Chapter 4 Electronics I - Diode models p n A K Fall 2017 talarico@gonzaga.edu 1
Effect of Temperature on I/V curves Source: Hu Figure 4.21 The IV curves of the silicon PN diode shift to lower voltages with increasing temperature talarico@gonzaga.edu 2
Effect of Temperature on I/V curves For a given we want to understand how the diode voltage varies with temperature The diode s I/V equation contains temperature in two places: V T = KT/q and I S = I S V e T 1 D 2 I S = qan p i + D n L p N d L n N a = qan µ 2 p V T i + µ V n T L p N d L n N a Inside the expression of I S the terms n i and μ also depends on temperature: µ T 3/2 where: n 2 KT i = N C N V e = A C A V T 3 KT e N v and N c are respectively the effective density of energy states in valence band and conduction band N C = A C T 3/2 and N V = A V T 3/2 talarico@gonzaga.edu 3
Effect of Temperature on I/V curves If for simplicity we ignore the weak temperature dependence of the effective mass density of the conduction band electrons (m n* ) and the valence band holes (m p* ) the values of A C and A V are about constant (sources: Pierret p. 51): n 2 i = A C A V T 3 KT e with : A C = 2 m * nk 2π! 2 A V = 2 m * pk 3/2 3/2 6.2 10 15 cm 3 K 3/2 3.52 10 15 cm 3 K 3/2 n 2 i = A C A V T 3 KT e 2π! 2 k = Boltzmann s constant = 8.62 10 5 ev/k = 1.38 10 23 Joule/K ħ= Reduced Planck s constant = 1.055 10 34 Joule-sec m n* = 1.18 m 0 m p* = 0.81 m 0 m 0 = electron rest mass = 9.11 10 31 Kg n i =! A C A V T 3/2 2KT e = B T 3/2 e 2KT talarico@gonzaga.edu 4
Effect of Temperature on I/V curves Therefore : n i = B T 3/2 2KT e with B 5 10 15 cm 3 K 3/2 If for simplicity we ignore the weak temperature dependence of the band gap energy: source: Plummer E G = E G0 4.73 10 4 T 2 where E G0 is the band gap energy at 0K (E G0 1.17 ev) and the temperature T is expressed in K We can finally write I S as follows: T + 636 E G0 3 10 3 T 1.16 3 10 () T I S = qab 2 T 3 kt e κ µ, p T 1.5 kt / q L p N d + κ µ,n T 1.5 kt / q L n N a ς T 2.5 e kt talarico@gonzaga.edu 5
Effect of Temperature on I/V curves First pass: I S = qab 2 T 3 kt e κ µ, p T 1.5 kt / q L p N d + κ µ,nt 1.5 kt / q L n N a ς T 2.5 e kt the temperature dependence caused by the polynomial term is weak compared to the one caused by the exponential term, so we will ignore it I S = ς T 2.5 kt e = ζ kt ζ e Let s get back to our objective: given a fixed value of current we want to find out the voltage variation occurring Δ when there is a temperature variation ΔT occurring TC d dt @ID =const talarico@gonzaga.edu 6
Effect of Temperature on I/V curves Let s assume forward bias and flip the I/V eq.: I S d dt = d dt k q kt q T E G qt V e T I S V = e T V T ln I S kt q ln I D = k I S q ln I D + kt I S q + kt q d dt ln lnζ + E G kt T + kt q constant constant = kt q ln I D I S d ( dt ln I ln S ) d dt E G kt And let s keep in mind a couple of useful math rules: d ( uv) = v du dx + u du dx dx d dx lnu d dx x N ( ) = 1 u du dx ( ) = N x N 1 E G and k are constants take them out of the derivative talarico@gonzaga.edu 7
Effect of Temperature on I/V curves TC d dt T E G / q T Assuming =0.5V and T=300K: 0.5-1.12 TC = 0.5 1.1 = = 2mV -2.1 mv/k / K = -2.1 mv/degree 300 At room temperature, a diode s forward voltage-drop has a thermal coefficient of about 2mV per degree I diode T 1 > T 0 ΔV/ΔT = (1 0)/(T 1 T 0 ) 2mV/degree V diode If we know the diode voltage at some reference temperature T 0 we can estimate the diode voltage at any other temperature T as follows: (T ) (T 0 ) 2mV/degree (T T 0 ) = (T 0 )+TC (T T 0 ) talarico@gonzaga.edu 8
