PN Junction PDF

## Unbiased P-N Junction ###### Immediately after formation * A diagram shows a P-N junction with holes 'h' on the left side and electrons 'e' on the right side. ###### Few moments later * A diagram shows a P-N junction with a depletion region. * The depletion region is marked by '+' signs on th...

## Unbiased P-N Junction ###### Immediately after formation * A diagram shows a P-N junction with holes 'h' on the left side and electrons 'e' on the right side. ###### Few moments later * A diagram shows a P-N junction with a depletion region. * The depletion region is marked by '+' signs on the P side and '-' signs on the N side. * An electric field marked 'E' is generated. ## Depletion layer - A diagram shows a P-N junction with a depletion layer. - X1 and X2 mark the depletion region on the P and N sides, respectively. - The depletion layer is marked by '+' signs on the P side and '-' signs on the N side. - An electric field marked 'E' is generated from the N side to the P side. - A graph shows the electric field (E) vs position (x). - The electric field is constant in depletion region. ## Formulas - $N_A$ = Acceptor Dopant concentration - $N_D$ = Donor "" - -$ρ_1$ = Volume charge density of Acceptor dopant in depletion layer - $ρ_2$ = Volume change density of donor dopant in depletion layer - $ρ_1 = N_Ae $ ; $ρ_2 = N_De$ - $ρ_1 Δx_1 = ρ_2 Δx_2$ => $ρ_1x_1 = ρ_2x_2$ - $E_o = ρ_1 Δx_1 + ρ_2 Δx_2 / 2Aε $ - $ = ρ_1x_1 + ρ_2x_2 / 2ε $ - $ = ρ_1x_1 / ε $ = $ ρ_2x_2 / ε $ ## Electric field (E) - A graph shows the electric field (E) vs position (x). - E = ( - $E_o (1 + x/x_1) ; -x_1 ≤ x ≤ 0$ - E = ( - $E_o (1 - x/x_1) ; 0 ≤ x ≤ x_2 $ - Let V = 0 at x = 0 - $V(x_2 ) = -∫ E dx = -∫ ( - E_o (1 - x/x_1) ) dx = E_ox_2 / 2$ - $V(-x_1) = -∫ E dx = -∫ ( E_o (1 + x/x_1) ) dx = E_o(-x_1) / 2$ - $V_B = V(x_2) - V(-x_1) = E_o(x_1 + x_2) / 2$ - $V_B = ρ_1x_1 + ρ_2x_2 / 2ε$ ## Barrier Potential (V_B) - $V_B = ρ_1x_1 + ρ_2x_2 / 2ε$ - $V_B = (ρ_1x_1 + ρ_2x_2) x^2 / 2ε(ρ_1x_1 + ρ_2x_2 )^2$ - $V_B = (ρ_1ρ_2) x^2 / 2ε(ρ_1x_1 + ρ_2x_2)$ - $x = √2εV_B / (ρ_1 + ρ_2)$ - $x = √2εV_B / (N_A + N_B)$ - $ρ_1x_1 = ρ_2x_2$ ; - $x = x_1 + x_2$ - $x_1 = (ρ_2 / (ρ_1 + ρ_2))x$ ; $x_2 = (ρ_1 / (ρ_1 + ρ_2))x$ - $ρ_1 = N_Ae$ ; $ρ_2 = N_De$ - x = width of depletion layer - V_B = Barrier potential ## Current in unbiased P-N junction - A diagram shows a P-N junction with a depletion layer. - An electric field marked 'E' is generated from the N side to the P side. - I diffusion is marked as an arrow pointing right. - I drift is marked as an arrow pointing left. - $I_{diff}$ = $I_{drift}$ = $I_o$ - $I_{int}$ = $I{diff} - I_{drift}$ = 0 - $I_o$ is of the order of a few microamps. ## Bias in p-n Junction ### Forward Biasing - A diagram shows a p-n junction with a battery, '+' on the P side, '-' on the N side. - An arrow pointing right is marked as a diode symbol. - I_diff = $I_o e^{eV/kT}$ - I_drift = $I_o$ - I_int = $I_o (e^{eV/kT} - 1)$ Ideal diode equation - A graph shows the current (I) vs the applied voltage (V). - V_K - Knee voltage or Cut off voltage. - V_R = { 0.7V for Si { 0.3V for Ge ### Reverse Biasing - A diagram shows a p-n junction with a battery, '-' on the P side, '+' on the N side. - An arrow pointing left is marked as a diode symbol. - I_diff = 0 - I_drift ≈ $I_o $ - I_int = I_diff - I_drift - I_int = -$I_o$ - A graph shows the current (I) vs the applied voltage (V). - The current is in the order of a few microamps. ## Diode Approximation ### Ideal diode - A diagram shows a diode with a barrier potential. - No barrier potential. - Zero resistance in forward bias. - ∞ resistance in reverse bias. - A diagram shows two series circuit with an ideal diode in each. - In the forward bias circuit, the ideal diode is replaced by a wire. - In the reverse bias circuit, the ideal diode is removed. ### Non-ideal diode - A diagram shows a diode with cut-in voltage V_c. - Non-ideal diode (first approx) ### Contact potential - In a diode, contact potential is equal to the barrier potential of an unbiased diode. - Barrier potential changes with biasing but contact potential does not. - As forward biasing increases, Barrier potential decreases. - As reverse biasing increases, Barrier potential increases. ## P-N Junction under Bias ### Unbiased - A diagram shows a P-N junction. - A diagram shows a P-N junction with contact potential and barrier potential. - 0.7V is identified on both sides. ### Forward Bias - A diode symbol with a battery is represented. - '+' is on the P side and '-' is on the N side. - A P-N junction with 0.7V potential difference is represented. ### Reverse Bias - A circuit with a battery is represented with '+' on the P side, '-' on the N side. - A p-n junction with 0.7V potential difference is represented. - 1. 