+ Post New Thread
Results 1 to 18 of 18
  1. #1
    Junior Member level 1
    Points: 284, Level: 3

    Join Date
    Sep 2017
    Posts
    18
    Helped
    0 / 0
    Points
    284
    Level
    3

    Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Hi

    I am trying to estimate the shot-noise from ambient light sources (sunlight and Fluorescent lamps).
    I expect to shield my photodiode as good as possible - so no direct light is hitting it.
    Second, what is the relationship between photodiode area size and noise from ambient light - is it proportional like: Id(shot-current) = Area*responsivity*irradiance.


    Thank you.

  2. #2
    Super Moderator
    Points: 77,854, Level: 68
    Achievements:
    7 years registered
    Awards:
    Most Frequent Poster 3rd Helpful Member

    Join Date
    Apr 2014
    Posts
    15,785
    Helped
    3592 / 3592
    Points
    77,854
    Level
    68

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Hi,

    Shot noise should be caused by the semiconductor of the photodiode. It should be independent of light source.
    You need to measure the current, not the voltage. An amplifier will introduce additional noise.

    The light source will be noisy, too. But I have no values.
    I assume the noise of sunlight is very low.
    But for sure if you run an FFT on sunlight intensity on earth you will get two major frequency spots:
    * 1/day
    * 1/year
    * i assume there will be a third one: 1/week, caused by pollution
    The rest of it will be very random, mainly cased by clouds.
    I don't expect much high frequency noise.

    For flourecscent light the dominant frequencies will be much higher..it depends on how the driving voltage is generated.
    I'd say it is impossible to give a general answer on this.

    Klaus



  3. #3
    Super Moderator
    Points: 260,279, Level: 100
    Awards:
    1st Helpful Member

    Join Date
    Jan 2008
    Location
    Bochum, Germany
    Posts
    45,454
    Helped
    13829 / 13829
    Points
    260,279
    Level
    100

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Light sources don't involve shot noise, received photons and charge carriers in a photo detector however do.

    Respectively you get a certain amount of shot noise produced in the photo detector which is not originated from intensity fluctuation of the optical source. Although SNR is increasing with light intensity, the absolute noise level is still increasing with square root of intensity. In so far a high level of constant ambient light affects the measurement of a small AC signal.

    Fluorescent lamps have strong 100/120 Hz and in case of inverter operated devices also multi 10 kHz components that can be quite problematic for all applications using modulated light. Sunlight has almost clean continuous intensity.



  4. #4
    Junior Member level 1
    Points: 284, Level: 3

    Join Date
    Sep 2017
    Posts
    18
    Helped
    0 / 0
    Points
    284
    Level
    3

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Okay, thank u.
    Sorry – for not be being precise, I meant the shot-noise (from the photodiode) generated by the ambient light ��.
    I have been trying to calculate the shot-noise current. I found out that room lighting is approximately 500 lux – which corresponds to 3-5 Watts/m^2.
    With a photodiode active area of 0.5mm^2 – The shot-noise Id is 1.5ĩA (using the eq. Area*responsivity(0.6)*irradiance.).
    This number sounds quite big in my ear – this is of course without any shielding.



    •   AltAdvertisement

        
       

  5. #5
    Super Moderator
    Points: 260,279, Level: 100
    Awards:
    1st Helpful Member

    Join Date
    Jan 2008
    Location
    Bochum, Germany
    Posts
    45,454
    Helped
    13829 / 13829
    Points
    260,279
    Level
    100

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Shot noise in regular photo diodes can be easily calculated.

    Noise density is in = √(2 q Id) [A/√Hz]
    with q being the elementary charge and Id the diode current. See https://en.wikipedia.org/wiki/Shot_noise



  6. #6
    Junior Member level 1
    Points: 284, Level: 3

    Join Date
    Sep 2017
    Posts
    18
    Helped
    0 / 0
    Points
    284
    Level
    3

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Quote Originally Posted by FvM View Post
    Shot noise in regular photo diodes can be easily calculated.

    Noise density is in = √(2 q Id) [A/√Hz]
    with q being the elementary charge and Id the diode current. See https://en.wikipedia.org/wiki/Shot_noise
    How do I find Id?



  7. #7
    Super Moderator
    Points: 260,279, Level: 100
    Awards:
    1st Helpful Member

    Join Date
    Jan 2008
    Location
    Bochum, Germany
    Posts
    45,454
    Helped
    13829 / 13829
    Points
    260,279
    Level
    100

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Basically as calculated in your previous post, light intensity multiply diode sensitivity. Spectral light intensity and diode sensitivity curve must be considered however.

