Lesson04:Bipolar Junction Transistors
• Explore the physical structure of bipolar transistor• Study terminal characteristics of BJT.• Explore differences between npn and pnp transistors.• Develop the Transport Model for bipolar devices.• Define four operation regions of the BJT.• Explore model simplifications for the forward active
region.• Understand the origin and modeling of the Early effect.• Present a PSPICE model for the bipolar transistor. Discuss
bipolar current sources and the current mirror.
• The BJT consists of 3 alternating layers of n- and p-type semiconductor called emitter (E), base (B) and collector (C).
• The majority of current enters collector, crosses the base region and exits through the emitter. A small current also enters the base terminal, crosses the base-emitter junction and exits through the emitter.
• Carrier transport in the active base region directly beneath the heavily doped (n+) emitter dominates the i-vcharacteristics of the BJT.
Transport Model for the npn Transistor
• The narrow width of the base region causes a coupling between the two back to back pn junctions.
• The emitter injects electrons into base region; almost all of them travel across narrow base and are removed by collector.
• Base-emitter voltage vBE and base-collector voltage vBCdetermine the currents in the transistor and are said to be positive when they forward-bias their respective pn junctions.
• The terminal currents are the collector current(iC ), the base current (iB) and the emitter current (iE).
• The primary difference between the BJT and the FET is that iB is significant, while iG = 0.
npn Transistor: Forward Characteristics
Forward transport current is
IS is saturation current
VT = kT/q =0.025 V at room temperature
Base current is given by
20 F 500 is forward current gain
Emitter current is given by
In this forward active operation region,
npn Transistor: Reverse Characteristics
Reverse transport current is
Emitter current is given by
is reverse current gain
Base current is given by
Base currents in forward and reverse modes are different due to asymmetric doping levels in the emitter and collector regions.
npn Transistor: Complete Transport Model Equations for Any Bias
The first term in both the emitter and collector current expressions gives the current transported completely across the base region.
Symmetry exists between base-emitter and base-collector voltages in establishing the dominant current in the bipolar transistor.
pnp Transistor: Operation
• The voltages vEB and vCB are positive when they forward bias their respective pn junctions.
• Collector current and base current exit the transistor terminals and emitter current enters the device.
pnp Transistor: Forward Characteristics
Forward transport current is:
Base current is given by:
Emitter current is given by:
pnp Transistor: Reverse Characteristics
Reverse transport current is:
Base current is given by:
Emitter current is given by:
pnp Transistor: Complete Transport Model Equations for Any Bias
Circuit Representation for Transport Models
In the npn transistor (expressions analogous for the pnp transistors), total current traversing the base is modeled by a current source given by:
Diode currents correspond directly to the 2 components of base current.
Operation Regions of the Bipolar Transistor
Base-emitter junction Base-collector junctionReverse Bias Forward Bias
Forward Bias Forward active region(Normal active region)
Saturation region(Not same as FETsaturation region)(Closed switch)
Reverse Bias Cutoff region(Open switch)
Reverse-active region(Inverse active region)
i-v Characteristics Bipolar Transistor: Common-Emitter Output Characteristics
For iB=0, the transistor is cutoff. If iB >0, iCalso increases.
For vCE > vBE, the npn transistor is in the forward active region, iC = F iB is independent of vCE..
For vCE< vBE, the transistor is in saturation.
For vCE< 0, the roles of collector and emitter are reversed.
i-v Characteristics of Bipolar Transistor: Common-Emitter Transfer Characteristic
This characteristic defines the relation between collector current and base-emitter voltage of the transistor.
It is almost identical to the transfer characteristic of a pn junction diode.
Setting vBC =0 in the collector-current expression:
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 these doping differences, the base-emitter diode has a relatively low breakdown voltage (3 to 10 V). The collector-base diode is typically designed to break down at much larger voltages.
• Transistors must therefore be selected in accordance with the possible reverse voltages in circuit.
