5 Chapter Bipolar Junction Transistors
Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock
Chap 5 - 1
Chapter Goals
Explore physical structure of bipolar transistor Understand bipolar transistor action and importance of carrier transport across base region Study terminal characteristics of BJT. Explore differences between npn and pnp transistors. Develop Transport model for the bipolar device. Define four operation regions of BJT. Explore model simplifications for each operation region. Understand origin and modeling of Early effect. Present SPICE model for bipolar transistor. Provide examples of worst-case and Monte Carlo analysis of bias circuits.
Chap 5 - 2
Bipolar Transistor: Physical Structure
Consists of 3 alternating layers of n- and p-type semiconductor called emitter (E), base (B) and collector (C). Majority of current enters collector, crosses base region and exits through emitter. A small current also enters base terminal, crosses base-emitter junction and exits through emitter. Carrier transport in the active base region directly beneath the heavily doped (n+) emitter dominates i-v characteristics of BJT.
Chap 5 - 3
Transport Model for the npn Transistor
Base-emitter voltage vBE and base-collector voltage vBC determine currents in transistor and are said to be positive when they forward-bias their respective pn junctions. The terminal currents are collector current(iC ), base current (iB) and emitter current (iE). Primary difference between BJT and FET is that iB is significant while iG = 0.
Chap 5 - 4
Narrow width of the base region causes coupling between the two back-to-back pn junctions. Emitter injects electrons into base region, almost all of them travel across narrow base and are removed by collector
npn Transistor: Forward Characteristics
Base current is given by
F F
iF I S vBE iB = = exp 1 VT
20 500 is forward common-emitter F
current gain Emitter current is given by
F
Forward transport current is
v iC =F = exp iI V
S BE T
I S vBE iE =C +B = ii exp 1 VT
1
0.95 F = F 1.0 is forward common+ 1 base current gain F
In this forward active operation region,
iC iB = F iE iC =F
Chap 5 - 5
IS is saturation current
18 10 A IS 10 9 A
VT = kT/q =0.025 V at room temperature
npn Transistor: Reverse Characteristics
current gain Base currents in forward and reverse modes are different due to asymmetric doping levels in emitter and collector regions. Emitter current is given by Reverse transport current is
v iR = iE = exp I V
S BC T
0 R is reverse common-emitter 20
1
I S vBC iC = exp 1 VT
R
Base current is given by
R R
0 R =
iR I S vBC iB = = exp 1 VT
R 0.95 + 1 R
is reverse common-base current gain
Chap 5 - 6
npn Transistor: Complete Transport Model Equations for Any Bias
v v v BC 1 iC =I exp exp exp V V V T v I v v iE =IS exp BE exp BC + exp BE 1 V V V T T T v IS IS vBE 1 + exp BC 1 iB = exp F VT R VT
S BE T BC T
IS
R S F
First term in both emitter and collector current expressions gives current transported completely across base region. Symmetry exists between base-emitter and base-collector voltages in establishing dominant current in bipolar transistor.
Chap 5 - 7
Transport Model Calculations: Example
Evaluating the expressions for terminal currents,
IC =1.07mA I E =1.09mA
Problem: Find terminal voltages and currents. Given data: VBB = 0.75 V, VCC = 5.0 V, IS = 10-16 A, F = 50, R = 1 Assumptions: Room temperature operation, VT = 25.0 mV. Analysis: VBE = 0.75 V, VBC = VBB - VCC = 0.75 V-5.00V = -4.25 V
IB =21.4A
1.07mA = =50 I B 0.0214mA I = C =1.07mA =0.982 F I E 1.09mA
= F
IC
Chap 5 - 8
pnp Transistor: Structure
Voltages vEB and vCB are positive when they forward bias their respective pn junctions. Collector current and base current exit transistor terminals and emitter current enters the device.
