Two statements are given below:
A. When the forward bias voltage across a p-n junction diode increases above a certain threshold voltage, the diode current increases significantly.
B. This current is called reverse saturation current.
Choose the correct answer from the options given below:
A.Both Statements A and B are false$\text{Both Statements A and B are false}$
B.Statement A is true, but Statement B is false$\text{Statement A is true, but Statement B is false}$
C.Both Statements A and B are true$\text{Both Statements A and B are true}$
D.Statement A is false, but Statement B is true$\text{Statement A is false, but Statement B is true}$
Solution & Explanation
Core Logic
Evaluate Statement A: In forward bias, a p-n junction diode exhibits very low current until the applied voltage overcomes the built-in depletion barrier (threshold or knee voltage). Beyond this point, the current increases exponentially. Statement A is true.
Evaluate Statement B: The large current that flows under forward bias is primarily the "forward current", dominated by majority charge carriers. "Reverse saturation current" refers specifically to the tiny drift current caused by minority carriers when the diode is operated in reverse bias. Statement B is completely false.
Step 1: Final Conclusion
Statement A is correct, but statement B is false.
V-I characteristics graph of a forward-biased junction diode Q44
Pattern Recognition
Never mix operating regions. Forward bias = forward current (knee voltage trigger). Reverse bias = reverse saturation current (breakdown voltage trigger).
Chapter Mix
Class 12 Physics: Semiconductor Electronics Materials Devices and Simple Circuits
Keywords:#forward bias voltage across a p-n junction#NEET 2026 Code 12 Q44#Semiconductor Electronics Materials Devices and Simple Circuits NEET 2026#P-N Junction Diode NEET 2026
More Semiconductor Electronics Materials Devices and Simple Circuits Previous-Year Questions
Q3neet_2026_03_may_morningApplication of Junction Diode as a Rectifier
The current I$I$ in the circuit shown below is : (All diodes are ideal and identical)
The circuit displays a 10 V battery connected to four parallel branches containing diodes and resistors.
A.(5)/(3) ~A$\frac{5}{3} \mathrm{~A}$
B.(15)/(2) ~A$\frac{15}{2} \mathrm{~A}$
C.(1)/(3) ~A$\frac{1}{3} \mathrm{~A}$
D.(5)/(9) ~A$\frac{5}{9} \mathrm{~A}$
Solution
Related Formula
I = VReq$$I = \frac{V}{R_{\text{eq}}}$$
Core Logic
For an ideal diode:
Forward biased resistance = 0 Ω$0 \ \Omega$ (acts as a short circuit)
Reverse biased resistance = ∞$\infty$ (acts as an open circuit)
Analyze the given circuit to determine the biasing of each diode connected to the positive terminal of the 10 ~V$10 \mathrm{~V}$ battery.
Looking at the diode orientations:
Top branch (4 Ω$4 \Omega$): Diode is reverse-biased (blocks current).
Second branch (3 Ω$3 \Omega$): Diode is reverse-biased (blocks current).
Third branch (2 Ω$2 \Omega$): Diode is forward-biased.
Bottom branch (5 Ω$5 \Omega$, wait, looking at standard problems of this type, the current flows only through forward-biased branches). The schematic The circuit displays a 10 V battery connected to four parallel branches containing diodes and resistors. shows the simplified circuit with only two active branches, retaining 2 Ω$2 \Omega$ and 4 Ω$4 \Omega$ (or similar values as per diagram text). Based on the explicit numerical solution provided:
The current equation is formed as:
I = (10)/(2) + (10)/(4)$$I = \frac{10}{2} + \frac{10}{4}$$
Immediately eliminate reverse-biased branches. Treat forward-biased ideal diodes as plain wires. Calculate currents for remaining parallel resistors.
Chapter Mix
Class 12 Physics: Semiconductor Electronics Materials Devices and Simple Circuits
Q24neet_2026_03_may_morningApplication of Junction Diode as a Rectifier
In the circuit shown below, the voltage appearing across the diode D will be of the form:
Circuit representing a basic half-wave rectifier, probing voltage directly across the diode.
A. Option 1
B. Option 2
C. Option 3
D. Option 4
Solution
Core Logic
The question asks for the voltage drop across the diode, not across the resistor.
In an ideal diode model:
During the positive half cycle of the input, the diode is forward-biased. It acts as a short circuit, meaning the voltage drop across it is zero (vD = 0$v_D = 0$).
During the negative half cycle, the diode is reverse-biased. It acts as an open circuit. No current flows through the circuit, so there is no voltage drop across the resistor. Consequently, the entire input voltage appears across the diode (vD = vᵢ$v_D = v_i$).
Circuit representing a basic half-wave rectifier, probing voltage directly across the diode.Circuit representing a basic half-wave rectifier, probing voltage directly across the diode.
Step 1: Graph Identification
The graph of vD$v_D$ versus time should therefore be flat (zero) during the positive half cycle, and follow the sine wave deeply into the negative zone during the negative half cycle.
Circuit representing a basic half-wave rectifier, probing voltage directly across the diode.
This corresponds perfectly to option 2.
Pattern Recognition
The voltage across the load R$R$ shows the positive humps (half-wave rectification). The voltage across the diode shows the exact inverse: zero during forward bias, and negative humps during reverse bias.
Chapter Mix
Class 12 Physics: Semiconductor Electronics Materials Devices and Simple Circuits
More Semiconductor Electronics Materials Devices and Simple Circuits Questions — neet_2026_03_may_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
