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Simple Photodiode question

  1. May 7, 2016 #1
    Hi all,

    I'm having a little bit of trouble understanding photodiodes in reverse bias. Say for instance I have the circuit shown, If I'm asked for a certain voltage required to switch on RA I understand that this corresponds to a certain current being allowed through the diode which corresponds to a certain light intensity from the graph. But it's the voltage that I'm unclear about. Say for instance the RA requires a Voltage drop of 6V to operate. This corresponds to the diode allowing 2mA of current through. Does this mean that the voltage drop across the diode in this case is 9V ? And subsequently if the Emf was reduced to say 10V does that mean that the RA would still operate with a V drop of 6v at the same light intensity but the voltage drop across the diode would now be 4V.

    Many thanks in advance.

    Attached Files:

  2. jcsd
  3. May 7, 2016 #2

    Paul Colby

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    For reverse bias the current is proportional to the charge carriers per second generated in the diode. This number depends on the light generating particle hole pairs and is pretty independent of the applied reverse bias as shown by the V-I curve. For very large reverse bias the diode will begin to break down. Break down occurs when the charge carriers gain enough energy to generate more charge carriers. In this limit the diode will act somewhat like a geiger counter for photons.

    The role of ##R_A## in the circuit is to turn the current (which is proportional to the light) into a voltage. The greater the resistance the greater the voltage drop for a given light level. Hope this helps.
  4. May 7, 2016 #3

    Paul Colby

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    Ah, you're asking what the role of the reverse bias is. The larger the reverse bias the lower the junction capacitance of the diode. This will effect the speed or bandwidth of the diode to a time varying optical signal. Clearly the ##R_A C## time constant where ##C## is the diode junction capacitance will limit the circuit response time.
  5. May 7, 2016 #4
    I'm interested in the voltage, In the picture I attached there is a supply voltage of 15V and for a given question let's say the RA needs 6V to work, when the current through the photodiode is sufficient to give 6V to the RA does this mean that the voltage drop across the diode is 15-6 = 9 ?
  6. May 8, 2016 #5

    Paul Colby

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    Well, I don't understand what you mean by "needs 6V to work" in the above. For the most part resistors always "work" and there will be a a voltage drop given by Ohm's law, ##V=IR##. The voltage drop across the diode is that part of 15V that isn't dropped across the resistor. If current through the resistor is ##I=I_\text{leak}+I_\text{photo}## the voltage drop across ##R_A## is this current times ##R_A##. All this holds in the DC limit.

    So, if sufficient current is passing through the diode to make the voltage drop 6V across the resistor then the voltage drop across the diode will be ##9 = 15 - 6## volts. It might be best to look at the diode as a current source and not think in terms of voltages. For typical photodiodes ##R_A## would have to be very very large to drop 6V for the typical photocurrent.

    (By photodiode you don't mean solar cell do you? In that limit 6V might make some sense. You could help us out by sharing what you're trying to do)
    Last edited: May 8, 2016
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