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Chap 3 Diodes Ideal Diode Terminal Characteristics Semiconductors Physical Operation Open Ckt pn junction Reverse bias pn junction Forward bias pn junction Diode Circuits Small signal model Rectifiers Clipping and clamping The SPICE diode C. Hutchens Chap 3 ECEN 3313 Handouts 1 Simple Diode I-V Representation DC or Capacitance free Diode I-V models. At higher frequencies one most include Cs C reverse CDep or CJ “Junction” C forward CD C. Hutchens Chap 3 ECEN 3313 Handouts 2 Diode Symbols and Ckt Representation Ideal Math model I D I s ( V / nVt 1) where n is the emission coifficient and = 1 for Si, Is is the saturation current, and Vt is the thermal voltage and = 26mV at room temp For V> 4 n Vt 100mV VZ or Avalanche I D I s V / nVt For V>> 4 n Vt short ckt with a voltage drop of 0.5 to 1.0V and for V< 4Vt 100mV ID Is 0 C. Hutchens Chap 3 ECEN 3313 Handouts 3 Diode SPICE parameters .MODEL D1N914/125C D + IS = 2.25833E-15 + RS = 1.799439 + N = 1.067043 + TT = 1.46E-7 + CJO = 3.0177E-12 + VJ = 0.4 + M = 0.2147523 + EG = 1.11 + XTI = 4.799594 + KF = 0 + AF = 1 + FC = 0.764906 + BV = 94 + IBV = 5E-6 +) C. Hutchens Chap 3 ECEN 3313 Handouts ( Saturation Current Contact Resistance Ideality factor Fwd Transit time Zero Bias Junction C Built in potential doping grading Coeff. Breakdown Voltage 4 Simple Diode Rectifer EX 2 Find the fraction of each cycle the diode conducts and the peak current if vin = 24 COS 2 f t . Note Typically f will be in the 10 to 1000s of Hertz for this application. Assume vin = 24Cos 2 f t and V = 12 V, and R =100 ohms. 24 Cos = 12 Cos =1/2 Cos-1 1/2 = 600 Fraction = 2 x 60/360 = 33% Ipeak = (24-12)/100 Other items you must consisder. If Rs = 10 ohms, and the resistance and all diode parameters vary by +/-20% what is the worst case peak current, peak and continious power dissipationa? What if vin varies by 20%. C. Hutchens Chap 3 ECEN 3313 Handouts 5 Limiters/Clippers Observations VO > V1 VO <V2 V2 < VO < V1 Vo > V1 D1 is a short circuit else o.c. Vo < V2 D1 is a short circuit else o.c. The polarity and magnitude of V 1 and V2 shift the conduction point. Ex vin = 20V peak to peak Sinusoid, V1 = 10 V and V2 = -8 V C. Hutchens Chap 3 ECEN 3313 Handouts 6 LOG COMPUTATION AND THE DIODE I D I s ( V / nVt 1) where n is the emission = 1 Both IS and VT are temperature dependent Valid for as many as 7+ Decades V / nV 1) and I D 2 I S 2 ( V / nV 1) Consider I D1 I S 2 ( t t I D1 / I D2 (V2 V1 ) / nVT V2 V1 V nVT Ln(I D1/ I D2 ) 2.3nVT Log 10 (I D1/ I D2 ) = 60mV /Decade at Rm Temp Every Decade Increase in current results in a 60mV increase in diode voltage. Typical Values of Diode on voltage are 100 to 800mV. C. Hutchens Chap 3 ECEN 3313 Handouts 7 COMPUTATION AND THE DIODE Now ID is a function of diode area or A JS where JS is the current density Rewriting the previous equation V 2.3nVT Log 10 (J S A D1 / J S A D2 ) 2.3nVT Log 10 (A D1 / A D2 ) If two diodes are fabricated side by side and care the same current but are of unequal areas the voltage difference will be a function of the Log of the ratio their areas. See EX 3.3 For a single diode the FORWARD voltage shift is approx. -2mV/C0 Use Temperature measurement Reverse bias current -IS "Area dependent Doubles every 50C." C. Hutchens Chap 3 ECEN 3313 Handouts 8 Basic Semiconductor Concepts-Review Conductivity is controlled by carrier concentration Intrinsic carrier concentration ni = p =n = 1.5x1010 carriers/cm3 at Room Temp. ni BT 3 EG / kT where B is a material dependent para 8.62x10-5eV/K, EG is the bandgap and for Si 1.12 eV, T is Temp. K0 equal 2730 at room temp. Si has 5x1022 atms/cm3 vs. ni = 1.5x1010 C. Hutchens Chap 3 ECEN 3313 Handouts 9 Basic Semiconductor Concepts Physical factors controlling pn junction