Semiconductor Conductivity Math (Kankon)

ece-1109 ECE-1109 Introduction to ECE

1. Semiconductor Conductivity and Resistivity

Formulae Used:

• Conductivity ($\sigma$): $\sigma =(n{\mu }_n+p{\mu }_p)e$
• Resistivity ($\rho$): $\rho =1/\sigma$
• Mass Action Law: $np=n_i^2$
• Current Density ($J$): $J=\sigma E$

Problem 1: Intrinsic Conductivity

• Question: Calculate intrinsic conductivity given mobilities ${\mu }_n=0.38m^2/Vs$, ${\mu }_p=0.18m^2/Vs$, and intrinsic carrier density $n_i=2.5\times 10^{14}/m^3$.
• Process: Since it is intrinsic, $n=p=n_i$. $\sigma =(n{\mu }_n+p{\mu }_p)e=n_i({\mu }_n+{\mu }_p)e$ $\sigma =(2.5\times 10^{14})\times (0.38+0.18)\times (1.6\times 10^{-19})$
• Answer: $\sigma =2.24\times 10^{-5}{\Omega }^{-1}{m}^{-1}$

Problem 2: Doped Resistivity

• Question: A donor impurity is added to Germanium (1 part per $10^8$ atoms). $N_{Ge}=4.41\times 10^{22}/c{m}^3$, ${\mu }_n=3800c{m}^2/Vs$, $n_i=2.5\times 10^{13}/c{m}^3$. Show resistivity drops to $3.7\Omega \text{cm}$.
• Process:
  1. Find Donor concentration ($N_D$): $\frac{4.41\times 10^{22}}{10^8}=4.41\times 10^{14}/c{m}^3$.
  2. Assume $n\approx N_D$ and find $p$: $p=n_i^2/n=(2.5\times 10^{13}{)}^2/(4.41\times 10^{14})=1.41\times 10^{12}/c{m}^3$.
  3. Calculate Resistivity: $\rho =\frac{1}{(n{\mu }_n+p{\mu }_p)e}$.
  4. Since $n\gg p$, approximate $\sigma \approx n{\mu }_{n}e$. $\rho \approx \frac{1}{(4.41\times 10^{14}\times 3800)\times 1.6\times 10^{-19}}$
• Answer: $\rho =3.724\Omega \text{cm}$,

Problem 3: Current Density in Silicon Bar

• Question: Si bar, cross-section $5c{m}^2$, length $0.5cm$, applied field $20V/cm$. Calculate electron/hole current density components and total current. Given: $n_i=1.5\times 10^{10}/c{m}^3$, ${\mu }_n=1400$, ${\mu }_p=460$.
• Process:
  1. Current Density ($J$): $J_n=nq{\mu }_{n}E$ and $J_p=pq{\mu }_{p}E$.
     • $J_n=1.5\times 10^{10}\times 1.6\times 10^{-19}\times 1400\times 20$
     • $J_p=1.5\times 10^{10}\times 1.6\times 10^{-19}\times 460\times 20$
  2. Total Current ($i$): $i=(J_n+J_p)\times A$.
• Answers:
  • $J_n=6.72\times 10^{-5}A/c{m}^2$
  • $J_p=2.21\times 10^{-5}A/c{m}^2$
  • Total Current $i=4.465\times 10^{-4}A$,

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2. PN Junction Parameters

Formulae Used:

• Barrier Potential: $V_b=\frac{kT}{e}\ln (\frac{N_A{N}_D}{n_i^2})$
• Depletion Width: $W_d=\sqrt{\frac{2ϵ}{q}[\frac{1}{N_A}+\frac{1}{N_D}]V_b}$

Problem 1: Calculate Barrier Potential ($V_b$)

• Question: Si junction at 300K. $N_A=10^{16}/c{m}^2$, $N_D=10^{17}/c{m}^2$, $n_i=1.5\times 10^{10}$.
• Process: $V_b=\frac{1.38\times 10^{-23}\times 300}{1.6\times 10^{-19}}\ln (\frac{10^{16}\times 10^{17}}{(1.5\times 10^{10}{)}^2})$
• Answer: $V_b=0.757\text{V}$

Problem 2: Calculate Depletion Width ($W_d$)

