Infra Red Emitters

[Paul]

I'd like to produce a beam of IR, to trip a remote alarm when the beam
is broken.

Lots of IR Emitter LED's (and frequency matching IR Phototransistors)
are available, but I have no idea how '90mW/Sr' translates to real
life. (In fact, I don't know what 'Sr' stands for!)

I would like to span about 3 meters outdoors - so rain, fog, etc must
be catered for.

Anyone any ideas about what I need, please?

Thanks.

Paul.

 

[John Popelish]

Paul wrote:

I'd like to produce a beam of IR, to trip a remote alarm when the beam
is broken.

Lots of IR Emitter LED's (and frequency matching IR Phototransistors)
are available, but I have no idea how '90mW/Sr' translates to real
life. (In fact, I don't know what 'Sr' stands for!)

I would like to span about 3 meters outdoors - so rain, fog, etc must
be catered for.

Anyone any ideas about what I need, please?

Thanks.

Paul.

Steradian is a measure of cone angular shape. A cone of 4*pi
steradian is so wide that it folds back on itself and includes all
directions. An angle of 1 radian is about 57 degrees. A steradian is
sort of square radians, since it involves angle over one more
dimension (the shape of a cone, instead of an angle) than an angle on
a flat surface.
http://www.physlink.com/Education/AskExperts/ae174.cfm
Specifically, just as a radian is the angle you get if you form an
angle from the center of a circle through the ends of a radius long
section of the circumference of the circle. a steradian is the cone
you get if you form it to pass from the center of the sphere through
the circumference of a circle on the sphere that has an area of the
radius squared.
http://www.schorsch.com/kbase/glossary/solid_angle.html

For long distances, you need a high energy in a small cone high watts
per steradian). This doesn't mean that the total power out equals
this number or that the light covers a steradian. But if the light
did cover a cone this wide, and the intensity remained as bright as it
is in the center of the actual beam, then the total power would equal
the specified power.

I like the TSAL6100 (940 nano meter wavelength, 10 degree beam) and
TSFF5210 (870 nm wavelength and 10 degree beam) and similar devices
from Vishay.
http://www.vishay.com/docs/81009/81009.pdf
http://www.vishay.com/docs/81090/81090.pdf
http://www.vishay.com/ir-emitting-diodes/
--
John Popelish

 

[Paul]

John Popelish <jpopelish@rica.net> wrote in message news:<4083DBA9.E983F075@rica.net>...

Steradian is a measure of cone angular shape. A cone of 4*pi
steradian is so wide that it folds back on itself and includes all
directions. An angle of 1 radian is about 57 degrees. A steradian is
sort of square radians, since it involves angle over one more
dimension (the shape of a cone, instead of an angle) than an angle on
a flat surface.
http://www.physlink.com/Education/AskExperts/ae174.cfm
Specifically, just as a radian is the angle you get if you form an
angle from the center of a circle through the ends of a radius long
section of the circumference of the circle. a steradian is the cone
you get if you form it to pass from the center of the sphere through
the circumference of a circle on the sphere that has an area of the
radius squared.
http://www.schorsch.com/kbase/glossary/solid_angle.html

For long distances, you need a high energy in a small cone high watts
per steradian). This doesn't mean that the total power out equals
this number or that the light covers a steradian. But if the light
did cover a cone this wide, and the intensity remained as bright as it
is in the center of the actual beam, then the total power would equal
the specified power.

I like the TSAL6100 (940 nano meter wavelength, 10 degree beam) and
TSFF5210 (870 nm wavelength and 10 degree beam) and similar devices
from Vishay.
http://www.vishay.com/docs/81009/81009.pdf
http://www.vishay.com/docs/81090/81090.pdf
http://www.vishay.com/ir-emitting-diodes/

Thanks for that John. Just a couple of questions:

In order to cope with rain, fog etc, what kind of power would I need
to span 3 metres?
Am I better off with a shorter or longer wavelength for this
application, please?

Thanks again.

Paul.

 

[John Popelish]

Paul wrote:

John Popelish <jpopelish@rica.net> wrote in message news:<4083DBA9.E983F075@rica.net>...

Steradian is a measure of cone angular shape. A cone of 4*pi
steradian is so wide that it folds back on itself and includes all
directions. An angle of 1 radian is about 57 degrees. A steradian is
sort of square radians, since it involves angle over one more
dimension (the shape of a cone, instead of an angle) than an angle on
a flat surface.
http://www.physlink.com/Education/AskExperts/ae174.cfm
Specifically, just as a radian is the angle you get if you form an
angle from the center of a circle through the ends of a radius long
section of the circumference of the circle. a steradian is the cone
you get if you form it to pass from the center of the sphere through
the circumference of a circle on the sphere that has an area of the
radius squared.
http://www.schorsch.com/kbase/glossary/solid_angle.html

For long distances, you need a high energy in a small cone high watts
per steradian). This doesn't mean that the total power out equals
this number or that the light covers a steradian. But if the light
did cover a cone this wide, and the intensity remained as bright as it
is in the center of the actual beam, then the total power would equal
the specified power.

