Question

In the circuit shown assume the diode to be ideal. When ${{V}_{i}}$ increases from $2V$ to $6V$, the change in current is (in $mA$)
(A). zero
(B). $20$
(C). $\dfrac{80}{3}$
(D). $40$

Answer 1

Hint: In the figure shown is a circuit containing a forward biased diode. In forward bias, the current is allowed to flow through the diode. As the potential across its terminals changes, the state of the diode changes from forward to reverse when the potential applied to the positive terminal is less than the potential applied to the negative terminal. Based on this and applying Kirchhoff’s loops, we can calculate change in current.
Formulas used:
${{V}_{i}}-150\times I=3$

Complete answer:

In the given figure, the diode is forward biased which means it will let the current flow through it.
Applying loop in the given figure we get,
$\begin{align}
  & {{V}_{i}}-150\times I=3 \\
 & \Rightarrow {{V}_{i}}-150I=3 \\
\end{align}$
When ${{V}_{i}}=2V$,
From the figure above, for any
${{V}_{i}}<3V$ The diode will be reverse biased and hence will not allow the flow of current through the circuit Therefore, the value of current will be 0.
When ${{V}_{i}}=6V$,
$\begin{align}
  & 6-150I=3 \\
 & \Rightarrow 3=150I \\
 & \therefore I=0.02A \\
\end{align}$
The value of current will be $0.02A$.
The change in current will be-
$\begin{align}
  & \Delta I={{I}_{2}}-{{I}_{1}} \\
 & \Rightarrow \Delta I=0.02-0 \\
 & \therefore \Delta I=0.02A=20mA \\
\end{align}$
The difference in the value of current will be $20mA$.
Therefore, the change in current when potential difference across a diode changes is $20mA$.

Hence, the correct option is (B).

Note:
The potential across an ideal diode is zero, this means that it has zero resistance. When the negative terminal is at a higher potential, the diode becomes reverse biased. Forward biased means that the positive terminal of the battery is connected to the p-terminal and the negative terminal is connected to the n-terminal of the diode and the opposite is true for reverse biased.