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Chapter 5 Bipolar Junction Transistors. Chapter Goals. 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. - PowerPoint PPT Presentation

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Chapter 5 Bipolar Junction Transistors

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Chapter 5Bipolar Junction Transistors

Chapter Goals

• 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.

Physical Structure

• 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-v characteristics 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 vBC determine 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

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−== 1expTV

BEvS

IFiCi

A910A1810 −≤≤−SI

VT = kT/q =0.025 V at room temperature

Base current is given by

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−== 1expTV

BEv

F

SI

F

Fi

Bi ββ

€

20≤βF

≤ 500 is forward current gain

Emitter current is given by

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=+= 1expTV

BEv

F

SI

BiCiEi α

0.11

95.0 ≤+

=≤F

FF β

βα

In this forward active operation region,

FB

iCi

β= FE

iCi

α=

npn Transistor: Reverse Characteristics

Reverse transport current is

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=−= 1expTV

BCv

SIEiRi

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−== 1expTV

BCv

R

SI

R

Ri

Bi ββ

200 ≤≤ Rβ

Emitter current is given by

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−= 1expTV

BCv

R

SI

Ci

α

95.01

0 ≤+

=≤R

RR β

βα

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

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−−= 1expexpexp

TVBC

v

R

SI

TVBC

v

TVBEv

SI

Ci

β

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−= 1expexpexp

TVBE

v

F

SI

TVBC

v

TVBEv

SIEi β

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−= 1exp1exp

TVBC

v

R

SI

TVBEv

F

SI

Bi ββ

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:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−== 1expTV

EBvS

IFiCi

Base current is given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−== 1expTV

EBv

F

SI

F

Fi

Bi ββ

Emitter current is given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−+=+= 1exp11TV

EBv

FS

IBiCiEi β

pnp Transistor: Reverse Characteristics

Reverse transport current is:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=−= 1expTV

CBv

SIEiRi

Base current is given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−== 1expTV

CBv

R

SI

R

Fi

Bi ββ

Emitter current is given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−+−= 1exp11TV

CBv

RS

ICi

β

pnp Transistor: Complete Transport Model Equations for Any Bias

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−−= 1expexpexp

TVCB

v

R

SI

TVCB

v

TVEBv

SI

Ci

β

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−= 1expexpexp

TVEB

v

F

SI

TVCB

v

TVEBv

SIEi β

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−= 1exp1exp

TVCB

v

R

SI

TVEBv

F

SI

Bi ββ

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:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=−=TV

BCv

TVBEv

SIRiFiTi expexp

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−+−= 1exp1expTV

BCv

R

SI

TVBEv

F

SI

Bi ββ

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)

(Good Amplifier)

Saturation region(Not same as FETsaturation region)(Closed switch)

Reverse Bias Cutoff region(Open switch)

Reverse-active region(Inverse active region)

(Poor amplifier)

i-v Characteristics Bipolar Transistor: Common-Emitter Output Characteristics

For iB=0, the transistor is cutoff. If iB >0, iC also increases.

For vCE > vBE, the npn transistor is in the forward active region, iC = βF 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:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−= 1expTV

BEvS

ICi

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 =>qkT

BEv V1.04 −=−<qkT

BCv

€

iC≅ I

Sexp

vBEV

T

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟+

IS

βR

≅ IS

expvBEV

T

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

iE≅

IS

αF

expvBEV

T

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟+

IS

βF

≅IS

αF

expvBEV

T

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

iB≅

IS

βF

expvBEV

T

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟−

IS

βF

−IS

βR

≅IS

βF

expvBEV

T

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ BiFEi

BiFCi

EiFCi

)1( +=

=

=

ββα

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 βF 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: βF = 50, VBC =VB - VC= -9 V• Assumptions: Forward-active region of operation, VBE = 0.7 V• Analysis:

€

VBE

+8200IE

+(−VEE

)=0

∴IE

= 8.3V8200Ω

=1.01mA

IB

=IE

βF

+1=1.02mA

51=19.8μA

IC

=βF

IB

=0.990mA

VCE

=−(−VEE

)+VCC

−VR

∴VCE

=9+9−8.3=9.7V

Note: 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

21

1RR

R

CCV

EQV

+=

21

21RR

RR

EQR

+=

EIERBEVBIEQ

REQ

V ++=

€

4 =12,000IB

+0.7+16,000(βF

+1)IB

∴IB

= 4V-0.7V

1.23×106Ω=2.68μA

€

IC

=βF

IB

=201μA

€

IE

=(βF

+1)IB

=204μA

€

VCE

=VCC

−RC

IC

−RE

IE

=VCC

− RC

+R

Eα

F

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

IC

= 4.32V

Q-point is (4.32 V, 201 A)

€

βF = 75

BE Loop

CE Loop

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

+R

Fα

F

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

IC

=12−38,200IC

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 choosing

This ensures that power dissipation in base bias resistors is < 17 % of the total quiescent power consumed by the circuit, while I2 >> IB.

