In an unbiased $P-N$ junction,holes diffuse from the $p-$ region to $n-$ region because

  • A
    free $e^-$ in the $n-$ region attract them
  • B
    They move across the junction by the potential difference
  • C
    Hole concentration in $p-$ region is more as compared to $n-$ region
  • D
    All of these

Explore More

Similar Questions

$A$ source voltage of $8V$ drives the diode in the figure through a current-limiting resistor of $100 \Omega$. Then the magnitude of the slope of the load line on the $V-I$ characteristics of the diode is

The band gap in a semiconductor is $0.6 eV$. The maximum wavelength of electromagnetic radiation which can create a hole-electron pair in this semiconductor is equal to [Use $hc = 1242 eV-nm$]. (in $nm$)

For a $p-n$ junction,the intensity of the electric field is $1 \times 10^{6} \text{ V/m}$ and the width of the depletion region is $5000 \text{ Å}$. The value of the potential barrier is $\dots \text{ V}$.

$A$ diode is connected in parallel with a resistance $R$ as shown in the figure. The most probable current $(I)$ - voltage $(V)$ characteristic is:

In the circuit, all three diodes $D_1, D_2, D_3$ have a forward resistance of $50 \text{ } \Omega$ each and infinite backward resistance. If the battery voltage is $5 \text{ V}$, find the current through the $100 \text{ } \Omega$ resistance. (in $\text{mA}$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo