Consider a circuit consisting of a capacitor (20 μF), resistor \((100 \Omega)\) and two identical diodes as shown in figure. The resistance of diode under forward biasing condition is 10 Ω. The time constant of the circuit is \(\alpha \times 10^{-3}\) s. The value of \(\alpha\) is ________.
Step 1: Understanding the Concept:
The circuit consists of a capacitor, a resistor, and a parallel combination of two anti-directional diodes connected in series with a voltage source. The time constant of an RC circuit determines how quickly the capacitor charges or discharges. Step 2: Key Formula or Approach:
The time constant \(\tau\) of an RC circuit is given by the product of the equivalent resistance and the capacitance:
\(\tau = R_{\text{eq}} \times C\).
Because the two diodes are in parallel and point in opposite directions, whichever direction the current flows, exactly one diode will be forward-biased (conducting) while the other will be reverse-biased (blocking). Step 3: Detailed Explanation:
The reverse-biased diode acts as an open circuit (infinite resistance), and the forward-biased diode acts as a resistor with \(R_d = 10 \, \Omega\).
Therefore, the equivalent resistance of the diode pair branch is simply \(10 \, \Omega\).
This diode branch is connected in series with the main resistor \(R = 100 \, \Omega\).
The total equivalent resistance of the circuit is:
\(R_{\text{eq}} = R + R_d = 100 + 10 = 110 \, \Omega\).
The capacitance is given as \(C = 20 \text{ \mu F} = 20 \times 10^{-6} \text{ F}\).
Now, calculate the time constant:
\(\tau = R_{\text{eq}} \times C = 110 \times 20 \times 10^{-6} \text{ s}\)
\(\tau = 2200 \times 10^{-6} \text{ s} = 2.2 \times 10^{-3} \text{ s}\). Step 4: Final Answer:
Comparing this result with the given format \(\alpha \times 10^{-3} \text{ s}\), we find that \(\alpha = 2.2\).