€

IB

=V

EQ−V

BER

EQ+(β

F+1)R

E

≅V

EQ−V

BE(β

F+1)R

E

€

REQ

<<(βF

+1)REfor

5/2 C

II ≤

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 =αFRC

• If REQ<<(βF+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<<(βF+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

2VCC

⎛

⎝ ⎜

⎞

⎠ ⎟5VCC

IC

⎛

⎝ ⎜

⎞

⎠ ⎟= 5

VCC −VCE + 2VBE

2IC

⎛

⎝ ⎜

⎞

⎠ ⎟

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: βF = 50, VCC = 9 V• Assumptions: Forward-active region operation with VEB = 0.7 V• Analysis:

€

9=VEB

+18,000IB

+1000(IC

+ IB

)

∴9=VEB

+18,000IB

+1000(51)IB

∴IB

= 9V−0.7V69,000Ω

=120μA

IC

=50IB

=6.01mA

VEC

=9−1000(IC

+ IB

)=2.87V

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 RC on 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 A

Forward current gain = 100

Reverse current gain = 0.5

Forward Early voltage = 75 V

Base resistance = 250 Collector Resistance = 50 Emitter Resistance = 1 Forward transit time = 0.15 ns

Reverse 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:

where

NAB = the doping concentration in the base

ni2 = the intrinsic carrier concentration (1010/cm3)

nbo = ni2 / NAB

Dn = the diffusivity = (kT/q)n

• Saturation current for the PNP transistor is:

• Due to the higher mobility () of electrons compared to holes, the npn transistor conducts higher current than the pnp for equivalent doping and applied voltages.

BW

ABN

innqAD

BW

bon

nqADS

I2

==

BW

DBN

inpqAD

BW

bop

pqADS

I2

==

Diffusion Capacitance

• 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,

FT

VT

I

TVBEvB

Wbo

qAn

TV

poQBEdv

dQDC τ==

−

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

exp2

1

int

FT

VC

I

DC τ=

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

expvBEV

T

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥

1+vCEV

A

⎡

⎣

⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥

=βF

IB

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+=A

VCE

v

FOF 1ββ⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

TVBE

v

FO

SI

Bi expβ

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, βFO, and VA,.

211 BI

BI

CI

RBE

VBB

V

REFI ++=−

=

BJT Current Mirror (continued)

With an infinite βFO 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.

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛++

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

TVBE

V

FO

SI

AVCE

V

TVBE

V

SIREFI exp211exp

β

FOAVBE

VA

VCE

V

REFI

AVCE

V

TVBE

V

SI

CI

β21

2121exp2

++

+

=+⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=∴

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

€

MR=IO

IREF

=

1+V

CE2V

A

1+V

BEV

A

+ 2β

FO

is the "Mirror Ratio".

BJT Current Mirror: Example

• Problem: Find output current for given current mirror

• Given data: βFO = 75, VA = 50 V

• Assumptions: Forward-active operation region, VBE = 0.7 V

• Analysis:

€

IREF

=V

BB−V

BER

=12V−0.7V56kΩ

=202μA

IO

= MR×IREF

=(202μA)1+12

751+ 0.7

75+ 2

50

=223μA

VBE = 6.7333e-01IC2 = 5.3317e-04IC21 = 5.3317e-04

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.

AE

A

SOI

SI = 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.

FOAVBE

VA

VCE

V

REFInO

I

β21

21

.++

+

=1

2

EAE

An=

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≡∂io∂vo Q− pt

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟

−1

=IC2

VA

+VCE

⎡

⎣

⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥

−1

≅V

AIO

€

iO

= iC2

= IREF

1+V

CE2+v

ce2V

A

1+V

BEV

A

+ 2β

FO

= IREF

1+V

CE+vo

VA

1+V

BEV

A

+ 2β

FO

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