BJT Bipolar Junction Transistor

pnp transistor

Large current
Operation of npn transistor

Modes of operation of a BJT transistor
Mode BE junction BC junction
cutoff reverse biased reverse biased
linear(active) forward biased reverse biased
saturation forward biased forward biased

Summary of npn transistor behavior

Summary of pnp transistor behavior

Summary of equations for a BJT
IE ≈ IC
IC = βIB
β is the current gain of the transistor ≈ 100
VBE = 0.7V(npn)
VBE = -0.7V(pnp)

Graphical representation of transistor characteristics
IB
IC
IE
Output
circuit
Input
circuit

Input characteristics
Acts as a diode
VBE ≈0.7V
IB
IB
VBE
0.7V

Output characteristics
IC
IC
VCE
IB = 10μA
IB = 20μA
IB = 30μA
IB = 40μA
Cutoff
region
At a fixed IB, IC is not dependent on VCE
Slope of output characteristics in linear region is near 0 (scale exaggerated)
Early voltage

Biasing a transistor
We must operate the transistor in the linear region.
A transistor’s operating point (Q-point) is defined by
IC, VCE, and IB.

A small ac signal vbe is superimposed on the DC voltage VBE. It gives rise to a collector signal current ic, superimposed on the dc current IC.
Transconductance
ac input signal
(DC input signal 0.7V)
ac output signal
DC output signal
IB
The slope of the ic - vBE curve at the bias point Q is the transconductance gm: the amount of ac current produced by an ac voltage.

Analysis of transistor circuits at DC
For all circuits: assume transistor operates in linear region
write B-E voltage loop
write C-E voltage loop
Example -1
B-E junction acts like a diode
VE = VB - VBE = 4V - 0.7V = 3.3V
IE
IC
IE = (VE - 0)/RE = 3.3/3.3K = 1mA
IC ≈ IE = 1mA
VC = 10 - ICRC = 10 - 1(4.7) = 5.3V

Example-2
B-E Voltage loop
5 = IBRB + VBE, solve for IB
IB = (5 - VBE)/RB = (5-.7)/100k = 0.043mA
IC = βIB = (100)0.043mA = 4.3mA
VC = 10 - ICRC = 10 - 4.3(2) = 1.4V
IE
IC
IB
β = 100

Exercise-3
VE = 0 - .7 = -0.7V
IE
IC
IB
β = 50
IE = (VE - -10)/RE = (-.7 +10)/10K = 0.93mA
IC ≈ IE = 0.93mA
IB = IB/β = .93mΑ/50 = 18.6μΑ
VC = 10 - ICRC = 10 - .93(5) = 5.35V

Prob.
Use a voltage divider, RB1 and RB2 to bias VB to avoid two power supplies.
Make the current in the voltage divider about 10 times IB to simplify the analysis. Use VB = 3V and I = 0.2mA.
IB
I
(a) RB1 and RB2 form a voltage divider.
Assume I >> IB I = VCC/(RB1 + RB2)
.2mA = 9 /(RB1 + RB2)
AND
VB = VCC[RB2/(RB1 + RB2)]
3 = 9 [RB2/(RB1 + RB2)], Solve for RB1 and RB2.
RB1 = 30KΩ, and RB2 = 15KΩ.

Prob.
Find the operating point
Use the Thevenin equivalent circuit for the base
Makes the circuit simpler
VBB = VB = 3V
RBB is measured with voltage sources grounded
RBB = RB1|| RB2 = 30KΩ || 15KΩ = . 10KΩ

Prob.
Write B-E loop and C-E loop
B-E
loop
C-E loop
B-E loop
VBB = IBRBB + VBE +IERE
IE =2.09 mA
C-E loop
VCC = ICRC + VCE +IERE
VCE =4.8 V
This is how all DC circuits are analyzed
and designed!