Simplified Forward-Active Region Model
In the forward-active region, the base-emitter junction is forward-biased and the base-collector junction is reverse-biased. vBE > 0, vBC < 0If we assume that
then the transport model terminal current equations simplify to:V1.04
BEv V1.04 qkT
iC IS expvBEVT
R IS exp
The BJT is often considered a current-controlled current source, although fundamental forward active behavior suggests a voltage-controlled current source.
Simplified Circuit Model for Forward-Active Region
• Current in the base-emitter diode is amplified by the common-emitter current gain F and appears at the collector
• The base and collector currents are exponentially related to the base-emitter voltage.
• The base-emitter diode is often replaced by a constant voltage drop model (VBE = 0.7 V), since it is forward-biased in the forward-active region.
Simplified Forward-Active Region Model (Analysis Example)
• Problem: Find Q-point• Given data: F = 50, VBC =VB - VC= -9 V• Assumptions: Forward-active region of operation, VBE = 0.7 V• Analysis:
VBE 8200 IE (VEE ) 0
IB I E
F 1 1.02 mA
IC F IB 0.990 mAVCE (VEE )VCC VRVCE 9 9 8.3 9.7VNote : VR IE R here.
Biasing for BJT
• The goal of biasing is to establish a known Q-point, which in turn establishes the initial operating region of transistor.
• In BJT circuits, the Q-point is represented by (VCE, IC) for the npn transistor or (VEC, IC) for the pnp transistor.
• In general, during circuit analysis, we use a simplified mathematical relationships derived for the specified operation region of the transistor.
• The practical biasing circuits used with BJTs are:– The Four-Resistor Bias network– The Two-Resistor Bias network
Four-Resistor Bias Network for BJT
2.68A IC FIB201A
IE (F 1)IB204A
VCE VCC RC IC RE IE
VCC RC RE F
IC 4.32 V
Q-point is (4.32 V, 201 A)
Four-Resistor Bias Network for BJT (Check Analysis)
• All calculated currents > 0, VBC = VBE - VCE = 0.7 - 4.32 = - 3.62 V• Hence, the base-collector junction is reverse-biased and the assumption
of forward-active region operation is correct.• The load-line for the circuit is:
VCE VCC RC
IC 12 38 ,200 IC
The two points needed to plot the load line are (0, 12 V) and (314 A, 0). The resulting load line is plotted on the common-emitter output characteristics for IB= 2.7 A.
The intersection of the corresponding characteristic with the load line determines the Q-point.
Four-Resistor Bias Network for BJT: Design Objectives
• From the BE loop analysis, we know that
• This will imply that IB << I2 so that I1 = I2 to good approximation in the base voltage divider. Then the base current doesn’t disturb the voltage divider action, and the Q-point will be approximately independent of base voltage divider current.
• Also, VEQ is designed to be large enough that small variations in the assumed value of VBE won’t have a significant effect on IB.
• Base voltage divider current is limited by choosingThis ensures that power dissipation in base bias resistors is < 17 % of the total quiescent power consumed by the circuit, while I2 >> IB.
EREQ (F 1)REfor
Four-Resistor Bias Network for BJT: Design Guidelines
• Choose I2 = IC/5. This means that (R1+R2) = 5VCC/IC .• Let ICRC =IERE = (VCC - VCE)/2. Then RC = (VCC - VCE)/2IC; RE =FRC• If REQ<<(F+1)RE, then IERE = VEQ - VBE.• Then (VCC - VCE)/2 = VEQ - VBE, or VEQ = (VCC - VCE + VBE)/2.• Since VEQ = VCCR1/(R1 +R2) and (R1+R2) = 5VCC/IC,
• Then R2 = 5VCC/IC - R1.• Check that REQ<<(F+1)RE. If not, adjust bullets 1 and 2 above.• Note: In the LabVIEW bias circuit design VI (NPNBias.vi), bullet 1
is called the “Base Margin” and bullet 2 is called the “C-E V(oltage) Drops”.
R1 VCC VCE 2VBE
5 VCC VCE 2VBE
Problem 5.87 4-R Bias Circuit Design
Two-Resistor Bias Network for BJT: Example
• Problem: Find the Q-point for the pnp transistor in the 2-resistor bias circuit shown below.
