Question

For the series diode configuration of the given figure, determine \(V_D\) and \(I_D\).

• A. \(0.4 \text{ V and } 0 \text{ A}\)
• B. \(0.7 \text{ V and } 0.85 \text{ A}\)
• C. \(-0.7 \text{ V and } 0.23 \text{ A}\)
• D. \(0 \text{ V and } 0.31 \text{ A}\)

Hint:

For silicon diodes: \[ V_{\gamma}\approx0.7V \] If the applied forward voltage is less than \(0.7V\), the diode remains OFF and: \[ I_D=0 \] In OFF condition, the diode behaves like an open circuit and the entire supply voltage appears across the diode itself.

Answer 1

The Correct Option is A

Concept: A silicon diode conducts current only when it is forward biased. For a silicon diode: \[ V_D \approx 0.7V \] when the diode is ON (forward biased). If the diode is reverse biased or the applied voltage is insufficient to turn it ON, then: \[ I_D = 0 \] and the diode behaves like an open circuit. In a practical circuit:
• If the applied voltage across the diode is less than \(0.7V\), the silicon diode does not conduct.
• When the diode is OFF, current through the entire series branch becomes zero.
• Since no current flows through the resistor, the voltage drop across the resistor also becomes zero. These observations are extremely important while analyzing diode circuits.

Step 1: Identify the applied source voltage. From the given figure, the applied voltage is: \[ 0.4V \] The circuit consists of:
• A silicon diode
• A \(1.3k\Omega\) resistor
• A DC supply of \(0.4V\)

Step 2: Compare the applied voltage with the cut-in voltage of silicon diode. For a silicon diode to conduct properly, the forward voltage should approximately be: \[ V_{\gamma} \approx 0.7V \] However, the available source voltage is only: \[ 0.4V \] Clearly, \[ 0.4V < 0.7V \] Hence, the applied voltage is not sufficient to forward bias the silicon diode. Therefore, the diode remains in OFF state.

Step 3: Determine the diode current. Since the diode is OFF, it behaves like an open circuit. An open circuit does not allow current flow. Therefore, \[ I_D = 0A \]

Step 4: Determine the voltage across the resistor. The resistor voltage is: \[ V_R = I_D R \] Since: \[ I_D=0 \] therefore, \[ V_R = 0 \times 1.3k\Omega \] \[ V_R=0V \] Thus, no voltage is dropped across the resistor.

Step 5: Determine the diode voltage \(V_D\). Because the resistor drop is zero, the entire applied source voltage appears across the diode. Hence, \[ V_D = 0.4V \]

Step 6: Write the final answer. Thus, \[ V_D = 0.4V \] and \[ I_D = 0A \] Therefore, the correct option is: \[ \boxed{(A)\ 0.4\text{ V and }0\text{ A}} \]