I like the TSAL6100 (940 nano meter wavelength, 10 degree beam) and
TSFF5210 (870 nm wavelength and 10 degree beam) and similar devices
from Vishay.
http://www.vishay.com/docs/81009/81009.pdf
http://www.vishay.com/docs/81090/81090.pdf
http://www.vishay.com/ir-emitting-diodes/

Thanks for that John. Just a couple of questions:

In order to cope with rain, fog etc, what kind of power would I need
to span 3 metres?
Am I better off with a shorter or longer wavelength for this
application, please?

Thanks again.

Paul.

I'm not sure which wavelength would be best. The shorter wavelength
is closer to the peak response of most silicon detectors (at 940 nm,
silicon is getting to be pretty transparent) and these are also happen
to be slightly more powerful emitters. But from a signal to noise
standpoint, you might get an advantage in using the longer wavelength
ones if you use a silicon detector filtered to exclude all light that
is much shorter wavelength than 900 nm, since the silicon has so
little response left longer than about 1000 nm. In other words, the
longer wavelength filtered silicon sees very little daylight between
the filter cutting off the short wavelengths and the silicon cutting
off the longer wavelengths. The longer wavelength should do a bit
better in fog, also. But I doubt there is a 2:1 advantage to either.
I would just run the LED at rated peak power, and possibly add a small
plano convex lens (perhaps a plastic fresnel) to more nearly collimate
the beam. If the whole thing operates in the dark, you may be able to
get by with a DC excitation, but if there is any significant
interfering light, you will need to modulate the emitter with some
carrier frequency (40 kHz is common) and detect energy only in that
band.

Does you design allow a synchronous detection scheme (electrical
signals shared by emitter and detector) or are you forced to a power
detector following a bandpass filter, only? The nice thing about the
latter is that they are available as an integrated optical long pass
filter, detector, amplifier, bandpass filter, rectifier and comparator
to produce a binary go/no go output from a modulated beam. (used for
infrared data communication) But the synchronous detector has an
advantage of higher ultimate signal to noise ratio capability if there
is interference near the modulation frequency.

Have you picked out a detector yet?

--
John Popelish

 

[Anand Dhuru]

Thanks for that John. Just a couple of questions:

In order to cope with rain, fog etc, what kind of power would I need
to span 3 metres?
Am I better off with a shorter or longer wavelength for this
application, please?

Thanks again.

Paul.

Hi Paul,

To achieve a long range, this is paramount; you need to *modulate* the
IR beam, and then again modulate the modulated signal! This might
sound complex but is really very easy to achieve.

Design 2 oscillators, one at 38KHz, the other at around 500Hz; you
could use a pair of 555s for this. Connect your transmitter LED to the
outputs of these; anode to the 500Hz, cathode to the 38KHz.

At the receiver end, use one of the 3 pin IR receiver/demodulator
modules, at the output of which you will get your 500Hz signal. You
could then use a 555 based missing-pulse-detector to detect the
absense of this pulse train.

This arrangement is essentially what is used in consumer IR (TVs, DVDs
etc.); it will give you ample range, and is quite immune to ambient
conditions.

Of course, in place of 2 555s you could use a 556, or a PIC
microcontroller to reduce the component count.

Hope this helps!

Regards,

Anand Dhuru

 

[Paul]

John Popelish <jpopelish@rica.net> wrote in message news:<40867B2B.97AA32B0@rica.net>...

Thanks again for your help, John.

I'm not sure which wavelength would be best. The shorter wavelength
is closer to the peak response of most silicon detectors (at 940 nm,
silicon is getting to be pretty transparent) and these are also happen
to be slightly more powerful emitters. But from a signal to noise
standpoint, you might get an advantage in using the longer wavelength
ones if you use a silicon detector filtered to exclude all light that
is much shorter wavelength than 900 nm, since the silicon has so
little response left longer than about 1000 nm. In other words, the
longer wavelength filtered silicon sees very little daylight between
the filter cutting off the short wavelengths and the silicon cutting
off the longer wavelengths. The longer wavelength should do a bit
better in fog, also. But I doubt there is a 2:1 advantage to either.
I would just run the LED at rated peak power, and possibly add a small
plano convex lens (perhaps a plastic fresnel) to more nearly collimate
the beam.

That's a good idea - thanks.

If the whole thing operates in the dark, you may be able to
get by with a DC excitation, but if there is any significant
interfering light, you will need to modulate the emitter with some
carrier frequency (40 kHz is common) and detect energy only in that
band.

This will be operating in daylight too. Anand also suggests this
approach.

Does you design allow a synchronous detection scheme (electrical
signals shared by emitter and detector) or are you forced to a power
detector following a bandpass filter, only?

No, it won't be synchronous - it would merely detect any absence of a
signal greater than 100 ms.

The nice thing about the
latter is that they are available as an integrated optical long pass
filter, detector, amplifier, bandpass filter, rectifier and comparator
to produce a binary go/no go output from a modulated beam. (used for
infrared data communication) But the synchronous detector has an
advantage of higher ultimate signal to noise ratio capability if there
is interference near the modulation frequency.

Have you picked out a detector yet?

No, not yet John - I'm still trying to understand as much as possible.
I started out on this by thinking it would be quite simple - I now see
that if this is going to work with any degree of reliability, a lot
more knowledge is required.

Paul.

 

[Paul]

ardhuru@vsnl.com (Anand Dhuru) wrote in message news:<8b81210b.0404212202.63f82118@posting.google.com>...

Thanks for that Anand. Alas, I fear your assumption regarding my level
of knowledge is somewhat more than reality . Sorry to ask what are
probably very basic questions:

To achieve a long range, this is paramount; you need to *modulate* the
IR beam, and then again modulate the modulated signal! This might
sound complex but is really very easy to achieve.

Why is such a signal desirable? Does it make it less prone to
interference?

Design 2 oscillators, one at 38KHz, the other at around 500Hz; you
could use a pair of 555s for this. Connect your transmitter LED to the
outputs of these; anode to the 500Hz, cathode to the 38KHz.

So - the transmitted signal is a pair of square waves, one about 70
times the frequency of the other, superimposed?

At the receiver end, use one of the 3 pin IR receiver/demodulator
modules,

Do you, by any chance, have an example mfctr no. of such a device?

at the output of which you will get your 500Hz signal.

An ordinary 500Hz square wave signal? Because the demodulator
'removes' the 38kHz element? (I'm assuming a 555 only produces a
square wave, BTW - this assumption may be incorrect of course!)

You
could then use a 555 based missing-pulse-detector to detect the
absense of this pulse train.

OK, I understand that.

This arrangement is essentially what is used in consumer IR (TVs, DVDs
etc.); it will give you ample range, and is quite immune to ambient
conditions.

Of course, in place of 2 555s you could use a 556, or a PIC
microcontroller to reduce the component count.

Hope this helps!

Yes, thank you.

 

[Sřren]

Hi Paul,

Why is such a signal desirable? Does it make it less prone to
interference?

The integrated IR-receivers demodulate the signal (removes the carrier
wave) but they need it to differentiate between the ambient light and the
IR signal.

If you used only the carrier wave to modulate the IR-LED, the AGC
(Automatic Gain Control) circuit in the receiver would assume it was some
kind of static "noise" and turn down its gain.

By making the carrier intermittent (in pulses), the receiver keep the gain
high.

The signal you need looks something like:

__||||____||||____||||__

If you want to save on the power, you can use shorter bursts of the
carrier wave, with longer delays in between, making it look something
like:

__||______||______||__


Most IR-detectors needs at least 6..12 periods of the carrier for AGC
adjustment, sĺ with eg. a 38kHz carrier, so use bursts of at least 0.5ms.

A person breaking the beam will do that for at least 60ms (and they have
to be speedy to accomplish this), as IR tends to "creep" around objects a
slight bit, perhaps 50ms is a better number to be on the safe side.
That means, that you would have to use at least 2 bursts each 50ms and the
lowest frequency is as such 40 Hz.

Your average power consumption is then very low, as illustrated by this:
______|__________________________________________________|_______

Only the pulses use any real power.

Design 2 oscillators, one at 38KHz, the other at around 500Hz; you
could use a pair of 555s for this. Connect your transmitter LED to
the outputs of these; anode to the 500Hz, cathode to the 38KHz.

So - the transmitted signal is a pair of square waves, one about 70
times the frequency of the other, superimposed?

It is better to let one oscillator gate the other, like this:

+-----------------------------+---O +12V
_ | _ |
+-| \ _ Vdd +-| \ [120]
| | )O--------+-----| \ | | )O--+ _|_
+-|_/ | | )O---------+---+-|_/ | _\_/_ // IR-LED
| | +-|_/ | | | |
+----[680k]----+ | | | _ | |
| | +--[36k]--[4k7]-+ +-| \ | b|/c
+--[18k]--|<|--+ | ^ | | | )O--+--[3k3]--| BC337
_|_ 1N914 _|_ |___| +-|_/ |\e
--- 47nF or equ. --- 1nF |
| | IC=4093 |
+------------------+----------------------------------------+---O Gnd

Adjust the 4k7 trimmer for 38kHz (and get 38kHz receivers


--
Venlig hilsen,
Sřren
* If it puzzles you dear... Reverse engineer *
LM317-PSU-Designer v1,0b <http://www.ElektronikTeknolog.dk/cgi-bin/LM317/>

 

[Paul]

Hi Soren,

Thanks very much for that, and the excellent ASCII circuit diagram.

I think I'm going to have to do a lot more background study on this
subject, and then have a play.

Anyway, the answers have given me a lot of useful information.

Many thanks to you and others who have helped with my understanding.

Regards,

Paul.