Question

The Zener diode has reverse breakdown voltage \(V_{Z}\). The current through the load resistance \(R_{L}\) is \(I_{L}\). The current through Zener diode is
(\(V_{0}\) = Input voltage , \(R_{S}\) = Series resistance)

• A. \(\frac{(V_{0}-V_{Z})}{R_{S}}\)
• B. \(\frac{(V_{0}-V_{Z})}{R_{L}}\)
• C. \(\frac{V_{Z}}{R_{L}}\)
• D. \(\left(\frac{V_{0}-V_{Z}}{R_{S}}\right)-I_{L}\)

Hint:

The current through the series resistance splits into the load current and the Zener current.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
In a Zener regulator the series resistance \(R_S\) carries the total current, which divides between the load and the Zener diode.

Step 2: Total current
The Zener holds the output at \(V_Z\), so the voltage across \(R_S\) is \(V_0-V_Z\).
\(I_S=\frac{V_0-V_Z}{R_S}\).

Step 3: Zener current
\(I_S=I_Z+I_L\), so \(I_Z=\frac{V_0-V_Z}{R_S}-I_L\).
This is option (D).

Final Answer:
The Zener current equals the series current minus the load current. \[ \boxed{\text{(D) }\left(\dfrac{V_{0}-V_{Z}}{R_{S}}\right)-I_{L}} \]