ECE 1071 Basic Electronics PDF

ECE_1071 BASIC ELECTRONICS Basics of Diode Dr. Balachandra Achar HV 9481360288 Dept. of ECE, MIT Bengaluru 1  Basics of PN junction diode Syllabus PN junction diodes Introduction PN junction un...

ECE_1071 BASIC ELECTRONICS Basics of Diode Dr. Balachandra Achar HV 9481360288 Dept. of ECE, MIT Bengaluru 1  Basics of PN junction diode Syllabus PN junction diodes Introduction PN junction under no-bias, reverse-bias and forward-bias Diode IV characteristics and current equation Ideal and practical equivalent circuits of diode Reverse breakdown mechanism Reference Book: Boylestad and Nashelsky “Electronic Devices & Circuit Theory” 2  PN junction diode Prerequisites (Self-study) Atomic structure of semiconductors Intrinsic and Extrinsic semiconductors PN junction diode Diode is a 2-terminal, single junction semiconductor device made up of silicon or germanium or compound semiconductors such as GaAs. One side of diode is doped with acceptor impurities P-type semiconductor, Anode terminal Other side of diode is doped with donor impurities N-type semiconductor, Cathode terminal 3  PN junction diode under no-bias Bias means application of external voltage across the diode No-bias means zero voltage Leads to zero current through the diode P-type material contains majority holes, minority free-electrons and negative ions (formed by combination of hole and free-electron) N-type material contains majority free-electrons, minority holes and positive ions (formed by loss of one electron) At the junction, free electrons and holes will combine Results in lack of free carriers – called depletion region 4  PN junction diode under reverse bias External voltage is applied such that positive terminal of battery is connected to n-type (cathode) and negative terminal of battery is connected to p-type (anode) Voltage is considered to be negative (< 0) Majority free-electrons in N-type and majority holes in P-type are attracted away rom the junction Results in widening of depletion region Minority free-electrons in P-type and minority holes in N-type cross the junction Results in very small current called “reverse saturation current” Nano-amperes or pico-ampere current flows (negligible) Diode is considered to be open-switch 5  PN junction diode under forward bias External voltage is applied such that positive terminal of battery is connected to P-type (anode) and negative terminal of battery is connected to N-type (cathode) Voltage is considered to be positive (> 0) Majority free-electrons in N-type and majority holes in P-type cross the junction Results in narrowing of depletion region Current flows in the forward direction If forward bias is increased, depletion region disappears, current increases exponentially Electrons and holes can freely cross the junction Results in large current (in milli amperes) 6  Diode current equation Diode current is calculated by using Shockley’s equation, for both forward and reverse bias conditions: 𝐼𝐷 = 𝐼𝑂 (𝑒 (𝑉𝐷/𝜂𝑉𝑇) −1) ID = current through the diode IO = reverse saturation current VD = voltage across the diode (positive for forward bias, negative for reverse bias) η = ideality factor (Assumed 1 for germanium and 2 for silicon) VT = thermal voltage 𝑘𝑇 𝑇 𝑉𝑇 = = 𝑞 11600 T = temperature in kelvin Reverse saturation current doubles for every 10° rise in temperature 𝐼𝑂2 = 𝐼𝑂1 2(𝑇2−𝑇1)/10 IO1 = reverse saturation current at temperature T1 IO2 = reverse saturation current at temperature T2 7  Problem on diode current equation Determine the current through the silicon diode at 27° C if the reverse saturation current is 40 nA and applied bias voltage is (i) 0.5 V, (ii) 0.8 V, (iii) -5 V Sol: Given: T = 27° C = (27 + 273)K = 300 K η = 2 (for silicon) IO = 40 nA = 40⨯10–9 A We have: 𝐼𝐷 = 𝐼𝑂 (𝑒 (𝑉𝐷/𝜂𝑉𝑇) −1) Case (i) VD = 0.5 V 0.5×11600 ( ) 𝐼𝐷 = 40 × 10 −9 × (𝑒 2×300 −1) = 0.000631 A = 0.631 mA Case (ii) VD = 0.8 V 0.8×11600 𝐼𝐷 = 40 × 10−9 × (𝑒 ( 2×300 ) −1)= 0.209 A = 209 mA Case (iii) VD = –5 V (reverse bias) −5×11600 𝐼𝐷 = 40 × 10−9 × (𝑒 ( 2×300 ) −1) = 40 × 10−9 A = 40 nA (same as IO) 8  Problem on diode current equation Determine the current through the silicon diode at 47° C if the reverse saturation current is 40 nA at 27° C and applied bias voltage is 0.7 V To find IO at 47°C Given: T1 = 27° C, T2 = 47° C, IO1 = 40⨯10–9 A We have: 𝐼𝑂2 = 𝐼𝑂1 2(𝑇2−𝑇1)/10 𝐼𝑂2 = 40 × 10−9 × 2(47−27)/10 = 160⨯10–9 A = 160 nA To find the diode current: Given: η =2, T = 47+273 = 320, IO = 160 nA, VD = 0.7 V We have: 𝐼𝐷 = 𝐼𝑂 (𝑒 (𝑉𝐷/𝜂𝑉𝑇) −1) 0.7×11600 𝐼𝐷 = 160 × 10−9 × (𝑒 ( 2×320 ) −1) = 0.0518 A = 51.8 mA 9  Problem on diode current equation A germanium diode passes 80 mA current when it is forward biased by 0.5 V at 25°C temperature. Determine the reverse saturation current at the given temperature. 