behavior. Diffusion - Thermal agitation coupled with conc. (carriers/cm3) Drift - E field associated carrier motion. n-type equilibrium carrier conc. nn0 ND i.e. P atms/cm3 free electons p-type equilibrium carrier conc. pp0 NA i.e. B atms/cm3 free holes nn0 pp0 = ni2 then C. Hutchens Chap 3 ECEN 3313 Handouts pp0 = ni2/ND and nn0 = ni2/NA 10 Doping Example Given ND = 1017/cm3 of P find nn0 pp0 at 300 K0. ni BT 3 E / kT where B is a material dependent para 8.62x10-5eV/K, EG is the bandgap and for Si 1.12 eV, T is Temp. K0 equal 2730 at room temp. G ni 2 = (1.5x 1010 /cm3)2 nn0 ND = 1017/cm3 pp0 = ni2/ND =2.25 x 103 C. Hutchens Chap 3 ECEN 3313 Handouts 11 Diffusion Cathode Anode x Consider bars of Si with the following hole and electron concentrations Cp = -mp x and Cn = mn x "Linear" the current (A/cm^2) can be written as Jp = q Dp dp/dx = -qDp mp for pos. current in the x dir. Jn = q Dn dn/dx = -q Dn mn for electrons where D is the diffusion constant for holes ( 12cm2/s ) and electrons ( 34cm2/s) and q is the charge on an electron of hole. C. Hutchens Chap 3 ECEN 3313 Handouts 12 Drift vdrift = E where is the mobility of a charge and E is the electric field Then considering a Si crystal with a charge density qp coulomb/cm3 Jp-drift = q pp E hole drift current Jn-drift = q pn E electron drift current (neg charge in neg x dir. results in positive current) Jtotal drift = q E(pp + pn) or using ohms Law and R= L/A = 1/( q (pp + pn)) Note V/I = R = L/A ; = (V/L)(A/I) = E/J x C. Hutchens Chap 3 ECEN 3313 Handouts 13 Open Ckt pn Junction Vo = VJ, n = N Drift (IS) and Diffusion (ID) Currents are in Equilbrium (IS = ID) for an o.c. diode and for charge neutrality qx p AN A qxn AN D x p N A xn N D where xp and xn are the respective depletion depths in the n and p type materials respectively. x n x p Wdep 2 Si q 1 / N A 1 / N D Vo defines the depletion width. C. Hutchens Chap 3 ECEN 3313 Handouts 14 Open Ckt pn Junction The pn work function given by; N N Vo Vt Ln A D ni Ex Given NA = 1017 , ND = 1016 , T = 300K0 and using ni = 1.5x1010/cm3 and Si = 11.70 Fd/cm (0 = 1.04x10-12). Find Vo, W det and its componets. Vo = 738mV W det = 0.32um xp = 0.03um xn = 0.29um C. Hutchens Chap 3 ECEN 3313 Handouts 15 Rev. pn Junction-Depletion Cap. Note C = q/RVR = VQ or Cj = A/t = D/W dep Cj Cj0 1 VR / Vo where 2 N A N D C j 0 A Si qVo N A N D No Applied Reverse Voltage C. Hutchens Chap 3 ECEN 3313 Handouts Estimating Cj - This is done by using Cj0/2 to Cj0 i.e. VD = is 0 to 0.5V Cj = Cj0 VD = is 0 to 5V Cj = Cj0/2 Engineers to be, must be thinking and make judgement calls Leave the details to the simulator 16 Reverse Bias pn Junction Applied Reverse Voltage Depletion Capacitance Junction charges can be expressed as a function of either depletion region as: qJ qn qx n AN D and using x p N A xn N D and qJ qn qxn A xn x p Wdep NDN A ND N A where xp and xn are the respective depletion depths in the n and p type materials respectively WHERE VR IS THE REVERSE VOLTAGE CONTROLING W dep, Wdep 2 Si q 1 / N A 1 / N D Vo VR . C. Hutchens Chap 3 ECEN 3313 Handouts 17 Fwd. pn Junction on the Small Signal Forward bias small signal transconductance or resistance is defined by the slope of the line at Qpt or the derivative of the I vs. V characteristics evaluated at the operating point. gd I 1 I V I I rd nVt Q C. Hutchens Chap 3 ECEN 3313 Handouts The idea of a QPT is "Very Critical" ! 