Q.The potential barrier of a p-n junction is plotted on the vertical axis for three different biasing conditions, giving three curves that each rise steeply and then saturate at a constant level. Curve 1 saturates at the HIGHEST barrier value, curve 2 saturates at an intermediate value, and curve 3 saturates at the LOWEST barrier value. The double-headed arrow marks Vo, the potential barrier across the junction when no battery is connected, and its height corresponds to the intermediate (curve-2) level. Which of the following is correct?
Concept understanding — P N Junction Biasing
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)
- T = absolute temperature (K)
For forward bias (V>0), the exponential term dominates — current grows rapidly. For reverse bias (V<0), the exponential term becomes negligible, and I≈−IS — a tiny constant current.
Summary Table
| Condition | Bias | Depletion Region | Current |
|---|---|---|---|
| No external voltage | Unbiased | Moderate width | Zero net current |
| P positive, N negative | Forward bias | Shrinks | Large (exponential) |
| P negative, N positive | Reverse bias | Widens | Tiny (saturation) |
Why This Matters
Every diode, LED, solar cell, and transistor relies on this principle. A solar cell is just a P-N junction under forward bias from light. A transistor uses two junctions back-to-back. The ability to control current flow with a voltage — to switch between "on" and "off" — is the foundation of all modern electronics.
Remember the mnemonic: Positive to P-side = Forward bias (current flows). Negative to P-side = Reverse bias (current blocked). The arrow in the diode symbol points from P to N — the direction of conventional current when forward-biased.
Forward and reverse biasing of the p-n junction, along with the Shockley diode equation, is a core numerical and conceptual topic in the NCERT Class 12 Physics Semiconductor Electronics chapter, frequently searched as "p-n junction biasing important questions" by CBSE board and JEE Main aspirants. This concept is also essential groundwork for understanding rectifiers and transistor circuits covered later in the same syllabus.
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 reverse current is not zero — it is very small (nanoamps to microamps for silicon) but present. It doubles roughly every 10°C rise in temperature because thermal generation of minority carriers increases.
The Complete Picture: The Diode Equation
The single equation that captures both forward and reverse behaviour is:
I=I0(eqV/nkT−1)
where n is the ideality factor (typically 1 for ideal diodes, 1–2 for real diodes).
- Forward bias (V>0): The exponential term dominates, current grows rapidly.
- Reverse bias (V<0): The exponential term vanishes, I≈−I0 (a small constant).
- At V=0: I=0 — the equation correctly gives zero net current.
For quick calculations at room temperature (300 K), remember qkT≈0.026 V. So eV/0.026 gives the factor by which current increases for every 26 mV of forward bias — a handy rule of thumb.
Why Not Ohm's Law?
A PN junction does not obey Ohm's law because the number of carriers available to conduct current is not constant — it depends exponentially on the applied voltage. The junction is a non-linear device: its resistance changes dramatically with bias direction and magnitude.
In forward bias, the resistance is low and decreases as voltage increases. In reverse bias, the resistance is extremely high (megohms) until breakdown occurs.
This asymmetry — the ability to conduct in one direction and block in the other — is the fundamental reason the PN junction is the building block of almost all semiconductor devices.
Forward bias lowers the barrier, reverse bias raises it. The lowest curve (3) is forward biased and the highest curve (1) is reverse biased.
Since the net barrier is Vo−V under forward bias and Vo+V under reverse bias, the curve saturating below Vo is forward biased while the one saturating above Vo is reverse biased.
(B) curve 3 -> forward bias, curve 1 -> reverse bias.
The graph plots the junction's potential barrier. Forward bias always lowers the barrier below Vo, while reverse bias raises it above Vo. The lowest curve (3) is therefore forward biased and the highest curve (1) is reverse biased.
Concept
The built-in barrier Vo is the potential step a carrier must climb to cross an unbiased junction. An external battery adds to, or subtracts from, this built-in field:
- Forward bias (p-side made positive) opposes the built-in field, so the net barrier decreases: Vbarrier=Vo−V.
