Question

For the given circuit, the power across zener diode is __________ mW. 

 

Hint:

In Zener diode circuits, first check if the diode is "on" (i.e., V\(_{in}\)>V\(_Z\)). If it is, the voltage across the parallel load is fixed at V\(_Z\). Then, analyze the rest of the circuit using Ohm's law and Kirchhoff's laws.

Answer 1

Correct Answer: 120

Step 1: Understanding the Question:
We are given a voltage regulator circuit using a Zener diode. We need to calculate the power dissipated by the Zener diode.
Step 2: Key Formula or Approach:
1. A Zener diode, when operating in the breakdown region, maintains a constant voltage across it, which is the Zener voltage V\(_Z\).
2. The voltage across the load resistor R\(_L\) will be equal to V\(_Z\) since they are in parallel.
3. Calculate the current through the load resistor (I\(_L\)) and the total current (I) flowing through the series resistor.
4. The current through the Zener diode (I\(_Z\)) is the difference between the total current and the load current (I = I\(_Z\) + I\(_L\)).
5. The power dissipated by the Zener diode is P\(_Z\) = V\(_Z\) \(\times\) I\(_Z\).
Step 3: Detailed Explanation:
Given values:
Input Voltage, V\(_{in}\) = 24 V
Series Resistance, R\(_s\) = 1 k\( \Omega \) = 1000 \( \Omega \)
Zener Voltage, V\(_Z\) = 10 V
Load Resistance, R\(_L\) = 5 k\( \Omega \) = 5000 \( \Omega \)
Since the input voltage (24 V) is greater than the Zener voltage (10 V), the Zener diode is in its breakdown region and will regulate the voltage. The voltage across the load resistor is V\(_L\) = V\(_Z\) = 10 V.
Calculate the current through the load resistor, I\(_L\):
\[ I_L = \frac{V_L}{R_L} = \frac{10 \text{ V}}{5000 \, \Omega} = 0.002 \text{ A} = 2 \text{ mA} \] The voltage drop across the series resistor R\(_s\) is:
\[ V_s = V_{in} - V_Z = 24 \text{ V} - 10 \text{ V} = 14 \text{ V} \] Calculate the total current I flowing from the source:
\[ I = \frac{V_s}{R_s} = \frac{14 \text{ V}}{1000 \, \Omega} = 0.014 \text{ A} = 14 \text{ mA} \] This total current splits between the Zener diode and the load resistor. By Kirchhoff's current law:
\[ I = I_Z + I_L \] \[ I_Z = I - I_L = 14 \text{ mA} - 2 \text{ mA} = 12 \text{ mA} \] Finally, calculate the power dissipated by the Zener diode:
\[ P_Z = V_Z \times I_Z = 10 \text{ V} \times 12 \text{ mA} = 10 \text{ V} \times (12 \times 10^{-3} \text{ A}) = 120 \times 10^{-3} \text{ W} = 120 \text{ mW} \] Step 4: Final Answer:
The power across the Zener diode is 120 mW.