A diode has Zener voltage of 10 V and maximum power dissipation of 0.5 W, then the minimum resistance to be

Question

A diode has Zener voltage of 10 V and maximum power dissipation of 0.5 W, then the minimum resistance to be used in series with this diode for safety when it is connected to a 25 V power supply is _____ $Ω$ . [2026]

Smart Achievers · Accepted Answer

(300)

$P_{Max} = V_{Z} I_{Z}$

$\Rightarrow I_{Z} = \frac{P_{Max}}{V_{Z}} = \frac{0.5}{10} = 0.05 A$

$V_{R} = V_{s} - V_{z} = 25 - 10 = 15 V$

$R_{\min} = \frac{V_{R}}{I_{Z}} = \frac{15.0}{0.05} = 300 Ω$

Solution

Q.
A diode has Zener voltage of 10 V and maximum power dissipation of 0.5 W, then the minimum resistance to be used in series with this diode for safety when it is connected to a 25 V power supply is _____ $Ω$ . [2026]

Ans.
(300)

$P_{Max} = V_{Z} I_{Z}$

$\Rightarrow I_{Z} = \frac{P_{Max}}{V_{Z}} = \frac{0.5}{10} = 0.05 A$

$V_{R} = V_{s} - V_{z} = 25 - 10 = 15 V$

$R_{\min} = \frac{V_{R}}{I_{Z}} = \frac{15.0}{0.05} = 300 Ω$

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Solution

Q. A diode has Zener voltage of 10 V and maximum power dissipation of 0.5 W, then the minimum resistance to be used in series with this diode for safety when it is connected to a 25 V power supply is _____ Ω. [2026]

Ans. (300) PMax=VZIZ ⇒IZ=PMaxVZ=0.510=0.05 A VR=Vs-Vz=25-10=15 V Rmin=VRIZ=15.00.05=300 Ω

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Q.
A diode has Zener voltage of 10 V and maximum power dissipation of 0.5 W, then the minimum resistance to be used in series with this diode for safety when it is connected to a 25 V power supply is _____ $Ω$ . [2026]

Ans.
(300)

$P_{Max} = V_{Z} I_{Z}$

$\Rightarrow I_{Z} = \frac{P_{Max}}{V_{Z}} = \frac{0.5}{10} = 0.05 A$

$V_{R} = V_{s} - V_{z} = 25 - 10 = 15 V$

$R_{\min} = \frac{V_{R}}{I_{Z}} = \frac{15.0}{0.05} = 300 Ω$