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Photodiode problem help

  1. May 6, 2008 #1
    Hey guys, just another question regarding photonics. From looking at an I - V characteristic of a typical photodiode:

    http://www.rp-photonics.com/img/photodiode.png

    I dont understand what it means by there being a negative voltage and a negative current. I suspect the negative in the current simply means that the charged particles move in the opposite direction mean while i dont understand what the negative voltage would mean. Is the negative voltage there because a supply of charge carriers are entering the diode from the power supply?
     
    Last edited: May 6, 2008
  2. jcsd
  3. May 6, 2008 #2

    Hootenanny

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    If the charge particles are moving in the opposite direction as you say, what does that tell you about the potential (voltage)?
     
  4. May 6, 2008 #3
    Hmmmm i guess what im really getting at is how can u have a negative voltage. Does the negative voltage simple coincide with the direction that it is travelling?
     
  5. May 6, 2008 #4

    Hootenanny

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    Indeed it does. Consider the gravitational analogy:
    Your stood on the ground at the bottom of a ladder, here we define your (gravitational) potential energy to be zero. Now you climb 5 meters up the ladder, hence you gain 5mg joules of potential energy. Ergo your change in potential energy is +5mg. Now, you climb back down the ladder to ground level. At the top of the ladder you have 5mg of potential energy and now at the bottom you have 0 potential energy. Hence, you have lost 5mg of potential energy. Therefore, your change in potential energy is -5mg.​

    Does that make sense?
     
  6. May 6, 2008 #5
    That seems to make perfect sense :smile:

    I cant help thinking now with a reverse-biased diode, you obviously have some small leakage current that is defined as negative (due to the direction it is traveling). If we consider the movement of electrons, from negative to positive, does that mean that the tiny amount of electrons that are moving are infact moving to positive to negative? This is because the direction of current is opposite? Does that make sense??? :confused:

    ALSO

    What does it mean if something is said to have a reverse biase voltage of 4?
     
    Last edited: May 6, 2008
  7. May 6, 2008 #6

    Hootenanny

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    No, electrons will always move from a lower voltage to a higher one. A reverse bias simply means that a potential difference is applied across the diod in the opposite direction to which current would normally flow. Hence, you get a small amount of current leakage in the opposite direction to normal, but the electrons are still moving from a lower voltage to a higher one, it just happens that the voltage is applied in the opposite direction.
     
  8. May 6, 2008 #7
    I really really dun understand this. Its so annoying cause ive looked at so many sources and yet it still wont sick in. When you apply a current to a reverse-biased diode, can you please explain what exactly happens. I understand the analogy given before but i dont understand how you can have negative voltage in the diode. I understand the voltage in forward biased, as it is needed to 'open' the diode. What is the point of the reverse voltage now!!! :mad:
     
  9. May 6, 2008 #8
    Okay, so a more reasonable approach:

    Before i go to bed, let me just point out some things i dont understand

    1. No matter if the diode is reverse biased or forward biased, the direction the electrons travel will still be the same?

    2. If this is the case then why do we refer to negative current?

    3. How can we have negative voltage?
     
  10. May 6, 2008 #9

    Redbelly98

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    Most of the time, yes, the photodiode current will be negative. Look at the I-V curve again. If enough forward voltage is applied to the photodiode, the current will become positive. But this requires connecting a battery or other voltage source to it, with a positive polarity in the connection. Also, there is really no useful reason to put a positive voltage on a photodiode, as then the signal would be very insensitive to incident light level

    If you simply connect the photodiode to a resistor only, the current will always be negative -- or zero, if there is no light hitting it.

    If there is no light hitting the photodiode, it's I-V curve will behave just like a regular diode. There is a standard convention for positive and negative current directions in regular diodes, and it makes sense to keep that convention for all diodes: photodiodes, LED's, and laser diodes too.

    It happens that in a photodiode, the current generated by incident light goes in the opposite direction to the normal conduction direction of a regular diode. So this current is negative, by agreed convention.

    By connecting a battery to the photodiode, it is possible to have a negative voltage across it. This is a useful way of getting a linear signal (i.e., one that does not saturate) over a wide range of light levels.
     
  11. May 6, 2008 #10

    Redbelly98

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    By the way, I've attached the two circuits used to generate the dashed-line traces shown in the photodiode I-V curve from message #1 of this thread:

    http://www.rp-photonics.com/img/photodiode.png

    Perhaps this will help to better understand the I-V curves.
     

    Attached Files:

  12. May 7, 2008 #11
    So even if a battery is not connected, there will be movement of charged particles and hence a voltage?
     
    Last edited: May 7, 2008
  13. May 7, 2008 #12

    Redbelly98

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    No. With no battery, and no light, there will be zero current and zero voltage.

    Just like like when there is no battery connected to a regular diode.
     
  14. May 7, 2008 #13
    How about a solar cell? They dont require a battery do they?
     
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