Aside: second pass If we want to consider also the temperature dependence caused by the polynomial term, all we have to do is a little more work taking the derivative: d dt = d dt T + kt q T 2.5V T T kt q ln I D = k I S q ln I D + kt I S q d dt ln I lnζ lnt 2.5 D + E G kt E G / q T d ( dt ln ln I S ) I S = ς T 2.5 kt e d ( dt lnt m ) = 1 d ( T m dt T m ) = m T T m 1 = m m T The term /T results from the temperature dependence on V T. The negative terms results from the temperature dependence of I S, and does not depend on the voltage across the diode Assuming =0.5V and T=300K: -1.12 TC = 0.5 2.5 26 10 3 1.1 300 2.2mV -2.3 mv/degree So for conservative design it is common to assume: TC = d dt 2.5mV -2.5 mv/degree talarico@gonzaga.edu 9
Effect of Temperature on I/V curves In reverse bias we find out that the variation of I S with T is about 14.6% percent/degree: Recalling: d ( dx lnu ) = 1 u du dx d ( dt ln I S ) = 1 di S I S dt ( ) di S dt = I S d ln I S dt d dt ln ς T 2.5 e =I S kt = d dt lnς + lnt 2.5 E G = 2.5 kt T + E G kt = 2.5 2 T + E G q kt 2 q = 2.5 T + E / q G T V T di S dt = I S 2.5 T + E G / q T V T At room temperature: di S dt = I S 2.5 300 + 1.12 = I 300 26 10 3 S 14.6 15.2 100 [A/degree] Since (1.152) 5 2 we conclude that the saturation current approximately doubles for every 5 degrees rise in temperature talarico@gonzaga.edu 10
Effect of Temperature on I/V curves In reverse bias: I S (T ) = I S (T 0 ) 2 (T T 0 )/5 I diode T 1 > T 0 V diode Although still quite small, real diodes exhibit reverse currents that are much larger than I S. A large part of the reverse current is due to leakage effects. These leakage effects are proportional to the junction area A just as I S is. As a rule of thumb I R doubles for every 10 degree rise in temperature I R (T ) = I R (T 0 ) 2 (T T 0 )/10 talarico@gonzaga.edu 11
Different Models d d = g D V γ = 0.7V V γ V γ0 γ rd=1/gd V γ V γ0 VD Figure 5 23 Streetman approximations of junction diode characteristics: (a) the ideal diode; (b) ideal diode with an offset voltage; (c) ideal diode with an offset voltage and a resistance to account for slope in the forward characteristic. talarico@gonzaga.edu 12
Ideal diode with voltage offset source: Razavi Constant voltage model: the constant voltage is called cut-in voltage, turn-on voltage, or threshold voltage and is usually denoted as,on or V γ 1N4148 Forward continuous current: IF = 300 ma This model is based on the observation that a forward-conducting diode has a voltage drop that varies in a relatively narrow range (e.g. 0.6 to 0.8). We ll assume V γ 0.7V Below V Υ the current is very small (less than 1% of the maximum rated value). T A = 25 C for VF=0.7V, IF 3 ma for T > 25! C P D,max = 500 1.68 (T 25) [ mw ] T J = T A + R ΘJA P D talarico@gonzaga.edu 13
Ideal diode with voltage offset source: Neamen The diode current increases a factor of 10 every 60 mv increase in voltage 0.2 0.4 0.6 0.8 I/V characteristic of a theoretical diode with I S =10 14 A = 10 fa Forward-bias part of the characteristics with current plotted on a log scale NOTE: real diodes exhibit reverse currents that are considerably larger than I S (this is mainly due to holes and electrons being generated within the space charge region). A typical value of reverse-bias current maybe 1nA (still small and negligible in most cases) talarico@gonzaga.edu 14