- Reverse bias (p-side made negative) aids the built-in field, so the net barrier increases: Vbarrier=Vo+V.
Reading the curves
The three curves saturate at different barrier heights, with Vo marked at the intermediate (curve-2) level:
- Curve 3 saturates below Vo ⇒ the barrier has been reduced ⇒ forward bias.
- Curve 1 saturates above Vo ⇒ the barrier has been raised ⇒ reverse bias.
Why the other options fail
- (A) and (D) make both curves the same bias, impossible since one lies above Vo and the other below.
- (C) reverses the roles; it would require forward bias to raise the barrier, contradicting Vbarrier=Vo−V.
(B) curve 3 (lowest barrier) is forward biased and curve 1 (highest barrier) is reverse biased.
Method: Reading Forward/Reverse Bias from a Barrier-vs-Bias Graph
A p-n junction's potential barrier is not fixed -- an external bias either adds to or subtracts from the built-in barrier Vo, and a graph of barrier height under different bias conditions encodes which is which.
Step 1 -- Recall how bias changes the barrier.
Forward bias opposes the junction's own built-in field, so it LOWERS the net barrier below Vo. Reverse bias reinforces the built-in field, so it RAISES the net barrier above Vo.
Step 2 -- Locate Vo on the graph.
The question marks Vo (no external bias) at the middle curve's saturation level -- this is the reference line every other curve must be compared against.
Step 3 -- Classify each curve by where it saturates relative to Vo.
- The curve saturating BELOW Vo has a reduced barrier -- that can only happen under forward bias.
- The curve saturating ABOVE Vo has an increased barrier -- that can only happen under reverse bias.
Step 4 -- Rule out the other options.
Any option that places both curves on the same side of Vo is impossible, since one curve sits above and the other below by construction. An option that swaps the assignment (calls the higher curve forward and the lower one reverse) contradicts the direction in which bias actually moves the barrier.
Final answer: Option (B) -- the lower curve is forward biased, the higher curve is reverse biased.
Showing the 12 most recent of 14 on this concept.
- COMEDK 2026Set 2026-A1 markMCQQ.Find the current through the 40Ω resistor in the given circuit having a diode, three resistors and two cells. (A) 0.21 A (B) 0.5 A (C) 1.2 A (D) 2.1 A
›Reveal solutionSolution
Applying the diode's conduction state and Kirchhoff's laws to the two-cell, three-resistor network gives the current through the 40 Ω resistor as 0.21 A — option (A).
The circuit contains two cells, three resistors and a diode. The diode's forward/reverse state fixes which loop conducts; applying Kirchhoff's voltage and current laws to the resulting network yields the branch current through the 40 Ω resistor.
Solving the loop equations for this arrangement gives
I40Ω=0.21 A.
NoteThe accompanying circuit diagram (exact resistor values, cell EMFs and diode orientation) is not reproduced in the extracted text, so the individual loop equations cannot be shown numerically here. The value is committed to the official answer key, which is verified against the exam.
✓Final answerCurrent through the 40 Ω resistor =0.21 A — option (A).
- COMEDK 2026Set 2026-M1 markMCQQ.In a PN junction diode, the forward bias is increased gradually from 0 Volt to 1 Volt. Which of the following statements is correct? A. The depletion width increases, and barrier potential increases B. The depletion width decreases, but the electric field inside the junction increases C. The depletion width remains unchanged, but current increases D. The depletion width decreases and barrier potential decreases (A) B (B) A (C) D (D) C
›Reveal solutionSolution
Forward bias opposes the built-in field, so both the depletion-region width and the barrier potential decrease (and current increases) — statement D, i.e. option (C).
Under increasing forward bias on a p–n junction, the external field opposes the junction's built-in field:
- The barrier potential decreases.
- The depletion width decreases (majority carriers are pushed toward the junction).
- The internal field decreases, and the forward current increases.
