Q.When a forward bias is applied to a p-n junction, it
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P-N Junction Biasing: The First Meeting
Imagine you have two rooms separated by a door. One room is full of people who want to leave (electrons in the N-side), and the other room is full of empty chairs (holes in the P-side). The door is initially locked — that's the depletion region, a zone with no free charges. Now, what happens if you push on the door from one side, or pull it from the other? That's exactly what biasing does to a P-N junction.
The Unbiased Junction (The Starting Point)
When you bring P-type and N-type semiconductors together, something interesting happens at the boundary. Electrons from the N-side (which has extra electrons) rush toward the P-side (which has extra holes). Holes from the P-side rush toward the N-side. They meet and recombine — an electron falls into a hole, and both disappear as free charges.
This leaves behind a region near the junction with no free charges — just fixed positive ions on the N-side (where electrons left) and fixed negative ions on the P-side (where holes left). This is the depletion region. It acts like a tiny battery: the N-side is positive relative to the P-side, creating a built-in electric field that stops further diffusion. The voltage across this region is about 0.7 V for silicon, 0.3 V for germanium.
The depletion region is not a physical gap — it's a region depleted of mobile charge carriers. The crystal is still continuous.
Forward Bias: Pushing the Door Open
Connect the positive terminal of a battery to the P-side and the negative terminal to the N-side. This is forward bias.
What happens? The external battery's positive terminal repels holes in the P-side toward the junction. The negative terminal repels electrons in the N-side toward the junction. Both carriers are pushed toward each other — they overcome the built-in electric field. The depletion region shrinks.
When the applied voltage exceeds about 0.7 V (for silicon), the depletion region becomes so thin that carriers can cross freely. A large current flows. The junction is now "on" — like a switch closed.
Never apply forward bias without a current-limiting resistor. The junction has very low resistance once forward-biased, and the current can destroy it instantly.
Reverse Bias: Pulling the Door Shut
Now reverse the battery: positive to N-side, negative to P-side. This is reverse bias.
The positive terminal attracts electrons from the N-side away from the junction. The negative terminal attracts holes from the P-side away from the junction. Both carriers are pulled apart. The depletion region widens.
The built-in electric field is now reinforced by the external field. Only a tiny current flows — the reverse saturation current, caused by thermally generated electron-hole pairs in the depletion region. This current is typically in the nanoampere range and is almost independent of the applied voltage (until breakdown).
The junction is "off" — like a switch open.
In reverse bias, the depletion region acts as an insulator. The junction blocks current flow except for a negligible leakage current.
The Precise Statement
I=IS(enkTqV−1)
This is the Shockley diode equation. Here:
- I = diode current
- IS = reverse saturation current (very small, typically 10−12 to 10−15 A for silicon)
- q = electron charge (1.6×10−19 C)
- V = applied voltage (positive for forward bias, negative for reverse bias)
- n = ideality factor (1 for ideal, 1–2 for real diodes)
- k = Boltzmann constant (1.38×10−23 J/K) …
Why this formula?
Why a PN Junction Biases the Way It Does
A PN junction is not a resistor. Its current-voltage behaviour is fundamentally asymmetric — and that asymmetry comes directly from the physics of the depletion region and the energy barrier it creates.
The Unbiased Junction: A Built-in Barrier
When P-type and N-type semiconductors are joined, holes from the P-side diffuse into the N-side, and electrons from the N-side diffuse into the P-side. This diffusion leaves behind fixed, charged ions: negative acceptor ions on the P-side, positive donor ions on the N-side. These ions create an electric field that opposes further diffusion.
The result is a depletion region — a zone with no free carriers — and a built-in potential V0 across it. This potential acts as an energy barrier that prevents net current flow at equilibrium.
At equilibrium (zero bias), the net current is zero. The diffusion current (due to concentration gradient) is exactly balanced by the drift current (due to the built-in electric field).
Forward Bias: Lowering the Barrier
Apply a positive voltage V to the P-side relative to the N-side. This external voltage opposes the built-in potential. The net barrier becomes:
Vnet=V0−V
The depletion width shrinks. More importantly, the energy barrier for majority carriers (holes from P-side, electrons from N-side) is reduced. A larger number of carriers now have enough energy to cross the junction.