Is the model good enough? Should we worry about the fact that the diode has resistance? r diode = di D d 1 = d I S e dv D V T 1 1 = I S V e T V T 1 = I Se V T I S + I S V T 1 = V T + I S Forward Region: r diode = V T + I S V T V T / is only a few Ω >> I S Example: I S 10 fa => I S exp( /V T ) = 10 14 exp(0.7/26m) 4.9 ma => V T / = 26/4.9 5.3 Ω Reverse Region: r diode = V T + I S [Ω] I S talarico@gonzaga.edu 15
Ideal Diode source: Razavi > 0 =0 a short (R=0) in forward bias < 0 = 0, an open (R= ) in reverse bias In applications that involve voltages much greater than the diode voltage drop (0.6V-0.8V) we may neglect the diode voltage drop altogether. talarico@gonzaga.edu 16
Piecewise linear model This model is useful when there is a small varying signal superimposed to the biasing voltage Let s say we want to forward bias a diode so that it operates at a given value of, we need to find the corresponding (Q = operating point = (, ) = D R = D R R KVL = topological eq. V I S e T constitutive eq. of device This is the eq, of a line ( vs. ) with slope 1/R talarico@gonzaga.edu 17
NOTE: ln x = α 5 Log x ln e = 1 = α 5 Log e = α 5 0.43 α 2.3 Piecewise linear model Example We have D =5V and the diode at 0 =0.7V has a current of 0 =1mA. We want 4.3mA. Let s assume 0.7V (and iterate until we find the right value corresponding to ) R = D 5 0.7 = 4.3m =1KΩ A voltage increase of 60 mv corresponds to a 10 x current increase 1 0 = 2.3V T Log I D1 1 = 0 + 60mV Log I D1 = 0.7+ 60mV Log 4.3m 0.738V 1m 0 0 (iteration 1) 2 = D 1 R = 5 0.738 1K = 4.262mA 2 = 1 + 60mV Log I D2 0.738+ 60mV Log 4.262m 4.3m 0.738V 1 (iteration 2) No further iterations are necessary. The circuit used gives 4.262mA and 0.738V talarico@gonzaga.edu 18
Piecewise linear model aa We have a signal superimposed to the bias voltage : v D = + v d v d = v D Δ v D = + Δ v D V i D I S e T +Δ V = I S e T Δ V = I S e T V e T Δ V = e T V γ0 If the max excursion of the signal is small (that is Δ /V T <<1): Δ V i D = e T 1+ Δ Δ V T <<1 V T e x =1+ x + x2 2! + x3 3! +... for x<<1: e x 1+ x i D + I = g d = i d D Δ = + g d Δ = + Δ V T we find out that the response of the diode is about linear and it make sense to approximate the diode characteristic with the tangent line at Q: 1 r d = g d = Δ Δ = i d v d = di D = Δ dv D @ talarico@gonzaga.edu 19 *
Piecewise linear model Equation of line given two points: y y 0 = m( x x 0 ) m = y y 1 0 = Δy x 1 x 0 Δx i D slope: g d Q=(, ) i D = I S V e T 1 I + g D d ( v D) = + slope=δi/δv V γ 0 ( v D ) v D g d = V T V T = V γ 0 V γ 0 = V T alternatively: y = mx + n i D v D r d V γ 0 r d y(x) 0 = mx + n n = mx V γ0 V γ * How small is a small signal? e x 1+ x The error of the approx. should be 10% Error = e x 1 x 0.1 Solving numerically we se that for x 0.4 the error is 0.09 Δ V T 0.4 Δ 26mV 0.4 10.4mV talarico@gonzaga.edu 20
Diode Circuits Finally let s start building some circuit Applications: Rectifiers Limiting Circuits (a.k.a. Clippers) Level Shifters (a.k.a. Clampers) Detectors Voltage doublers Regulators Switches talarico@gonzaga.edu 21