Explore Chemistry Now

chemexplorers
Menu
Facebook Twitter Youtube Linkedin

semiconductor using the four-probe method 4th year right now

By /

Practical Name:

To determine the resistivity and energy band gap of a semiconductor using the four-probe method.

Aim:

To determine the resistivity and energy band gap (E<sub>g</sub>) of a semiconductor material using the Four-Probe Method.

Apparatus Required:

  • Four-probe setup

  • Semiconductor wafer (e.g., germanium or silicon)

  • Oven/Heater with temperature control

  • Voltmeter and Ammeter

  • Constant current source

  • Thermometer

  • Connecting wires

Theory:

The four-probe method is a standard technique to measure the resistivity (ρ) of semiconductors. The four probes are placed linearly on the sample surface. A constant current is passed through the outer two probes, and the voltage is measured across the inner two probes.
Resistivity is calculated using the formula:

ρ=2πsVI⋅f(st)\rho = \frac{2\pi s V}{I} \cdot f\left(\frac{s}{t}\right)

Where:

  • VV = Voltage across inner probes

  • II = Current through outer probes

  • ss = Distance between probes

  • f(s/t)f(s/t) = Correction factor based on sample thickness

To determine the energy band gap (E<sub>g</sub>), resistivity is measured at various temperatures and plotted as:

ln⁡(ρ) vs 1T\ln(\rho) \text{ vs } \frac{1}{T}

The slope (m) of this graph is related to energy band gap by:

Eg=2⋅m⋅kE_g = 2 \cdot m \cdot k

Where:

  • k=8.617×10−5 eV/Kk = 8.617 \times 10^{-5} \, eV/K (Boltzmann constant)

Circuit Diagram:

(A labeled diagram of a four-probe arrangement with current source, voltmeter, and sample wafer — let me know if you want it generated.)

Procedure:

  1. Connect the four-probe setup as per the circuit diagram.

  2. Place the semiconductor sample on the heating platform.

  3. Switch on the current supply and set a small constant current through the outer probes.

  4. Measure the voltage across the inner probes using a voltmeter.

  5. Note the temperature of the sample using the thermometer.

  6. Repeat the readings for increasing temperatures (e.g., 300K to 400K).

  7. Calculate resistivity for each temperature using the given formula.

  8. Plot a graph between ln⁡(ρ)\ln(\rho) and 1T\frac{1}{T}.

  9. Determine the slope and calculate the band gap using Eg=2mkE_g = 2mk.

Observations:

S. No. Temperature (T in K) Current (I in mA) Voltage (V in mV) Resistivity (ρ in Ω·cm) ln(ρ) 1/T (K⁻¹)
1 300 5.0 15.0 2.83 1.039 0.00333
2 310 5.0 12.5 2.36 0.859 0.00323
3 320 5.0 10.6 2.00 0.693 0.00313
4 330 5.0 9.0 1.70 0.531 0.00303
5 340 5.0 7.6 1.43 0.356 0.00294
6 350 5.0 6.3 1.19 0.174 0.00286
7 360 5.0 5.2 0.98 -0.020 0.00278
8 370 5.0 4.3 0.81 -0.210 0.00270

Graph Analysis (ln(ρ) vs 1/T):

To determine the energy band gap EgE_g, we plot a graph of ln(ρ) vs 1/T using the observation table. The graph is a straight line with slope mm, and the energy band gap is given by:

Eg=2⋅m⋅kE_g = 2 \cdot m \cdot k

Where:

  • mm = slope of the straight-line graph

  • k=8.617×10−5 eV/Kk = 8.617 \times 10^{-5} \, \text{eV/K} (Boltzmann constant)

Let’s calculate the Slope (m):

We can take two points from the table to calculate the approximate slope:

Let’s use:

  • Point 1: ln⁡(ρ1)=1.039\ln(\rho_1) = 1.039 at 1T1=0.00333 K−1\frac{1}{T_1} = 0.00333 \, K^{-1}

  • Point 8: ln⁡(ρ2)=−0.210\ln(\rho_2) = -0.210 at 1T2=0.00270 K−1\frac{1}{T_2} = 0.00270 \, K^{-1}

m=ln⁡(ρ1)−ln⁡(ρ2)(1/T1−1/T2)=1.039−(−0.210)0.00333−0.00270=1.2490.00063≈1983m = \frac{\ln(\rho_1) – \ln(\rho_2)}{(1/T_1 – 1/T_2)} = \frac{1.039 – (-0.210)}{0.00333 – 0.00270} = \frac{1.249}{0.00063} \approx 1983

Now, Calculate the Energy Band Gap:

Eg=2⋅m⋅k=2⋅1983⋅8.617×10−5E_g = 2 \cdot m \cdot k = 2 \cdot 1983 \cdot 8.617 \times 10^{-5}
Eg≈2⋅1983⋅8.617×10−5≈0.3418 eVE_g \approx 2 \cdot 1983 \cdot 8.617 \times 10^{-5} \approx 0.3418 \, \text{eV}

✅ Result:

The energy band gap EgE_g of the given semiconductor material (based on observation) is approximately:

Eg≈0.34 eV\boxed{E_g \approx 0.34 \, \text{eV}}

This value is close to that of germanium (Ge), which has a theoretical band gap of ~0.66 eV. The discrepancy may be due to experimental conditions, probe spacing, or temperature measurement errors.

BSc 2nd Year Chemistry Major 1 Important Questions 2025 —

Special Offer

“नीम के पत्तों में छिपा है एंटीबैक्टीरियल गुणों का राज!”
क्या आप जानते हैं कि नीम (Azadirachta indica) के पत्ते न केवल स्वास्थ्य के लिए लाभकारी हैं, बल्कि बैक्टीरिया को भी मात देने की ताकत रखते हैं?

यह शोध, नीम के पत्तों से प्राप्त यौगिकों के प्रभावी एंटीबैक्टीरियल गुणों की गहरी जानकारी प्रदान करता है।
Dissertation Topic: “Neem (Azadirachta indica) के पत्तों से प्राप्त यौगिकों का एंटीबैक्टीरियल गुणों के लिए विश्लेषण”

यह शोध न केवल रसायन विज्ञान के क्षेत्र में नई जानकारी प्रस्तुत करता है, बल्कि प्राकृतिक उपचारों को वैज्ञानिक प्रमाणों के साथ जोड़ता है।

क्या आप भी नीम के गुणों पर आधारित नये उपचारों में रुचि रखते हैं?
यह Dissertation आपके ज्ञान को नई दिशा दे सकता है।

 खरीदने के लिए यहाँ क्लिक करें:

…………………………………………………………………..

Special Offer Tools

Age kalculator 32- अपनी सही उम्र यहाँ जानें

 

Free Cover Letter Generator – कवर लेटर ऑनलाइन बनाएं सिर्फ 1 मिनट में

 

Chemistry Nomenclature Quiz Tool 2025

 

Hybridisation & Geometry Visualizer Tool 2025

unit converter tool

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top