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A $p$ - $n$ junction diode can withstand upto $20 \mathrm{~mA}$ current under forward bias. The diode has a potential difference of $0.5 \mathrm{~V}$ across it, which is assumed to be independent of current. What is the maximum voltage of the battery used to forward bias the diode when a resistance of $125 \Omega$ is connected in series with it?
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Verified Answer
The correct answer is:
$3.0 \mathrm{~V}$
Given, potential difference across diode,
$V_D=0.5 \mathrm{~V}$
Maximum current in forward bias,
$i=20 \mathrm{~mA}=20 \times 10^{-3} \mathrm{~A}$
$\Rightarrow$ Resistance of diode,
$R_D=\frac{V_D}{i}=\frac{0.5}{20 \times 10^{-3}}=25 \Omega$
Now, another resistance of $125 \Omega$ is connected with this diode in series.
So, effective resistance,
$R=25+125=150 \Omega$
Clearly, maximum voltage to be used during forward bias,
$V=i R=20 \mathrm{~mA} \times 150 \Omega$
$=3000 \mathrm{mV}=3.0 \mathrm{~V}$
$V_D=0.5 \mathrm{~V}$
Maximum current in forward bias,
$i=20 \mathrm{~mA}=20 \times 10^{-3} \mathrm{~A}$
$\Rightarrow$ Resistance of diode,
$R_D=\frac{V_D}{i}=\frac{0.5}{20 \times 10^{-3}}=25 \Omega$
Now, another resistance of $125 \Omega$ is connected with this diode in series.
So, effective resistance,
$R=25+125=150 \Omega$
Clearly, maximum voltage to be used during forward bias,
$V=i R=20 \mathrm{~mA} \times 150 \Omega$
$=3000 \mathrm{mV}=3.0 \mathrm{~V}$
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