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):
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- Ideal diode conducts only when \( V_i>1V \).
- When \( V_i<1V \), no current flows as the diode is reverse biased.
- Apply Ohm’s Law for current calculation when the diode is forward biased.
Step 1: Understanding the Circuit {- The circuit consists of a series resistor (\(250 \Omega\)) and an ideal diode} - The output side of the diode is fixed at +1V. - The diode allows current to flow only when the input voltage \( V_i \) is greater than 1V (i.e., it is forward biased). Step 2: Analyzing Current Flow The current through the resistor is given by Ohm’s Law: \[ I = \frac{V_{{in}} - V_{{diode}}}{R} \] where: - \( R = 250\Omega \), - \( V_{{diode}} = 1V \) (diode voltage when conducting). For different values of \( V_i \): 1. When \( V_i = -2V \): \[ I = \frac{-2V - 1V}{250\Omega} = \frac{-3}{250} = -12 { mA} \] Since the current is negative, the diode is reverse biased, meaning no current flows. Thus, for \( V_i \leq 1V \), current is zero. 2. When \( V_i = 6V \): \[ I = \frac{6V - 1V}{250 \Omega} = \frac{5}{250} = 20 { mA} \] Step 3: Change in Current - Initial current at \( V_i = -2V \) is 0 mA. - Final current at \( V_i = 6V \) is 20 mA. - The change in current is: \[ \Delta I = 20 - 0 = 20 { mA} \] Thus, the correct answer is: \[ \Delta I = 20 { mA} \]
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