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Author Topic: Measuring PAR (photosynthetic active radiation)  (Read 2291 times)
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Hello everyone,

For a new project (actually a add-on for an old one -> link) I'm building a PAR meter. The difference between measuring PAR and Lumen, for example is the spectrum being analysed is different. Since PAR sensors are quite expensive I've found a way to do it with led's using 'Mims discovery/effect'...
There's also a book called 'Atmospheric Monitoring With Arduino' where one can build a simple photometer with some LED's and use this to measure the PAR spectrum (400nm - 700nm), but I don't know if I'm allowed to post the schematic and/or code from this project here...correct me if I'm wrong smiley-wink

When I attach a led to my multimeter and set this to measure mV it measures about 80mV when I turn on a light, so the LED is producing a voltage...Yeah... Thank you Forrest!! smiley
When connected to the arduino the output starts to swing going from 230 to 0 and then back up...I think it has something to do with the amount of mV the arduino can register --> 5V/1023 = 0.004887586V = 4mV per step. So I've read some threads on this forum and others and started thinking about it.
First I have to step up the voltage with a linear rail-to-rail op-amp and second stabelise the reading by using a capacitor. Now my electronic skill set isn't much to write home about and this is why I need some help...
I've found several diagrams which would do the trick, but none of them tell me which kind of op-amp and capacitor I should use...


What do you guys/galls think?

With kind reagards,

Billie

p.s. If I forgot to mention some information I sincerely apologise smiley-wink
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Manchester (England England)
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I think that using a 320M resistor is very stupid. In effect the amplifier is working at what is known as open loop gain and it will be very suceptable to picking up noise.
That is not a practical circuit especially for beginners.

Your problems have nothing to do with step side and everything to do with picking up mains interference.
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Belgium
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I think that using a 320M resistor is very stupid.
Thank you for pointing that out...

Your problems have nothing to do with step side and everything to do with picking up mains interference.
Could you tell me how I can smooth out the 'mains interference'? Is this why the value's jump up and down?
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Well you could try a filter with a resistor and capacitor but it would be good to use a proper sensor and a realistic value of feed back resistor.
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Hello sir, I want to buy a resistor of 320 Mega ohm.
- So you want an insulator ?
No, a resistor.
- Okay, keep you hands up with a small gap in between.
Now what ?
- Can you see the resistor ?
No.
- It's there.
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Denmark
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I just found this.

5 minutes reading

http://dkc1.digikey.com/us/en/tod/ADI/Top5Problems_NoAudio/Top5Problems_NoAudio.html
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Belgium
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Well you could try a filter with a resistor and capacitor but it would be good to use a proper sensor and a realistic value of feed back resistor.
I know a decent sensor would be better, but a PAR sensor is in the price range of 180$. I thought the geniuses here would come up with a nice diagram that I could actually use...
Knowing that the feedback resistor is way to much, but since you guys love gobbling up those diagrams, and this was one I could find in a whim...I posted this one, not expecting the funny replies...Please give Erdin a dollar cause he's SOOO funny  smiley-razz
Any other ideas how I could measure PAR without the op-amp?

Thank you for the useful link. Saw it earlier this weekend, but didn't know what to make of all those diagrams and explanations... Still don't actually, that's why I came here to learn something...So just for the heck of it I'm going to try my best to decipher the 5min course (it already took me a whole lot longer smiley-wink ) and come back with what i've learned. And then please comment on this, even if it's just to make fun! Because that just means one has read my jibberjabber and understands it better then I do smiley-wink

Thank you all for reading and like I said earlier on this forum... I'm not going to give up that easy!
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Belgium
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Ok, I've read the document...a few times, hehe. And came to this conclusion...
I should use a JFET Op-amp because of the low noise, which is rail-to-rail and through hole (because I don't have a reflow oven). It also has to have a single voltage supply (5V) so I chose the AD820 for this...
Also made a diagram, instead of picking one from the internet (won't be doing that again smiley-wink )
The feedback resistor value and capacitor value are still a guess, but I think the filter is in the right place...Please correct me if I'm wrong.
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Denmark
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I just tried to measure on a photodiode i have.

I can measure about 300mV when i put it direct under my desptop lamp.

So i tried hooking it up to an op amp, like the schematic below, which gave med readings from 0 to 5V, whit just a little light.

So the 10Mohm resitor is far to big, but I didn't have a 1Mohm.



