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Simple Photodiode Reverse Bias Circuit - Current?

Discussion in 'General Electronics Discussion' started by Corwin Schwarzchild, May 3, 2015.

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  1. Corwin Schwarzchild

    Corwin Schwarzchild

    May 3, 2015
    Hi there,

    I am an EE student studying for an upcoming final. My question pertains to a very simple circuit containing a resistor, a voltage source, and a photodiode.

    The photodiode is receiving an input intensity of 500 mW/m^2. The PD has a responsivity of 0.4 A/W. Assume the area of the detector is 1mm^2. The load resistor we'll call 100 0hms. See image.

    We're going to model the PD IV curve with a simple piecewise linear function, see image. Rd we'll call 10 ohms and V0 we'll call 0.5 A.

    If I make the bias voltage so that the photodiode is in reverse mode, say Vb = -5V, what current am I going to see in the circuit? How can I figure this out? Please explain thoroughly because I am confused.

    Corwin S

    Attached Files:

    • IV.PNG
      File size:
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    • PD.PNG
      File size:
      17.6 KB
  2. hevans1944

    hevans1944 Hop - AC8NS

    Jun 21, 2012
    Your diagram shows the photodiode with forward bias, which is incorrect. Photodiodes are always reverse-biased when used as light sensors. OTOH, if this is a power-producing photocell, disregard the discussion below.

    What is your question? What is VO? Where is the piece-wise linear approximation you said would be attached? Why do you want a piece-wise linear approximation to a forward biased "photodiode"?

    You would normally connect the photodiode anode to a negative bias voltage and connect its cathode to the summing junction of an op-amp with a largeish feedback resistor to convert the photodiode reverse current to an output voltage linearly proportional to light intensity. No piece-wise linear approximation is necessary: the photodiode current is linear with intensity over at least four decades of intensity.

    I would also include another photodiode, reverse-biased with a positive bias voltage applied to its cathode and its anode connected to the summing junction of the op-amp. Keep this second photodiode in the dark, but in good thermal contact with the first (active) photodiode so they both are at the same temperature. By doing this you will effectively cancel the dark-current of both photodiodes and extend the range to very low intensities. The op-amp should be powered from equal positive and negative rails, typically ±12 to ±15 V DC. An offset potentiometer may be necessary to nuil the output when the light measuring photodiode is in the dark. The feedback resistor may need to be smaller to accommodate very high light levels without saturating the op-amp output.

    The reverse current of an un-illuminated photodiode is called the dark current. It is virtually independent of the bias voltage but is very much a function of the junction temperature. Hence the need for dark current temperature compensation as described above. See also the diode equation for details.

    The photodiode current, which is in the same direction as the dark current since it is produced by the same reverse bias voltage, is given by the product of the diode sensitivity (0.4 A/W) and the optical power input to the photodiode. The optical power input is the product of Illiumination power density (500 mW/m²) and the active photodiode area (1 mm²). So do the math.

    This thread should be in the homework section since you are an EE student.
    Last edited: May 4, 2015
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