In the circuit shown, diode has \(20\Omega\) forward resistance. When \(V_i\) increases from \(8V\) to \(12V\), change in current is \(x\,mA\). Find \(x\).
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Always evaluate current separately in OFF and ON regions of diode.
Step 1: Understanding the Concept:
A diode only conducts in forward bias. In this circuit, the diode is connected to a \(+10\text{ V}\) supply at the cathode. For the diode to be forward biased, the input voltage \(V_{i}\) at the anode must exceed the cathode potential. Step 2: Key Formula or Approach:
Net Voltage \(\Delta V = V_{i} - V_{\text{bias}}\) (when \(V_{i}>V_{\text{bias}}\)).
Current \(I = \frac{\Delta V}{R_{\text{total}}}\). : Detailed Explanation:
1. Case 1: \(V_{i} = 8\text{ V}\).
Since \(8\text{ V}<10\text{ V}\), the diode is reverse biased. No current flows.
\[ I_{1} = 0\text{ mA} \]
2. Case 2: \(V_{i} = 12\text{ V}\).
Since \(12\text{ V}>10\text{ V}\), the diode is forward biased.
Effective voltage \(= 12 - 10 = 2\text{ V}\).
Total resistance \(= R_{\text{series}} + R_{\text{diode}} = 200 + 20 = 220\text{ }\Omega\).
\[ I_{2} = \frac{2}{220} = \frac{1}{110}\text{ A} \approx 0.00909\text{ A} = 9.09\text{ mA} \]
3. Change in current:
\[ \Delta I = I_{2} - I_{1} = 9.09 - 0 \approx 9\text{ mA} \] Step 3: Final Answer:
The value of \(x\) is \(9\).