Chap 5 - 9
pnp Transistor: Forward Characteristics
Base current is given by
iF I S vEB iB = = exp 1 VT
F F
Emitter current is given by Forward transport current is
iC =F = exp iI
S EB T
v V
1
v iE =C +B = 1+ iiI exp V 1
S
F
EB T
1
Chap 5 - 10
pnp Transistor: Reverse Characteristics
Base current is given by
iF I S vCB iB = = exp 1 VT
R R
Emitter current is given by Reverse transport current is
iR = iE = exp I
S CB T
v V
1
v iC = I 1+ exp V 1
S
R
CB T
1
Chap 5 - 11
pnp Transistor: Complete Transport Model Equations for Any Bias
v I v v CB 1 iC =I exp exp exp V V V T v I vEB expvCB EB 1 iE =IS exp exp V V VT T T v I v IS iB = exp EB 1+ S exp CB 1 F VT R VT
S EB T CB T S R S F
First term in both emitter and collector current expressions gives current transported completely across base region. Symmetry exists between base-emitter and base-collector voltages in establishing dominant current in bipolar transistor.
Chap 5 - 12
Circuit Representations for the Transport Models
In npn transistor (expressions analogous for pnp transistors), total current traversing base is modeled by a current source given by:
v iT = = exp iF iR I V
S BE T
exp
BC T
v V
Diode currents correspond directly to the two components of base current.
I S vBE I S vBC iB = exp 1 + exp 1 VT VT
F R
Chap 5 - 13
Operation Regions of Bipolar Transistors
Base-Emitter Junction Base-Collector Junction
Reverse Bias
Forward Bias
Forward Bias
Forward-Active Saturation Region Region (Closed Switch) (Good Amplifier) Cutoff Region (Open Switch) Reverse-Active Region (Poor Amplifier)
Binary Logic States
Reverse Bias
Chap 5 - 14
i-v Characteristics of Bipolar Transistor: Common-Emitter Output Characteristics
For iB = 0, transistor is cutoff. If iB > 0, iC also increases. For vCE > vBE, npn transistor is in forward-active region, iC = F iB is independent of vCE. For vCE < vBE, transistor is in saturation. For vCE < 0, roles of collector and emitter reverse.
Chap 5 - 15
i-v Characteristics of Bipolar Transistor: Common-Base Output Characteristics
For vCB > 0, npn transistor is in forward-active region, iC = iE is independent of and vCE. For vCB < 0, base-collector diode becomes forward-biased and iC grows exponentially (in negative direction) as base-collector diode begins to conduct.
Chap 5 - 16
i-v Characteristics of Bipolar Transistor: Common-Emitter Transfer Characteristic
Defines relation between collector current and base-emitter voltage of transistor. Almost identical to transfer characteristic of pn junction diode Setting vBC = 0 in the collector-current expression yields
vBE iC = I exp 1 nVT
S
Collector current expression has the same form as that of the diode equation
Chap 5 - 17
Simplified Cutoff Region Model
If we assume that
4kT 4kT vBE < andvBC < q q
where -4kT/q = -0.1 V, then the transport model terminal current equations simplify to IS iC =+ In cutoff region both junctions are reverse-biased, transistor is said to be in off state vBE < 0, vBC < 0
R
IS
iE = iB =
F
IS
F
R
Chap 5 - 18
IS
Simplified Cutoff Region Model (Example)
Problem: Estimate terminal currents using simplified transport model Given data: IS = 10-16 A, F = 0.95, R = 0.25, VBE = 0 V, VBC = -5 V Assumptions: Simplified transport model assumptions Analysis: From given voltages, we know that transistor is in cutoff.
IC =I 1+
S
1
I E =IS =10 A IS IB = = 1016 A 3
R 16
=
R
IS
=4 16 A 10
R
For practical purposes, all three currents are essentially zero.
