Question:
For a step-down Type A chopper, the waveforms of input current I and output current \(\text{I}_\text{o}\) respectively are
For a step-down Type A chopper, the waveforms of input current I and output current \(\text{I}_\text{o}\) respectively are
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Resistors have no memory or energy storage!
For a purely resistive load, the current waveform always looks exactly like the voltage waveform. Since a chopper outputs pulsed, discontinuous voltage blocks, the resulting current loops must be discontinuous and discontinuous.
Updated On: Jun 25, 2026
- \( \text{discontinuous and continuous for a resistive load} \)
- \( \text{discontinuous and discontinuous for a resistive load} \)
- \( \text{continuous and continuous for an inductive load} \)
- \( \text{continuous and discontinuous for an inductive load} \)
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The Correct Option is B
Solution and Explanation
Concept:
A Type-A step-down chopper uses a high-speed switch connected in series between the DC input source and the output load, along with a parallel freewheeling diode.
The behavior of the current waveforms depends heavily on the type of load connected to the output:
• Purely Resistive Load (\(R\)): A resistor has no energy storage capacity. According to Ohm's Law ($i = \frac{v}{R}$), the current waveform perfectly tracks the voltage waveform. When the switch turns off, the voltage drops to zero instantly, causing the current to drop to zero instantly as well.
• Inductive Load (\(R-L\)): An inductor stores magnetic energy and prevents sudden changes in current. During the switch off-time, the inductor discharges its stored energy through the freewheeling diode, keeping the output current flowing continuously.
Step 1: Analyzing behavior with a purely resistive load.
Let's look at the circuit states when driving a purely resistive load $R$:
• During \(T_{\text{ON}}\) phase: The switch is closed. The output voltage equals the source voltage ($V_0 = V_s$). The current flowing from the source into the resistor is: \[ I = I_0 = \frac{V_s}{R} \]
• During \(T_{\text{OFF}}\) phase: The switch opens. The output voltage drops to zero instantly ($V_0 = 0$). Since there is no inductor to maintain current flow, the output current drops to zero instantly ($I_0 = 0$). No current is drawn from the source ($I = 0$). Because both the input current $I$ and output current $I_0$ drop to zero and stay at zero during the entire switch off-time, both waveforms are discontinuous.
Step 2: Verification of the options.
Let's check our finding against the choices:
• Option (2) states that the waveforms are "discontinuous and discontinuous for a resistive load", which perfectly describes this behavior.
• For an inductive load, the output current flows continuously through the freewheeling diode, but the input current from the source still drops to zero when the switch opens. This means the waveforms would be discontinuous for the input and continuous for the output, which does not match options (3) or (4). Hence, the correct choice is option (2).
• Purely Resistive Load (\(R\)): A resistor has no energy storage capacity. According to Ohm's Law ($i = \frac{v}{R}$), the current waveform perfectly tracks the voltage waveform. When the switch turns off, the voltage drops to zero instantly, causing the current to drop to zero instantly as well.
• Inductive Load (\(R-L\)): An inductor stores magnetic energy and prevents sudden changes in current. During the switch off-time, the inductor discharges its stored energy through the freewheeling diode, keeping the output current flowing continuously.
Step 1: Analyzing behavior with a purely resistive load.
Let's look at the circuit states when driving a purely resistive load $R$:
• During \(T_{\text{ON}}\) phase: The switch is closed. The output voltage equals the source voltage ($V_0 = V_s$). The current flowing from the source into the resistor is: \[ I = I_0 = \frac{V_s}{R} \]
• During \(T_{\text{OFF}}\) phase: The switch opens. The output voltage drops to zero instantly ($V_0 = 0$). Since there is no inductor to maintain current flow, the output current drops to zero instantly ($I_0 = 0$). No current is drawn from the source ($I = 0$). Because both the input current $I$ and output current $I_0$ drop to zero and stay at zero during the entire switch off-time, both waveforms are discontinuous.
Step 2: Verification of the options.
Let's check our finding against the choices:
• Option (2) states that the waveforms are "discontinuous and discontinuous for a resistive load", which perfectly describes this behavior.
• For an inductive load, the output current flows continuously through the freewheeling diode, but the input current from the source still drops to zero when the switch opens. This means the waveforms would be discontinuous for the input and continuous for the output, which does not match options (3) or (4). Hence, the correct choice is option (2).
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