    Unless you have an exactly defined exposition scenario, it's probably easier to measure the actual diode current with a reference setup.


    1 members found this post helpful.

    •   AltAdvertisement

        
       

  8. #8
    Junior Member level 1
    Points: 284, Level: 3

    Join Date
    Sep 2017
    Posts
    18
    Helped
    0 / 0
    Points
    284
    Level
    3

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Okay thank you.

    I am using TIA for a current to voltage conversion. The photocurrent is 600nA, the question is therefore should I use an input (in series with the photodiode) resistor to get Vin (how big should this resistor be)– since the gain is A = Vout/Vin?
    The input signal is modulated (the LED) with 600 Hz square wave with a duty on 1% - this gives a frequency of 60KHz. How do I decide the following parameters 1. The gain A 2) The Bandwidth 3) Settling time (how is this affect by the duty-cycle)



    •   AltAdvertisement

        
       

  9. #9
    Super Moderator
    Points: 77,854, Level: 68
    Achievements:
    7 years registered
    Awards:
    Most Frequent Poster 3rd Helpful Member

    Join Date
    Apr 2014
    Posts
    15,785
    Helped
    3592 / 3592
    Points
    77,854
    Level
    68

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Hi,

    confusing.

    With a TIA usually the voltage across the photodiode is constant.
    And with a TIA your formula "A = Vout/Vin" is not valid.

    The input signal is modulated (the LED) with 600 Hz square wave with a duty on 1% - this gives a frequency of 60KHz
    This statement makes no sense to me.

    Gain of a TIA is given in V per A = V/A

    Bandwidth of a TIA is determined by the feedback resistor and the feedback capacitor. fc = 1/ (2 x Pi x R x C)
    Settling time depends on OPAMP - and the cutoff frequency fc. and is not affected by the duty cycle.

    Klaus



  10. #10
    Super Moderator
    Points: 260,279, Level: 100
    Awards:
    1st Helpful Member

    Join Date
    Jan 2008
    Location
    Bochum, Germany
    Posts
    45,454
    Helped
    13829 / 13829
    Points
    260,279
    Level
    100

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Series resistor makes no sense for TIA.
    The input signal is modulated (the LED) with 600 Hz square wave with a duty on 1% - this gives a frequency of 60KHz.
    I guess you want to estimate the useful signal bandwidth. 60 kHz gives at least an order of magnitude for an amplifier bandwidth required to reconstruct the input waveform. The useful bandwidth depends however on the signal detector properties. You need to specify the detection algorithm to determine an optimal bandwidth related to best SNR.



  11. #11
    Junior Member level 1
    Points: 284, Level: 3

    Join Date
    Sep 2017
    Posts
    18
    Helped
    0 / 0
    Points
    284
    Level
    3

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Hmm. I am not quite sure, I am understanding you. My 3db cutoff (bandwidth) should be at least 60 KHz (fp) Right?
    Using the eq. fp = 1/(2pi*CF*RF). Which parameters should I tweak on. I know Cf creates a pole that counteracts the zero from beta, but how should I chose Rf. (Voutmax-Voutmin/Idmax)?.
    For choosing an appropriate amplifier, I should calculate the Gain bandwidth product – fGBW = Cin+CF/2piRfCF^2 right?.
    I am also confused about how I should chose my open loop Gain? The current signal is 600nA.



  12. #12
    Super Moderator
    Points: 77,854, Level: 68
    Achievements:
    7 years registered
    Awards:
    Most Frequent Poster 3rd Helpful Member

    Join Date
    Apr 2014
    Posts
    15,785
    Helped
    3592 / 3592
    Points
    77,854
    Level
    68

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Hi,

    My 3db cutoff (bandwidth) should be at least 60 KHz (fp) Right?
    We don't know.
    You give informations piece by piece therefore after each post we another situation.

    Which parameters should I tweak on
    First calculate Rf, since it is the gain factor. Rf = Vout / Iin
    Then calculate Cf according your desired cutoff frequency.

    GBW of the amplifier needs to be higher than your desired fc. I recommend at least ten times.

    Why you want to calculate open loop gain?
    Do you want to use the amplifier in open loop?
    What is it good for?
    At which frequency? It is about GBW / f.

    Klaus



  13. #13
    Junior Member level 1
    Points: 284, Level: 3

    Join Date
    Sep 2017
    Posts
    18
    Helped
    0 / 0
    Points
    284
    Level
    3

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Click image for larger version. 