• Given data: F = 50, VCC = 9 V• Assumptions: Forward-active region operation with VEB = 0.7 V• Analysis:
9 VEB 18 ,000 IB 1000 (IC IB ) 9 VEB 18 ,000 IB 1000 (51)IB
IB 9V 0.7V69 ,000
IC 50 IB 6.01 mA
VEC 9 1000 (IC IB ) 2.87 V
Q-point is : (6.01 mA, 2.87 V)
PNP Transistor Switch Circuit Design
Emitter Current for PNP Switch Design
BJT PSPICE Model• Besides the capacitances which are associated with the physical structure, additional model components are: diode current iS, capacitance CJS, related to the large area pn junction that isolates the collector from the substrate and one transistor from the next.
• RB is the resistance between external base contact and intrinsic base region.
• Collector current must pass through RCon its way to the active region of the collector-base junction.
• RE models any extrinsic emitter resistance in the device.
BJT PSPICE Model -- Typical Values
Saturation Current = 3 e-17 AForward current gain = 100Reverse current gain = 0.5Forward Early voltage = 75 VBase resistance = 250 Collector Resistance = 50 Emitter Resistance = 1 Forward transit time = 0.15 nsReverse transit time = 15 ns
Minority Carrier Transport in Base Region
• With a narrow base region, minority carrier density decreases linearly across the base, and the Saturation Current (NPN) is:
NAB = the doping concentration in the base ni
2 = the intrinsic carrier concentration (1010/cm3)nbo = ni
2 / NABDn = the diffusivity = (kT/q)n
• Saturation current for the PNP transistor is:
• Due to the higher mobility () of electrons compared to holes, the npntransistor conducts higher current than the pnp for equivalent doping and applied voltages.
• For vBE and hence iC to change, charge stored in the base region must also change.
• Diffusion capacitance in parallel with the forward-biased base-emitter diode produces a good model for the change in charge with vBE.
• Since transport current normally represents collector current in the forward-active region,
Early Effect and Early Voltage• As reverse-bias across the collector-base junction increases, the width of
the collector-base depletion layer increases and the effective width of base decreases. This is called “base-width modulation”.
• In a practical BJT, the output characteristics have a positive slope in the forward-active region, so that collector current is not independent of vCE.
• “Early” effect: When the output characteristics are extrapolated back to where the iC curves intersect at common point, vCE = -VA (Early voltage), which lies between 15 V and 150 V.
• Simplified F.A.R. equations, which include the Early effect, are:
iC IS expv
BJT Current Mirror
• The collector terminal of a BJT in the forward-active region mimics the behavior of a current source.
• Output current is independent of VCC as long as VCC ≥ 0.8 V. This puts the BJT in the forward-active region, since VBC ≤ - 0.1 V.
• Q1 and Q2 are assumed to be a “matched” pair with identical IS, FO, and VA,.
BJT Current Mirror (continued)
With an infinite FO and VA (ideal device), the mirror ratio is unity. Finite current gain and Early voltage introduce a mismatch between the output and reference currents of the mirror.
is the " Mirror Ratio" .
BJT Current Mirror: Example
• Problem: Find output current for given current mirror• Given data: FO = 75, VA = 50 V• Assumptions: Forward-active operation region, VBE = 0.7 V• Analysis:
IREF V BB V BE
R 12 V 0.7V
56 k 202 A
IO MR IREF (202 A)1 12
BJT Current Mirror: Altering the Mirror Ratio
The Mirror Ratio of a BJT current mirror can be changed by simply changing the relative sizes of the emitters in the transistors. For the “ideal” case, the Mirror Ratio is determined only by the ratio of the two emitter areas.
SOISI where ISO is the saturation current of a BJT with one unit of emitter area: AE =1(A). The actual dimensions of A are technology-dependent.
BJT Current Mirror: Output Resistance• A current source using BJTs doesn’t have an output current that is
completely independent of the terminal voltage across it, due to the finite value of Early voltage. The current source seems to have a resistive component in series with it.
• Ro is defined as the “small signal” output resistance of the current mirror.
Ro iovo Q pt
iO iC2 IREF
CE 2 v