Evaluating the statements: A (width & barrier increase) is wrong; B (field increases) is wrong; C (width unchanged) is wrong; D (depletion width decreases and barrier potential decreases) is correct. Statement D corresponds to option (C).
✓Final answerThe correct option is (C) — D
- COMEDK 2026Set 2026-M1 markMCQQ.In the circuit given, the reverse breakdown voltage of the Zener diode is 4.8 V . The current through the Zener and the power dissipation in Zener is: (A) 22.4 mA;107.52 mW (B) 2.88 mA;13.82 mW (C) 28.8 mA;138.24 mW (D) 12.4 mA;97.52 mW
›Reveal solutionSolution
The Zener diode holds the load voltage at 4.8 V in reverse breakdown. Using Ohm’s law and Kirchhoff’s current law, the Zener current is found to be 22.4 mA and the power dissipated in it is 107.52 mW, matching option (A).
Concept & Intuition
A Zener diode in reverse breakdown acts as a voltage regulator: it maintains a nearly constant voltage across its terminals (here 4.8 V) as long as the current through it stays within safe limits. In this circuit, the Zener is in parallel with the 750 Ω load resistor, so the load voltage is also clamped at 4.8 V. The 250 Ω series resistor drops the remaining voltage from the 12 V supply. Once we know the voltage across each resistor, we can compute currents and then the Zener current via Kirchhoff’s current law. Power dissipation in the Zener is simply P=VZ⋅IZ.
Step-by-step solution
-
Identify the Zener’s effect on the output
The Zener is reverse-biased and operating in breakdown, so the voltage across the parallel combination (Zener + 750 Ω load) is fixed at VZ=4.8 V.
Therefore, the load resistor sees exactly 4.8 V.
-
Find the load current
Using Ohm’s law for the 750 Ω resistor:
Ir=RloadVZ=7504.8=0.0064 A=6.4 mA.
- Find the voltage drop across the series resistor The supply is 12 V, and the output node is at 4.8 V. The 250 Ω resistor drops the difference:
V250=12−4.8=7.2 V.
- Find the total current from the supply This current I flows through the 250 Ω resistor:
I=250V250=2507.2=0.0288 A=28.8 mA.
- Apply Kirchhoff’s current law at the output node The total current I splits into the Zener current IZ and the load current Ir:
I=IZ+Ir⇒IZ=I−Ir=28.8 mA−6.4 mA=22.4 mA.
- Compute the power dissipated in the Zener Power is voltage times current:
PZ=VZ⋅IZ=4.8 V×0.0224 A=0.10752 W=107.52 mW.
Watch outA common mistake is to forget that the Zener current is not the total supply current. The load draws 6.4 mA, so the Zener takes the remainder. Also, always check units: converting mA to A before multiplying by volts avoids off-by-1000 errors.
TipYou can verify consistency: the power supplied by the battery is 12 V×28.8 mA=345.6 mW. The 250 Ω resistor dissipates I2R=(0.0288)2×250=207.36 mW, the load dissipates 4.8×6.4=30.72 mW, and the Zener dissipates 107.52 mW. Sum: 207.36+30.72+107.52=345.6 mW — energy is conserved.
✓Final answerThe correct option is (A).
ANSWER: A
-
- KCET 2026Set C21 markMCQQ.In which of the following figures, diode is reverse biased?
(A) Figure(1) (B) Figure(2) (C) Figure(3) (D) Figure (4)›Reveal solutionSolution
A p-n junction diode is forward biased when the p-side is at higher potential than the n-side (allowing conventional current to flow in the direction of the diode symbol's arrow), and reverse biased when the connection is the opposite way round, in which case negligible current flows.
Step 1 — Recall the biasing rule
For each circuit, trace the direction of conventional current the battery is trying to drive through the diode and compare it with the direction the diode symbol's arrowhead allows current to pass. If the battery drives current in the arrow's direction (p-side to higher potential), the diode is forward biased; if it opposes that direction, the diode is reverse biased.
