Question 209: Semiconductors and Logic Gates

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 $\_\_\_\_$ $\Omega$.

Accepted Answer

For safety, the current through the Zener diode must not exceed the value corresponding to its maximum power.
Given:
Zener voltage: $V_Z = 10\,	ext{V}$
Maximum power: $P_{\max} = 0.5\,	ext{W}$
Supply voltage: $V = 25\,	ext{V}$
First, find the maximum safe current through the Zener diode:
$$ P_{\max} = V_Z I_{\max} $$
$$ I_{\max} = \frac{P_{\max}}{V_Z} = \frac{0.5}{10} = 0.05\,	ext{A} $$
Now, the series resistor must drop the remaining voltage:
$$ V_R = 25 - 10 = 15\,	ext{V} $$
Using OhmÃ¢ÂÂs law:
$$ R_{\min} = \frac{V_R}{I_{\max}} = \frac{15}{0.05} = 300\,\Omega $$
Therefore, the minimum resistance required is
$$ \boxed{300\,\Omega} $$

Semiconductors and Logic Gates

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 \_\_\_\_ \Omega.

Integer / Numeric Answer Type