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

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Answer 1

Given:
$V_{D} = 0.5 V$
$V_{B} = 1.5 V$
Excess current
$I_{E} = 10 m A$
Required current
$I = 5 m A = 5 \times 1 0^{- 3} A$
In order to use a $1.5 V$ battery to forward bias the diode, the resistor used in series with the diode so that the maximum current does not exceed $5 m A$ is

$R = \frac{V _{B} - V _{D}}{I}$

$R = \frac{1 . 5 - 0 . 5}{5 \times 1 0 ^{- 3}}$

$R = \frac{1}{5 \times 1 0 ^{- 3}}$

$R = 0.2 \times 1 0^{3} = 2 \times 1 0^{2} Ω$

Answer 2

Given:

V_{D} = 0.5 V

V_{B} = 1.5 V

Excess current

I_{E} = 10 m A

Required current

I = 5 m A = 5 \times 1 0^{- 3} A

In order to use a

1.5 V

battery to forward bias the diode, the resistor used in series with the diode so that the maximum current does not exceed

5 m A

is

R = \frac{V _{B} - V _{D}}{I}

R = \frac{1 . 5 - 0 . 5}{5 \times 1 0 ^{- 3}}

R = \frac{1}{5 \times 1 0 ^{- 3}}

R = 0.2 \times 1 0^{3} = 2 \times 1 0^{2} Ω