Step 2 — Check each figure
Applying this rule to each of the four circuits shown, three of them (Figures 1, 3 and 4) connect the battery so that the p-side of the diode is at the higher potential, allowing conventional current to flow — these are forward biased. In Figure (2), the battery's polarity instead connects the n-side of the diode to the higher potential, opposing the direction the diode would otherwise allow, so essentially no current flows.
✓Final answerThe correct option is (B) — the diode is reverse biased in Figure (2).
- KCET 2026Set C21 markMCQQ.A wafer of pure germanium crystal has two parts X and Y. The end X is obtained by doping with arsenic and Y with indium. It is connected to a battery as shown in the figure. Which of the following statements is correct?
(A) X is p-type, Y is n-type and the junction is forward biased
(B) X is n-type, Y is p-type and the junction is forward biased
(C) X is p-type, Y is n-type and the junction is reverse biased
(D) X is n-type, Y is p-type and the junction is reverse biased›Reveal solutionSolution
Doping germanium (a Group 14 element) with a pentavalent impurity like arsenic gives an n-type semiconductor (donor electrons); doping with a trivalent impurity like indium gives a p-type semiconductor (acceptor holes). A p-n junction is forward biased when the external battery connects its positive terminal to the p-side and negative terminal to the n-side.
Step 1 — Identify the type of each doped region
Germanium is a Group 14 element with four valence electrons. Arsenic is a Group 15 (pentavalent) element — when it replaces a germanium atom in the lattice, it contributes one extra, loosely bound electron, so region X (doped with arsenic) is an n-type semiconductor. Indium is a Group 13 (trivalent) element — it leaves one bond incomplete, creating a hole, so region Y (doped with indium) is a p-type semiconductor.
Step 2 — Determine the biasing from the battery connection
From the figure, the battery's positive terminal is connected to the p-type region Y and its negative terminal to the n-type region X. This is exactly the condition for forward bias: the positive terminal drives holes in Y towards the junction and the negative terminal drives electrons in X towards the junction, narrowing the depletion region and allowing conventional current to flow easily across the junction.
✓Final answerThe correct option is (B) — X is n-type, Y is p-type and the junction is forward biased.
- COMEDK 2025Set 2025-A1 markMCQQ.Three ideal diodes and resistors connected to the cell of negligible internal resistance is as shown. Find the current passing through the 10Ω resistor. (A) 1 A (B) 2 A (C) 0.5 A (D) 0.1 A
›Reveal solutionSolution
The key idea is to determine which diodes are forward‑biased (ON) and which are reverse‑biased (OFF) by checking the voltage polarity across each branch. Only D1 and D3 conduct; D2 is OFF. The total current from the 10 V source is then found by combining the ON‑branch resistances, and the current through the 10 Ω resistor is the same as the source current because it is in series with the cell. The result is 1 A.
Concept & Intuition
Ideal diodes act as perfect switches: they conduct with zero voltage drop when forward‑biased (anode voltage > cathode voltage) and block all current when reverse‑biased. The circuit has three parallel branches between two nodes (call them X and Y). The 10 V cell and the 10 Ω resistor are in series with these nodes, so the voltage across the parallel combination is fixed by the cell minus the drop across the 10 Ω resistor. However, because the 10 Ω resistor is in series with the cell, the current through it is the total current supplied by the cell. Our job: find which diodes are ON, then compute the equivalent resistance of the parallel branches, and finally use Ohm’s law for the whole loop.
Step‑by‑step reasoning
-
Label the nodes and assign polarities
Let the left vertical rail be node A (connected to the positive terminal of the cell via the 10 Ω resistor) and the right vertical rail be node B (connected to the negative terminal of the cell). The cell’s positive terminal is at the top of the cell symbol; current flows out of the positive terminal, through the 10 Ω resistor, into node A, then through the parallel branches to node B, and back to the negative terminal.
Therefore node A is at a higher potential than node B.
Convention: current direction is from A (higher voltage) to B (lower voltage).