Chap 5 - 19
Simplified Forward-Active Region Model
In forward-active region, emitter-base junction is forward-biased and collector-base junction is reverse-biased. vBE > 0, vBC < 0 kT kT If we assume that v 4 = 0.1Vandv 4 = 0.1V
BE
then the transport model terminal current equations simplify to
v iC =IS exp V IS
BE T
q
BC
q
+
IS IS v vBE iE = exp += exp V F F F VT IS I I I vBE S S = S expvBE iB = exp V VT T BE T F F R F
R
IS
=IS exp
BE T
v V
iC =F iE iC =F iB iE =(F +1)iB
BJT is often considered a current-controlled device, though fundamental forward-active behavior suggests a voltage- controlled current source.
Chap 5 - 20
Simplified Forward-Active Region Model (Example 1) Problem: Estimate terminal currents and base-emitter voltage
Given data: IS =10-16 A, F = 0.95, VBC = VB - VC = -5 V, IE = 100 A Assumptions: Simplified transport model assumptions, room temperature operation, VT = 25.0 mV Analysis: Current source forward-biases base-emitter diode, VBE > 0, VBC < 0, we know that transistor is in forward-active operation region.
IC =F I E =0.95100A =95A F 0.95 F = = =19 1F 10.95 IE 100A IB = = =5A F +1 20
VBE =VT ln
F I E
IS
=0.69V
Chap 5 - 21
Simplified Forward-Active Region Model (Example 2)
Problem: Estimate terminal currents, base-emitter and base-collector voltages. Given data: IS = 10-16 A, F = 0.95, VC = +5 V, IB = 100 A Assumptions: Simplified transport model assumptions, room temperature operation, VT = 25.0 mV Analysis: Current source causes base current to forward-bias baseemitter diode, VBE > 0, VBC <0, we know that transistor is in forward-active operation region. IC = IB =19 100 =1.90mA A F I E =( + B =20 1)I 100 =2.00mA A F
VBE =VT ln C =0.764V I S VBC =VB VC =VBE VC = 4.24V
Chap 5 - 22
I
Simplified Circuit Model for ForwardActive Region
Current in base-emitter diode is amplified by common-emitter current gain F and appears at collector; base and collector currents are exponentially related to base-emitter voltage. Base-emitter diode is replaced by constant voltage drop model (VBE = 0.7 V) since it is forward-biased in forward-active region. Dc base and emitter voltages differ by 0.7-V diode voltage drop in forwardactive region.
Chap 5 - 23
Simplified Forward-Active Region Model (Example 3)
Problem: Find Q-point Given data: F = 50, R = 1 VBC = VB - VC = -9 V Assumptions: Forward-active region of operation, VBE = 0.7 V Analysis:
VBE +8200I E V EE =0 8.3V E= I =1.01mA 8200 I 1.02mA IB = E = =19.8A F +1 51 IC =F IB =0.990mA VCE =VCC I R ( BE )= V CC 90.99mA(4.3K )+0.7 =5.44V
Chap 5 - 24
Simplified Reverse-Active Region Model
In reverse-active region, basecollector diode is forward-biased and base-emitter diode is reversebiased. Simplified equations are:
iC =
R
IS
exp
BC T
v V
iE =IS exp iB =
R
IS
exp
BC T BC T
v V
v V
iE = iC R iE = RiB
Chap 5 - 25
Simplified Reverse-Active Region Model
Problem: Find Q-point Given data: F = 50, R = 1 VBE = VB - VE = -9 V. Combination of R and the voltage source forward biases base-collector junction. Assumptions: Reverse-active region of operation, VBC = 0.7 V Analysis:
0.7V-(-9V) C = I = 1.01mA 8200 IC 1.01mA IB = = =0.505mA R + 1 2 E =IB =0.505mA I
Chap 5 - 26
Simplified Saturation Region Model
In saturation region, both junctions are forward-biased, and the transistor operates with a small voltage between collector and emitter. vCESAT is the saturation voltage for the npn BJT.
v iC =S exp I V
BE T IS IS v v IS v BC BE BC exp iB = exp + exp VT VT VT R C R B C FB F R
i 1+ 1)i 1 ( + CESAT = T ln v V 1 i R i
foriB F
iC
Simplified >
Model
No simplified expressions exist for terminal currents other than iC + iB = iE.
