Maximum power delivered by a solar cell to a resistive load

In summary: It's often a good approximation for an LED as well.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. It's often a good approximation for an LED as well.In summary, the conversation discusses the current-voltage characteristic of a photovoltaic energy converter, which can be approximated by a given formula. It then asks for the optimum value of a resistor in
  • #1
pc2-brazil
205
3

Homework Statement



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?

Homework Equations



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:

[tex]v+Ri=0[/tex]

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.
 

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  • #2
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)
 
  • #3
Simon Bridge said:
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)

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?
 
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  • #4
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.
 
  • #5
rude man said:
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.

OK, thank you for confirming it and for providing an alternative approach.
 
  • #6
pc2-brazil said:
OK, thank you for confirming it and for providing an alternative approach.

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.
 

FAQ: Maximum power delivered by a solar cell to a resistive load

1. What is the maximum power point of a solar cell?

The maximum power point of a solar cell is the point at which the solar cell is producing the maximum amount of power. This occurs when the voltage and current output of the solar cell are balanced, resulting in the highest possible power output.

2. How is the maximum power point of a solar cell determined?

The maximum power point of a solar cell is determined by the characteristics of the solar cell and the external load connected to it. It is typically found by plotting the voltage and current outputs of the solar cell on a graph and identifying the point where the curve reaches its peak.

3. What factors affect the maximum power point of a solar cell?

The maximum power point of a solar cell can be affected by various factors, including the intensity of sunlight, temperature, and the characteristics of the external load. Changes in any of these factors can shift the maximum power point of the solar cell.

4. Can the maximum power point of a solar cell change over time?

Yes, the maximum power point of a solar cell can change over time. Factors such as changes in the sunlight intensity or temperature can cause the maximum power point to shift. Additionally, as a solar cell ages, its maximum power point may decrease due to wear and tear.

5. How is the maximum power delivered by a solar cell to a resistive load calculated?

The maximum power delivered by a solar cell to a resistive load is calculated using the equation Pmax = Vmax x Imax, where Pmax is the maximum power, Vmax is the voltage at the maximum power point, and Imax is the current at the maximum power point. This equation takes into account the voltage and current outputs of the solar cell at the maximum power point.

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