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  1. Problems
  2. /A resistor is cut from n-type silicon in which ne = 1.0 × 10²¹ m⁻³…
HardPhysicsHigh School (11–12)Electronics › Semiconductors and DiodesNumericalJEE MainBoard ExamOlympiadAEEE (Amrita)AMUEEEAP EAPCET
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A resistor is cut from n-type silicon in which ne = 1.0 × 10²¹ m⁻³…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A resistor is cut from n-type silicon in which ne=1.0×1021 m−3n_e = 1.0 \times 10^{21}\ \text{m}^{-3}ne​=1.0×1021 m−3. The holes can be left out. The bar is 1.01.01.0 cm long and has a cross-section of 2.0 mm22.0\ \text{mm}^22.0 mm2. A battery of 5.05.05.0 V is joined across its ends. Take μe=0.135 m2/V s\mu_e = 0.135\ \text{m}^2/\text{V s}μe​=0.135 m2/V s and e=1.602×10−19 Ce = 1.602 \times 10^{-19}\ \text{C}e=1.602×10−19 C. The drift speed of a carrier is vd=μEv_d = \mu Evd​=μE, where EEE is the electric field in the bar. Find the electric field in the bar, the drift speed of the electrons, the resistance of the bar and the current. Check your current by finding how many electrons cross a section of the bar each second. Silicon doped like this is common in integrated circuits, where a bar acts as a resistor.

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V/m
m/s
ohm
mA
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The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. L=0.010L = 0.010L=0.010 m, A=2.0×10−6 m2A = 2.0 \times 10^{-6}\ \text{m}^2A=2.0×10−6 m2, V=5.0V = 5.0V=5.0 V.

Step 1. The field. E=VL=5.0 V0.010 m=500 V/mE = \dfrac{V}{L} = \dfrac{5.0\ \text{V}}{0.010\ \text{m}} = 500\ \text{V/m}E=LV​=0.010 m5.0 V​=500 V/m.

Step 2. The drift speed. vd=μeE=0.135 m2/V s×500 V/m=67.5 m/sv_d = \mu_e E = 0.135\ \text{m}^2/\text{V s} \times 500\ \text{V/m} = 67.5\ \text{m/s}vd​=μe​E=0.135 m2/V s×500 V/m=67.5 m/s.

Step 3. The resistance. The conductivity is σ=e neμe=1.602×10−19 C×1.0×1021 m−3×0.135 m2/V s=21.6 S/m\sigma = e\, n_e \mu_e = 1.602 \times 10^{-19}\ \text{C} \times 1.0 \times 10^{21}\ \text{m}^{-3} \times 0.135\ \text{m}^2/\text{V s} = 21.6\ \text{S/m}σ=ene​μe​=1.602×10−19 C×1.0×1021 m−3×0.135 m2/V s=21.6 S/m. So ρ=0.0462 Ω m\rho = 0.0462\ \Omega\ \text{m}ρ=0.0462 Ω m and R=ρLA=0.0462 Ω m×0.010 m2.0×10−6 m2=231 ΩR = \dfrac{\rho L}{A} = \dfrac{0.0462\ \Omega\ \text{m} \times 0.010\ \text{m}}{2.0 \times 10^{-6}\ \text{m}^2} = 231\ \OmegaR=AρL​=2.0×10−6 m20.0462 Ω m×0.010 m​=231 Ω.

Step 4. The current. I=VR=5.0 V231 Ω=21.6I = \dfrac{V}{R} = \dfrac{5.0\ \text{V}}{231\ \Omega} = 21.6I=RV​=231 Ω5.0 V​=21.6 mA.

Answer: 500500500 V/m, 67.567.567.5 m/s, 231 Ω231\ \Omega231 Ω and 21.621.621.6 mA.

Check. I=neeAvd=1.0×1021 m−3×1.602×10−19 C×2.0×10−6 m2×67.5 m/s=21.6I = n_e e A v_d = 1.0 \times 10^{21}\ \text{m}^{-3} \times 1.602 \times 10^{-19}\ \text{C} \times 2.0 \times 10^{-6}\ \text{m}^2 \times 67.5\ \text{m/s} = 21.6I=ne​eAvd​=1.0×1021 m−3×1.602×10−19 C×2.0×10−6 m2×67.5 m/s=21.6 mA. About 1.35×10171.35 \times 10^{17}1.35×1017 electrons cross a section each second. The two methods agree.

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A donor gives one electron, so ne≈NDn_e \approx N_Dne​≈ND​. An acceptor gives one hole, so nh≈NAn_h \approx N_Anh​≈NA​. The law of mass action is ne nh=ni2n_e\, n_h = n_i^2ne​nh​=ni2​, with ni=1.0×1016 m−3n_i = 1.0 \times 10^{16}\ \text{m}^{-3}ni​=1.0×1016 m−3 for silicon at 300 K. Conductivity is σ=e (neμe+nhμh)\sigma = e\,(n_e \mu_e + n_h \mu_h)σ=e(ne​μe​+nh​μh​), with e=1.602×10−19 Ce = 1.602 \times 10^{-19}\ \text{C}e=1.602×10−19 C.

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