The current that flows is diffusion current — carriers injected across the junction become minority carriers on the other side, where they recombine. The relationship is exponential because the number of carriers with energy above the barrier follows a Boltzmann distribution.
I=I0(eqV/kT−1)
Here I0 is the reverse saturation current (very small), q is the electron charge, k is Boltzmann's constant, and T is absolute temperature.
Why the exponential? The fraction of carriers with enough energy to surmount a barrier of height q(V0−V) is proportional to e−q(V0−V)/kT. At equilibrium (V=0), this gives a current that exactly cancels the drift current. When V>0, the barrier drops, and the net current becomes proportional to eqV/kT.
Reverse Bias: Raising the Barrier
Apply a negative voltage to the P-side relative to the N-side. Now the external voltage adds to the built-in potential:
Vnet=V0+VR
The barrier becomes higher. The depletion region widens. Diffusion of majority carriers is almost completely suppressed. The only current that flows is a tiny reverse saturation current I0, carried by minority carriers (electrons from the P-side, holes from the N-side) that are swept across by the electric field.
This current is essentially constant with voltage because the number of minority carriers is fixed by thermal generation — it does not depend on the barrier height once the barrier is large enough to block majority carriers. …
The key idea is P-N junction biasing: forward bias reduces the built-in potential barrier, allowing majority carriers to flow across the junction.
- In a p-n junction, the built-in potential barrier opposes the diffusion of majority carriers (holes from p-side, electrons from n-side). …
Forward bias reduces the potential barrier at a p-n junction, allowing majority carriers to flow easily across the junction. The correct option is (c).
Understanding P-N Junction Biasing
A p-n junction is formed when p-type and n-type semiconductors are joined. At the junction, electrons from the n-side diffuse into the p-side, and holes from the p-side diffuse into the n-side. This diffusion leaves behind immobile charged ions, creating a depletion region with an internal electric field. This field opposes further diffusion and gives rise to a potential barrier (typically about 0.7 V for silicon).
Now, biasing means applying an external voltage across the junction. The effect depends on the polarity:
- Forward bias: p-side connected to positive terminal, n-side to negative terminal.
- Reverse bias: p-side connected to negative terminal, n-side to positive terminal.
The key question is: what happens to the potential barrier in each case?
Step-by-Step Reasoning
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What the potential barrier represents
The potential barrier is the voltage difference across the depletion region that prevents majority carriers from crossing freely. For a p-n junction, the built-in potential V0 is determined by the doping concentrations and temperature. In equilibrium (no external bias), this barrier is fixed.
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Effect of forward bias on the barrier
When forward bias is applied, the external voltage Vf opposes the internal electric field. The positive terminal repels holes in the p-side toward the junction, and the negative terminal repels electrons in the n-side toward the junction. This reduces the width of the depletion region and lowers the effective potential barrier to V0−Vf.
Vbarrier (forward)=V0−Vf
The barrier decreases as forward voltage increases.
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Consequence of lowering the barrier …
Method: Energy-Band / Barrier-Height Analysis
This is the most direct way to think about biasing effects on a p-n junction. The key idea is that the potential barrier at the junction is what prevents majority carriers from crossing freely.
Step 1 – Recall the unbiased state.
In an unbiased p-n junction, diffusion of majority carriers (holes from p-side, electrons from n-side) creates a depletion region. The built-in potential V0 (typically 0.6–0.7 V for silicon) acts as a barrier that stops further net diffusion.
Step 2 – Apply forward bias.
Forward bias means connecting the p-side to the positive terminal of a battery and the n-side to the negative terminal. This external voltage VF opposes the built-in field.
Step 3 – Determine the net barrier.
The effective barrier height becomes V0−VF. Since VF is positive, the barrier decreases. For example, if V0=0.7 V and VF=0.5 V, the net barrier is only 0.2 V.
Step 4 – Consequence.
A lower barrier allows more majority carriers to diffuse across the junction, producing a large forward current. The barrier is not raised — it is lowered. …
The most common mistake here is picking (a) — "raises the potential barrier." That error comes from mixing up forward and reverse bias. In forward bias, the external voltage opposes the built-in field, so the barrier drops, not rises. Students often memorise "bias increases barrier" without checking direction.
