Try the solar panel

Efficiency of the solar module

$$\text{Efficiency}\eta =\frac{\text{electrical output}}{\text{irradiated power}}=\frac{{\mathrm{P.}}_{the\; end}}{{\mathrm{P.}}_a}$$

The efficiency $\eta$ is de fi ned as the ratio of the radiated power P._a and the electric power output from the solar cell P._{the end} at the point of maximum performance.

The solar cell emits the maximum electrical power in the Maximum Power Point. The value for P._{the end} is equivalent to P._MPP, is already known (in this example P._{the end} = $\mathrm{0,311}\text{W.}$).

Determination with radiation power meter

With a radiation power meter, the power of the irradiated light per area (irradiance E.) measured. For determining the performance P._athat hits the solar cell, this value must be multiplied by the effective area of ​​the solar module.

$${\mathrm{P.}}_a=\mathrm{E.}·\mathrm{A.}$$

Legend

${\mathrm{P.}}_a$	-	Power of the light that hits the solar module in $\text{W.}$
$\mathrm{E.}$	-	Irradiance in $\text{W.}{\text{m}}^{-2}$
$\mathrm{A.}$	-	effective area of ​​the solar module in ${\text{m}}^2$

The efficiency can now be adjusted with $\eta =\frac{{\mathrm{P.}}_{the\; end}}{{\mathrm{P.}}_a}$ to calculate.

Determination without a radiation power meter

If no measuring device is available to measure the radiated power, the multimeter can be used to estimate the radiated light power. The fact that the short-circuit current (maximum photocurrent) is proportional to the number of photons (radiation) that hit the solar cell is used for this. The short-circuit current is therefore proportional to the radiated light power. For this, the factor F. used (proportionality factor).

$${\mathrm{P.}}_a=\mathrm{F.}·{\mathrm{I.}}_K·\mathrm{A.}$$

Legend

${\mathrm{P.}}_a$	-	Power of the light that hits the solar module in $\text{W.}$
$\mathrm{F.}$	-	factor F. (see below)
${\mathrm{I.}}_K$	-	Short circuit current in $\text{mA}$
$\mathrm{A.}$	-	effective area of ​​the solar module in ${\text{m}}^2$

factor $\mathrm{F.}$

The open circuit voltage is characteristic of the semiconductor material from which the solar cell is made. It is not proportional to the incident light and can therefore not be used for this measurement. In order to obtain a quantitative statement about the light output and to use the multimeter as a measuring device for the light output, the short-circuit current displayed on the multimeter must have a factor F. be multiplied. This factor depends on the maximum value of the short-circuit current of the solar cell.

The maximum output of the incident light per area in sunshine in summer is approx. $1.000\text{W.}{\text{m}}^{-2}$. The characteristics of the solar modules refer to the standard test conditions of $1.000\text{W.}{\text{m}}^{-2}$ Exposure to sunlight $25\text{° C}$ Cell temperature. The maximum value specified by the manufacturer for the short-circuit current is achieved with this irradiation power.

Assume the maximum short circuit current is $600\text{mA}$ under standard test conditions. The factor $\mathrm{F.}$ is then calculated using the following formula:

$$\mathrm{F.}=\frac{{\mathrm{P.}}_{Max}/\mathrm{A.}}{{\mathrm{I.}}_{KMax}}=\frac{1.000\text{W.}{\text{m}}^{-2}}{600\text{mA}}=\mathrm{1,67}\frac{\text{W.}}{{\text{m}}^2·\text{mA}}$$

If the short-circuit current displayed on the multimeter is now with the factor

$$\mathrm{F.}=\mathrm{1,67}\frac{\text{W.}}{{\text{m}}^2·\text{mA}}$$

multiplied, one has the approximate value of the area-related radiation output that hits the solar module.

In order to calculate the light output radiated onto the solar module, the effective area of ​​the solar module must be measured and multiplied by the area-related radiation output.

example

Solar panel area: 4 cells each $26\text{mm}$ x $77\text{mm}$, $\mathrm{A.}=8·{10}^{-3}{\text{m}}^2$Short circuit current: ${\mathrm{I.}}_K=200\text{mA}$

$${\mathrm{P.}}_a=\mathrm{F.}·{\mathrm{I.}}_K·\mathrm{A.}=\mathrm{1,67}\frac{\text{W.}}{{\text{m}}^2·\text{mA}}·200\text{mA}·8·{10}^{-3}{\text{m}}^2=\mathrm{2,67}\text{W.}$$

As already determined in this example, the maximum electrical power output is$\mathrm{0,311}\text{W.}$ (with an irradiated power of $\mathrm{2,67}\text{W.}$).

The efficiency can now be determined using the above equation.

$$\eta =\frac{{\mathrm{P.}}_{the\; end}}{{\mathrm{P.}}_a}=\frac{\mathrm{0,311}}{\mathrm{2,67}}=\mathrm{0,116}\approx \mathrm{11,6}\%$$

Legend

$\mathrm{A.}$	-	effective area of ​​the solar module in ${\text{m}}^2$
${\mathrm{I.}}_K$	-	Short circuit current in $\text{mA}$
${\mathrm{P.}}_a$	-	Power of the light that hits the solar module in $\text{W.}$
${\mathrm{P.}}_{the\; end}$	-	maximum output in $\text{W.}$
$\eta$	-	Efficiency of the solar module