# Engineering Experience 4: Design a Small Solar Vehicle/Nl/2013: Team PM5/Uitmeten zonnepanelen pm5
## Measure Solar

When one wishes to use a solar panel, it is important to know what the current-voltage characteristic of the panel. We need to know how the current is in proportion to the current.
To find these characteristics we must first determine the two extremes, this is the current at a voltage of 0V or the short-circuit current and voltage at a current of 0A or the open circuit voltage. Through these two values, which we know our chart will be. To determine the other values ​​of the graph, we proceed as follows. An adjustable resistor in series placed with the solar panel, an ammeter is also placed in series with the resistance and the solar panel. Parallel across the solar panel is a voltmeter placed.
By increasing the variable resistor, the voltage will increase about it, and may reduce the value for the current through the circuit reading of the ampere meter. We obtained the following results:
| Voltage U(V) | Current I(A) |
| 0,09 | 0,60 |
| 1,50 | 0,60 |
| 5,67 | 0,60 |
| 7,10 | 0,58 |
| 7,34 | 0,53 |
| 7,48 | 0,47 |
| 7,51 | 0,44 |
| 7,53 | 0,42 |
| 7,61 | 0,37 |
| 7,61 | 0,35 |
| 7,63 | 0,33 |
| 7,69 | 0,28 |
| 7,72 | 0,25 |
| 7,75 | 0,21 |
| 7,79 | 0,17 |
| 7,81 | 0,14 |
| 7,82 | 0,13 |
| 7,82 | 0,11 |
| 7,82 | 0,10 |
| 7,83 | 0,08 |
| 7,85 | 0,05 |
| 7,84 | 0,04 |
When we plot these results we obtain the following graph:
Since the voltage and current are measured with the same measuring device with the same precision, they both have the same error. The current values is an error of 0.01 A and the voltage values ​​of a 0.01 V.
So we see that at the beginning of the current remains constant at varying voltage but that the current will decrease with increasing voltage. quickly from a certain tension We can see from these data the voltage-output characteristic obtained, this should always do once the voltage on the power. this gives us the following values​​:
| Voltage U(V) | Current I(A) | Power P(W) |
| 0,09 | 0,60 | 0,054 |
| 1,50 | 0,60 | 0,9 |
| 5,67 | 0,60 | 3,402 |
| 7,10 | 0,58 | 4,118 |
| 7,34 | 0,53 | 3,8902 |
| 7,48 | 0,47 | 3,5156 |
| 7,51 | 0,44 | 3,3044 |
| 7,53 | 0,42 | 3,1626 |
| 7,61 | 0,37 | 2,8157 |
| 7,61 | 0,35 | 2,6635 |
| 7,63 | 0,33 | 2,5179 |
| 7,69 | 0,28 | 2,1532 |
| 7,72 | 0,25 | 1,93 |
| 7,75 | 0,21 | 1,6275 |
| 7,79 | 0,17 | 1,3243 |
| 7,81 | 0,14 | 1,0934 |
| 7,82 | 0,13 | 1,0166 |
| 7,82 | 0,11 | 0,8602 |
| 7,82 | 0,10 | 0,782 |
| 7,83 | 0,08 | 0,6264 |
| 7,85 | 0,05 | 0,3925 |
| 7,84 | 0,04 | 0,3136 |
Plotting these values ​​gives us the following graph:
The determination of the error is done using the following formula:
    $\left(U\pm 0,01\right)\left(I\pm 0,01\right)=\left(U*I\right)\pm \left(|U*I|*\left({\frac {0,01}{|U|}}+{\frac {0,01}{|I|}}\right)\right)\,$
When we apply it to all values ​​we obtained following table:
| Spanning U(V) | Stroom I(A) | Vermogen P(W) | Error on the ability (W) |
| 0,09 | 0,60 | 0,054 | 0,0069 |
| 1,50 | 0,60 | 0,9 | 0,021 |
| 5,67 | 0,60 | 3,402 | 0,0627 |
| 7,10 | 0,58 | 4,118 | 0,0768 |
| 7,34 | 0,53 | 3,8902 | 0,0787 |
| 7,48 | 0,47 | 3,5156 | 0,0795 |
| 7,51 | 0,44 | 3,3044 | 0,0795 |
| 7,53 | 0,42 | 3,1626 | 0,0795 |
| 7,61 | 0,37 | 2,8157 | 0,0798 |
| 7,61 | 0,35 | 2,6635 | 0,0796 |
| 7,63 | 0,33 | 2,5179 | 0,0796 |
| 7,69 | 0,28 | 2,1532 | 0,0797 |
| 7,72 | 0,25 | 1,93 | 0,0797 |
| 7,75 | 0,21 | 1,6275 | 0,0796 |
| 7,79 | 0,17 | 1,3243 | 0,0796 |
| 7,81 | 0,14 | 1,0934 | 0,0795 |
| 7,82 | 0,13 | 1,0166 | 0,0795 |
| 7,82 | 0,11 | 0,8602 | 0,0793 |
