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USING OPT101 VALUE WITH LIGHT FREQUENCY

Debasish Banerjee
Intellectual 560 points
Other Parts Discussed in Thread: OPT101

OK, now I have understood again that I was foolish at the last time my posts were removed, now I know that when I am using g2553 with an opto101, i can calculate the output as OPTO101_value =((OPTO101*3.67)-7.5)/1000, where 3.67 is my vcc(reference voltage this is equal to 1023*(vin/vr), 7.5 is the dark current correction for 7.5mV, and since the result is in mV, hence the /1000 is there. Now the problem is since the photodiode gives output as 0.45A/W(Ampere/Watt) at 650 nm visible red and so the OPT101 operating with a single power supply(as in my case) and using the internal 1MΩ internal feedback resistor for a response of 0.45V/μW at 650nm, all I am having now is only the light intensity, but how to get the wavelength  as well? Also can I filter the input of the photodiode without any cap or lense, I mean so that it only sees a certain wavelength and gives the output?

Sorry, just saw that the internal 1 mega Ohms feedback resistor, which I am using, is laser trimmed to give the aforesaid output at 650nm, then I hope all the values of the photodiode output is already being calculated for 650nm wavelength, no matter which light is in the ambience, right? Then how come it is not giving the output as near zero when the room is dark, no lights are on(assuming that visible red at 650nm should also not be there in this case?)? 

Thank you all, and though this is irrelevant, LM35CAZ basic circuit also works with 3.67 vcc, but the result starts as 0mV=2 Degree Celsius, and I overlooked that! 

 

  • 0
    Guru 227245 points

    Debasish Banerjee said:
    how to get the wavelength  as well?

    You have more variables than equations. This means that there is an infinite number of solutions.

    if you don't know neither wavelength nor intensity, the output of the sensor means absolutely nothing. If you at least know that the wavelength is constant, the sensor output can tell you the relative change in intensity, but not the absolute intensity.

    Usually, thsi sensor is used to calibrate intensity of a light source of known frequency or known frequency spectrum distribution. Such as LED, Laser or CCFL. (on normal light bulbs, spectral distribution changes with intensity).

    Debasish Banerjee said:
    Also can I filter the input of the photodiode without any cap or lense, I mean so that it only sees a certain wavelength and gives the output?

    Only by applying some sort of filter, like a colored plastic film.

    BTW: I already answered yesterday to your other thread, but due to your ban, the answer didn't show up and now the thread is gone too. (actually, this is the first time I noticed a ban in the MSP forum, so you are maybe holding the record now :) )
    Unfornately I already deleted the notification message from my mail account, so it is gone.

  • 0 in reply to Jens-Michael Gross
    Intellectual 560 points

    Jens-Michael Gross said:
    actually, this is the first time I noticed a ban in the MSP forum, so you are maybe holding the record now :)

    Thank you Jens, but I am not very fond of making this kind of records! May God bless me :\

    Jens-Michael Gross said:
    if you don't know neither wavelength nor intensity, the output of the sensor means absolutely nothing.

    I know the relative intensity for sure, as it comes up as the output for the above formula used, and I mentioned that I saw in datasheet that while using the internal 1 MOhms resistor, output is 0.45v/microwatts at 650nm visible red, does that mean it usually ignores all other wavelengths, no matter what is out there? If yes, then why it doesnot show near zero output when the room is completely(ok, atleast to a human eye) dark?

    It is written in the Page 8 of 15, in case if you need 2313.opt101.pdf

    output is ~6.13 micro watts(/0.45) under 40 watts white fluorescent tube-light , which increases upon further focusing a white single led torch into it, but still remains >6 microwatts when all lights are off. connected opto101 as per datasheet , p1(vs) to (3.67vcc), p3(v-) and p8(common) both to ground, p4(1MΩ Feedback) and p5(Output) to p1.7 adc channel, reading is fine, except it always above that 6 microwatts, why its so?

  • 0 in reply to Debasish Banerjee
    Guru 227245 points

    Debasish Banerjee said:
    I know the relative intensity for sure

    Only if you know htat the frequency is constant. If you measure a light bulb, lower power means not only lower intensity but also a frequency shift.
    And I covered this case just two sentences below.

    Debasish Banerjee said:
    I saw in datasheet that while using the internal 1 MOhms resistor, output is 0.45v/microwatts at 650nm visible red, does that mean it usually ignores all other wavelengths, no matter what is out there?

    No. I covers all wavelengths with varying weight.

    So if you measure 0.6V, you don't know whether it is 1µW@850nm or 2µW@550nm or a mixture of both. Or something completely else.

    You have two variables (frequency and intensity) and one equation : Vout = F(f,i). For F(x,y) you have the transfer courve. However, for any given (measured) Vout, there are multiple x,y pairs possible. So without knowing anything about f or i, Vout is meaningless. Knowing something about f or i (e.g. that f is constant), gives Vout some meaning. Knowing f or i and Vout, you can exactly determine the third one.