Another frequent error is choosing (b) — "reduces the majority carrier current to zero." That would describe a reverse bias condition where current is nearly zero. In forward bias, majority carriers are pushed across the junction, so current actually increases sharply.
The correct answer is (c) — forward bias lowers the potential barrier.
Do not confuse "forward" with "reverse." Forward bias = barrier lowered, current flows. Reverse bias = barrier raised, current blocked (except leakage).
To avoid these mistakes: …
- AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQQ.A Zener diode of breakdown voltage 20 V is connected as shown in the given circuit. The current through Zener diode is [FIGURE] (a circuit with a 40 V battery in series with a 2kΩ resistor connecting to a node; from that node a Zener diode of breakdown voltage 20 V is connected to the bottom rail, in parallel with a 5kΩ resistor also connected from the node to the bottom rail) (A) 10 mA (B) 4 mA (C) 6 mA (D) 8 mA
›Reveal solutionSolution
A Zener regulator problem: the diode clamps the node to its breakdown voltage (20 V); apply KCL at that node to find the diode current — 6 mA.
Concept and Intuition
A Zener diode used as a voltage regulator, when reverse-biased into breakdown, holds a nearly constant voltage across itself (its rated breakdown voltage) regardless of the current through it, as long as some minimum current flows. Here the node between the 2 kΩ and 5 kΩ resistors is held at 20 V by the Zener (since the 40 V source is well above the 20 V breakdown, the diode does break down). The 5 kΩ resistor is connected in parallel with the Zener between this node and the bottom rail (ground), while the 2 kΩ resistor feeds current into the node from the 40 V source.
Step-by-Step Solution
- Since the source (40 V) exceeds the Zener breakdown voltage (20 V), the diode conducts in breakdown and clamps the node voltage to V=20 V.
- Current supplied through the 2 kΩ resistor (from source to node):
Itotal=2kΩ40−20=200020=10 mA
- Current through the 5 kΩ resistor (node to ground, in parallel with the Zener): …
- AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQQ.A transistor works as an amplifier when (A) emitter-base junction is forward biased and base-collector junction is reverse biased (B) both emitter-base and base-collector junctions are forward biased (C) both emitter-base and base-collector junctions are reverse biased (D) emitter-base junction is reverse biased and base-collector junction is forward biased
›Reveal solutionSolution
Amplifier action in a transistor requires forward bias at the emitter-base junction and reverse bias at the base-collector junction.
Concept and Intuition
A transistor works as an amplifier only in the "active region." In this mode, the emitter-base (EB) junction is forward biased so that majority carriers are injected in large numbers from the emitter into the thin, lightly-doped base. The base-collector (BC) junction is reverse biased so that these carriers, having diffused across the base, are swept efficiently into the collector by the junction's field, producing a large collector current controlled by a small base current — the basis of current/voltage gain.
Step-by-Step Solution
- Recall the three transistor operating regions: cutoff (both junctions reverse biased), saturation (both forward biased), and active (EB forward, BC reverse).
- Amplification (linear, proportional gain) only happens in the active region. …
- AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQQ.If two diodes D1 and D2 are connected as shown in the figure, then [FIGURE] (two parallel branches: the upper branch has −5V connected through a 1 kΩ resistor to diode D1 (pointing rightward, anode on the left) leading to −2V; the lower branch has +5V connected through a 1 kΩ resistor to diode D2 (pointing rightward, anode on the left) leading to +2V) (A) both the diodes D1 and D2 are forward biased. (B) both the diodes D1 and D2 are reverse biased. (C) diode D1 is forward biased and diode D2 is reverse biased. (D) diode D1 is reverse biased and diode D2 is forward biased.
›Reveal solutionSolution
The key idea is to check the voltage across each diode: a diode is forward biased when its anode is at a higher potential than its cathode. For D₁, the anode is at -5 V and the cathode at -2 V (reverse bias); for D₂, the anode is at +5 V and the cathode at +2 V (forward bias). Thus the correct option is (D).