Chap 5 - 27
Nonideal BJT Behavior: Junction Breakdown Voltages
If reverse voltage across either of the two pn junctions in the transistor is too large, the corresponding diode will break down. The emitter is the most heavily-doped region, and the collector is the most lightly doped region. Due to doping differences, the base-emitter diode has a relatively low breakdown voltage (3 to 10 V). The collector-base diode can be designed to break down at much larger voltages. Transistors must be selected in accordance with possible reverse voltages in circuit.
Chap 5 - 28
Nonideal BJT Behavior: Minority Carrier Transport in the Base Region
BJT current dominated by diffusion of minority carriers (electrons in npn and holes in pnp transistors) across base region. Base current consists of hole injection back into emitter and collector and a small additional current to replenish holes lost to recombination with electrons in base. Minority carrier concentrations at the two ends of the base region are:
v n(0)=bo exp n V
BE T
nbo is equilibrium electron density in the p-type base region.
andn(WB )=bo exp n
BC T
v V
Chap 5 - 29
Minority Carrier Transport in the Base Region (cont.)
For narrow base devices, minority carrier density decreases linearly across the base, and the diffusion current in the base is:
qADn ni 2 IS = qADn = WB N ABWB nbo
NAB = doping concentration in base ni2 = intrinsic carrier concentration (1010/cm3) nbo = ni2 / NAB Saturation current for the pnp transistor is
qAD p ni 2 IS = qADp = WB N DBWB pbo
Due to higher mobility of electrons than holes, the npn transistor conducts higher current than the pnp for a given set of applied voltages.
Chap 5 - 30
Nonideal BJT Behavior: Base Transit Time
Forward transit time is the time constant associated with storing minority-carrier charge Q required to establish career gradient in base region. vBE 1WB Q =qAnbo exp 2 VT qADn vBE 1 iT = nboexp W VT
WB 2 Q WB 2 F = = = I T 2Dn 2VT n
B
Base transit time places upper limit on useful operating frequency of transistor.
Chap 5 - 31
Nonideal BJT Behavior: Diffusion Capacitance
For vBE and hence iC to change, charge stored in base region must also change. Diffusion capacitance in parallel with forward-biased base-emitter diode models the change in charge with vBE.
dQ CD = dv BE
vI 1 qAnboWB BE = T = exp F VT 2 VT V T
Q point
Since transport current normally represents collector current in forward-active region,
I CD = C F VT
Chap 5 - 32
-Cutoff Frequency, Transconductance and Transit Time
Forward-biased diffusion and reverse-biased pn junction capacitances of BJT cause current gain to be frequency-dependent. Unity gain frequency is frequency at which current gain is unity F F f )= ( = 2 2 f = fT / F is -cutoff frequency f f F
1+
fT
1+
f
B
Transconductance is defined by:
gm = dVBE diC
Q point
d = dV
S BE
VBE I I exp =C V VT T Q point
Transit time is given by: =C D F
gm
Chap 5 - 33
Early Effect and Early Voltage
As reverse-bias across collector-base junction increases, width of the collector-base depletion layer increases and width of the base decreases (base-width modulation). In a practical BJT, output characteristics have a positive slope in forwardactive region; collector current is not independent of vCE. Early effect: When output characteristics are extrapolated back to point of zero iC, curves intersect (approximately) at a common point vCE = -VA which lies between 15 V and 150 V. (VA is named the Early voltage) Simplified equations (including Early effect):
v v BE vCE I S BE v iC = exp 1+ CE I = FO 1+ iB = exp F V V V VA T T A FO S
Chap 5 - 34
BJT SPICE Model
Besides capacitances associated with the physical structure, additional components are: diode current iS and substrate capacitance CJS related to the large area pn junction that isolates the collector from the substrate and one transistor from the next. RB is resistance between external base contact and intrinsic base region. Collector current must pass through RC on its way to active region of collectorbase junction. RE models any extrinsic emitter resistance in device.
