Elenco Deluxe Solar Educational Kit - Manual (Page 8)

8
EXAMPLE 2
CHART B, ITEM NO.5 SOLAR MOTOR 1.5V 450mW
Solar motor requires the power of 1.5V 450mW, then we need to use the following steps to
calculate the current (that is I) of the motor:
Calculate the current (I) and then determine the connecting system.
1. I =
2. I =
3. I =
4. I = 0.3A
5. I = 300mA
6. After calculate of the above, we know that a solar motor requires a 1.5V 300mA solar panel.
7. Then we have to determine the connecting system to make the current of the solar panel
for 300mA.
8. The parallel connecting system will increase the current (amperage I).
9. Each row of solar cells output is:
Voc: 1.5V
± 0.1V (each row cells)
Isc: 90mA
± 10mA (each row cells)
See Figure 13, 14 & 15
10. Therefore, we select the parallel connecting system, which will bring the solar module with
1.5V 300mA output.
11. To increase the current (amperage I), connect each row of cells in parallel (negative point
connect to negative point, positive point connect to positive point).
To increase the current (amperage I), formula as follows:
I total = I
1
+ I
2
+ I
3
e.g.:(I
1
)100mA + (I
2
)100mA + (I
3
)100mA = 300mA
E total = E
1
= E
2
= E
3
But the voltage (V) remains constant at 1.5V
12. How many rows of solar cells are needed to make a solar panel work for a solar motor of
1.5V 300mA.
Each row of cells are 1.5V 100mA, so 3 rows of cells are needed to connect in parallel
system to increase current:
I total = 3 rows of cells x 100mA = 300mA
But the voltage(E) remains constant
Or another calculation method:
I total = I
1
+ I
2
+ I
3
= (I
1
)100mA + (I
2
)100mA + (I
3
) 100mA
= 300mA
* In parallel connecting method, see Figure 15, 18, 20, 21, 22, 23 & 24
P
E
450mW
1.5V
0.450W
1.5V