In a pure silicon, what is the time for an electron to drift $1\mu m$ in an electric field of 100 V/cm? Assume electron mobility of $1350 \text{ cm}^2/V-s$
Silicon Electron Drift Time Calculation
To find the time an electron takes to drift across a distance under an electric field, we use the concepts of drift velocity and mobility.
Drift Velocity Calculation
The drift velocity ($v_d$) of an electron in a semiconductor is directly proportional to the applied electric field ($E$) and the electron mobility ($\mu_n$). The formula is:
$v_d = \mu_n E$
Drift Time Determination
Drift time ($t$) is the time taken to travel a distance ($d$) at the drift velocity ($v_d$). The relationship is:
$t = \frac{d}{v_d}$
Step-by-Step Calculation
First, ensure all units are consistent. We need to convert the drift distance to centimeters to match the mobility units.
- Given values:
- Drift Distance, $d = 1 \mu m = 1 \times 10^{-6}$ m $= 1 \times 10^{-4}$ cm
- Electric Field, $E = 100$ V/cm
- Electron Mobility, $\mu_n = 1350 \text{ cm}^2/V-s$
- Calculate Drift Velocity ($v_d$):
Substitute the given values into the drift velocity formula:
$v_d = (1350 \text{ cm}^2/V-s) \times (100 \text{ V/cm})$
$v_d = 135000 \text{ cm/s}$
- Calculate Drift Time ($t$):
Now, use the drift velocity and distance to find the time:
$t = \frac{d}{v_d} = \frac{1 \times 10^{-4} \text{ cm}}{135000 \text{ cm/s}}$
$t = \frac{10^{-4}}{1.35 \times 10^5} \text{ s}$
$t \approx 0.7407 \times 10^{-9} \text{ s}$
Converting this to nanoseconds (ns), where $1 \text{ ns} = 10^{-9} \text{ s}$:
$t \approx 0.74 \text{ ns}$
Conclusion
The calculated drift time is approximately $0.74$ ns, which matches Option A.
Important Questions from Carriers in Semiconductors
A. An impurity scattering cause decrease in mobility with decrease in temperature.
B. A slow moving carrier is likely to scatter more strongly than carrier with great momentum
C. Drift velocity of an electron increases with decrease in mobility
D. Drift velocity of an electron decrease with decrease in mobility
E. At critical electric field ($=10^5V/cm$) drift velocity tends to saturate.
Choose the correct answer from the options given below:
An n-type GaAs Gunn diode has the following parameters drift velocity (Vd) = $10^7$ cm/s, Doping concentration = $2 \times 10^{14}$ cm$^{-3}$, and electric field = 3200 V/cm. Choose the correct option:
A. Current density = 320 A/cm²
B. Electron mobility = 3100 cm²/v-s
C. Current density = 120 A/cm²
D. Electron mobility = 1450 cm²/v-s
E. Electron mobility = 1500 cm²/v-s
Choose the correct answer from the options given below:
Consider the following statements :
If an electric field is applied to an n-type semiconductor bar, the electrons and holes move in opposite directions due to their opposite charges. The net current is
1. due to both electrons and holes with electrons as majority carriers.
2. the sum of electron and hole currents.
3. the difference between electron and hole current.Which of these statements is/are correct ?
Consider the following :
1. In semiconductors, the mobility of electrons is more than that of holes.
2. In semiconductors, the resistivity increases with the increase in temperature.
3. Metal has positive TCR.
4. In metals, thermal conductivity is inversely proportional to electronic conductivity at a particular temperature.Which of the following statements are correct ?
The Debye length is a characteristic length for semiconductor. Arrange the doping density given below in such a way that Debye length changes from maximum to minimum.
(A) N = 1015 cm–3
(B) N = 1017 cm–3
(C) N = 3 × 1015 cm–3
(D) N = 1016 cm–3
(E) N = 5 × 1016 cm–3Choose the most appropriate answer from the options given below :
