Junction And External Potential In Solar Cell

2lai

15th July 2009 - 01:19 PM

Question about measured I-V characteristic.

According to PVCDROM, the IV curve of a solar cell is the superposition of the IV curve of the solar cell diode in the dark with the light-generated current. Illumination light shifts the IV curve down into the fourth quadrant.

If there is a series resistance Rs, the junction potential will be V+I*Rs where V is the external measured voltage. Considering the negative sign of current (I), is the junction potential always greater or lower than the measured voltage?

Enthalpy

26th July 2009 - 02:05 AM

Fourth quadrant? Sign? What's that bad joke?

Computers can use signs. Humans make mistakes with signs. Smart engineers use only positive numbers.

Or could it be that this is a homework...? Looks like.

Anyway, and with positive voltages: the available voltage is lower than the junction potential.

Then, add if you wish that it is less negative than more positive but is measured upside down from the back through a mirror but with Spice's convention which is opposed to the normal convention but only for generators.

Confused2

26th July 2009 - 08:04 PM

QUOTE (2lai+)

Considering the negative sign of current (I), is the junction potential always greater or lower than the measured voltage?

The junction potential must always be greater than (or equal to) the measured voltage by an amount equal to the voltage drop across the source resistance ( I * Rs ). If no current is flowing then clearly no voltage drop across the source resistance so V_measured = V_junction

-C2.

rpenner

26th July 2009 - 10:45 PM

Why is this confusing???

Slope of a curve is Δy/Δx. The first quadrant of a graph is where x is positive and y is positive. The fourth quadrant of a graph is where x is positive and y is negative.

A V-I diagram plots V on the x axis and I on the y axis.
Most V-I diagrams slope upward like this: /
The inverse of the slope of the line is the resistance. Thus ΔV = ΔI×R(V) or R(V) = ΔV/ΔI which is the inverse of the slope ΔI/ΔV.
And for non-active materials, V = 0 means I = 0, all non-active materials are expected to be solely in the first quadrant.
For linear, non-active materials, R(V) is a constant function and the first equation simplifies to V = I×R.

Now a solar cell in the dark is non-active (but not-necessarily linear).
But in the light, it exhibits a small tendency to have current. So if you hook it up one way you will measure plus current and the other way minus current. So if you add external voltage to it you will pass through the fourth quadrant (with negative current and (small) positive external voltages) to the first (with both positive).

Confused2

27th July 2009 - 12:07 AM

QUOTE

Now a solar cell in the dark is non-active (but not-necessarily linear).
But in the light, it exhibits a small tendency to have current.

2lai, please comment, your course is more up to date than mine..

Essence of solar cell is (light dependent) voltage source with high source resistance?

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