Practical No. 08: Determination of static and dynamic resistance of given P N junction diode.
XIII. Interpretation of Results
- The forward resistance (both static and dynamic) was low, while the reverse resistance was very high. The knee voltage was identified, and the values of static and dynamic resistance were calculated from the graph.
XIV. Conclusions and Recommendations
- It is concluded that a PN junction diode conducts easily in forward bias and offers very high resistance in reverse bias.
- It is recommended to carefully limit the current in forward bias and the voltage in reverse bias to avoid damaging the diode.
XV. Practical Related Questions & Answers
Q1. Write the ideal value of knee voltage for the diode used.
- Answer: For a Silicon diode (e.g., IN4007), the knee voltage is approximately 0.7 V. For a Germanium diode, it is approximately 0.3 V.
Q2. Write the steps to identify the p and n terminals of diode using multimeter.
- Answer:
- Set the multimeter to diode test mode.
- Connect the red probe to one terminal and the black probe to the other.
- If the meter shows a reading (between 0.5V to 0.8V for Si), the terminal connected to the red probe is the P-side (Anode) and the black is the N-side (Cathode).
- If the meter shows “OL” or “1”, the connections are reversed.
Q3. Give reasons for using micro ammeter in a reverse bias mode.
- Answer: The current in reverse bias is very small (in microamperes or even less). Using an ammeter with a higher range (e.g., milliammeter) would not give a measurable or accurate reading for such a small current.
Q4. State the specifications of diode used in this lab experience.
- Answer: The diode used was IN4007. Its specifications are:
- Maximum Average Forward Current: 1 A
- Peak Repetitive Reverse Voltage: 1000 V
- Forward Voltage Drop: ~1.1 V at 1 A
Q5. Calculate the current in the given circuit.
- Answer:
- If forward biased: The diode is ON. Assuming a knee voltage (V_k = 0.7V for Si) drops across it, the current is:
- Formula: I = (V_supply – V_k) / R
- Given circuit simple calculation (as per manual text short note): I = V / R = 5 / 1000 = 0.005 A
- If reverse biased: The diode is OFF. The current is approximately zero (only a very small leakage current flows).
- If forward biased: The diode is ON. Assuming a knee voltage (V_k = 0.7V for Si) drops across it, the current is:
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