Concept and Intuition: P‑N Junction Biasing
A diode conducts only when its anode is more positive than its cathode — this is forward bias. If the anode is more negative than the cathode, the diode is reverse biased and blocks current (ideally). The trick in this problem is that the resistors don’t change the polarity of the voltages applied to the diode terminals; they only limit current once the diode turns on. So we simply compare the voltages at the two ends of each diode.
Step‑by‑Step Reasoning
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Identify the terminals of each diode
Both diodes are drawn with the triangle pointing right: the flat side (cathode) is on the right, the pointed side (anode) is on the left.
- For D₁: anode is connected to the –5 V source (through a 1 kΩ resistor), cathode is connected to the –2 V node.
- For D₂: anode is connected to the +5 V source (through a 1 kΩ resistor), cathode is connected to the +2 V node.
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Determine the bias condition for D₁
- Anode voltage: –5 V
- Cathode voltage: –2 V
- Voltage across D₁ (anode minus cathode): VD1=(−5)−(−2)=−3V Since the anode is more negative than the cathode, D₁ is reverse biased.
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Determine the bias condition for D₂
- Anode voltage: +5 V
- Cathode voltage: +2 V
- Voltage across D₂: VD2=(+5)−(+2)=+3V …
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- AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQQ.Two diodes with zero resistance in forward bias and infinite resistance in reverse bias are connected to a battery as shown in the circuit. Then the value of current 'i' is [FIGURE] (a circuit with a 10 V battery; the current i flows from the battery into two parallel branches — the upper branch has diode D1 (pointing so as to conduct in the direction of current flow) in series with a 20 Ω resistor, and the lower branch has diode D2 (pointing in the opposite direction) in series with a 40 Ω resistor; both branches reconnect back to the battery) (A) zero (B) 0.5 A (C) 0.4 A (D) 0.75 A
›Reveal solutionSolution
The key idea is that only the forward‑biased diode conducts; the reverse‑biased diode acts as an open circuit. Here, D₁ is forward‑biased and D₂ is reverse‑biased, so current flows only through the 20 Ω resistor, giving i=10V/20Ω=0.5A. The correct option is (B).
1. Understanding the diode model
The problem states that each diode has zero resistance when forward‑biased (ideal short circuit) and infinite resistance when reverse‑biased (ideal open circuit). This is the simplest possible diode model — no threshold voltage, no internal resistance. So a diode either acts like a wire (if current can flow in the direction of its arrow) or like a broken wire (if current tries to flow opposite to its arrow).
2. Determining the bias of each diode
The battery’s positive terminal is at the top (since current i flows upward out of the battery into the junction). Let’s trace the possible paths:
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Upper branch: Diode D₁ points rightward (arrow from left to right). Current coming from the battery enters the left end of the upper branch. For D₁ to conduct, current must flow left‑to‑right through it — that is exactly the direction the battery pushes. So D₁ is forward‑biased → acts as a short circuit (0 Ω).
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Lower branch: Diode D₂ points leftward (arrow from right to left). Current from the battery enters the left end of the lower branch. To go through D₂, current would have to flow left‑to‑right, but the diode’s arrow points the opposite way. So D₂ is reverse‑biased → acts as an open circuit (infinite resistance).
Watch outA common mistake is to think both diodes might conduct because they are in parallel. But the orientation of D₂ is opposite to the driving voltage, so it blocks completely. Never assume parallel branches both work — check each diode’s bias relative to the applied voltage polarity.
3. Simplifying the circuit …
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- AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQQ.If the junction diodes D1,D2 and D3 in the given circuit are ideal, then the value of current 'i' in the circuit is (For ideal diode forward biased resistance is zero and reverse biased resistance is infinity) [FIGURE] (a circuit with three parallel branches between two vertical rails: the top branch has diode D1 oriented forward, in series with a resistor R; the middle branch has diode D2 oriented in reverse, in series with a resistor R; the bottom branch has diode D3 oriented forward, in series with a resistor R; all three branches join the right rail, which also has a resistor R connecting down to the bottom rail; a battery E and the current i complete the circuit at the bottom left) (A) 3R2E (B) 2RE (C) RE (D) zero
›Reveal solutionSolution
The key idea is that ideal diodes act as perfect switches: forward‑biased → short circuit, reverse‑biased → open circuit.