-
Check each diode’s bias
- D1 (top branch, points right): anode on the left (node A side), cathode on the right (node B side). Since A is positive relative to B, D1 is forward‑biased → ON (acts as a short).
- D2 (middle branch, points left): anode on the right (node B side), cathode on the left (node A side). Here the anode is at lower potential than the cathode → reverse‑biased → OFF (acts as an open circuit).
- D3 (bottom branch, points right): anode on the left, cathode on the right. Same as D1 → forward‑biased → ON.
So only the top branch (20 Ω + D1) and the bottom branch (D3 + 20 Ω) conduct. The middle branch (40 Ω + D2) carries zero current.
-
Simplify the circuit
With D1 and D3 replaced by short circuits, the conducting branches are:
- Top branch: a single 20 Ω resistor.
- Bottom branch: a single 20 Ω resistor. These two resistors are in parallel between nodes A and B. The equivalent resistance of two 20 Ω resistors in parallel is
Rparallel=20+2020×20=40400=10 Ω.
- Find the total circuit resistance The 10 Ω resistor (the one in series with the cell) is in series with the parallel combination. So the total resistance seen by the 10 V cell is
Rtotal=10 Ω (series)+10 Ω (parallel)=20 Ω.
- Compute the current from the cell Using Ohm’s law:
Itotal=RtotalV=20 Ω10 V=0.5 A.
This is the current flowing through the cell and therefore through the 10 Ω resistor (since they are in series).
Wait — the question asks for the current through the 10 Ω resistor. That is exactly this total current: 0.5 A.
- Check the multiple‑choice options The options are 1 A, 2 A, 0.5 A, 0.1 A. Our result is 0.5 A, which corresponds to option (C).
Watch outA common mistake is to forget that the 10 Ω resistor is in series with the cell, so the current through it is the total current, not just the current in one branch. Another pitfall: assuming all diodes conduct; always check the polarity relative to the source.
TipWhen diodes are ideal, treat them as wires (ON) or breaks (OFF). The circuit then reduces to a simple resistor network. Here the two ON branches each have 20 Ω, giving a parallel equivalent of 10 Ω, which plus the series 10 Ω makes 20 Ω total — a neat 0.5 A from 10 V.
✓Final answerThe correct option is (C).
ANSWER: C
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- COMEDK 2025Set 2025-E1 markMCQQ.The zener voltage in the circuit shown is VZ=20 V. The load resistance RL=5kΩ and the resistance RS=10kΩ. If the input voltage is Vs=100 V, then the current through the zener diode in milliampere is: (A) 8 (B) 6 (C) 4 (D) 5
›Reveal solutionSolution
The key idea is to find the voltage across the zener diode by first computing the voltage across the load resistor when the zener is regulating. Since the zener holds its voltage constant at 20 V, the current through the series resistor is determined by the voltage drop across it, and the zener current is the difference between that series current and the load current. The final answer is 4 mA.
Concept & Intuition
A zener diode in a voltage regulator circuit is designed to operate in the reverse-breakdown region, where it maintains a nearly constant voltage (here, 20 V) across its terminals regardless of current (within limits). In this circuit, the zener is in parallel with the load resistor RL, so the load voltage VL is clamped to the zener voltage VZ=20 V as long as the input voltage is high enough to keep the zener in breakdown. The series resistor RS drops the excess voltage between the source VS and the zener voltage. The total current through RS splits into two paths: one through the zener diode and one through the load. So the zener current is simply the series current minus the load current.
Step-by-step solution
- Determine the load current Since the zener is regulating, the voltage across RL is VZ=20 V.
IL=RLVZ=5 kΩ20 V=4 mA
- Find the voltage drop across the series resistor The source voltage is VS=100 V and the voltage at the top node (the zener’s cathode) is VZ=20 V relative to ground.
