A zener diode of power rating 2 W is to be used as a voltage regulator. If the zener diode has a breakdown of 10 V and it has to regulate voltage fluctuated between 6 V and 14 V, the value of R s for safe operation should be Ω.

Question

A zener diode of power rating 2 W is to be used as a voltage regulator. If the zener diode has a breakdown of 10 V and it has to regulate voltage fluctuated between 6 V and 14 V, the value of R s for safe operation should be Ω. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

When unregulated voltage is 14 V voltage across zener diode must be 10 V So potential difference across resistor Δ V Rs =4 V and Pz e n e r=2 W VI =2 I =(2/10)=0.2 A Δ VRs = IRs 4 × 0.2 Rs ⇒ Rs = (40/2) = 20 Ω

Q. A zener diode of power rating $2 W$ is to be used as a voltage regulator. If the zener diode has a breakdown of $10 V$ and it has to regulate voltage fluctuated between $6 V$ and $14 V$ , the value of $R_{s}$ for safe operation should be _____ $Ω$ .

When unregulated voltage is $14 V$ voltage across zener diode must be $10 V$ So potential difference across resistor $Δ V_{R s} = 4 V$ and $P_{ze n er} = 2 W$
$V I = 2$
$I = \frac{2}{10} = 0.2 A$
$Δ V_{R s} = I R_{s}$
$4 \times 0.2 R_{s} \Rightarrow R_{s} = \frac{40}{2} = 20 Ω$

Q. A zener diode of power rating 2W is to be used as a voltage regulator. If the zener diode has a breakdown of 10V and it has to regulate voltage fluctuated between 6V and 14V, the value of Rs​ for safe operation should be _____Ω.

When unregulated voltage is 14V voltage across zener diode must be 10V So potential difference across resistor ΔVRs​=4V and Pzener​=2W VI=2 I=102​=0.2A ΔVRs​=IRs​ 4×0.2Rs​⇒Rs​=240​=20Ω

Q. A zener diode of power rating $2 W$ is to be used as a voltage regulator. If the zener diode has a breakdown of $10 V$ and it has to regulate voltage fluctuated between $6 V$ and $14 V$ , the value of $R_{s}$ for safe operation should be _____ $Ω$ .

When unregulated voltage is $14 V$ voltage across zener diode must be $10 V$ So potential difference across resistor $Δ V_{R s} = 4 V$ and $P_{ze n er} = 2 W$
$V I = 2$
$I = \frac{2}{10} = 0.2 A$
$Δ V_{R s} = I R_{s}$
$4 \times 0.2 R_{s} \Rightarrow R_{s} = \frac{40}{2} = 20 Ω$