What is the first step in analysing the Zener network?

Remove the Zener and find the voltage across the load; compare with VZ

Assume the Zener is ON and VL = 10 V

For RL = 3 kΩ, what is VL and the logic?

Open-circuit VL = 12 V > 10 V so Zener ON and VL = VZ = 10 V

For RL = 3 kΩ, what are Iz and Pz, and is it safe?

Iz = 2.667 mA, Pz = 26.67 mW < 30 mW, safe

Iz = 3.333 mA, Pz = 33.3 mW, unsafe

2081 Baishakh (new course) · 6 marks

Zener network with two load values

Short answer

For RL=1.2kΩ, VL=8.73V (<10V), Zener off, so VR=7.27V, IZ=0, PZ=0. For RL=3kΩ, VL=10V, Zener on, VR=6V, IZ=2.667mA, PZ=26.67mW.

For the Zener diode network shown below, determine $V_L$ , $V_R$ , $I_Z$ and $P_Z$ for $R_L = 1.2\ k\Omega$ and $R_L = 3\ k\Omega$ . [Figure: $V_i = 16$ V source in series with $R = 1\ k\Omega$ ; Zener diode with $V_Z = 10$ V and $P_{ZM} = 30$ mW in parallel with load $R_L$ ; $V_R$ across R, $I_Z$ through the Zener.]

Given: $V_i=16$ V, $R=1\ \text{k}\Omega$ , $V_Z=10$ V, $P_{ZM}=30$ mW.

Method. First assume the Zener is OFF (removed) and find the voltage across the load. If it is at least $V_Z$ , the Zener is ON and $V_L=V_Z$ .

Case 1: $R_L=1.2\ \text{k}\Omega$

V_L=\frac{R_LV_i}{R+R_L}=\frac{1.2\times16}{1+1.2}=8.727\ \text{V}

Since $8.727\ \text{V}<10\ \text{V}$ , the Zener is OFF (not in breakdown).

V_L=8.73\ \text{V},\quad V_R=16-8.727=7.27\ \text{V},\quad I_Z=0,\quad P_Z=0

Case 2: $R_L=3\ \text{k}\Omega$

V_L=\frac{3\times16}{1+3}=12\ \text{V}>10\ \text{V}

The Zener is ON, so $V_L=V_Z=10$ V.

\begin{aligned} V_R&=16-10=6\ \text{V}\\ I_R&=\frac{6}{1\ \text{k}\Omega}=6\ \text{mA}\\ I_L&=\frac{10}{3\ \text{k}\Omega}=3.333\ \text{mA}\\ I_Z&=I_R-I_L=6-3.333=2.667\ \text{mA}\\ P_Z&=V_ZI_Z=10\times2.667=26.67\ \text{mW} \end{aligned}

$P_Z=26.67\ \text{mW}<P_{ZM}=30$ mW, so the Zener is safe.

$R_L$	$V_L$	$V_R$	$I_Z$	$P_Z$
1.2 kΩ	8.73 V	7.27 V	0	0
3 kΩ	10 V	6 V	2.667 mA	26.67 mW

✓ Answer checked by Claude Opus 5.5 (high effort)

Zener circuit output voltage and current · Basic Electrical and Electronics Engineering, chapter 3
Zener diode as voltage regulator · Basic Electrical and Electronics Engineering, chapter 3
Mesh analysis with 3 A source · Basic Electrical and Electronics Engineering, chapter 1
Series AC circuit element parameters · Basic Electrical and Electronics Engineering, chapter 1
Parallel RL and RC branch currents, power · Basic Electrical and Electronics Engineering, chapter 1
Two-wattmeter method star load · Basic Electrical and Electronics Engineering, chapter 1
Zener current for two load values · Electronics Devices and Circuit, chapter 6
Zener input range, 91 Ω, 8 V, 400 mW · Electronics Devices and Circuit, chapter 6
Zener network: VL, IL, Iz, IR · Basic Electronics Engineering, chapter 2
Zener: range of load for VL = 5 V · Electronics Devices and Circuit, chapter 6

All Basic Electrical and Electronics Engineering chapter 3 questions →

Zener network with two load values

Case 1: RL=1.2 kΩR_L=1.2\ \text{k}\OmegaRL​=1.2 kΩ

Case 2: RL=3 kΩR_L=3\ \text{k}\OmegaRL​=3 kΩ

Related questions

Case 1: $R_L=1.2\ \text{k}\Omega$

Case 2: $R_L=3\ \text{k}\Omega$