Exercise-4
(a) Find VC, VB, and VE, given: β = 100, VA = 100V
IE = 1 mA
IB ≈ IE/β = 0.01mA
VB = 0 - IB10K = -0.1V
VE = VB - VBE = -0.1 - 0.7 = -0.8V
VC = 10V - IC8K = 10 - 1(8) = 2V
VB

Example-5
2-stage amplifier, 1st stage has an npn transistor; 2nd stage has an pnp transistor.
IC = βIB
IC ≈ IE
VBE = 0.7(npn) = -0.7(pnp)
β = 100
Find IC1, IC2, VCE1, VCE2
Use Thevenin circuits.

Example -5
RBB1 = RB1||RB2 = 33K
VBB1 = VCC[RB2/(RB1+RB2)]
VBB1 = 15[50K/150K] = 5V
Stage 1
B-E loop
VBB1 = IB1RBB1 + VBE +IE1RE1
Use IB1 ≈ IE1/ β
5 = IE133K / 100 + .7 + IE13K
IE1 = 1.3mA
IB1
IE1

Example -5
C-E loop
neglect IB2 because it is IB2 << IC1
IE1
IC1
VCC = IC1RC1 + VCE1 +IE1RE1
15 = 1.3(5) + VCE1 +1.3(3)
VCE1= 4.87V

Example-5
Stage 2
B-E loop
IB2
IE2
VCC = IE2RE2 + VEB +IB2RBB2 + VBB2
15 = IE2(2K) + .7 +IB2 (5K) + 4.87 + 1.3(3)
Use IB2 ≈ IE2/ β, solve for IE2
IE2 = 2.8mA

Example-5
Stage 2
C-E loop
IE2
IC2
VCC = IE2RE2 + VEC2 +IC2RC2
15 = 2.8(2) + VEC2 + 2.8 (2.7)
solve for VEC2
VCE2 = 1.84V

Summary of DC problem
Bias transistors so that they operate in the linear region B-E junction forward biased, C-E junction reversed biased
Use VBE = 0.7 (npn), IC ≈ IE, IC = βIB
Represent base portion of circuit by the Thevenin circuit
Write B-E, and C-E voltage loops.
For analysis, solve for IC, and VCE.
For design, solve for resistor values (IC and VCE specified).

Summary of npn transistor behavior
npn
collector
emitter
base
IB
IE
IC
small
current
large current
+
VBE
-

Transistor as an amplifier
Transistor circuits are analyzed and designed in terms of DC and ac versions of the same circuit.
An ac signal is usually superimposed on the DC circuit.
The location of the operating point (values of IC and VCE) of the transistor affects the ac operation of the circuit.
There are at least two ac parameters determined from DC quantities.

A small ac signal vbe is superimposed on the DC voltage VBE. It gives rise to a collector signal current ic, superimposed on the dc current IC.
Transconductance
ac input signal
(DC input signal 0.7V)
ac output signal
DC output signal
IB
The slope of the ic - vBE curve at the bias point Q is the transconductance gm: the amount of ac current produced by an ac voltage.

Transconductance = slope at Q point
gm = dic/dvBE|ic = ICQ
where IC = IS[exp(-VBE/VT)-1]; the equation for a diode.
Transconductance
ac input signal
DC input signal (0.7V)
ac output signal
DC output signal
gm = ISexp(-VBE/VT) (1/VT)
gm ≈ IC/VT (A/V)

ac input resistance ∝ 1/slope at Q point
rπ = dvBE/dib|ic = ICQ
rπ ≈ VT /IB
re ≈ VT /IE
ac input resistance of transistor
ac input signal
DC input signal (0.7V)
ac output signal
DC output signal
IB

Small-signal equivalent circuit models
ac model
Hybrid-π model
They are equivalent
Works in linear region only

Steps to analyze a transistor circuit
1 DC problem
Set ac sources to zero, solve for DC quantities, IC and VCE.
2 Determine ac quantities from DC parameters
Find gm, rπ, and re.
3 ac problem
Set DC sources to zero, replace transistor by hybrid-π model, find ac quantites, Rin, Rout, Av, and Ai.

