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I-V Characteristic Graphs

  1. May 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Using the diagram for the I-V characteristic of a semi-conductor (attached). Answer the following questions:
    1. Why is the resistance of an I-V characteristic not determined by the gradient of the graph
    2. Explain why the temperature of semiconductor increases
    3. Explain why the resistivity goes down.

    I am mainly stuck with question 2.

    2. Relevant equations
    p = RA/l (p = resistivity, R = resistance, A = area, I = length.
    P= I^2 R

    3. The attempt at a solution
    1. I don't know whether they are trying to get at - is it because it is the inverse as it is current versus voltage or is it that the gradient is the rate of change which would be wrong as resistance is just the ratio of voltage to current (i.e. if its a curve you would not take the tangent)?
    2. This is the bit I am stuck with. I checked the answer and it says that you must use P = I^2 R to justify the increase in temperature and cannot justify it in terms of voltage increase. This is what confuses me. The graph shows the voltage increases and voltage is proportional to power, so I don't understand why you must use P = I^ R?
    3. Resistivity goes down because area and length are constant but resistance goes down so therefore resistivity must go down.

    Many thanks for any help.
     

    Attached Files:

  2. jcsd
  3. May 20, 2015 #2

    BvU

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    Hi Jim,
    (1) Current versus voltage is indeed not so good: with V = I R for a resistor (i.e. a constant resistance R) 1/R is indeed the slope of a current versus voltage curve. But I don't think that's what they mean. Here there is no single resistance: the slope changes with the voltage. The best you get from the gradient (bad word here, slope is much better) is a kind of local resistance ##\Delta V = R(V) \Delta I##

    (2) Apperently there is some kind of context for the question here, because if you just leave a chunk of semiconductor laying around, there is no reason for its temperature to increase spontaneously

    (3) is circular reasoning in your answer, so you'll have to find something better!
     
  4. May 20, 2015 #3
    Thanks for the help. My answers for parts 1 and 3 are correct as far as I can tell form the markscheme. I still don't know what you mean for question 2. I still have no idea why you must use P = I^2 R and why the markscheme says you cannot talk about the increase in power coming from voltage. The graph clearly shows voltage increases and a bigger voltage gives more power so why can't you argue this way?
     
  5. May 20, 2015 #4
    Just to add to what I said. If you take two points on the graph, you get the following:

    at 4V the current is 0.008A giving a power of 0.032W and a resistance of 500 Ohms
    at 8V the current is 0.026A giving a power of 0.208W and the resistance is 307 Ohm's

    If we kept the voltage the same (4V) but lowered the resistance to 307 Ohm's the current instead would be 0.013A giving a power of 0.052W. Therefore you can clearly see that the increase in power (and therefore temperature) surely comes form the increase in voltage?
     
  6. May 20, 2015 #5

    BvU

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    I really can't help much with this kind of question/marking scheme.

    In part (1) it's made clear that there is no such thing as a single-value resistance here, and then the marking scheme wants you to use it ? Apparently in (2) and in (3) as well ?

    Power dissipated is I x V is what I learned. For a resistor (constant R) Ohm's law lets you write I2R or V2/R equally well.

    And in (3), if the marking scheme is happy with "resistance" down ##\Rightarrow## resistivity down, and you are happy with the marking scheme, then who am I to claim that doesn't explain why ?
     
  7. May 20, 2015 #6
    Thanks again. I'm happy with part 3 its just part 2 as I would explain the power increase to the semiconductor through an increase in voltage. Here is how I would put it - the voltage across the semiconductor increases therefore more power is dissipated and the temperature increases. This causes more electrons to be freed increasing the current. So the part about increase temperature I would simply explain by P = V I, both V and I will increase to give a larger power.

    I have attached the full question with the mark scheme and the notes from the examiners as we are told when revising past questions to read the examiners notes as they contain useful information about understanding what they want. It is 1c I am unhappy with. From the examiners notes on question 1 c it seems to me that they are saying that it is incorrect thinking to link voltage to power dissipated as they say "it was common for students to link temperature rise to voltage rather than current".
     

    Attached Files:

  8. May 20, 2015 #7
    I think I've found some information which has provided an answer (I am unsure about how reliable for source is). Apparently, although an increase in voltage is associated with an increase in power, it is not usually associated with an increase in something called "Joule Heating". Joule heating is strictly power loss through heating which has a strong dependence on current rather than voltage. So I guess what they are getting at in the question is that the temperature rise comes from Joule heating which depends on current rather than voltage. Does anyone know why this is - i.e. why the flow of charge (current) is a strong indicator of heat loss in a conductor rather the energy the charges have (voltage)? I guess it would have some microscopic explanation?
     
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