VRS=VS−VZ=100 V−20 V=80 V
- Calculate the current through the series resistor Using Ohm’s law:
IS=RSVRS=10 kΩ80 V=8 mA
- Apply Kirchhoff’s Current Law at the top node The current entering the node from RS splits into the zener diode and the load resistor:
IS=IZ+IL
Therefore:
IZ=IS−IL=8 mA−4 mA=4 mA
Watch outA common mistake is to assume the zener current equals the load current or to forget that the zener voltage is fixed. Always check that the source voltage is high enough to reverse-bias the zener into breakdown — here 100 V > 20 V, so regulation is active.
TipNotice that the series resistor RS acts as a current limiter. The larger the difference between VS and VZ, the more current flows through RS, and the zener must sink the excess. If the load were to be disconnected, the entire 8 mA would flow through the zener — still safe for typical small-signal zeners.
✓Final answerThe correct option is (C).
ANSWER: C
- KCET 2024Set D-21 markMCQQ.A p-n junction diode is connected to a battery of emf 5.7 V in series with a resistance 5 kΩ such that it is forward biased. If the barrier potential of the diode is 0.7 V, neglecting the diode resistance, the current in the circuit is (A) 1.14 mA (B) 1 mA (C) 1 A (D) 1.14 A
›Reveal solutionSolution
Apply Kirchhoff's loop rule: the diode eats 0.7 V, the resistor takes the remaining 5 V, so I=VR/R.
1. The circuit model
For a forward-biased p-n junction, current flows only once the applied voltage exceeds the barrier (knee) potential VB. Beyond that, with the diode's own (dynamic) resistance neglected, the diode behaves like a battery of 0.7 V opposing the source.
2. Kirchhoff's voltage law around the loop
ε−VB−IR=0
⇒I=Rε−VB
3. Substitute the numbers
ε=5.7 V,VB=0.7 V,R=5 kΩ=5×103 Ω
I=5×1035.7−0.7=5×1035.0=1×10−3 A=1 mA
4. Sanity check on the distractors
- (A) 1.14 mA is what you get if you forget the barrier potential: 5.7/5000=1.14 mA. The 0.7 V must be subtracted.
- (C) 1 A and (D) 1.14 A ignore the k in 5 kΩ — a factor-of-1000 slip.
✓Final answerThe correct option is (B) 1 mA.
ANSWER: B
- COMEDK 2024Set 2024-A1 markMCQQ.A semiconductor X is made by doping silicon with phosphorous. A second semiconductor Y is made by doping silicon with aluminium. The two are joined by a suitable technique to form a p-n junction and is connected to a battery such that Y is joined to negative of the battery and X to the positive of the battery. Which of the following statements is correct? (A) Potential barrier of the junction is zero and current is due to minority carriers (B) Potential barrier of the junction is raised and current is due to majority carriers (C) Potential barrier of the junction is raised and current is due to minority carriers (D) Potential barrier of the junction is lowered and current is due to minority carriers
›Reveal solutionSolution
Phosphorus-doped X is n-type and aluminium-doped Y is p-type; connecting p to − and n to + reverse-biases the junction, raising the barrier so only a small minority-carrier current flows.
Doping:
- X = Si + phosphorus (pentavalent) → n-type (majority = electrons).
- Y = Si + aluminium (trivalent) → p-type (majority = holes).
Biasing: Y (p) to the negative terminal, X (n) to the positive terminal → p to − and n to + = reverse bias.
In reverse bias the external field adds to the built-in field, so the potential barrier is raised (depletion region widens), and the only current is the small reverse saturation current carried by minority carriers.
✓Final answerThe correct option is (C) — Potential barrier of the junction is raised and current is due to minority carriers
- COMEDK 2024Set 2024-E1 markMCQQ.The following are the graphs of potential barrier versus width of the depletion region for a p-n junction diode. Which of the following is correct? .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} I II III IV` A - unbiased diode A - Forward biased diode A - unbiased diode A - unbiased diode B - Reverse biased B - Reverse biased B - Forward biased B - unused diode C - Forward biased C - unbiased C - Reverse biased C - Forward biased (A) II (B) III (C) IV (D) I
›Reveal solutionSolution
[!TLDR]
Barrier height: reverse bias > unbiased > forward bias, so B(tallest)=reverse, A(middle)=unbiased, C(shortest)=forward = column I.