* 662.jpg (58.28 KB, 664x532 - viewed 33 times.)
« Last Edit: April 02, 2013, 12:12:22 pm by Erni » Logged

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Thank you so much for providing me with a schematic! I'm still off with mine I guess... smiley-confuse
I've got a couple of new questions now, hehe...
The 10MΩ is the feedback resistor. The greater the value, the higher the output you get? right?
Where are the 1KΩ and 100KΩ resistors for?
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Manchester (England England)
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I know a decent sensor would be better, but a PAR sensor is in the price range of 180$
There is a reason for that price, it actually measures what you want to measure. An LED is like any semiconductor junction it just measures photons. There is no discrimination that would match it to a PAR.  So while we could give you a circuit it will just be a general light meter and not measure what you want.
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BillieBricks

I corrected the schematic, but you are right: the gain is  10/0.1 = 100 , which is too much when using my photodiode, it should be about 15, so a 1.5 Mohm resistor should be suitable.
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Hey Mike,

I'm sorry to contradict you, but LED's do transmit and absorb in a specific bandwidth... This is why I got interested in using them in the first place...

Quote from: Atmospheric Monitoring with Arduino
Mims reasoned that since LEDs take in
electricity and emit light (along that narrow emission band, remember), they
should also take in light and give off electricity. A series of experiments proved
him right, and by 1973 he had formalized what is now known as the Mims
Effect: LEDs will absorb light along a relatively narrow band of color, and emit
a small amount of electricity.

So when one uses different kinds of led each with their specific bandwidth, one can make a photometer to measure them.
The next thing is to figure out how to extrapolate the actual PAR value from the total of all the leds, but I know someone who uses a PAR meter professionally and he's willing to let me use it to calibrate my device to his...But first need to get mine working smiley-wink



BillieBricks

I corrected the schematic, but you are right: the gain is  10/0.1 = 100 , which is too much when using my photodiode, it should be about 15, so a 1.5 Mohm resistor should be suitable.
Again. Thank you for taking the time and even test the theory/schematic.
I know I've got lots to learn, but you make it fun!!! smiley-grin
« Last Edit: April 02, 2013, 12:35:08 pm by BillieBricks » Logged

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Quote
Mims reasoned that since LEDs take in ....
Is this the same person who suggested using a 320M resistor in the feedback of an op amp?
I think his credibility is a bit blown by that.

So I looked at the book Atmospheric Monitoring with Arduino on Amazon and in the preview section I came across this diagram (attached).
Any one who thinks that is the way to connect an LED to an arduino without damaging the arduino knows absolutely nothing about arduinos and electronics. As this book according to Amazon has not been released how did you get a copy?

This is the bandwidth you are looking for:-
http://www.licor.com/env/products/light/quantum_sensors/
Now do you actually think that can be replaced by an LED?

LEDs detect photons by creating electron hole pairs in the intrinsic region of a PN junction. These then get swept up by the potential barrier on each side of the intrinsic region and that represents a current.
In order to create the electron hole pair the photon energy E which is hv (h = planks constant v = electromagnetic frequency) has to be greater than the gap between the valance band and the conduction band in the material of the junction. However, this is only the minimum energy, so while an LED will react to light at the same frequency that it emits and not to lower frequencies, it will also generate a photo current for photons of higher energy / frequency.

I ask again how meaningful will measurements from an LED be?
Well good luck.


* Arduino & LED.png (107 KB, 559x372 - viewed 28 times.)
« Last Edit: April 02, 2013, 02:42:00 pm by Grumpy_Mike » Logged

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Quote
Mims reasoned that since LEDs take in ....
Is this the same person who suggested using a 320M resistor in the feedback of an op amp?
I think his credibility is a bit blown by that.
No it is not, but I think you already knew that smiley-wink

So I looked at the book Atmospheric Monitoring with Arduino on Amazon and in the preview section I came across this diagram (attached).
Any one who thinks that is the way to connect an LED to an arduino without damaging the arduino knows absolutely nothing about arduinos and electronics.
I agree with that. He's probably using old diagrams...A shame actually, but it is the first edition of the book. You can probably score some points with them pointing that out in your charismatic ways...


As this book according to Amazon has not been released how did you get a copy?
Maybe you are looking in the wrong place. Here, for you. I know how you love links I send you

This is the bandwidth you are looking for:-
http://www.licor.com/env/products/light/quantum_sensors/
Thanks for the link, but have you got any idea how expensive these sensors are?

Now do you actually think that can be replaced by an LED?
I hope so!

LEDs detect photons by creating electron hole pairs in the intrinsic region of a PN junction. These then get swept up by the potential barrier on each side of the intrinsic region and that represents a current.
In order to create the electron hole pair the photon energy E which is hv (h = planks constant v = electromagnetic frequency) has to be greater than the gap between the valance band and the conduction band in the material of the junction. However, this is only the minimum energy, so while an LED will react to light at the same frequency that it emits and not to lower frequencies, it will also generate a photo current for photons of higher energy / frequency.

I ask again how meaningful will measurements from an LED be?
Well good luck.

Think out of the box I would say. wouldn't it be cool if it where to work?
I've got a link for you... These people use a led for taking scientific measurement...

With kind regards,


Billie
« Last Edit: April 04, 2013, 05:26:36 am by BillieBricks » Logged

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