Zener Diode Protective Resistance Calculation
The problem asks for the approximate value of the protective resistance ($R_s$) in a Zener diode voltage regulating circuit.
Given Parameters:
- Zener breakdown voltage ($V_z$) = 10 V
- Maximum power dissipation ($P_{z,max}$) = 0.4 W
- Input voltage ($V_{in}$) = 15 V
Calculation Steps:
- Calculate Maximum Zener Current ($I_{z,max}$): The maximum power dissipation is related to the Zener voltage and the maximum Zener current by the formula: $P_{z,max} = V_z \times I_{z,max}$ Substituting the given values: $0.4 \text{ W} = 10 \text{ V} \times I_{z,max}$ Solving for $I_{z,max}$: $I_{z,max} = \frac{0.4 \text{ W}}{10 \text{ V}} = 0.04 \text{ A}$ So, the maximum allowable Zener current is 0.04 A (or 40 mA).
- Calculate Voltage Across Series Resistor ($V_{Rs}$): The series resistor ($R_s$) must drop the excess voltage, which is the difference between the input voltage and the Zener voltage: $V_{Rs} = V_{in} - V_z$ $V_{Rs} = 15 \text{ V} - 10 \text{ V} = 5 \text{ V}$
- Calculate Protective Resistance ($R_s$): The protective resistance ($R_s$) is determined by the voltage across it ($V_{Rs}$) and the maximum current it must handle without damaging the Zener diode. Assuming the load current is minimal for calculating the minimum required resistance, the current through $R_s$ is approximately $I_{z,max}$. $R_s = \frac{V_{Rs}}{I_{z,max}}$ $R_s = \frac{5 \text{ V}}{0.04 \text{ A}} = 125 \text{ } \Omega$
Therefore, the approximate value of the protective resistance ($R_s$) required is 125 $\Omega$. This value ensures the Zener current does not exceed the maximum allowable limit.
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Important Questions from Modern Physics
The binding energy for the following nuclear reactions are expressed in MeV.
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($h = 6.63 \times 10^{-34} \text{ J.s}, e = 1.6 \times 10^{-19} \text{ C}, c = 3 \times 10^8 \text{ m/s}$) - A particle having electric charge $3 \times 10^{-19}\text{ C}$ and mass $6 \times 10^{-27}\text{ kg}$ is accelerated by applying an electric potential of $1.21\text{ V}$. Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12}\text{ m}$. The value of $\alpha$ is __________. (Take Planck's constant $= 6.6 \times 10^{-34}\text{ J.s}$)
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