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Diodes Lesson #6 Chapter 3 Diodes Lesson #6 Chapter 3 BME 372 Electronics I – 180 J.Schesser Diodes • Typical Diode VI Characteristics – Forward Bias Region i – Reverse Bias Region d – Reverse Breakdown Region + - vd – Forward bias Threshold 5 i d 4 3 2 1 v d 0 -7-5-3-11357 -1 Reverse -2 breakdown Reverse bias Forward bias region -3 region region -4 -5 VI stands for Voltage Current BME 372 Electronics I – 181 J.Schesser Zener Diodes • Operated in the breakdown region. • Used for maintain a constant output voltage BME 372 Electronics I – 182 J.Schesser Load Line Analysis • Let’s see how to use a diode in a circuit. R Load Line 39 i D amps 35 + + 31 iD 27 VSS vD 23 -- -- 19 15 11 7 3 • Use KVL for this circuit -1 -0. -0. -0. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 3 2 1 v D volts Vss = RiD + vD • This equation is plotted on the same graph as the diode VI characteristics. BME 372 Electronics I – 183 J.Schesser Load Line Analysis i D amps Vss = RiD + vD 39 Operating Point Vss =1.6V, R=50kΩ 35 Q-point vD=0; 31 27 Load Line iD= Vss / R 23 Highest Voltage =32amps 19 vD=Vss Highest Current 15 =1.6V; 11 7 iD=0 3 -1 -0. -0. -0. 0 0.10.20.30.40.50.60.70.80.91 1.11.21.31.41.51.61.71.8 3 2 1 BME 372 Electronics I – 184 J.Schesserv D volts Ideal Diode • Basically, a switch – Forward Bias: any current allowed, diode on – Reverse Bias: zero current, diode off 5 – No reverse breakdown region i d 4 3 2 Diode on Diode off 1 v d 0 -7-5-3-11357 -1 -2 -3 -4 -5 BME 372 Electronics I – 185 J.Schesser How do we Analysis a Circuit with an Ideal Diode • For a real diode we use load line (graphical analysis) • For an ideal diode, we use a deductive method: 1. Assume a set of states for the diodes 2. Solve the circuit to find the currents, iD, of diodes assumed to ON and the voltages, vD, of the diodes assume to be OFF 3. Check to see if iD is positive for all diodes assumed to be ON and vD is negative for all diodes assumed to be OFF 4. If this is true, then the solution is complete; otherwise return to step 1 by assuming a different set of states for the diodes. BME 372 Electronics I – 186 J.Schesser Example 4k D1 D2 + + 6k 10V 3V -- -- 1. Assume D1 OFF and D2 ON D 2. Solve the circuit to find the currents, iD, of 4k D1 2 iD2 diodes assumed to ON and the voltages, vD, + vD1 -- of the diodes assume to be OFF + + 6k iD2 = 3/6k = 0.5mA OK POSITIVE 10V 3V vD1 = 10 - 3 = 7 V NOT OK SHOULD BE -- -- NEGATIVE 3. NEED TO START OVER WITH ANOTHER ASSUMPTION BME 372 Electronics I – 187 J.Schesser Example 4k D1 D2 + + 6k 10V 3V -- -- 1. Assume D1 ON and D2 OFF 2. Solve the circuit to find the currents, iD, of D D2 4k 1 diodes assumed to ON and the voltages, v , iD1 D -- vD2 + of the diodes assume to be OFF + + 6k iD1 = 10/10k = 1mA OK POSITIVE 10V 3V v = 3 - 6 = -3 V OK NEGATIVE -- D2 -- 3. SOLUTION FOUND!!!! BME 372 Electronics I – 188 J.Schesser How Do We Use Diodes • Rectifier circuits – Half-wave: only one (positive or negative) side of a waveform is passed – Full-wave: waveform is made single sided • Wave Shaping – Clipping Circuits: waveforms are limited in amplitude – Clamping Circuits: the extreme values of a waveform is clamped to a set value • Logic Circuits – AND and OR gates • Voltage Regulators BME 372 Electronics I – 189 J.Schesser Rectifier Circuits • Half Wave Rectifier 1.5 1 Vin(t) = Vm sin(ωt) 0.5 0 0 5 10 15 20 25 30 -0.5 -1 -1.5 1.5 + + 1 Vin(t) -- RL vo 0.5 -- 0 0 5 10 15 20 25 30 -0.5 -1 -1.5 BME 372 Electronics I – 190 J.Schesser Rectifier Circuits • Half Wave Rectifier 1.5 1 with a smoothing 0.5 0 0 5 10 15 20 25 30 Capacitor -0.5 -1 • Peak Detector -1.5 • Envelop Detector 1.5 1 0.5 0 + + 0 5 10 15 20 25 30 -0.5 Vin(t) -- vo -1 -- -1.5 BME 372 Electronics I – 191 J.Schesser Rectifier Circuits • Full Wave Rectifier 1.5 1 Vin(t) = Vm sin(ωt) 0.5 0 0 5 10 15 20 25 30 -0.5 -1 -1.5 1.5 + + 1 Vin(t) -- vo 0.5 -- 0 0 5 10 15 20 25 30 -0.5 -1 -1.5 R BME 372 Electronics I – L 192 J.Schesser Wave Shaping Circuits • Clipper Circuits 20 10 V (t) = V sin(ωt) in m 0 R 0 5 10 15 20 -10 - + V -20 D 2 + 1 + vo Vin(t) -- + V 20 1 D2 - -- 10 v 0 o 0 5 10 15 20 V D is ON 1 1 -10 vin -20 D1 & D2 are OFF D is ON V BME 372 Electronics I – 193 2 2 J.Schesser Clamping Circuit Clamping Circuit Vin 15 Capacitor 10 Vout • Clamping Circuits 5 0 0 5 10 15 20 25 Vin(t) = Vm sin(ωt) ‐5 ‐10 C ‐15 ‐20 ‐25 Time (seconds) D + Vo + 1 - Vin(t) -- • Note at t = 0, the voltage across the capacitor is zero and when the diode is forward biased, the capacitor charges up to Vm during the time the diode conducts which is when Vin(t) > 0 Vo(t) = vd = Vin(t) – VC • Thereafter the diode remains reverse Note VC charges to VM biased since vd = Vm sin(ωt) – Vm will never be positive and the capacitor Vo(t) = vd = Vm sin(ωt) – Vm can’t discharge and Vo is always < 0 BME 372 Electronics I – 194 J.Schesser Wave Shaping Circuits Clamping Circuit Vin+VI • Clamping Circuits 20 Capacitor 15 Vdiode 10 Vout Vin(t) = Vm sin(ωt) 5 0 0 5 10 15 20 25 ‐5 ‐10 + ‐15 D ‐20 + 1 V ‐25 Vin(t) -- - o ‐30 Time (seconds) V1 + - • The capacitor charges up to Vm + V1 during the time the diode conducts which Vo(t) = Vin(t) – VC is when Vin(t) +V1 > 0 v = V (t) – V +V d in C 1 • Thereafter the diode remains reverse Note VC charges to VM +V1 biased since vd = Vm sin(ωt) – Vm will Vo(t) = Vm sin(ωt) – (Vm+ V1) never be positive the capacitor can’t vd = Vmsin(ωt) – (Vm+ V1)+V1 discharge and Vo is always < -V1 BME 372 Electronics I – = Vm sin(ωt) – Vm 195 J.Schesser Wave Shaping Circuits Clamping Circuit Vdiode 20 Capacitor 15 Vin+VI • Clamping Circuits 10 Vout 5 Vin(t) = Vm sin(ωt) 0 024681012 ‐5 C ‐10 ‐15 + ‐20 D ‐25 + 1 V ‐30 Time (seconds) Vin(t) -- - o V 1 • A resistor is added to the + - circuit to allow for changes in input voltage so that the Vo(t) = Vin(t) – VC capacitor can discharge if Vin drops below V