LC 3908, Physical lectronics, Lecture 17 Bipolar Transistor njection Models
Lecture Outline Last lecture looked at qualitative operation of the BJT, now want to develop a quantitative model to predict terminal currents as functions of terminal voltages and physical parameters (as for the diode) The injection model developed in this lecture applies to dc operation small signal model will be considered in a later lecture - switching behavior is outside scope of 3908 Development is by region in the transistor, and basic form is the same as for the two regions of the diode (lecture 8) base region is the biggest added complexity Most important insight is the relationship between terms in the equations and physical components of current flow Page 17-2
Bipolar njection Models The most basic method for modelling current transport in the bipolar transistor is an injection model, which uses the currents injected across each junction to determine the terminal currents Not the most accurate method, but useful at low current densities and easy to relate terms to current components Page 17-3
Forward Active Current Components Consider forward active region for derivation - equations apply to all operating conditions because based on the general gradients and general boundary conditions at depletion region edges Goal is to determine the current components shown in terms of injection across each pn-junction Page 17-4
Review of Diffusion Current Conduction Recall that conduction by diffusion is modelled in terms of a diffusion coefficient and a carrier gradient n, diff p, diff qad dn ( x ) = n dx qad dp ( x ) = p dx Page 17-5
Collector Neutral Region njection n the neutral collector, p C (x) is p Co at the contact and p Co e qvbc/ at the depletion edge The gradient of hole density is therefore Using qvbc dpc( x) pcoe p = dx The corresponding current term is therefore pc p, diff C qad dp ( x ) = p dx Co qad pcpco = 1 C qvbc ( e ) Page 17-6
mitter Neutral Region njection n the neutral emitter, p (x) is p o at the contact and p o e qvbc/ at the depletion edge The gradient of hole density is therefore dp( x) po poe = dx qv B Using p, diff qad dp ( x ) = p dx The corresponding current term is therefore p qad ppo = 1 qvb ( e ) Page 17-7
Base Neutral Region njection n the neutral base, n B (x) is n Bo e qvbc/ on the C side and n Bo e qvb/ on the side The gradient of electron density is therefore qvb dnb( x) nboe nboe = dx Using n, diff The corresponding current term is therefore nb B qv qad dn ( x ) = n dx qadnbnbo = B BC qvb qvbc ( e e ) Page 17-8
njection Components in the BJT Structure Have now defined the basic injection components flowing through the BJT structure Page 17-9
Positive Current Conventions The current terms so far carry a sign indicating direction relative to the +ve x axis For circuit modelling applications, want to choose a set of positive directions relative to the device, not the x axis Page 17-10
Terminal Currents and njection Components Once the positive convention is chosen, there is a correspondence between the injection components and the terminal currents C = pc + nb B = pc + p = nb + p Page 17-11
Final Collector Current xpression Substituting for the components found earlier (note that in forward active V BC < 0, giving a positive collector hole term as shown) C qad pcpco qa D n = + C 14444 244443 144444 B 2444443 collector hole injection ( qvbc nb Bo qvb qvbc ( e 1) ( e e ) pc ) linking (electron) current ( ) nb Page 17-12
Final mitter Current xpression Substituting for the components found earlier = qa D n qvb qvbc qad ppo qvb ( ) + ( 1) 1444442444443 14444244443 nb Bo e e e B linking (electron) current ( ) emitter hole injection ( ) nb p Page 17-13
Final Base Current xpression Substituting for the components found earlier (note in forward active operation V BC < 0 gives a negative collector term as shown) = qa D p + qa D p 14444244443 14444244443 qvbc qvb ( 1) ( 1) pc Co p o B e e C collector hole injection ( ) pc emitter hole injection ( ) p Page 17-14
