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shot noise source

Discussion in 'Electronic Design' started by [email protected], Sep 19, 2005.

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  1. Guest


    I need to build a white noise source from 1 Hz to 100 kHz (the more the
    and came across a photodiode shot noise source which seems fine because
    it can
    be calculated and I don't have the facilities to calibrate a zener
    noise source.

    Message from Phil Hobbs

    Now I need a concrete design idea:

    a) use an inverting amplifier with a big feedback resistor (1 G)
    Illuminate the diode for 1 nA Photocurrent.
    This gives me 1.8e-14 A/sqrt(Hz) current noise,
    with 10 KHz bandwidth 1.8 mV after the amplifier.
    LF411 has around 10 fA/sqrt(Hz) noise, this is not negligible with
    respect to the shot noise.
    In addition, johnson noise from the feedback resistor is to large,
    the voltage is below 50 mV (2kT/e)

    b) use 1 mA photocurrent and see how to get rid of the DC before
    Use the active load circuit from Zetex application note AN3, figure 6.

    ------|------- +Vee
    |1k5 |
    ..-. |33n
    | | ---
    | | --- ___
    '-' | ||___||
    | | | |
    | | 330k | |
    |---|-|___|--------------|-\ | ||
    /| | | >------||-
    | | GND-|+/ ||
    |-------------- |/| out
    ------------ -Vcc
    (created by AACircuit v1.28.6 beta 04/19/05

    Probably I would use a current source to sink the current from the
    and use negative feedback from the output of the OPA to the light
    source to have
    a better response at low frequencies

    Do you have any comments?

    Many thanks for your feedback

  2. wrote...
    Silly man, doing all that work, getting only 1nA. Improve the
    optical pathway. Turn up that LED current! Use 1mA, use 20mA,
  3. 1

    1 Guest

    Do I understand it correctly that increasing the current (and therefore
    decreasing Rfeedback) I will never be able to reach shot noise limit
    2kT/e, since Uout = sqrt(2 e I) * Rfeedback = sqrt(2 e I) * Ucc/I < 2kt/e
    (DC coupling)
    So you would go with the current sink?


  4. 1 wrote...
    Think carefully, if you choose a current-sense resistor high
    enough so the DC photodiode current drops significantly more
    than 50mV dc, such as 1V, etc., the noise developed across
    that resistor will be entirely shot noise. You can then ac
    couple and amplify to get your desired output noise signal.
  5. 1

    1 Guest

    ok, my mistake was to assume the _noise_ current had to be larger than
    50 mV. With 1 mA photocurrent I have 1.7 µV behind the amplifier (1k
    feedback) over 10 kHz bandwidth and this can be AC-coupled with a second

    many thanks

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