Example-6
Find vout/vin, (β = 100)
DC problem
Short vi, determine IC and VCE
B-E voltage loop
3 = IBRB + VBE
IB = (3 - .7)/RB = 0.023mA
C-E voltage loop
VCE = 10 - ICRC
VCE = 10 - (2.3)(3)
VCE = 3.1V
Q point: VCE = 3.1V, IC = 2.3mA

Example -6
ac problem
Short DC sources, input and output circuits are separate, only coupled mathematically
gm = IC/VT = 2.3mA/25mV = 92mA/V
rπ = VT/ IB = 25mV/.023mA = 1.1K
vbe= vi [rπ / (100K + rπ)] = 0.011vi
vout = - gm vbeRC
vout = - 92 (0.011vi)3K
vout/vi = -3.04
+
vout
-
e
b
c
+
vbe
-

Exercise-7
Find gm, rπ, and r0, given: β = 100, VA = 100V,IC=1 mA
gm = IC/VT = 1 mA/25mV = 40 mA/V
rπ = VT/ IB = 25mV/.01mA = 2.5K
r0 = output resistance of transistor
r0 = 1/slope of transistor output characteristics
r0 = | VA|/IC = 100K

Summary of transistor analysis
Transistor circuits are analyzed and designed in terms of DC and ac versions of the same circuit.
An ac signal is usually superimposed on the DC circuit.
The location of the operating point (values of IC and VCE) of the transistor affects the ac operation of the circuit.
There are at least two ac parameters determined from DC quantities.

Steps to analyze a transistor circuit
1 DC Analysis Set ac sources to zero, solve for DC quantities, IC and VCE.
2 Determine ac quantities from DC parameters
Find gm, rπ, and ro.
3 AC Analysis
Set DC sources to zero, replace transistor by hybrid-π model, find ac quantities, Rin, Rout, Av, and Ai.
ro

Circuit
IE = 1 mA VC = 10V - IC8K = 10 - 1(8) = 2V
IB ≈ IE/β = 0.01mA gm = IC/VT = 1 mA/25mV = 40 mA/V
VB = 0 - IB10K = -0.1V rπ = VT/ IB = 25mV/.01mA = 2.5K
VE = VB - VBE = -0.1 - 0.7 = -0.8V
+
Vout
-

ac equivalent circuit
b
e
c
vbe = (Rb||Rpi)/ [(Rb||Rpi) +Rs]vi
vbe = 0.5vi
vout = -(gmvbe)||(Ro||Rc ||RL)
vout = -154vbe
Av = vout/vi = - 77
+
vout
-
Neglecting Ro
vout = -(gmvbe)||(Rc ||RL)
Av = vout/vi = - 80

Prob.
+
Vout
-
β=100

Prob.
+
Vout
-
(a) Find Rin
Rin = Rpi = VT/IB = (25mV)100/.1 = 2.5KΩ
(c) Find Rout
Rout = Rc = 47KΩ
Rin
Rout
(b) Find Av = vout/vin
vout = - βib Rc
vin = ib (R + Rpi)
Av = vout/vin = - βib Rc/ ib (R + Rpi
= - βRc/(R + Rpi)
= - 100(47K)/(100K + 2.5K)
= - 37.6
= βib
b
e
c
ib

Graphical analysis
Input circuit
B-E voltage loop
VBB = IBRB +VBE
IB = (VBB - VBE)/RB

Graphical construction of IB and VBE
IB = (VBB - VBE)/RB
If VBE = 0, IB = VBB/RB
If IB = 0, VBE = VBB
VBB/RB

Load line
Output circuit
C-E voltage loop
VCC = ICRC +VCE
IC = (VCC - VCE)/RC

Graphical construction of IC and VCE
VCC/RC
IC = (VCC - VCE)/RC
If VCE = 0, IC = VCC/RC
If IC = 0, VCE = VCC