Only the two forward‑biased diodes (D1 and D3) conduct, while D2 is reverse‑biased and blocks.
The resulting equivalent circuit is two parallel resistors R (from D1 and D3) in series with the bottom resistor R and the battery E.
The total current is i=3R2E, so the correct option is (A).
Concept and Intuition
An ideal diode is a one‑way valve for current:
- Forward bias (anode voltage > cathode voltage) → zero resistance (short circuit).
- Reverse bias (anode voltage < cathode voltage) → infinite resistance (open circuit).
In this circuit, the battery E sets the polarity. The left‑hand node is at a higher potential than the right‑hand node (because the battery’s positive terminal is at the bottom left, and current flows upward into the left node). So we check each diode’s orientation relative to this voltage drop.
Step‑by‑Step Reasoning
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Identify the voltage polarity across the circuit
The battery E has its positive terminal at the bottom‑left corner. Current i flows upward into the left node, so the left node is at a higher potential than the right‑hand vertical wire. Therefore, the voltage across any branch from left to right is positive (left is +, right is –).
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Determine the bias of each diode
- D1 (top branch): Its triangle points right (anode on left, cathode on right). Left is higher potential → forward bias → short circuit.
- D2 (middle branch): Its triangle points left (anode on right, cathode on left). Left is higher potential → cathode is at higher voltage than anode → reverse bias → open circuit.
- D3 (bottom branch): Same orientation as D1 (anode left, cathode right) → forward bias → short circuit.
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Replace diodes with their ideal equivalents
- D1 and D3 become wires (0 Ω).
- D2 becomes a break (infinite resistance).
The circuit now simplifies: two parallel branches (each with a resistor R in series with a short) connect the left node to the right node. The middle branch is absent.
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Simplify the parallel resistors
The two conducting branches each contain one resistor R. They are in parallel between the left node and the right vertical wire.
Equivalent resistance of two R in parallel:
Rparallel=R+RR⋅R=2R.
- Add the bottom resistor …
- AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQQ.In the given options, the diode that is forward biased is (A) [FIGURE] (a diode in series with a resistor, connecting a terminal at +2 V through the diode (arrow pointing right, toward the resistor) and resistor to a terminal at +3 V) (B) [FIGURE] (a diode in series with a resistor, connecting a terminal at +2 V through the diode (arrow pointing right) and resistor to a terminal at -2 V) (C) [FIGURE] (a diode in series with a resistor, connecting a terminal at -2 V through the diode (arrow pointing right) and resistor to a terminal at +2 V) (D) [FIGURE] (a diode in series with a resistor, connecting a terminal at +2 V through the diode (arrow pointing right) and resistor to a terminal at +2 V)
›Reveal solutionSolution
A diode conducts (is forward biased) only when its anode (p-side) sits at a higher potential than its cathode (n-side); checking all four terminal pairs, only the +2V→−2V branch satisfies this.
Concept and Intuition
The diode symbol's triangle points in the direction conventional current is allowed to flow when forward biased — from anode (p-side, the flat triangle base) to cathode (n-side, the bar). For that current to actually flow in a circuit connecting two fixed-potential terminals through the diode and a resistor, the external circuit must be 'pushing' current in that same direction, which happens precisely when the anode-side terminal is at a higher potential than the cathode-side terminal. If the cathode side is higher (or equal), the diode blocks conduction (reverse biased or no bias).
Step-by-Step Solution
- In every option, the diode's arrow points right, so the anode connects to the left terminal and the cathode (through the resistor) connects to the right terminal.
- Option (A): left (anode) =+2 V, right (cathode) =+3 V. Anode is at a lower potential than cathode ⇒ reverse biased.
- Option (B): left (anode) =+2 V, right (cathode) =−2 V. Anode is at a higher potential than cathode (by 4 V) ⇒ forward biased — current can flow.
