm
angle of half sensitivity = +-20 degrees

_________________________________________________________

Emitter has
Max current and voltage ratings
Reverse voltage 5V
Continous forward current 150mA
Forward voltage 1.3v typ. 1.7v max
radiant power output 13 - 15mW
Wavelength at peak emission 950nm

________________________________________________________

First , I would think that the forward current * forward voltage = the radiant power output?

But 1.3v * 0.15 A = 195 mW much >>> then 13mW - 15mW.
Is it lossing this much enery in the form of heat?

Wavelength at peak emission how did they calculate the peak emission?

And for the detector
peak sensitivity wavelength 850nm
spectral bandwidth range: 620 - 980nm
angle of half sensitivity = +-20 degrees

what is the angle of half sensitivity for? Does it mean if I put the detector at an angle of more then 20 degrees it would give a less then have strength of the signal?

spectral bandwidth range can I some how build a circuit so it output's any frequency I want with in the 620 - 980nm. If so what controlls this and how did they come up with this range?

And how do you know from these chracteristics what the max transmission can be? Like how far away can I put the detector from the emitter to have it still work.

And must their be a line of sight with infrared or can it pass thru stuff like wall's etc?

Thanks for any clarity on these questions.
 
Last edited:

beenthere

Joined Apr 20, 2004
15,819
Emitter:

I question the 150 ma figure - most I have seen have a rating for 50 ma. But IR diodes aren't all that efficient.

The peak emission was measure with a spectrophotometer. It's a device capable of measuring light energy at very narrow bandwidths.

Detector:

The little lens can't bring much IR to focus on the transistor with the source off axis.

The spectral bandwidth of the phototransistor is determined by photoelectric effect. If a photon with the correct energy falls on the sensitive surface, it will kick an electron or two loose. I strongly siuspect someone shined different frequencies of light on the phototransistor and determined the upper and lower boundaries of its response

Since the signal out of the phototransistor can be amplified, I don't think you can determine a range from these figures. If you modulate the source at, say, 38 KHz like most remotes, then it's easier to pick out the signal from a noisy background. All light sources but LED's tend to emit IR.

Gotta go up in frequency to pass through walls. You could go around corners with a mirror, I suppose.
 

Thread Starter

Mathematics!

Joined Jul 21, 2008
1,036
Ok

But now I am having trouble using it.

For example I connected a AA 1.5 volt battery to the transmitter leds and a switch.

Then I made another circuit

+|-------detector led --- regular red light led----
9V..............................................................|
-|---------------------------------------------|

I thought when I turn on the switch the transmitter led should transmit
and the detector should detected it and turn on the red light?

I had a dim red light when I put the circuits with in a half of an inch.
But now I cann't even get that dim light to go any more.

Does the current flow from emitter to collector or the other way around?
Because one way has max 5V the other 70V???

is the current going to slow to show the red light?


anode is on the + side of circuit
cathnode on the -

But still confused about where Emitter and Collector legs should go.
I was getting the dim light when I had Collector leg--emitter leg-- anode leg -- cathode leg ---

This is killing me I have been try to figure this out the whole day.
Any help would be great
Thanks again
 
Last edited:

beenthere

Joined Apr 20, 2004
15,819
The IRLED need 1.5 volts, on average, to go into conduction. Same for the red one.Having done so, there must be some device, like a resistor, to limit current. See our Ebook - http://www.allaboutcircuits.com/vol_3/chpt_3/12.html.

So you have either destroyed the PN junction in the red or IRLED, or drained the 9 volt cell. Those batteries are only meant to supply low currents like 15 ma or less.
 
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