Graphical analysis
Input
signal
Output
signal

Load-line A results in bias point QA which is too close to VCC and thus limits the positive swing of vCE.
Load-line B results in an operating point too close to the saturation region, thus limiting the negative swing of vCE.
Bias point location effects

Basic single-stage BJT amplifier configurations
We will study 3 types of BJT amplifiers
CE - common emitter, used for AV, Ai, and general purpose
CE with RE - common emitter with RE,
same as CE but more stable
CC common collector, used for Ai, low output resistance,
used as an output stage
CB common base (not covered)

Common emitter amplifier
ac equivalent circuit

Common emitter amplifier
Rin
Rout
+
Vout
-
Rin
(Does not include source)
Rin = Rpi
Rout
(Does not include load)
Rout = RC
AV
= Vout/Vin
Vout = - βibRC
Vin = ib(Rs + Rpi)
AV = - βRC/ (Rs + Rpi)
Ai
= iout/iin
iout = - βib
iin = ib
Ai = - β
ib
iout

Common emitter with RE amplifier
ac equivalent circuit

Common emitter with RE amplifier
Rin
Rout
+
Vout
-
Rin
Rin = V/ib
V = ib Rpi + (ib + βib)RE
Rin = Rpi + (1 + β)RE
(usually large)
Rout
Rout = RC
AV
= Vout/Vin
Vout = - βibRC
Vin = ib Rs + ib Rpi + (ib + βib)RE
AV = - βRC/ (Rs + Rpi + (1 + β)RE)
(less than CE, but less sensitive to β variations)
Ai
= iout/iin
iout = - βib
iin = ib
Ai = - β
ib
iout
ib + βib
+
V
-

Common collector (emitter follower) amplifier
b
c
e
+
vout
-
(vout at emitter)
ac equivalent circuit

Common collector amplifier
Rin
+
vout
-
Rin
Rin = V/ib
V = ib Rpi + (ib + βib)RL
Rin = Rpi + (1 + β)RL
AV
= vout/vs
vout = (ib + βib)RL
vs = ib Rs + ib Rpi + (ib + βib)RL
AV = (1+ β)RL/ (Rs + Rpi + (1 + β)RL)
(always < 1)
ib
ib + βib
+
V
-

Common collector amplifier
Rout
+
vout
-
Rout
(don’t include RL, set Vs = 0)
Rout = vout /- (ib + βib)
vout = -ib Rpi + -ibRs
Rout = (Rpi + Rs) / (1+ β)
(usually low)
Ai
= iout/iin
iout = ib + βib
iin = ib
Ai = β + 1
ib
ib + βib

Prob
ac
circuit
CE with RE
amp, because RE
is in ac circuit
Given
Rpi =VT/IB
= 25mV(50)/.2mA
= 6.25K

Prob.
(a) Find Rin
Rin = V/ib
V = ib Rpi + (ib + βib)RE
Rin = Rpi + (1 + β)RE
Rin = 6.25K + (1 + 50)125
Rin ≈ 12.62K
Rin
+
V
-
ib
ib + βib

Prob.
(b) Find AV = vout/vs
vout = - βib(RC||RL)
vs = ib Rs + ib Rpi + (ib + βib)RE
AV = - β (RC||RL) / (Rs + Rpi + (1 + β)RE)
AV = - 50 (10K||10K) /(10K + 6.25Ki + (1 + 50)125)
AV ≈ - 11
ib
ib + βib
+
vout
-
−βib

Prob.
(c) If vbe is limited to 5mV, what is the largest signal at input and output?
vbe = ib Rpi = 5mV
ib = vbe /Rpi = 5mV/6.25K = 0.8μA (ac value)
vs = ib Rs + ib Rpi + (ib + βib)RE
vs = (0.8μA)10K + (0.8μA) 6.25K + (0.8μA + (50)0.8μA )125
vs ≈ 18mV
ib
ib + βib
+
vout
-
+
vbe
-