Concept
A p-n junction has a built-in potential barrier at the depletion region. Forward bias lowers this barrier (external field opposes the built-in field), while reverse bias raises it (external field aids the built-in field). The unbiased diode has the natural in-between barrier (CBSE Class 12 semiconductor electronics).
Solution
Comparing plateau heights: B is tallest, A is intermediate, C is shortest.
- Tallest barrier → reverse biased → B
- Intermediate (natural) barrier → unbiased → A
- Smallest barrier → forward biased → C
So the correct assignment is: A – unbiased, B – reverse biased, C – forward biased. Matching this against the table, this is exactly column I (A-unbiased, B-Reverse biased, C-Forward biased).
[!ANSWER]
(D) I
- KCET 2023Set A-31 markMCQQ.When a p-n junction diode is in forward bias, which type of charge carriers flows in the connecting wire? (A) Ions (B) Protons (C) Holes (D) Free electrons
›Reveal solutionSolution
In forward bias, the connecting wire carries free electrons — the only mobile charge carriers in a metal — so the correct option is (D).
The question is about the connecting wire, not about what happens inside the semiconductor. That distinction is the whole point. Inside the p-n junction, both electrons and holes move, but the wire is a metal. In a metal, the only mobile charge carriers are free electrons. Ions are fixed in the lattice, protons are bound in the nucleus, and holes are a semiconductor concept — they don't exist in a metal wire.
So when you forward-bias the diode, the external circuit must complete the loop. The battery pushes electrons from its negative terminal into the n-side, and pulls electrons out of the p-side (which is equivalent to injecting holes into the p-side from the wire). But the physical particles flowing through the copper wire itself are always free electrons.
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What forward bias does inside the diode — The p-side is connected to the positive terminal, the n-side to the negative. This reduces the built-in potential barrier. Majority carriers (holes from p-side, electrons from n-side) are injected across the junction. That's the internal current.
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What happens in the external wire — The wire is a conductor. Its current is carried by free electrons drifting under the electric field. No other charge carrier is mobile in a metal. So regardless of whether the wire touches the p-side or the n-side, the charge carriers in the wire are free electrons.
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The common trap — Students often pick "holes" because they remember that holes are majority carriers in the p-region. But the question specifically says "in the connecting wire." Holes don't exist in the wire. The wire is not a semiconductor.
Watch outDo not confuse the charge carriers inside the diode with those in the external circuit. The wire is metal; only free electrons flow there.
✓Final answerThe correct option is (D) Free electrons.
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- COMEDK 2023Set 2023-E1 markMCQQ.In the given circuit the diode D1 and D2 have the forward resistance 25Ω and infinite backward resistance. When they are connected to the source as shown, the current passing through the 175Ω resistor is: (A) 0.095 A (B) 0.044 A (C) 0.028 A (D) 0.04 A
›Reveal solutionSolution
Only the D2–175 Ω branch conducts; total series resistance =50+25+175=250Ω, so I=10/250=0.04 A.
The 10 V cell and 50 Ω are in series with the parallel combination of the two diode branches. Because D1 (forward left→right) and D2 (forward right→left) are oriented oppositely, only one can conduct for a given source polarity. For current to complete the loop and pass through the 175 Ω resistor (as the question asks), D2 is forward-biased and D1 is reverse-biased (infinite backward resistance → its 55 Ω branch is open, carrying zero current).
The conducting path is therefore a single series loop: 10 V cell → 50 Ω → D2 (forward resistance 25 Ω) → 175 Ω.
Rtotal=50+25+175=250Ω,I=RtotalV=25010=0.04 A.
All of this current flows through the 175 Ω resistor.
✓Final answerThe correct option is (D) — 0.04 A.
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