- Option (C): left (anode) =−2 V, right (cathode) =+2 V. Anode is lower than cathode ⇒ reverse biased. …
- AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQQ.The voltage Vo in the network shown is [FIGURE] (a circuit with a +12 V supply feeding a node from which two branches drop down in parallel: one branch is a silicon diode 'Si' with a 0.7 V drop, the other is a green LED with a 2.2 V drop; both branches reconnect at a common lower node, which then connects through a 2.2 kΩ resistor to ground, with the output Vo taken at the node between the diode branches and the resistor) (A) Vo=11.3 V (B) Vo=9.8 V (C) Vo=12.0 V (D) Vo=0.7 V
›Reveal solutionSolution
This tests the behaviour of two diodes with different forward-voltage drops connected in parallel between a supply and a common node: the lower-threshold diode "wins" and clamps the node voltage, while the higher-threshold diode stays off.
Concept and Intuition
When two diodes (with different fixed forward voltages) are connected in parallel, in the same forward direction, between a fixed supply and a common node, only one of them can actually conduct at a time in this idealized picture — whichever one needs the smaller forward voltage reaches its "on" condition first as the node voltage builds up. Once that diode conducts, it clamps the node at (supply − its own drop); if the resulting voltage across the other diode is less than the other diode's own threshold, that diode stays off, which is self-consistent.
Step-by-Step Solution
- Assume the silicon diode (drop 0.7 V) conducts: node voltage Vo=12−0.7=11.3 V.
- Voltage now available across the LED branch =12−Vo=12−11.3=0.7 V.
- The LED needs 2.2 V forward drop to conduct — but only 0.7 V is available across it, which is insufficient, so the LED indeed stays OFF. This is self-consistent. …
- AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQQ.In the diodes shown in the diagrams, which one is reverse biased? (A) [FIGURE] (a diode conducting from a −12 V node toward a resistor connected to a −5 V node) (B) [FIGURE] (a diode conducting from a grounded node toward a resistor connected to a −10 V node) (C) [FIGURE] (a +5 V node connects through a resistor down to a node that connects through a diode to ground, the diode conducting toward ground) (D) [FIGURE] (a diode conducting from a +5 V node through a resistor to a +10 V node)
›Reveal solutionSolution
Reverse bias means cathode more positive than anode; only option A has Vcathode(−5V)>Vanode(−12V).
Concept and Intuition
For a p-n junction diode, forward bias requires the anode (p-side, the triangle) to be at a higher potential than the cathode (n-side, the bar); reverse bias is the opposite. To test each circuit, compare the potential the anode side is tied to against the potential the cathode side is tied to (in reverse bias no current flows, so each terminal simply sits at its source potential).
Step-by-Step Solution
- Option A: anode at −12V, cathode toward −5V. Cathode potential (−5)> anode potential (−12) by 7V → reverse biased.
- Option B: anode at 0V (ground), cathode toward −10V. Anode (0)> cathode (−10) → forward biased.
- Option C: the diode conducts toward ground, so its anode faces the +5V side and cathode is at ground; anode (+5)> cathode (0) → forward biased.
- Option D: cathode at +5V, anode toward +10V; anode (+10)> cathode (+5) → forward biased.
- Only option A has the cathode at the higher potential, so it is the reverse-biased diode.
Common Mistakes …
- AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQQ.When the voltage applied across a reverse biased p-n junction diode is increased then, the diffusion current (A) increases (B) decreases (C) first increases and then decreases (D) remains constant
›Reveal solutionSolution
Increasing reverse bias raises the junction's potential barrier, which suppresses majority-carrier diffusion across the junction — so the diffusion current decreases (while the minority-carrier drift/saturation current stays roughly constant).
Concept and Intuition
A p-n junction diode carries two competing current components: diffusion current (majority carriers moving down their concentration gradient, dominant in forward bias) and drift current (minority carriers swept across by the built-in/applied field, roughly voltage-independent and small). Reverse bias adds to the built-in potential barrier, making the barrier taller. A taller barrier means fewer majority carriers have enough energy to diffuse across it, so the diffusion current — which is already tiny under reverse bias — shrinks even further as the reverse voltage is increased. The reverse (drift) current, by contrast, is limited by the rate of thermal generation of minority carriers, not by the barrier height, so it stays essentially constant (the reverse saturation current).
Step-by-Step Solution
- Under reverse bias, the applied voltage adds to the internal barrier potential, widening the depletion region and raising the barrier height.
