Q.
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit ?
AIEEEAIEEE 2006Semiconductor Electronics: Materials Devices and Simple Circuits
Solution:
p-n Junction Diode offers negligible resistance in forward biasing and offers very large resistance in reverse biasing.
Step1: Identify the Diode in reverse biasing and remove that branch.
The diode $D _1$ is in reverse biasing thus that branch can be neglected. The diode $D _2$ is in forward biasing and it offers negligible resistance.
Step 2: Calculate net resistance of circuit and the find current. Equivalent resistance of the circuit is,
$
\begin{array}{l}
R _{\text {eq. }}=(4+2) \Omega \\
\Rightarrow R _{\text {eq. }}=6 \Omega
\end{array}
$
Thus current flowing in the circuit is, by Ohm's Law
$
\begin{array}{l}
I =\frac{ V }{ R _{\text {eq }}} \\
\Rightarrow I =\frac{12}{6} \\
\Rightarrow I =2 A
\end{array}
$
Thus, net current flowing through the circuit is $2 A$.