Chap 5 - 35
BJT SPICE Model Typical Values
Saturation Current IS = 3x10-17 A Forward current gain BF = 100 Reverse current gain BR = 0.5 Forward Early voltage VAF = 75 V Base resistance RB = 250 Collector Resistance RC = 50 Emitter Resistance RE = 1 Forward transit time TT = 0.15 ns Reverse transit time TR = 15 ns
Chap 5 - 36
BJT PARAMETER EXTRACTION
vBE iC = I exp 1 nVT
S
Determination of the Saturation Current IS
A straight-line t of the forward-bias ID (VPN ) characteristics on a log-linear plot for values of VPN > 0.1 V intercepts the y-axis at IS .
Determination of the Ideality Factor n
Select two values of the bias voltage VPN on the straight-line part of the current-voltage characteris-tics, e.g, VP N,1 and VP N,2. Determine the values of the corresponding currents ID(VPN,1) and ID(VP N,2).
Determination of the Ideality Factor n
Determination of the Series Resistance RE+RC
At bias voltages larger than 0.8 to 0.9 V, the eect of the diode series resistance starts aecting the ID (VPN ) characteristics as shown. Select a value of the diode current ID,1 in this bias range, and determine the dierence between the straight-line t and the diode characteristics. This dierence is ID,1RS . Knowing ID,1, determine RS . Repeat at dierent values of the current and calculate an average value of RS .
Forward Beta
Log(IB)
v iB = exp BE nV T IS
F
1
vBE iC = I exp 1 nVT
S
Log(IC) - Log(IB) = Log(F)
High Performance BJTs
Modern BJTs use combination of shallow and deep trench isolation processes to reduce device capacitances and transit times. Devices have polysilicon emitters, narrow bases, or SiGe base regions. SiGe transistors exhibit cutoff frequencies > 100 GHz.
Chap 5 - 43
Biasing for BJT
Goal of biasing is to establish known Q-point which in turn establishes initial operating region of the transistor. For a BJT, the Q-point is represented by (IC, VCE) for an npn transistor or (IC, VEC) for a pnp transistor. The Q-point controls values of diffusion capacitance, transconductance, input and output resistances. In general, during circuit analysis, we use simplified mathematical relationships derived for a specified operation region, and the Early voltage is assumed to be infinite. Two practical biasing circuits used for a BJT are:
Four-Resistor Bias network Two-Resistor Bias network
Chap 5 - 44
Four-Resistor Bias Network for BJT
R1 +R2 R1 +R2 VEQ =REQIB +VBE +RE I E 4 =12,000IB +0.7+ 16,000(F + B 1)I VEQ VBE 4V- 0.7V B = I = =2.68 A 6 R +( + 1)R 1.23 10 VEQ =VCC R1 REQ =R1 R2 = R1 R2
= 75 F
IC = IB =201 A F
EQ
F
E
IE =(F +1)IB =204A VCE =VCC RC IC RE I E
C
RF VCE =VCC R + IC =4.32V F F. A. region correct - Q-point is (201 A, 4.32 V)
Chap 5 - 45
Four-Resistor Bias Network for BJT (cont.)
All calculated currents > 0, VBC = VBE - VCE = 0.7 - 4.32 = - 3.62 V Hence, base-collector junction is reverse-biased, and assumption of forward-active region operation is correct. RF Load-line for the circuit is: VCE = CC RC + C = V I 12 38,200IC
F
The two points needed to plot the load line are (0, 12 V) and (314 A, 0). Resulting load line is plotted on common-emitter output characteristics. IB = 2.7 A, intersection of corresponding characteristic with load line gives Q-point.
Chap 5 - 46
Four-Resistor Bias Network for BJT: Design Objectives
We know that
V EQ BE EQ I B V EQ BE V R V IE = forREQIB < (V EQ BE ) < V RE RE
This implies that IB << I2, so that I1 = I2. So base current doesnt disturb voltage divider action. Thus, Q-point is independent of base current as well as current gain. Also, VEQ is designed to be large enough that small variations in the assumed value of VBE wont affect IE. Current in base voltage divider network is limited by choosing I2 IC/5. This ensures that power dissipation in bias resistors is < 17 % of total quiescent power consumed by circuit and I2 >> IB for > 50.