18 Fwd. pn Junction-Diffusion Cap. CD = T I/Vt = TT I/Vt diffusion cap EX IF a Diode is forward biased by a ID = 0.1mA current find the small signal ac model for a the diode the on voltage (VDQ) across the diode. Assume a Si diode with n=N=1, TT= 27.5nS, CJ0 = 8.3pFd and Is= IS= 2fA rd = nVt /I = 25mV/0.1mA = 250 ohms CD = (27.5nS) (0.1mA)/25mV 110pFd Ctotal = CD + CJ 110pFd + 8.3pFd = 118pFd Note we could just as well have ignored CJ VD = n Vt Ln (ID/IS) = 25mV Ln (0.1mA/2fA) = 0.640V C. Hutchens Chap 3 ECEN 3313 Handouts 19 Small Signal vs Large Signal Typically RS= RS should be ignoreable until VDQ exceeds 700 to 800mV. The Observations IDQ = (V-VDQ)/R, VDQ = n Vt Ln (IDQ/IS) Transindental functions criteria is rd << RS require iterations. Cblock is selected Large enough at time to appear as a short circuit in the frequecy band if intrest. Vbias DC voltage to set IDQ apperas as a short circuit. Depending on the value of R vs rd it may be possible to ignored. Node equation across diode vd vd (G + gd + sCj + s Cd + (vd - vin ) sCblock= 0 vd/vin = sCblock/(G+ gd + s Ctotal) (Cblock/Ctotal){1/( 1 + gd/sCtotal)} C. Hutchens Chap 3 ECEN 3313 Handouts 20 DC vs AC (Small signal) Observations: C is selected large enough to appear as a short circuit in the frequency band of the ac generator. To get started quickly V is selected equal 0.6V and should be checked at times with VD = n Vt Ln (ID/IS) = 25mV Ln (4mA/2fA) = 0.710V. Note that rd can be ignored for all intent and purposes. What is more important is that I D not destroy the diode or its reverse voltage not be exceeced. C. Hutchens Chap 3 ECEN 3313 Handouts 21 Zener Regulation Example - Assume VZ = 5.1, Vsmax = 15V, Vsmin = 12V Assume Izmin = Izmax /10 Izmax = ILMAX (Vsmax - VZ)/ (VSMIN - 0.9 VZ -0.1 VSMAX) Note PDISS of R is very high at all times. Therefore zener regulators are very impractical. C. Hutchens Chap 3 ECEN 3313 Handouts 22 Full Wave Rectifer I = Cdv/dt & f = /2 = 1/T Given Ckt diagram and Vp = 169 V for 120Vrms VZ = 5V, IZ min = 5mA Analysis of operation Vxpeak = 169/14 12V @ RL = Ixmax = (Vxpeak -VZ)/R1 = (12-5)/220 = 31.8mA Sets zener power dissapation. (This should be increased by 10% for saftey purposes) Vxmin = (IL+ Izmin)R1 + VZ = (5/3.8K + 5mA )220 + 5 = 5.29 V Minimum allowable voltage for zener operation. Note proper selection of C insures that Vxmin will not be exceeded. C> Imax dt/dv from I = Cdv/dt where Imax = IL+ Izmin 6.25mA where worst case dv/dt = (Vxpeak - Vxmin)2f = (12-5.29)120 C > 6.25mA/{(12-5.29)(120)} 7.8uFd C. Hutchens Chap 3 ECEN 3313 Handouts 23 Full Wave Rectifer Observations: The diode conducts briefly for a small fraction of the period. See figure. Assuming an ideal Diode; conductions occurs during the rising edge of the Vripple signal. The capcitor supplies current during the falling portion of Vripple. (IC IL) Only the capacitor supplies load current for the balance of the period T = 1/2f i.e. 1/(2 x 60) = 83mS When properly designed Vripple is small compared to Vp or Vo. i.e. 5% IL Vp/RL or more accurately IL (Vp - Vripple/2)/RL IC CdVr/dt or after rearranging dVr = IC C dt and integration. Vr (IC /C) T IC /(C2f) (Vp/RL )/(C 2f) C. Hutchens Chap 3 ECEN 3313 Handouts 24 Voltage Doubler/Clamped Cap Neg 1/2 cycle D1 conducts Pos 1/2 cycle D2 conducts How much loss occures due to the diodes? How do we determine Pdiss require of the diodes? How many additional diodes may be added to increase the voltage further? Is ripple on the capacitor a problem and if so can we solve it? C. Hutchens Chap 3 ECEN 3313 Handouts 25 Diode Switching C. Hutchens Chap 3 ECEN 3313 Handouts 26 Diode SPICE parameters .MODEL D1N914/125C D + IS = 2.25833E-15 + RS = 1.799439 + N = 1.067043 + TT = 1.46E-7 + CJO = 3.0177E-12 + VJ = 0.4 + M = 0.2147523 + EG = 1.11 + XTI = 4.799594 + KF = 0 + AF = 1 + FC = 0.764906 + BV = 94 + IBV = 5E-6 +) C. Hutchens Chap 3 ECEN 3313 Handouts ( Saturation Current Contact Resistance Ideality factor Fwd Transit time Zero Bias Junction C Built in potential doping grading Coeff. Breakdown Voltage 27 Diode Summary Fwd Rev Ideal DC or Large Signal i.e. fullwave rectifier Zener Small signal Small signal dynamic C. Hutchens Chap 3 ECEN 3313 Handouts 28