A PN junction diode when forward biased has a drop of 0.5V which is assumed to be independent of current. The current in excess of 10mA through the diode produces large joule heating which damages the diode. If we want to use a 1.5V battery to forward bias the diode, the resistor used in series with the diode so that the maximum current does not exceed 5mA is

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2 × 10⁵Ω

2 × 10³Ω

2 × 10⁴Ω

2 × 10²Ω

answer is A.

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Detailed Solution

Given,

$V_{D} = 0.5 V V_{B} = 1.5 V Excess current = I_{E} = 10 mA$

$Required current = I = 5 mA = 5 \times 10^{- 3} A$

The resistor used in series with the diode so that the maximum current does not exceed 5 mA is necessary in order to forward bias the diode using a 1.5 V battery.

$R = \frac{V_{B} - V_{D}}{I} R = \frac{1.5 - 0.5}{5 \times 10^{- 3}} R = \frac{1}{5 \times 10^{- 3}} R = 0.2 \times 10^{3} R = 2 \times 10^{2} Ω$

Hence the correct answer is $2 \times 10^{2} Ω .$

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