For a Zener diode as shown in the circuit diagram below, the Zener voltage \(V_Z\) is 3.7 V. For a load resistance (\(R_L\)) of 1 k\(\Omega\), a current \(I_1\) flows through the load. If \(R_L\) is decreased to 500 \(\Omega\), the current changes to \(I_2\). The ratio \(\frac{I_2}{I_1}\) is \rule{1cm{0.15mm}. (up to two decimal places)}
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The most common mistake in Zener diode problems is assuming the diode is always regulating. Always perform the initial check: calculate the voltage across the load as if the Zener wasn't there. If this voltage is less than \(V_Z\), the Zener is off and has no effect on the circuit.
Step 1: Understanding the Concept:
A Zener diode is used for voltage regulation. When reverse biased, it maintains a constant voltage \(V_Z\) across its terminals, provided the input voltage is high enough to keep it in the "breakdown" region. If the conditions do not support breakdown, the Zener acts like an open circuit. We must check the operating condition for each load resistance. Step 2: Key Formula or Approach:
1. Determine if the Zener diode is in breakdown. This occurs if the voltage across the load (if the Zener were absent) is greater than \(V_Z\). This no-load voltage \(V_{NL}\) is found using the voltage divider rule: \(V_{NL} = V_S \frac{R_L}{R_S + R_L}\).
2. If \(V_{NL} \ge V_Z\), the Zener is ON, and the load voltage is regulated to \(V_L = V_Z\). The load current is \(I_L = V_Z / R_L\).
3. If \(V_{NL}<V_Z\), the Zener is OFF and acts as an open circuit. The load current is then determined by the simple series circuit: \(I_L = V_S / (R_S + R_L)\). Step 3: Detailed Explanation:
Given: Source voltage \(V_S = 10\) V, Series resistance \(R_S = 1\) k\(\Omega\), Zener voltage \(V_Z = 3.7\) V.
Case 1: \(R_L = R_{L1} = 1\) k\(\Omega\)
- Check the condition for breakdown:
\[ V_{NL} = 10 \text{ V} \times \frac{1 \text{ k}\Omega}{1 \text{ k}\Omega + 1 \text{ k}\Omega} = 10 \text{ V} \times \frac{1}{2} = 5 \text{ V} \]
- Since \(V_{NL} = 5 \text{ V}>V_Z = 3.7 \text{ V}\), the Zener diode is ON and regulating.
- The voltage across the load is \(V_{L1} = V_Z = 3.7\) V.
- The load current is \(I_1 = \frac{V_{L1}}{R_{L1}} = \frac{3.7 \text{ V}}{1 \text{ k}\Omega} = 3.7 \text{ mA}\).
Case 2: \(R_L = R_{L2} = 500 \, \Omega = 0.5\) k\(\Omega\)
- Check the condition for breakdown:
\[ V_{NL} = 10 \text{ V} \times \frac{500 \, \Omega}{1000 \, \Omega + 500 \, \Omega} = 10 \text{ V} \times \frac{500}{1500} = \frac{10}{3} \text{ V} \approx 3.33 \text{ V} \]
- Since \(V_{NL} \approx 3.33 \text{ V}<V_Z = 3.7 \text{ V}\), the Zener diode is OFF (not in breakdown).
- It acts as an open circuit. The circuit is a simple voltage divider with \(R_S\) and \(R_{L2}\).
- The load current \(I_2\) is the total current in this series circuit:
\[ I_2 = \frac{V_S}{R_S + R_{L2}} = \frac{10 \text{ V}}{1000 \, \Omega + 500 \, \Omega} = \frac{10 \text{ V}}{1500 \, \Omega} \approx 6.667 \times 10^{-3} \text{ A} = 6.667 \text{ mA} \]
Calculate the ratio \(\frac{I_2}{I_1}\):
\[ \frac{I_2}{I_1} = \frac{6.667 \text{ mA}}{3.7 \text{ mA}} \approx 1.80189 \]
Step 4: Final Answer:
The ratio \(\frac{I_2}{I_1}\), rounded to two decimal places, is 1.80.
