experiment 2 reverse bias diode this experiment examines the reverse leakage current in a silicon 1n4004 diode the reverse current through a pn junction is typically very small na due to the 74106

Experiment 2: Reverse Bias Diode This experiment examines the reverse leakage current in a silicon 1N4004 diode. The reverse current through a PN junction is typically very small (nA), due to the extremely high reverse resistance. This current is called the reverse leakage current. The current is so small that it cannot be measured with a normal DMM. It can only be calculated using Ohm's Law. Note that the diode is now installed in the circuit in reverse bias. The current passing through the diode and the series 470 k resistor will be very small. Since this is a series circuit, the current through the diode will be the same as the current through the resistor. The DMM will measure the very small voltage that develops across the resistor. Since the voltage of the resistor is known, and the value of the resistor is known, the current through the resistor can be calculated. Procedure: 8. Measure the exact value of your resistor. DO NOT TOUCH IT WHILE MEASURING. Enter this measured value in Table 2. TABLE 2 Resistance 1N4004 Use the Agilent meter Vs 30 V R 470 k 9. Build the circuit shown in Figure 5. Figure 5: Reverse Bias Diode 10. Adjust the Current Control setting on the Power Supply to 100 mA. 11. Adjust the voltage of the power supply to each of the Voltages listed in Table 3 and enter the measured Rz voltage for each step in Table 3. 12. Use the measured R2 resistance and measured R2 voltage to calculate the reverse current I for each step and enter your calculated value in Table 3. TABLE 3 Step # Power Supply Voltage 1 0V 2 5V 3 10V 4 20V 5 30V Resistor R Voltage Calculated Current IR

Q1. Shading Effects and Bypass Diode Shaded Cell LOAD Current Consider 12 ideal and identical solar cells connected in series, and one of the cells is fully shaded as shown above. Each cell has Voc = 0.6V and Isc = 1A. Find the following: i) Find the relation between V and V at short circuit condition. (10 points) ii) Estimate the voltage across the shaded cell at short circuit condition assuming no reverse breakdown. (10 points) iii) If shading reduces the short circuit current to 0.1A, estimate the power dissipation in the shaded cell. (10 points) iv) If the reverse breakdown voltage of each cell is -2V, will the shaded cell undergo reverse breakdown? Justify your answer. (10 points) v) Find V and V at short circuit with breakdown voltage = -2V (Assume that after breakdown, the diode becomes highly conductive). (10 points) vi) Find the power dissipation in the shaded cell under reverse breakdown at short circuit. (10 points) vii) Find the open circuit voltage and short circuit current for the module with one cell fully shaded in the presence of bypass diodes across each cell. (10 + 10 points) viii) Estimate the minimum number of bypass diodes that need to be used to prevent reverse breakdown in this 12-cell module. Assume bypass diode's built-in voltage = 0.5V. (20 points)

Adi S.

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