Vo(t) = Vm sin(ωt) – (Vm+ V1) m BME 372 Electronics I – 196 J.Schesser Wave Shaping Circuits • Logic Circuits • AND gate •OR gate + 5V v va a v vb b + + v vc c Vo Vo - - • Output is high when • Output high only when any input is high all inputs are high BME 372 Electronics I – 197 J.Schesser Voltage Regulation • We want to design a circuit such that its output voltage does not fluctuate due to changes in the load or source. • Source Regulation: Change in Output voltage due changes in Source Voltage V Source Regulation load 100% VSS • Load: Change in Output voltage due changes in Load V V Load Regulation noload fullload 100% V fullload BME 372 Electronics I – 198 J.Schesser Zener Diode Regulator Id 5 • Load Line: Vss = -RiD - vD 0 R -20 -15 -10 -5 0 5 - -5 + Vd iD vD VSS -10 -- + -15 • R=1k, VSS=15 -20 – vD = -10.5 • V =20 SS V .5 – v = -11 Source Regulation load 100% 100% 10% D VSS 5 BME 372 Electronics I – 199 J.Schesser Zener Diode Regulator • Load Line Using a Thevenin’s Equivalent: Id 5 V = -R i – v 0 T T D D -20 -15 -10 -5 0 5 • V = 24, R = 1.2 k, R = 6 k -5 ss L Vd R -10 - + -15 iD vD RL VSS Full Load: Load = RkL 6 -20 -- + RL 6 Thevenin's Equivalent: VVTSS24 20 RRL 61.2 R RRL 61.2 k T RkT 1 RRL 61.2 VT 20 - Load Line Intercepts vVDT 20; i D 20 m + RkT 1 iD vD vV11 VT D No Load: Load = RL + V 24 -- Load Line Intercepts vV 24; i SS 20 m DSSDRk1.2 Load: vVD 11.5 .5 Load Regulation: 100% 4.5% 11 BME 372 Electronics I – 200 J.Schesser PN Junctions - Shockley Equation Shockley Equation 15 14 i D amps 13 12 11 i 10 d 9 8 7 6 5 - 4 + v 3 d 2 1 0 -0.3 -0.2 -0.1-1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 v D volts vD nVT iD I S (e 1) where iD and vD are the diode current and voltage, I S is called the reverse bias saturation current, and VT is the called the thermal voltage and is kT V where k is the boltzman constant, 1.38 10-23 Joule/ Kelvin, T q T is the temperature of the junction in degrees Kelvin, and q is the magnitude of electric charge of an electron, 1.6010-19 coulombs n is called the emission coeficient as takes values between 1 and 2 BME 372 Electronics I – 201 J.Schesser Small Signal Equivalent Circuit for a Diode Using the Shockley Equation • We can represent a diode by a resistor if the current and voltage are small signals Shockley Equation 15 14 i D amps 13 Define a resistance near the Q - point : 12 11 10 9 di 8 D 7 ΔiD ΔvD 6 5 Q point dv 4 D Q 3 2 1 1 0 -0.3 -0.2 -0.1-1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 diD v D volts rd dv D Q vd id rd 1 1 Note that I DQis the value of iD v d vD 1 DQ nVT nVT r [I (e 1)] I e VDQ d S S dv nV nVT D Q T Q at the Q - point I S e 1 V 1 DQ 1 I e nVT S I nV nV DQ T T r d nV I where VDQ is the value of vD at the Q - point T DQ BME 372 Electronics I – 202 J.Schesser Basic Semiconductor Electronics • Atomic Structure of Valence-4 elements like Carbon, Silicon, Germanium, etc.
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