10  Problem on diode current equation A germanium diode passes 6 mA current at 27°C when it is forward biased by some voltage. Determine the forward bias if the reverse saturation current is 1 nA. 11  I-V characteristics of diodes Voltage on x-axis and Current on y-axis (current versus voltage) For small voltages, current is very small. After crossing the knee voltage (VK), current increases exponentially VK for Ge diode is 0.3 V VK for Si diode is 0.7 V VK for GaAs diode is 1.2 V Reverse saturation current in Si and GaAs diodes is very small (pico amperes). In germanium diode, it is more (micro or nano amperes.) Magnitude of reverse breakdown voltage is normally lesser for Ge diode and higher for Si and GaAs diodes. 12  Reverse Breakdown in diode As the reverse bias voltage across diode is increased, at certain voltage the current increases drastically. This voltage is called “reverse breakdown voltage” Two phenomena are explained Avalanche breakdown Occurs in lightly doped diodes (wide depletion region) at reverse voltages greater than 5 V. As voltage is increased, velocity and kinetic energy of minority free-electrons increase When the free-electrons collide with other atoms, they transfer their energy and release the valence electrons from bonds and become free-electrons These additional free-electrons in-turn travel at high velocity and release more valence electrons from bonds Process get multiplied and large reverse current flows 13  Reverse Breakdown in diode Zener breakdown This occurs in heavily doped diodes (narrow depletion region) at reverse voltages lesser than 5 V. As the reverse voltage is increased, electric field in the region of junction increases Strong electric field disrupts the bonding forces within the atom and generate free-electrons and holes This results in large reverse current The maximum reverse bias voltage that can be applied to the diode before it enters the breakdown region (either Avalanche breakdown or Zener breakdown), is called “peak inverse voltage” (PIV). Breakdown mechanism is used in voltage regulator application. Such diodes are called Zener diodes. 14  Diode resistance DC or Static resistance It is simply the ratio of voltage across the diode and current flowing through the diode at the operating point 𝑅𝐷 = 𝑉𝐷ൗ𝐼𝐷 For bias voltages greater than knee-voltage, static resistance will be low For bias voltages lesser than knee-voltage, static resistance will be high Under reverse-bias, static resistance will be very high For example, At Q1, RD = 0.8/(20⨯10–3) = 40 ohms At Q2, RD = 0.5/(2⨯10–3) = 250 ohms At Q3, RD = -10/(-1⨯10–6) = 10 mega ohms 15  Diode resistance AC or Dynamic resistance It is defined as the ratio of change in the diode-voltage to the corresponding change in the diode-current, where the changes are very small 𝑟𝑑 = 𝑑𝑉𝐷ൗ𝑑𝐼𝐷 16  Problem on diode resistances A silicon diode having IO = 5 nA is forward biased by 0.7 V at 27° C. Determine the diode current, static resistance and dynamic resistance. 17  Diode equivalent circuits Ideal equivalent circuit An ideal diode when forward biased Acts like a closed switch with zero resistance and zero knee voltage An ideal diode when reverse biased Acts like an open switch with infinite resistance Simplified equivalent circuit When forward biased, it appears like a closed switch in series with a battery of 0.7 V (for Si) or 0.3 V (for Ge), with no resistance When reverse biased, it acts like an open switch with infinite resistance 18  Diode equivalent circuits Piecewise Linear equivalent circuit Forward IV characteristic is considered as two straight lines – Horizontal line from 0 to 0.7 V Slant line starting from 0.7 V having slope Average resistance of diode is 1/slope When forward biased, diode is considered as closed switch in series with a battery of 0.7 V and a resistor whose resistance = 1/slope of IV characteristics When reverse biased, diode is considered as open switch with infinite resistance 19