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Calculate the density of donor ions (ND) using the given values for barrier potential, temperature, acceptor concentration, and intrinsic concentration.

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Answer

3.9×1021 m−3\text{3.9} \times 10^{21} \, \text{m}^{-3}

Step-by-step solution

3 steps

To determine the density of donor ions (NDN_D), we use the formula for the barrier potential (V0V_0) of a P-N junction:

V0=VTln⁡(NANDni2)V_0 = V_T \ln\left(\frac{N_A N_D}{n_i^2}\right)

where VTV_T is the thermal voltage, NAN_A is the acceptor concentration, and nin_i is the intrinsic carrier concentration.

Given values from the problem:

  • Absolute temperature, T=350 KT = 350 \, \text{K}
  • Barrier potential, V0=0.7 VV_0 = 0.7 \, \text{V}
  • Concentration of acceptor ions, NA=7.5×1015 m−3N_A = 7.5 \times 10^{15} \, \text{m}^{-3}
  • Intrinsic concentration, ni=5×1013 m−3n_i = 5 \times 10^{13} \, \text{m}^{-3}

Constants (from question 4b):

  • Boltzmann constant, k=1.38×10−23 J/Kk = 1.38 \times 10^{-23} \, \text{J/K}
  • Electronic charge, q=1.6×10−19 Cq = 1.6 \times 10^{-19} \, \text{C}

  1. Calculate the thermal voltage (VTV_T).

    The thermal voltage is given by VT=kTqV_T = \frac{kT}{q}.

    VT=(1.38×10−23 J/K)×(350 K)1.6×10−19 CV_T = \frac{(1.38 \times 10^{-23} \, \text{J/K}) \times (350 \, \text{K})}{1.6 \times 10^{-19} \, \text{C}} VT=4.83×10−21 J1.6×10−19 CV_T = \frac{4.83 \times 10^{-21} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}} VT=0.0301875 VV_T = 0.0301875 \, \text{V}

  2. Rearrange the barrier potential formula to solve for NDN_D.

    V0=VTln⁡(NANDni2)V_0 = V_T \ln\left(\frac{N_A N_D}{n_i^2}\right) V0VT=ln⁡(NANDni2)\frac{V_0}{V_T} = \ln\left(\frac{N_A N_D}{n_i^2}\right) eV0VT=NANDni2e^{\frac{V_0}{V_T}} = \frac{N_A N_D}{n_i^2} ND=ni2NAeV0VTN_D = \frac{n_i^2}{N_A} e^{\frac{V_0}{V_T}}

  3. Substitute the values and calculate NDN_D.

    First, calculate the exponent term:

    V0VT=0.7 V0.0301875 V≈23.186087\frac{V_0}{V_T} = \frac{0.7 \, \text{V}}{0.0301875 \, \text{V}} \approx 23.186087 eV0VT=e23.186087≈1.1706×1010e^{\frac{V_0}{V_T}} = e^{23.186087} \approx 1.1706 \times 10^{10}

    Next, calculate ni2n_i^2:

    ni2=(5×1013 m−3)2=25×1026 m−6=2.5×1027 m−6n_i^2 = (5 \times 10^{13} \, \text{m}^{-3})^2 = 25 \times 10^{26} \, \text{m}^{-6} = 2.5 \times 10^{27} \, \text{m}^{-6}

    Now, substitute all values into the equation for NDN_D:

    ND=2.5×1027 m−67.5×1015 m−3×(1.1706×1010)N_D = \frac{2.5 \times 10^{27} \, \text{m}^{-6}}{7.5 \times 10^{15} \, \text{m}^{-3}} \times (1.1706 \times 10^{10}) ND=(0.33333×1012 m−3)×(1.1706×1010)N_D = (0.33333 \times 10^{12} \, \text{m}^{-3}) \times (1.1706 \times 10^{10}) ND=3.902×1021 m−3N_D = 3.902 \times 10^{21} \, \text{m}^{-3}

    Rounding to two significant figures:

    ND≈3.9×1021 m−3N_D \approx \boxed{\text{3.9} \times 10^{21} \, \text{m}^{-3}}
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