Basic Bipolar njection Model These three equations together comprise the basic bipolar injection model, applicable to any operating region qa D n qa D p nb Bo = + B 14444424444431 4444 244443 qv qv p o B BC qvb ( e e ) ( e 1) linking (electron) current ( ) emitter hole injection ( ) nb p B = qa D pc p Co qvbc qa D p po qvb ( e 1) + ( e 1) C 144 44 24444314444 244443 collector hole injection ( pc ) emitter hole injection ( p ) C qad pcpco qa D n = + C 14444 244443 144444 B 2444443 collector hole injection ( qvbc nb Bo qvb qvbc ( e 1) ( e e ) pc ) linking (electron) current ( ) nb Page 17-15
xample 17.1: njection Model Calculation Calculate the emitter, base and collector currents for the device structure and bias condition shown below. The emitter is 1 μm wide by 10 μm long. (The neutral widths were calculated in an earlier example) Page 17-16
xample 17.1: Solution The emitter width is found from A = b l = 1 10 10 10 = 10 10 4 4 8 cm 2 The currents are found by substituting the necessary values, with the results C B = 212. 4 10 43 + 21. 28 42 10 43 2. 28 10 collector hole injection 17 4 4 linking current = 212. 4 1034 + 118 1. 42 1034 118. 10 collector hole injection 17 6 6 emitter hole injection 4 6-4 = 21. 28 42 10 43+ 118 1. 42 1034 2.29 10 A linking current emitter hole injection Note that C and are much larger than B, and that is slightly larger than B, as expected A A Page 17-17
xample 17.1: Solution (con t) The values for and B can be compared to the currents calculated in the example of Lecture 16, where it was found that for the base-emitter junction, J n 3 = 228. 10 A / cm J = 112. A / cm 2 2 p Since is primarily determined by the linking current and B is primarily determined by the back injection current, the previous results can be compared by multiplying by the emitter area J A 3 8 4 n 8 6 p = 112. 10 10 = 112. 10 A B J A = 228. 10 10 10 = 228. 10 A Page 17-18
quation Simplifications - Forward Active n the forward active region, for V B > 3/q, the terms involving the V B exponential dominate the expressions The full injection model equations can therefore be simplified to qa D n qa D p nb Bo = + B 14444424444431 4444 244443 qv qv p o B BC qvb ( e e ) ( e 1) linking (electron) current ( ) emitter hole injection ( ) qa D n qa D nb Bo + B nb qv p o B qvb ( e ) ( e ) p p B qadpcpco qa D p = + C 144442444431 4444 244443 qa D p qv p o BC qvb ( e 1) ( e 1) collector hole injection ( ) emitter hole injection ( ) p o p qvb ( e ) p Page 17-19
quation Simplifications - Forward Active (con t) C qad pcpco qa D n = + C 14444 244443 144444 B 2444443 qa D n collector hole injection ( nb Bo B qvbc nb Bo qvb qvbc ( e 1) ( e e ) pc qvb ( e ) ) linking (electron) current ( ) nb Note these approximations drop the (small) collector hole injection Page 17-20
Forward Active Current Gain Using the simplified current expressions, the forward active current gain for the bipolar can be expressed in terms of physical parameters as β F ( ) qa D n e qvb C nb Bo B = = qvb B fwd active qa D p e ( ) p o D N D N nb D p AB B Note that the current gain (in this simple model) is directly proportional to the ratio of emitter to base dopings Page 17-21
xample 17.2: Forward Current Gain Calculate the forward active current gain β F for the device of xample 17.1. Page 17-22
xample 17.2: Solution Using the dopings given and the previously calculated values of diffusion coefficients and neutral widths, D N = = D N β F nb D p AB B 19 4 34. 9 10 10 12. 2 10 14. 10 17 4 = 204 This agrees to within roundoff error with the ratio of the actual calculated currents from xample 17.1 C B = 228. 10 112. 10 4 6 = 204 Page 17-23
quation Simplifications - Reverse Active n the reverse active region, for V BC > 3/q, the terms involving the V BC exponential dominate the expressions B = qa D nb n Bo B 144444 2 linking (electron) current qv qa D p p qv o qv qa DnB n B BC B Bo qv BC ( e e ) + ( e 1) ( e ) 444443 ( nb ) 14444 2 emitter hole injection 44443 qadpcpco qa D p = + C 144442444431 4444 244443 ( qv p o qv pc Co BC B qvbc ( e 1) ( e 1) ( e ) collector hole injection ( ) emitter hole injection ( ) p p p ) qa D C B p C qad pcpco qa D n = + C 14444 244443 144444 B 2444443 collector hole injection ( qa D p pc Co C qvbc nb Bo qvb qvbc ( e 1) ( e e ) pc ) linking (electron) current ( ) qa D n qvbc nb Bo qvbc ( e ) ( e ) B nb Page 17-24
quation Simplifications - Reverse Active These simplifications neglect the component of reverse injection from the emitter (acting as a collector) Page 17-25