| 7,82 | 0,10 | 0,782 | 0,0792 |
| 7,83 | 0,08 | 0,6264 | 0,0791 |
| 7,85 | 0,05 | 0,3925 | 0,0790 |
| 7,84 | 0,04 | 0,3136 | 0,0788 |
When we obtain the average pack a rounded error of 0.08 W
For the calculation of the diode factor, the following formula is used:
    $I=I_{sc}-I_{s}\left(e^{\frac {U}{m*N*U_{r}}}-1\right)\,$
Hierin is: Isc = short circuit current (A)
Is = saturation current (A) :$\left({\frac {10^{-8}A}{m^{2}}}\right)\,$
U = output voltage (V)
Ur = thermical voltage (V): 25,7 mV bij 25°C
Ur = (k*T)/e
k = Boltzmann constant =:$1,38*10^{-23}{\frac {J}{K}}\,$
T = temperature (K)
E = charge of an electron :$(V)=1.6*10^{-19}A_{s}\,$
M = diode factor (range 1~5)
N = some sunbathing cells in series
We know that the surface of the cells is equal to: $3.6*10-2m^{2}\,</math>,thisgivesusasaturatiestroomIs3.6*10-10AIfwewanttocalculatetheresultingdiodefactorweneedtheformulatransformedtothefollowingform::<math>m={\frac {U}{N*U_{r}*\ln \left({\frac {I-I_{sc}-I_{s}}{-I_{s}}}\right)}}\,$
The values ​​used in our calculations are:
| open circuit spanning | (V) | 8,70 |
| short circuit current | Isc (A) | 0,60 |
| saturatiestroom | Is (A) | 3,6E-10 |
| thermical voltage bij 25 ° | Ur (V) | 0,0257 |
| Number of solar cells in series | N | 15 |
This gave us the following values ​​for the diode factor:
| Voltage U(V) | Current I(A) | Diode factor |
| 0,09 | 0,60 | I=Isc |
| 1,50 | 0,60 | I=Isc |
| 5,67 | 0,60 | I=Isc |
| 7,10 | 0,58 | 1,033 |
| 7,34 | 0,53 | 0,998 |
| 7,48 | 0,47 | 0,985 |
| 7,51 | 0,44 | 0,978 |
| 7,53 | 0,42 | 0,975 |
| 7,61 | 0,37 | 0,974 |
| 7,61 | 0,35 | 0,970 |
| 7,63 | 0,33 | 0,969 |
| 7,69 | 0,28 | 0,968 |
| 7,72 | 0,25 | 0,968 |
| 7,75 | 0,21 | 0,966 |
| 7,79 | 0,17 | 0,967 |
| 7,81 | 0,14 | 0,966 |
| 7,82 | 0,13 | 0,966 |
| 7,82 | 0,11 | 0,965 |
| 7,82 | 0,10 | 0,964 |
| 7,83 | 0,08 | 0,963 |
| 7,85 | 0,05 | 0,963 |
| 7,84 | 0,04 | 0,961 |
The error on the diode factor is calculated using the following formula:
    $\delta m={\sqrt {\left({\frac {\delta m}{\delta U}}\delta U\right)^{2}+\left({\frac {\delta m}{\delta I}}\delta I\right)^{2}}}\,$
Here is:
    ${\frac {\delta m}{\delta U}}={\frac {1}{N*U_{r}*\ln \left({\frac {I-I_{sc}-I_{s}}{-I_{s}}}\right)}}\,$
En
    ${\frac {\delta m}{\delta I}}={\frac {U}{N*U_{r}}}*\left({\frac {{\frac {I_{s}}{I-I_{sc}-I_{s}}}*{\frac {1}{-I_{s}}}}{\ln ^{2}\left({\frac {I-I_{sc}-I_{s}}{-I_{s}}}\right)}}\right)\,$
By completing the formulas following results are obtained:
| Voltage U(V) | Current I(A) | Diode factor | |
| 0,09 | 0,60 | I=Isc | I=Isc |
| 1,50 | 0,60 | I=Isc | I=Isc |
| 5,67 | 0,60 | I=Isc | I=Isc |
| 7,10 | 0,58 | 1,033 | 0,029 |
| 7,34 | 0,53 | 0,998 | 0,008 |
| 7,48 | 0,47 | 0,985 | 0,004 |
| 7,51 | 0,44 | 0,978 | 0,003 |
| 7,53 | 0,42 | 0,975 | 0,003 |
| 7,61 | 0,37 | 0,974 | 0,002 |
| 7,61 | 0,35 | 0,970 | 0,002 |
| 7,63 | 0,33 | 0,969 | 0,002 |
| 7,69 | 0,28 | 0,968 | 0,002 |
| 7,72 | 0,25 | 0,968 | 0,002 |
| 7,75 | 0,21 | 0,966 | 0,002 |
| 7,79 | 0,17 | 0,967 | 0,002 |
| 7,81 | 0,14 | 0,966 | 0,002 |
| 7,82 | 0,13 | 0,966 | 0,002 |
| 7,82 | 0,11 | 0,965 | 0,002 |
| 7,82 | 0,10 | 0,964 | 0,002 |
| 7,83 | 0,08 | 0,963 | 0,002 |
| 7,85 | 0,05 | 0,963 | 0,001 |
| 7,84 | 0,04 | 0,961 | 0,001 |
We get an average diode factor of 0.97 with a standard deviation of 0.02, so m = 0.97 ± 0.02
Note: Some of the values ​​are I = Isc, this means that we ln 1 hatching which is equal to 0, this ensures that we divide by zero, which is impossible

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Gantt chart pm5
WBS PM5
Engineering Experience 4: Design a Small Solar Vehicle/Nl/2013: Team PM5