Chap 5 - 47
Four-Resistor Bias Network for BJT: Design Guidelines
Choose Thvenin equivalent base voltage Select R1 to set I1 = 9IB. Select R2 to set I2 = 10IB.
VEQ R1 = 9I B VCC EQ V R2 = 10I B VEQ BE V RE IC
V CE V RC CC E R IC
Chap 5 - 48
VCC V EQ CC V 4 2
RE is determined by VEQ and desired IC.
RC is determined by desired VCE.
Four-Resistor Bias Network for BJT: Example
Problem: Design 4-resistor bias circuit with given parameters. Given data: IC = 750 A, F = 100, VCC = 15 V, VCE = 5 V Assumptions: Forward-active operation region, VBE = 0.7 V Analysis: Divide (VCC - VCE) equally between RE and RC. Thus, VE = 5 V and VC = 10 V VCC VC I2 =10IB = 75.0A
RC = RE = VE IC =6.67k =6.60k
I1 = 9IB = 67.5A R1 = R2 = 9I B VCC VB 10I B VB = 84.4k =124k
Chap 5 - 49
IE VB =VE +VBE =5.7V I IB = C =7.5A
F
Two-Resistor Bias Network for BJT: Example
Problem: Find Q-point for pnp transistor in 2-resistor bias circuit with given parameters. Given data: F = 50, VCC = 9 V Assumptions: Forward-active operation region, VEB = 0.7 V 9 =V EB +18,000IB +1000(IC +IB ) Analysis:
=VEB +18,000IB +1000(51)IB 9 B= I 9V0.7V =120A 69,000 IC =50IB =6.01mA
VEC =91000(IC +IB )=2.88V VBC =2.18V
Forward-active region operation is correct Q-point is : (6.01 mA, 2.88 V)
Chap 5 - 50
Tolerances - Worst-Case Analysis: Example
Problem: Find worst-case values of IC and VCE. Given data: FO = 75 with 50% tolerance, VA = 50 V, 5 % tolerance on V V VCC , 10% tolerance for each resistor. I = EQ BE I Analysis:
C E
RE
To maximize IC , VEQ should be maximized, RE should be minimized and opposite for minimizing IC. Extremes of RE are: 14.4 k and 17.6 k.
VEQ = CC V
To maximize VEQ, VCC and R1 should be maximized, R2 should be minimized and opposite for minimizing VEQ.
Chap 5 - 51
R1 + 2 R
R1
Tolerances - Worst-Case Analysis: Example (cont.)
Extremes of VEQ are: 4.78 V and 3.31 V. Using these values, extremes for IC are: 283 A and 148 A.
V BE V VCE = CC C IC E I E CC C IC EQ V R R V R RE RE VCE CC C IC EQ + BE V R V V
To maximize VCE , IC and RC should be minimized, and opposite for minimizing VEQ. Extremes of VCE are: 7.06 V (forward-active region) and 0.471 V (saturated, hence calculated values for VCE and IC actually not correct).
Chap 5 - 52
Tolerances - Monte Carlo Analysis
In real circuits, it is unlikely that various components will reach their extremes at the same time, instead they will have some statistical distribution. Hence worst-case analysis over-estimates extremes of circuit behavior. In Monte Carlo analysis, values of each circuit parameter are randomly selected from possible distributions of parameters and used to analyze the circuit. Random parameter sets are generated, and the statistical behavior of circuit is built up from the analysis of many test cases.
Chap 5 - 53
Tolerances - Monte Carlo Analysis: Example
Full results of Monte Carlo analysis of 500 cases of the 4-resistor bias circuit yields mean values of 207 A and 4.06 V for IC and VCE respectively which are close to values originally estimated from nominal circuit elements. Standard deviations are 19.6 A and 0.64 V respectively. The worst-case calculations lie well beyond the extremes of the distributions
Chap 5 - 54