    Debasish Banerjee said:
    It is written in the Page 8 of 15

    THis pages tells you "At a wavelength of 650nm" which has to be read as "if the wavelength is 650 nm and you know that it is"

    Debasish Banerjee said:
    output is ~6.13 micro watts(/0.45)

    No. Output is mV. Since I don't know how you calculated the µW from the oputput voltage (and considering the said thigns above), I cannot confirm this value, nor can I work with it. Any use for analysis of your problem is "lost in translation".

    However, there is a dark current offset:

    DARK ERRORS, RTO(2)
    Offset Voltage, Output +5 / +7.5 / +10 mV

    This is caused by diode lwakage independent of any light. It's not that much, but it's there.

    Then there is a high sensitivity to infrared radiation (near infrared). Even in a 'dark' room (to the eyes) there might be much infrared radiation.
    Note that a fluorescent light or a diode doesn't emit much infrared radiation, so the the light you used to test the device adds just a second energy source to the ambient infrared. So the relative output change you see will not necessarily equal the change in visible light exposure.

    YOu also should do some sanity checking:

    Your fluorescent light is 40W. Some of this is visible light, some is infrared. Now the sensor is how far away from it? Calculate a sphere (or cylinder, depending of the CCF shape) around the lamp with this distance. Calculate the surface. Then divide the 5.2mm²/0.008in² sensor opening by this surface and multiply it with the 40W. This is the power that reaches the sensor. Multiply with 300000 for average sensitivity and amplifier gain. This should give you a rough estimation of the output voltage in V. Very rough, as I don't know the speactral composition of the lamp. So it may be off by +-50%

  • 0 in reply to Jens-Michael Gross
    Intellectual 560 points

    Thank you so much Jens, for your valuable insight on this. But one thing, lets assume that the light source is a 650nm visible red, then the diode gives that 0.45A/W and the through the op-amp it becomes 0.45V for per micorwatt of received power. If this is correct, then  OPTO101_value =((((OPTO101(//reading)*3.67(//this is refv))-7.5(//this is dark current typical output) gives me the output in millivolts, then when I divide the result by 1000, its gives me the result in volts per microwatt. Now if I divide this result with 0.45 to get the microwatts the diode gets on its 5.2 Sq.mm. surface, I get the output in the microwatts, isn't it? and further applying the same assumption, I wrote to you that ~6.13 micro watts is the output, I don't really care about absolute intensity, all I care about how much light is there in the eyes of my diode at a selected wavelength i.e. 650nm. This is just to be sure that my perception upto this is correct and hence is this writing.

    Jens-Michael Gross said:
    Then there is a high sensitivity to infrared radiation (near infrared). Even in a 'dark' room (to the eyes) there might be much infrared radiation.
    Note that a fluorescent light or a diode doesn't emit much infrared radiation, so the the light you used to test the device adds just a second energy source to the ambient infrared. So the relative output change you see will not necessarily equal the change in visible light exposure.

    I was thinking of the same, but was confused about that 

    Jens-Michael Gross said:
    while using the internal 1 MOhms resistor, output is 0.45v/microwatts at 650nm visible red
    you make me sure that it is like the way it seems. Thanks for your input, I will just use a optical filter to only let the 650nm in, hope these filters are available commercially, then I should be able to calculate the relative intensity of that, which will tell me whether I should turn on the light or not (day or night!).

  • 0 in reply to Debasish Banerjee
    Guru 227245 points

    Debasish Banerjee said:
    I will just use a optical filter to only let the 650nm in, hope these filters are available commercially

    For experiments, you may use a red laser pointer, as these usually have 650nm wavelength (the red HeNe gas lasers had 628nm, which seems to be brighter to the human eye)

    About the calculation above, well, I'm just about to leave to office for a 1 week vacation. So I don't have the time to check the math now. And I'm not sure how much time I will have in the next week.

  • 0 in reply to Jens-Michael Gross
    Intellectual 560 points

    Ok, finally I have managed to get the correct data out of it, the dark current for mine seems to be 7.34mV in complete darkness (excluding the concerns for IR), the reason for the anomaly I told previously was because of some short circuit, which is now fixed. Thanks for your valuable input Jean.

  • 0 in reply to Jens-Michael Gross
    Prodigy 30 points

    photons are special. they have size and number per steradian. In other words, there is direction from a target and a limit of view angle light gathering of a detector. Light can come in from many angles as long as it bounces in the detector and finally tickles the germanium on the substrate. Light in the IR can even go through the back side of the germanium without grabbing a hole to produce charge, it can heat the substrate and cause the noise to rise dramatically. Both the incoming light and the sensor have bandwidths. just like the human eye. A field stop and filter are required to limit most surrounding and reflected sources. Then you can talk about the noise due to shot, resistor, and temperature and sensor noise. ( cant distinguish between them of course, they are noise by definition. However calculations do exist so see a good tech spec. Filters are significant and need to be purchased with their own spectral response curve. That way you can calculate the light input from a KNOWN source, light from a STANDARD lamp...For fun and to do a good job you should use a chopped signal  to measure the input and calibrate the response. A standard lamp has a responsivity vs frequency curve for calculating microwatts per cm2 per nm-1 per str-1 or some other units you prefer.....

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