0.7V V_B = (0.7 - 0.1 ) V = 0.6 V - No current in the circuit - V_R = 0 - 2. 0.7V V_B = (0.7 - 0.3) V = 0.4 V - No current in the circuit - V_R = 0 - 3. 0.7V V_B = (0.7 - 0.7 ) V = 0 V - No current in the circuit - V_R = 0 - 4. 0.7V V_B = (0.7 - 0.7 ) V = 0 V - I = V_R/ R - V_R = 2 - 0.7 = 1.3 V ## Example Problems ### Example 14 - A diagram shows multiple circuits with diodes. - Which of the diodes are forward biased? - 2, 4, 5. ### Example 15 - Two identical capacitors A and B. - They are charged to the same potential V. - A is connected to a circuit with a diode in forward bias. - B is connected to a circuit with a diode in reverse bias. - Find the charges on the capacitors at t = CR. - A - Current flows Q = CV e^-t/τ - B - No current flows Q = CV ### Example 16 - A circuit with a battery, a p-n junction and a resistor. - "-1V" is on the diode side while "-4V" is on the battery side. - Find the current in the circuit. - R.B. - 0 amp ### Example 22 - A circuit with a battery, two diodes (Ge and Si) and a resistor. - The battery is 12V. - Ge conducts at 0.3V and Si at 0.7V. - Find the potential of terminal Y. - (4) 11.7 V ### Example 25 - A potential barrier of 0.50 V exists across a p-n junction. - The depletion region is 5.0 x 10^-7 m wide. - Find the intensity of the electric field in this region. - E = V/d = 0.50 / 5 x 10^-7 = 10⁶ V/m - An electron with speed 5.0 x 10⁵ m/s approaches the p-n junction from the n-side. - Find it's speed on the p-side. - A diagram shows a P-N junction with an electron moving from the N side to the P side. - v² = v_i² - 2(e/m) d - v² = (5 x 10⁵)² - 2(1.6 x 10^-19/9.1 x 10^-31) 5 x 10^-7 - v = 4.9 x 10⁵ m/s ### Example 28 - A circuit with a battery, two zener diodes and resistors. - The battery is 12V. - The Zener diodes are identical. - When 12V is used as input, find the power dissipated in each diode. - The current is 0.01A - The power dissipated in each diode is 40 mW. ### Example 29 - A circuit with a battery and zener diode, two resistances. - The battery is 12V. - There is a zener diode with V_z - 10V - Find the current through the zener diode. - Assume the zener diode is at breakdown voltage. - I_Z + 2I₁ = I - I_Z = I - 2I₁ = 12 / 500 - 2 ( 12 / 1500) = -14 / 1500 - Since, I_Z is negative, therefore, the Zener cannot be at breakdown mode. - I_Z = 0 ### Photodiode - A photodiode is a reverse biased p-n junction made from a photosensitive semiconductor. - The junction is embedded in plastic. - The upper surface is open to light. - The entire unit is very small. - When no light is falling on the junction, the "dark" current is very small. - When light is falling on the junction, additional electron-hole pairs are generated near the junction. - These minority carriers cross the junction and contribute to the current. - The current increases almost linearly with the incident light flux. - Photodiodes are used in light detection systems, light-operate switches, electronic counters etc. - The p-n photodiodes can operate at frequencies of the order of 1MHZ. ### Avalanche Breakdown - Avalance breakdown occurs in lightly doped junctions. - When reverse bias is made very high, the minority carriers acquire kinetic energy enough to break the covalent bonds near the junction, liberating electron-hole pairs. - These charge carriers are accelerated and produce more electron-hole pairs. - The reverse current increases abruptly. - The reverse bias voltage at which the reverse current increases abruptly is called the breakdown voltage. ### Zener Breakdown - Zener breakdown occurs in heavily doped junctions. - Under a high reverse bias voltage, the p-n junction's depletion region expands leading to a high strength of electric field across the junction. - The electric field breaks the covalent bonds of the semiconductor atoms, liberating a large number of free minority carriers. - The sudden generation of carriers rapidly increases the reverse current. - The reverse bias voltage at which the reverse current increases abruptly is called the Zener breakdown voltage. ### Zener Diode - A diagram shows a Zener diode symbol.