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Visualized Solution
The Sigma Insight: P-N Junction Diode
Analyzing the Setup
Imagine you are looking at a simple yet incredibly powerful electronic circuit. We have an alternating current (AC) source connected in series with a - junction diode and a load resistor . The AC source provides an input voltage that continuously changes its polarity over time, oscillating like a sine wave. Our objective is to determine the shape of the current that flows through the resistor as a function of time.
The Positive Half Cycle
Let's break the AC signal down into its two fundamental parts. First, we consider the positive half cycle. During this phase, the polarity of the AC source is such that the -side of the diode is at a higher electrical potential compared to the -side.
Because the -side is positive relative to the -side, the diode becomes forward biased. In an ideal scenario, a forward-biased diode acts exactly like a closed switch. It offers zero resistance, allowing the current to flow freely through the circuit and across the resistor . According to Ohm's Law, the current will be directly proportional to the input voltage . Therefore, the graph of the current will perfectly trace the positive hump of the input sine wave.
The Negative Half Cycle
Now, the AC source flips its polarity, entering the negative half cycle. Suddenly, the -side of the diode finds itself at a lower potential than the -side.
This reversal of polarity forces the diode into a reverse biased state. An ideal reverse-biased diode behaves like an open switch, offering infinite resistance to the flow of charge. As a result, the current is completely blocked. The current drops to exactly zero and remains there for the entire duration of the negative half cycle.
The Final Output
This alternating behavior repeats endlessly for every cycle of the AC input. The diode acts as a strict gatekeeper: it allows the positive half of the signal to pass through undisturbed but completely clips off the negative half.
Because it only rectifies half of the input wave, this circuit is famously known as a Half-Wave Rectifier. If we look at the options provided, the graph that perfectly illustrates this phenomenon—a positive sine hump followed by a flat zero line, repeating periodically—is option (c).
Similar Questions
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If in a p-n junction diode, a square input signal of 10 V is applied as shown.
(A)
(B)
(C)
(D)
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A full wave rectifier circuit along with the output is shown in figure. The contribution (s) from the diode 1 is (are)
* Multiple Correct Options
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In the following, which one of the diodes is reverse biased?
(A)
(B)
(C)
(D)
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In the forward bias arrangement of a junction rectifier, the end is connected to the ....... terminal of the battery and the direction of the current is from ....... to ...... in the rectifier.
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In a - junction diode not connected to any circuit
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the potential is the same everywhere.
(B)
the -type side is at a higher potential than the -type side.
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there is an electric field at the junction directed from the -side to the -type side.
(D)
there is an electric field at the junction directed from the -type side to the -type side.
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A battery is connected across the points X and Y. Assume and to be normal silicon diodes. Find the current supplied by the battery, if the positive terminal of the battery is connected to point X.
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(C)
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The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?
(A)
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(B)
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2.31 A
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Take the breakdown voltage of the zener diode used in the given circuit as . For the input voltage shown in figure below, the time variation of the output voltage is (Graphs are drawn schematically and on not to scale)
(A)
Graph (a)
(B)
Graph (b)
(C)
Graph (c)
(D)
Graph (d)
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The reading of the ammeter for a silicon diode in the given circuit is
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(B)
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