Name:	EqSettlingtime.PNG 
Views:	2 
Size:	73.7 KB 
ID:	141851

    Okay, Super. With the settling time vs duty-cycle. I found this eq. Cd is the output capacitance of the photodiode, epilson is the settling accuracy, A is the open-loop gain. D is the duty-cycle.
    Have anybody seen these equations before, why does the settling time depend on the open loop gain? eq (14). And how does the duty-cycle and switching frequency come into the picture in eq.15?
    Thank you.



  14. #14
    Super Moderator
    Points: 77,854, Level: 68
    Achievements:
    7 years registered
    Awards:
    Most Frequent Poster 3rd Helpful Member

    Join Date
    Apr 2014
    Posts
    15,785
    Helped
    3592 / 3592
    Points
    77,854
    Level
    68

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Hi,

    sadly you posted only a snippet of a document... therefore itīs rather difficult to verify if it complies or not.

    Cd is the output capacitance of the photodiode
    In my eyes the photodiode capacitance plays not a big role when the photodiode is connected to a TIA.
    The TIA just compensates the photodiode current to zero at the inverting input node of the TIA.
    This means the voltage is kept constant. With a constant voltage there is no current in a capacitance. No current means no influence.
    This is true for an ideal TIA.

    And the big benefit of the TIA circuit is, that the photodiode capacitance has (about) no influence on the signal bandwidth.

    If Cd really is the photodiode capacitance, then I donīt think that R is the feedback resistor.

    Klaus



    •   AltAdvertisement

        
       

  15. #15
    Junior Member level 1
    Points: 284, Level: 3

    Join Date
    Sep 2017
    Posts
    18
    Helped
    0 / 0
    Points
    284
    Level
    3

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Are you sure?
    The doc is attached.



  16. #16
    Junior Member level 1
    Points: 284, Level: 3

    Join Date
    Sep 2017
    Posts
    18
    Helped
    0 / 0
    Points
    284
    Level
    3

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Click image for larger version. 

Name:	CircuitG.PNG 
Views:	2 
Size:	134.4 KB 
ID:	141859


    The transimpedance doesn’t present a small impedance, because a Gain error signal Vo/AOL remains across the diode and its capacitances, which gives us the time constant: RfCD/(AoL).. See attached picture.



  17. #17
    Super Moderator
    Points: 77,854, Level: 68
    Achievements:
    7 years registered
    Awards:
    Most Frequent Poster 3rd Helpful Member

    Join Date
    Apr 2014
    Posts
    15,785
    Helped
    3592 / 3592
    Points
    77,854
    Level
    68

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Hi,

    Iīve had a look through the document.
    So indeed R is the feedback resistor.. but itīs divided by the open loop gain of the amplifier, therefore R/A becomes very small.
    This is a consideration of a real amplifier and not an ideal amplifier.
    Sadly they donīt put a schematic or at least drafts in the document, where it becomes more clearely what he is writing about.

    Letīs do some calculations:
    So assuming you have a photodiode with about 70pF and you want a cutoff frequency of 60kHz
    letīs calculate an imaginary input TIA input R:
    fc = 1/2 x Pi x R x C
    R = 1 / 2 x Pi x R x fc = 38 kOhms

    Now if you want a feedback resitor of 1Mohms, then you need an amplifier with an open loop gain higher than:
    1 MOhms / 38 kOhms = 26.5 at 60kHz
    26.5 = 28dB at 60kHz

    Klaus


    1 members found this post helpful.

  18. #18
    Junior Member level 1
    Points: 284, Level: 3

    Join Date
    Sep 2017
    Posts
    18
    Helped
    0 / 0
    Points
    284
    Level
    3

    Re: Ambient light noise (sunlight & Fluorescent lamps) in photodetector

    Ah okay. So it basically puts a requirement to my open-loop gain.

    I have calculated my resistor to be 12Mohm ( Voutmax (5V)-Vmin(0.1V)/Inmax (400nA)).

    (2) Then I calculate the Feedback capacitor fc=1/(2pi*Cf*Rf) at 60kHz, which gives 221fF. (to obtain stability)

    (3) Then I calculate fz =1/(2pi(Cin+Cf)*Rf) (the zero), where Cin = Cphotodiode(11pf)+Camplifer (10pf), I am rewriting this to get the desired resistor R = 1/(2pi(Cin+Cf)*fc = 84.961KOhm. (By the way isn’t the 3db cutoff decided by eq. 2?, and why didn't you include Cf in your procedure? )

    So my open loop gain should be 12MOhm/84.961KOhm 43dB. In the datasheet, they usually put the AOL gain with a specific resistor like at 10kOhm the open loop gain is 132 db. How should I read these numbers?

    (4) I calculate the GBW= Cin+Cf/(2piRfC^2)

    - Thank you



--[[ ]]--