Chapter 5 Bipolar Junction Transistors
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Transcript of Chapter 5 Bipolar Junction Transistors
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