Reverse Active Current Gain Using the simplified reverse active current expressions, the reverse active current gain can be expressed in terms of physical parameters as ( ) qa D n e β = = qvbc nb Bo B DnBNDCC R qvbc B rev active ( qad pcpco C) e DpCNABB The reverse current gain (again in this simple model) is directly proportional to the ratio of collector to base dopings Since a usual (epi, e.g.) integrated structure has a lower doping in the collector than the base, βr is much lower Page 17-26
xample 17.3: Reverse Current Gain Calculate the reverse active current gain β R for the device of xample 17.1. Page 17-27
xample 17.3: Solution Using the dopings given and the previously calculated values of diffusion coefficients and neutral widths, D N = = D N β R nb DC C pc AB B 34. 9 5 10 34. 10 17 4 12. 2 10 14. 10 15 4 = 035. Page 17-28
quation Simplifications - Saturation n saturation operation, both junctions are forward biased so only the 1 terms can be neglected (for Vs > 3/q) C qadnbnbo + B B qa D qv qv p o B BC qvb ( e e ) ( e ) qadpcpco + C qa D qv p o BC qvb ( e ) ( e ) qadpcpco qa D n + C These equations suggest that all three current components are (potentially) important - note the possibility of the linking current term going to zero qvbc nb Bo qvb qvbc ( e ) ( e e ) B p p Page 17-29
quation Simplifications - Cutoff n cutoff operation, both junctions are reverse biased, so all potential dependencies are lost (Vs < -3/q so exp(qv/) 0) qa D n qa D p qa D nb Bo qv qv p o B BC qvb = ( e e ) + ( e 1) B 14444424444431 44442 44443 linking (electron) current ( ) emitter hole injection ( ) nb p p p o B qadpcpco qa D p qa D p qv p o qv pc Co BC B = ( e 1) + ( e 1) C 144442444431 44442 44443 C collector hole injection ( ) emitter hole injection ( ) p p qa D p p o C qad pcpco qv qa D n BC nb Bo qvb qvbc = ( e 1) + ( e e ) C 14444 244443 144444 B 2444443 collector hole injection ( pc ) linking (electron) current ( ) qa D The remaining terms are composed of the saturation currents across each junction nb p pc Co C Page 17-30
α F The bers-moll Model f the following (algebraic) definitions are made S qa D n qa D p qa D n qa D nb Bo p o nb Bo + CS + B qadnbn Bo B qa D n + qa D p nb Bo B p o The injection equations can be written α R B qadnbn Bo B qa D n + qa D p p pc Co nb Bo B pc Co C qvb qvbc S( 1) α R CS( 1) qvb qvbc ( 1 α ) ( 1) ( 1 αr) CS( 1) qvb qvbc α ( 1) ( 1) = e e = e + e B F S = e e C F S CS C Page 17-31
The bers-moll Model (con t) f the following further definitions are made qvb qvbc ( 1) ( 1) e e F S The bipolar injection model can then be represented using two diodes and two controlled current sources - this is useful for implementation in a circuit simulator since it uses existing circuit elements R CS Page 17-32
xample 17.4: bers-moll Coefficients Calculate the bers-moll coefficients for the device in xamples 17.1, and calculate the terminal currents using the bers-moll model. Page 17-33
xample 17.4: Solution The bers-moll model parameters are found using the expressions given and the necessary physical values S 18 17 = 8. 42 10 A CS = 325. 10 A α = 0. 995 α = 0. 258 F The currents R and F are then given by ( qvb ) ( qvbc ) = e 1 = 2. 25 10 F S = e 1 = 325. 10 R CS R 4 A 17 A Page 17-34
xample 17.4: Solution (con t) Mathematically or from the equivalent circuit, the terminal currents are determined by the model terms involving F, since F >> R ( α ) = 225. 10 F α = 0. 995 2. 25 10 = 2. 24 10 C F F 1 = 0. 005 2. 25 10 = 113. 10 B F F These values agree to within roundoff error with the previous results 4 A 4 4 A 4 4 A Page 17-35
Lecture Summary This lecture derived an injection model for the BJT in terms of terminal voltages and physical parameters quations look intimidating, but are actually only composed of three unique terms, each with a very clear correspondence to a physical current component For each region of operation, simplification is possible based on sign of terminal voltage component, and once again remaining terms correspond to important current components in that region of operation The bers-moll model is just a way of rewriting the injection equations to allow implementation as pairs of diodes and controlled current sources for circuit simulation Page 17-36