Answer
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Hint: When the positive end of the diode is connected to the positive terminal of the battery, the diode is said to be forward biased. When the positive end of a diode is connected to the negative terminal of the battery, the diode is said to be reverse biased. We will use the expression for the voltage drop across a diode. The expression relates the voltage drop across a diode with the current through the diode and the saturation current.
Formula used:
Voltage across diode,
$V=R\left( I+{{I}_{s}} \right)$
Complete step by step answer:
Diode is a two terminal device that has very low resistance at one end and very high resistance at the other end. The main function of a diode is to allow electric current to pass in one direction, while blocking in the opposite direction.
A germanium diode is a type of diode which is made using the element Germanium. When we have a diode in a circuit and we pass through it and measure the voltage before and after the diode, germanium diode would have dropped the voltage by $0.3V$ .
A diode can be connected in an electric circuit in two ways; positive biasing of diode and negative biasing of diode.
Types of biasing of diode in a circuit:
(I)Forward biasing – When the positive end of a diode is connected to the positive terminal of the battery, the diode is said to be forward biased.
(II)Reverse biasing – When the positive end of a diode is connected to the negative terminal of the battery, the diode is said to be reverse biased.
The potential barrier in a diode is the barrier in which the charge requires additional force for crossing that particular region. We can say that the barrier in which the charge carriers are stopped by the obstructive force is known as the potential barrier.
The saturation current is defined as the limiting current in the circuit such that further increase of voltage produces no further increase in current.
Given that:
The saturation current in Germanium diode is,
${{I}_{s}}=10\mu A$
That is,
$\begin{align}
& {{I}_{s}}=10\times {{10}^{-6}}A \\
& {{I}_{s}}={{10}^{-5}}A \\
\end{align}$
Potential barrier for Germanium diode is$0.3$
(I)Calculating forward voltage when diode current is $100mA$:
Voltage across diode is given as,
$V=R\left( I+{{I}_{s}} \right)$
Where,
$R$ is the potential barrier
$I$ is the current across diode
${{I}_{s}}$ is the saturated current
We have,
$\begin{align}
& R=0.3 \\
& I=100mA \\
& {{I}_{s}}={{10}^{-5}}A \\
\end{align}$
Therefore,
$\begin{align}
& V=0.3\left( 100\times {{10}^{-3}}+{{10}^{-5}} \right) \\
& V=0.3\times {{10}^{-1}} \\
& V=0.03V \\
\end{align}$
(II)Calculating diode current at backward biased voltage $0.01V$:
$V=R\left( I+{{I}_{s}} \right)$
We have,
$\begin{align}
& V=0.01V \\
& R=0.3 \\
& {{I}_{s}}={{10}^{-5}}A \\
\end{align}$
Therefore,
$\begin{align}
& 0.01=0.3\left[ I+{{10}^{-5}} \right] \\
& -{{10}^{-5}}+\dfrac{{{10}^{-2}}}{0.3}=I \\
& I=0.33 \\
\end{align}$
The diode current at backward biased voltage $0.01V$ is $0.33A$
Note:
The potential barrier in a circuit exists even if the device is not connected to any power source. The significance of a potential barrier in a circuit is that it opposes both the flow of holes and electrons across the junction and is therefore called a potential barrier.
Formula used:
Voltage across diode,
$V=R\left( I+{{I}_{s}} \right)$
Complete step by step answer:
Diode is a two terminal device that has very low resistance at one end and very high resistance at the other end. The main function of a diode is to allow electric current to pass in one direction, while blocking in the opposite direction.
A germanium diode is a type of diode which is made using the element Germanium. When we have a diode in a circuit and we pass through it and measure the voltage before and after the diode, germanium diode would have dropped the voltage by $0.3V$ .
A diode can be connected in an electric circuit in two ways; positive biasing of diode and negative biasing of diode.
Types of biasing of diode in a circuit:
(I)Forward biasing – When the positive end of a diode is connected to the positive terminal of the battery, the diode is said to be forward biased.
(II)Reverse biasing – When the positive end of a diode is connected to the negative terminal of the battery, the diode is said to be reverse biased.
The potential barrier in a diode is the barrier in which the charge requires additional force for crossing that particular region. We can say that the barrier in which the charge carriers are stopped by the obstructive force is known as the potential barrier.
The saturation current is defined as the limiting current in the circuit such that further increase of voltage produces no further increase in current.
Given that:
The saturation current in Germanium diode is,
${{I}_{s}}=10\mu A$
That is,
$\begin{align}
& {{I}_{s}}=10\times {{10}^{-6}}A \\
& {{I}_{s}}={{10}^{-5}}A \\
\end{align}$
Potential barrier for Germanium diode is$0.3$
(I)Calculating forward voltage when diode current is $100mA$:
Voltage across diode is given as,
$V=R\left( I+{{I}_{s}} \right)$
Where,
$R$ is the potential barrier
$I$ is the current across diode
${{I}_{s}}$ is the saturated current
We have,
$\begin{align}
& R=0.3 \\
& I=100mA \\
& {{I}_{s}}={{10}^{-5}}A \\
\end{align}$
Therefore,
$\begin{align}
& V=0.3\left( 100\times {{10}^{-3}}+{{10}^{-5}} \right) \\
& V=0.3\times {{10}^{-1}} \\
& V=0.03V \\
\end{align}$
(II)Calculating diode current at backward biased voltage $0.01V$:
$V=R\left( I+{{I}_{s}} \right)$
We have,
$\begin{align}
& V=0.01V \\
& R=0.3 \\
& {{I}_{s}}={{10}^{-5}}A \\
\end{align}$
Therefore,
$\begin{align}
& 0.01=0.3\left[ I+{{10}^{-5}} \right] \\
& -{{10}^{-5}}+\dfrac{{{10}^{-2}}}{0.3}=I \\
& I=0.33 \\
\end{align}$
The diode current at backward biased voltage $0.01V$ is $0.33A$
Note:
The potential barrier in a circuit exists even if the device is not connected to any power source. The significance of a potential barrier in a circuit is that it opposes both the flow of holes and electrons across the junction and is therefore called a potential barrier.
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