Diode Diode Chapter 1 2 A diode is a single p-n junction device with conductive contacts wire leads co each region. The n region is called cathode, p region is called anode. The a Diode packages Diode - mode of operation AMN in direction of conventional (opposite to electron flow) symbol points AMN 1 Above figures show how forward-biased reverse-biased are connected. Notic 3 4 positive terminal negative terminal connected to anode cathode. When t forward biased, it acts like a closed (on) for reverse biased it acts like an
as figures a) b) below. ure c) shows ideal -I characteristic curve graphically ideal operation. -biased reverse-biased are connected. Notice about nal connected to anode cathode. When diode is 2. 2.Ideal Idealdiode diodecharacteristics. characteristics. 2. Ideal diode 2. Ideal diode characteristics. biascharacteristics. bias on) for reverse biased it acts like an open (off) gure c) shows ideal -I characteristic curve graphically!!!! bias Bias bias Bias biasbias BiasBias (+)(+) (-)(-) (-)(-) (+)(+) (+) (+) (-) (-) (-) (-) (+) (+) I I F IFF IF ON OFF ON OFF ON ON OFFOFF Infinite A@to@K!! byby by by!!by!!!! A@to@K! 5 AMN Ex: Assuming diodes to be ideal, find output in following circuits. 6 ure 3.3 (a) ectifier circuit. (b) Input waveform. (c) circuit when vi! 0. (d) circuit when vi to toto to!to!!! Ex: Assuming diodes to be ideal, find fraction of each cycle during which diode conducts peak value of diode. vs 24 The conduction angle θ: 24cosθ = 12 The peak value of diode: Id = (24-12)/0 = 0.12A
Constant-oltage-Drop model Constant-oltage-Drop Model Ex: Assuming diodes to be ideal, find values of I in following circuits. ) < > 0 9 Constant-oltage-Drop model Constant-oltage-Drop Model Example Example Ex: Find output vo in Model following circuits using Constant-oltage-Drop constant--drop modelv with =. D 5 Constant-oltage-Drop model Ex: Calculate load in following figure using constant--drop model, assuming D =. vo vi vvoo = vo vi ) = 0 vo vo tt vi vi 11 vivi The Thevenin : TH = 12, TH = 2kΩ The load : IL = (12-)/3k = 3.mA The load : L = 3.mA x 1kΩ = 3. 12
Piecewise-linear model IS : reverse saturation. D : forward biased. n : emission coefficient (1 n 2 depending on material). T : rmal. kt T = 25m @ room temp q B Piecewise-linear model k = Boltzmann s constant = 1.3 x -23 joules/kevin T = absolute temperature in kelvins = 23 + temperature in oc q = magnitude of electronic charge = 1.60 x - coulomb circuit representation 13 14 DC Load Load Line DC Line I (ma) For i >> IS: IS exp or I nt S 1.5 1 25 Ω! D1 Diode characteristic 40 30 ID 20 F B Operating point Q 0 60S 501 DC load line Slope = -1/1 0.4 S S F 1 Load line line equation: Load eq.: I I== 1 = 0. D 1.2 1.5 S F () 1.5 F 25Ω 49 15 16
Small-Signal Model Small-Signal Model v d (t) D i D(total) v D(total) v D(total) = D + v d (t) i D(total) = I D + i d (t) D + v d (t) i D(total) = I S exp vd (t) = I D exp I D + I D v d (t) D = I S exp I D 1+ v d (t) vd (t) exp,v d (t) di D dv d = 1 r d = I D ) r d = I D 1 1 Small Signal Analysis Step 1: Eliminate AC sources, find DC solution (ID, D) of circuit (use constant--drop model, unless specified orwise). Step 2: Calculate small signal parameters (rd). Step 3: Eliminate DC sources, replace diode with its small-signal equivalent model find small-signal solution. erify assumptions (vd nt). Step 4: The total solution is (DC solution + AC solution). Small Signal Analysis 1k v S v S = 5 + 0.2 sin(!t), Find v O (t). n = 2 v O DC Solution: AC Solution: v o (t) = O = I D = 5 1k r d = I D = 50 m 4.3 ma = 11.6 = 4.3 ma 11.6 3 + 11.6 0.2 sin(!t) = 2.3 3 sin(!t) Total: v O (t) = O +v o (t) = + 2.3 3 sin(!t) 20