• Question: Using the values above and relative permittivity ${ϵ}_{si}=1.04\times 10^{-12}F/cm$.
• Process: $W_d=\sqrt{\frac{2\times 1.04\times 10^{-12}}{1.6\times 10^{-19}}[\frac{1}{10^{16}}+\frac{1}{10^{17}}]\times 0.757}$
• Answer: $W_d=3.29\times 10^{-5}\text{cm}$

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3. Diode Current and Temperature

Formulae Used:

• Diode Equation: $I_D=I_S(e^{\frac{V_D}{\eta V_T}}-1)$
• Temp Effect (Forward): $\Delta V_D=-2.5mV\times \Delta T$
• Temp Effect (Reverse): $I_S$ doubles for every $10^{\circ }C$ rise

Problem 1: Reverse Saturation Current

• Question: Ge diode, Forward Voltage $0.25V$ at $10mA$ current at 300K. Find $I_S$. (Assume $\eta =1$).
• Process: $10\times 10^{-3}=I_S(e^{\frac{11600\times 0.25}{1\times 300}}-1)$
• Answer: $I_S=6.34\times 10^{-7}A$

Problem 2: Temperature Shifts

• Question (Forward Bias): If $V_D=0.7V$ at $25^{\circ }C$, find $V_D$ at $125^{\circ }C$.
  • Process: Change in temp = $100^{\circ }C$. Shift = $100\times 2.5mV=0.25V$. New $V_D=0.7-0.25$.
  • Answer: $0.45V$
• Question (Reverse Bias): If $I_S=10nA$ at $25^{\circ }C$, find $I_S$ at $125^{\circ }C$.
  • Process: $\Delta T=100^{\circ }C$ (10 doublings). $I_S=10nA\times 2^{10}$.
  • Answer: $10.24\mu A$

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4. Diode Circuit Analysis (DC & Resistance)

Problem 1: Load Line & Q-point

• Question: Circuit with $10V$ source, $1k\Omega$ resistor, Si diode ($V_D=0.8V$ assumed for Q-pt calculation). Find Q-point and $V_R$.
• Process:
  1. KVL: $10-10^3{I}_D-V_D=0$.
  2. For Q-point: Intersection of load line ($I_D=10mA$ at $V_D=0$, $V_D=10V$ at $I_D=0$) and characteristic curve.
  3. Calculation: If $V_{DQ}=0.8V$, then $I_{DQ}=\frac{10-0.8}{1k}$.
• Answer: $Q(9.2mA,0.8V)$ and $V_R=9.2V$

Problem 2: DC Resistance Levels ($R_D$)

• Question: Find $R_D$ at (a) $2mA$ ($V_D=0.5V$), (b) $20mA$ ($V_D=0.8V$), (c) $-10V$ ($I_S=1\mu A$).
• Process: $R_D=V_D/I_D$.
• Answers:
  • (a) $250\Omega$
  • (b) $40\Omega$
  • (c) $10M\Omega$

Problem 3: Average AC Resistance ($R_{av}$)

• Question: $V_{min}=0.65V(2mA)$, $V_{max}=0.725V(17mA)$.
• Process: $R_{av}=\frac{\Delta V_d}{\Delta I_d}=\frac{0.725-0.65}{(17-2)\times 10^{-3}}$.
• Answer: $5\Omega$

Problem 4: Series/Parallel Configurations

• Series Example: $20V$ source, Si diode, $20\Omega$ resistor.
  • Process: $I=(20-0.7)/20$.
  • Answer: $I=0.965A$
• Parallel Example: Parallel diodes, $10V$ source, $0.33\Omega$ series resistor.
  • Process: $I_{total}=(10-0.7)/0.33$. Since diodes are identical, $I_{D1}=I_{total}/2$.
  • Answer: $I_1=28.18mA$, $I_{D1}=14.09mA$

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5. Rectifiers

Formulae Used (Full Wave Bridge):

• $V_{avg}=2V_m/\pi$
• Efficiency $\eta =81.1$ (theoretical max)
• $V_{rms}=V_m/\sqrt{2}$