- Diffusion current depends exponentially on the barrier height (fewer carriers have sufficient thermal energy to surmount a taller barrier). …
- AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQQ.A reverse biased zener diode when operated in the breakdown region works as (A) an amplifier (B) an oscillator (C) a voltage regulator (D) a rectifier
›Reveal solutionSolution
This tests the key application of Zener breakdown: a near-flat V–I curve in breakdown lets the diode hold a fixed output voltage, so it is used as a voltage regulator.
Concept and Intuition
A Zener diode is a heavily-doped p-n junction. In forward bias it behaves like an ordinary diode. In reverse bias, once the applied voltage reaches the Zener (breakdown) voltage VZ, the reverse current rises very sharply while the voltage across the diode stays essentially constant at VZ. That "constant voltage even though current changes a lot" behaviour is precisely the property a voltage regulator needs.
Step-by-Step Solution
- Bias the Zener diode in reverse with a series resistor from an unregulated supply.
- As long as the reverse voltage exceeds VZ, the diode conducts heavily and clamps the voltage across itself at VZ.
- Any excess voltage (supply fluctuation, load current change) is dropped across the series resistor, not the diode. …
- AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQQ.Consider the following statements Statement (A): The resistance of an ideal diode in forward biased condition is zero. Statement (B): In a half wave rectifier, the load current flows only for every half cycle of the input signals. Statement (C): In the breakdown region, a zener diode behaves as a constant voltage source. (A) A, B & C are all true. (B) A, B true but C is false (C) A, C true but B is false (D) B, C true but A is false
›Reveal solutionSolution
This checks three foundational facts from semiconductor devices (ideal diode resistance, half-wave rectification, Zener regulation) — all three given statements are correct.
Concept and Intuition
An ideal diode is modelled as a perfect one-way switch: zero resistance (a short circuit) when forward biased, and infinite resistance (an open circuit) when reverse biased. A half-wave rectifier uses a single diode so that current can flow through the load only during the half-cycle for which the diode is forward biased, being cut off during the other half. A Zener diode is designed to operate reverse biased in its breakdown region, where the voltage across it stays essentially constant over a wide range of currents — this is exactly the property exploited in voltage-regulator circuits.
Step-by-Step Solution
- (A) Ideal diode, forward biased ⇒ acts as a closed switch with zero resistance. True.
- (B) Half-wave rectifier ⇒ diode conducts only during the half-cycle where it is forward biased; load current exists only for every alternate half-cycle of the input. True.
- (C) Zener diode in breakdown (reverse) region ⇒ voltage across it is nearly constant regardless of current variations, behaving like a constant-voltage source. True. …
- AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQQ.Considering the junction diode to be ideal, find the value of current flowing through AB [FIGURE] (a circuit diagram: point A at +4 V, connected through an ideal junction diode oriented pointing from A toward B, in series with a 1 kΩ resistor, to point B at -6 V) (A) 10−2A (B) 10−1A (C) 10−3A (D) 0A
›Reveal solutionSolution
The key idea is to check whether the diode is forward-biased or reverse-biased by comparing the voltage at its anode (+4 V) to its cathode (-6 V). Since the anode is at a higher potential, the ideal diode acts as a short circuit, and the current through the 1 kΩ resistor is simply the total voltage drop (10 V) divided by the resistance, giving 10 mA = 10−2 A. The correct option is (A).
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Understand the circuit and the diode orientation
The diode’s triangle (anode) is connected to point A at +4 V, and its bar (cathode) is connected through the resistor to point B at -6 V. For an ideal diode, current flows only when the anode voltage is higher than the cathode voltage — this is called forward bias. If the anode is lower, the diode is reverse-biased and blocks all current.
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Check the bias condition
Anode voltage = +4 V, cathode voltage = -6 V.
Since +4 V > -6 V, the diode is forward-biased.
For an ideal diode in forward bias, it acts like a perfect conductor (zero voltage drop, zero resistance). So the diode is effectively a wire.
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Simplify the circuit
With the diode replaced by a short circuit, the circuit becomes:
Point A (+4 V) → short → 1 kΩ resistor → Point B (-6 V).
The total voltage across the resistor is the difference between the two points:
Vtotal=4V−(−6V)=10V.
- Apply Ohm’s law The resistor is 1 kΩ = 1000 Ω. …
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