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Homework Help: Maximum power delivered by a solar cell to a resistive load

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data

    The current-voltage characteristic of a photovoltaic energy converter (solar cell) shown in the attached figure can be approximated by:

    [tex]i = I_1(e^{v/V_{TH}} - 1) - I_2[/tex]

    where the first term characterizes the diode in the dark and I2 is a term that depends on light intensity.

    Assume [itex]I_1 = 10^{-9}[/itex] and assume light exposure such that [itex]I_2 = 10^{-3}\ A[/itex].

    If it is desired to maximize the power that the solar cell can deliver to a resistive load, determine the optimum value of the resistor. How much power can this cell deliver?

    2. Relevant equations

    3. The attempt at a solution

    This question doesn't provide any value for VTH, but the book mentions that diodes tipically have VTH = 0.025 V, so I assume it is the value to be used here.

    Applying KVL to the attached figure:


    The question asks for the value of R for which [itex]P=Ri^2[/itex] is a maximum, so I suppose I should differentiate P with respect to R. However, I first need to solve [itex]v+Ri=0[/itex] and [itex]i = I_1(e^{v/V_{TH}} - 1) - I_2[/itex] simultaneously in order to find i in terms of R, and it doesn't appear to be possible analytically.

    Any hint on how to continue?

    Thank you in advance.

    Attached Files:

    • fig.GIF
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  2. jcsd
  3. Feb 5, 2013 #2

    Simon Bridge

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    Your problem is that you want to do ##\frac{d}{dR}i(R)=0## on $$i(R)=I_1\left ( e^{-iR/V_{TH}} -1 \right ) -I_2$$ ... won't it do that implicitly by the chain rule?
    (could be wrong - in which case I'd look for approximations)
  4. Feb 6, 2013 #3
    Thank you for the answer.

    I think that I arrived at a solution. I tried to do it as follows:

    Since the voltage across the resistor is [itex]-v[/itex] (in which I put a minus sign to follow the associated variables convention) and the current through the resistor is [itex]i[/itex], the power consumed by the resistor is [itex]P = -vi[/itex].

    [tex]\frac{dP}{dR} = -v\frac{di}{dR} - i\frac{dv}{dR}[/tex]

    I look for a maximum when the above expression is zero:

    [tex]v\frac{di}{dR} = -i\frac{dv}{dR}[/tex]

    [itex]\frac{di}{dR}[/itex] can be expressed as:

    [tex]\frac{di}{dR} = \frac{di}{dv}\frac{dv}{dR} = \frac{I_1e^{v/V_{TH}}}{V_{th}}\frac{dv}{dR} [/tex]

    Substituting that into [itex]v\frac{di}{dR} = -i\frac{dv}{dR}[/itex], I get:

    [tex]v\frac{I_1e^{v/V_{TH}}}{V_{th}}\frac{dv}{dR} = -i\frac{dv}{dR}[/tex]

    So, the [itex]\frac{dv}{dR}[/itex] ends up cancelling out. Since i is expressed in terms of v, I got an expression that involves v only, which can be solved for v. Doing this, I get that v is approximately 0.28 V. Plugging this into the expression for i, I get that i is approximately -0.00092 A.

    Thus, from v = -iR, the resistance is approximately 307.64 ohms, and power consumed is P = -vi = +0.26 mW.

    Now I'm left with proving that this is actually the maximum value for P. But I guess that this is reasonable, because P = 0 for R = 0, and R is not bounded to the right, so it appears the maximum can't be at an extremum.

    Is this correct?
    Last edited: Feb 6, 2013
  5. Feb 6, 2013 #4

    rude man

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    Happy to corroborate your answer. Nice work! My approach was a bit different: I solved for R(i), then extremalized (i^2)R(i) using d(i^2)R(i)/dt = 0. Had to use Excel to arrive at your number for i (transcendental equation).

    It's a maximum for exactly the reason you gave.
  6. Feb 6, 2013 #5
    OK, thank you for confirming it and for providing an alternative approach.
  7. Feb 6, 2013 #6

    rude man

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    OK, and of course I meant to say (d[i2R(i)/dR = 0.

    Also BTW I used the approximation exp(v/VTh) - 1 ~ exp(v/VTh). In this case exp(0.28v/0.026V) = 47535 which certainly >> 1. You can almost always do that for a diode i-V characteristic.
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