Problem 1: Bridge Rectifier Analysis

• Question: 10:1 Transformer, Input $220V$ (RMS), $50Hz$, $R_L=25\Omega$. Determine DC output voltage, Efficiency, PIV, Output Frequency.
• Process:
  1. Secondary Voltage ($V_s$): $V_p/V_s=10/1\Rightarrow 220/V_s=10\Rightarrow V_s=22V(rms)$.
  2. Peak Voltage ($V_m$): $V_m=\sqrt{2}\times V_s=\sqrt{2}\times 22=31.11V$.
  3. DC Voltage: $V_{dc}=2V_m/\pi =(2\times 31.11)/\pi$.
  4. Efficiency: $\eta =\frac{P_{out}}{P_{in}}=\frac{V_{dc}^2/R}{V_{rms}^2/R}$.
  5. Output Frequency ($f_o$): For full wave, $f_o=2\times f_{in}$.
• Answers:
  • (a) $V_{dc}=19.81V$
  • (b) Efficiency $=81.08$
  • (c) $PIV\ge 31.11V$
  • (d) $f_o=100Hz$,

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6. Zener Diodes

Problem 1: Temperature Coefficient ($T_c$)

• Question: $V_Z=10V$ at $25^{\circ }C$, $T_c=0.072$. New temp $100^{\circ }C$. Find change.
• Process: $\Delta V_Z=\frac{T_c\times V_Z\times (T_1-T_0)}{100}=\frac{0.072\times 10\times 75}{100}$.
• Answer: Change = $0.54V$, New Potential = $10.54V$

Problem 2: Regulator (Fixed $V_i$, Variable $R_L$)

• Question: $V_i=20V,R=222\Omega ,V_Z=10V,P_{ZM}=400mW$. Find Min/Max $R_L$.
• Process:
  1. For $R_{Lmin}$: $V_{th}=V_Z$. $\frac{V_i{R}_L}{R+R_L}=10$. Solve for $R_L$.
  2. For $R_{Lmax}$: Find $I_{Zmax}=P_{ZM}/V_Z$. Then $I_{Lmin}=I_R-I_{Zmax}$. Finally $R_{Lmax}=V_Z/I_{Lmin}$.
• Answers:
  • $R_{Lmin}=222\Omega$
  • $R_{Lmax}=1.98k\Omega$,

Problem 3: Regulator (Variable $V_i$)

• Question: $V_Z=20V,R=220\Omega ,R_L=1.2k\Omega ,I_{Zmax}=60mA$. Find Range of Input Voltage.
• Process:
  1. $V_{min}$: $V_Z=\frac{V_{min}{R}_L}{R+R_L}\Rightarrow V_{min}=\frac{V_Z(R+R_L)}{R_L}$.
  2. $V_{max}$: $I_{Rmax}=I_{Zmax}+I_L$. $V_{max}=I_{Rmax}R+V_Z$.
• Answer: $V_i$ range = $23.67V$ to $36.87V$

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7. BJT Biasing Configurations

Formulae Used:

• $I_C=\beta I_B$
• $V_{CE}=V_{CC}-I_C{R}_C$ (Common Emitter Loop)

Problem 1: Fixed Bias Analysis

• Question: $I_B=40\mu A,\beta =80,V_{CC}=14V,V_{CE}=6V$. Find $I_C,R_C,R_B$.
• Process:
  1. $I_C=80\times 40\mu A=3.2mA$.
  2. $R_C$: $14-(3.2mA)R_C-6=0$.
  3. $R_B$: $14-(40\mu A)R_B-0.7=0$.
• Answers: $I_C=3.2mA$, $R_C=2.5k\Omega$, $R_B=332.5k\Omega$,

Problem 2: Emitter Bias Analysis

• Question: $12V$ Source, $V_{CE}=7.6V$ (Wait, note says $V_C=7.6V,V_E=2.4V$), $\beta =80$. Find $R_E,R_C,R_B$.
• Process:
  1. $V_{CE}=7.6-2.4=5.2V$.
  2. $I_C\approx I_E$. $I_C=3mA$ (given).
  3. $R_E=V_E/I_E=2.4/3.04mA$.
  4. $R_C$: $12-I_C{R}_C-V_C=0$.
  5. $R_B$: $12-I_B{R}_B-V_B=0$ (where $V_B=V_E+0.7$).
• Answers: $R_E=790\Omega$, $R_C=1.47k\Omega$, $R_B=237.33k\Omega$

Problem 3: Collector Feedback Bias

• Question: $\beta =100,V_{CC}=20V,R_C=1k,R_B=100k$. Find Operating Point ($I_C,V_{CE}$).
• Process:
  1. $I_B=\frac{20-0.7}{(101\times 1k)+100k}$.
  2. $I_C=\beta I_B$.
  3. $V_{CE}=20-(I_C+I_B)R_C$.
• Answer: Operating Point $(9.6mA,10.3V)$