Prob.
(c) If vbe is limited to 5mV, what is the largest signal at input and output?
vout = vs AV
vout = 17.4mV(-11)
vout ≈ -191mV (ac value)
ib
ib + βib
+
vout
-
+
vbe
-

Prob.
Using this circuit, design an amp
with:
IE = 2mA
AV = -8
current in voltage divider I = 0.2mA
(CE amp because RE is not in ac circuit)
β = 100
Voltage divider
Vcc/I = 9/0.2mA = 45K
45K = R1 + R2
Choose VB ≈ 1/3 Vcc to put operating point near the
center of the transistor characteristics
R2/(R1 + R2) = 3V
Combining gives, R1 = 30K, R2 = 15K

Prob.
β = 100
Find RE (input circuit)
Use Thevenin equivalent
B-E loop
VBB=IBRBB+VBE+IERE
using IB ≈ IE/β
RE = [VBB - VBE - (IE/β)RBB]/IE
RE = [3 - .7 - (2mA/100)10K]/2mA
RE = 1.05KΩ
+
VBE -
IE
IB

Prob.
Find Rc (ac circuit)
Rpi = VT/IB = 25mV(100)/2mA = 1.25K
Ro = VA/IC = 100/2mA = 50K
Av = vout/vin
vout = -gmvbe (Ro||Rc||RL)
vbe = 10K||1.2K / [10K+ 10K||1.2K]vi
Av = -gm(Ro||Rc||RL)(10K||1.2K) / [10K||1.2K +Rs]
Set Av = -8, and solve for Rc, Rc ≈ 2K
+
vout
-

CE amplifier

CE amplifier
Av ≈ -12.2

FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009)
DC COMPONENT = -1.226074E-01
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
1 1.000E+03 1.581E+00 1.000E+00 -1.795E+02 0.000E+00
2 2.000E+03 1.992E-01 1.260E-01 9.111E+01 2.706E+02
3 3.000E+03 2.171E-02 1.374E-02 -1.778E+02 1.668E+00
4 4.000E+03 3.376E-03 2.136E-03 -1.441E+02 3.533E+01
TOTAL HARMONIC DISTORTION = 1.267478E+01 PERCENT
CE amplifier

CE amplifier with RE

CE amplifier with RE
Av ≈ - 7.5

FOURIER COMPONENTS OF TRANSIENT RESPONSE V($N_0009)
DC COMPONENT = -1.353568E-02
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
1 1.000E+03 7.879E-01 1.000E+00 -1.794E+02 0.000E+00
2 2.000E+03 1.604E-02 2.036E-02 9.400E+01 2.734E+02
3 3.000E+03 5.210E-03 6.612E-03 -1.389E+02 4.056E+01
4 4.000E+03 3.824E-03 4.854E-03 -1.171E+02 6.231E+01
TOTAL HARMONIC DISTORTION = 2.194882E+00 PERCENT

Summary
Av THD
CE -12.2 12.7%
CE w/RE (RE = 100) -7.5 2.19%

Prob.
+
vout
-
2 stage amplifier (a) Find IC and VC of each transistor
Both stages are the same (same for each stage)
Capacitively coupled
β = 100
RL=2K
Rc=6.8K

Prob.
(a) Find IC and VC of each transistor
(same for each stage)
B-E voltage loop
VBB = IBRBB + VBE + IERE
where RBB = R1||R2 = 32K
VBB = VCCR2/(R1+R2) = 4.8V, and
IB ≈ IE/β
IE = [VBB - VBE ]/[RBB/β + RE]
IE = 0.97mA
VC = VCC - ICRC
VC = 15 - .97(6.8)
VC = 8.39V
+
VC
-

Prob.
b
c
e
+
vout
-
(b) find ac circuit
b
c
e
RBB = R1||R2 = 100K||47K = 32KΩ
Rpi = VT/IB = 25mV(100)/.97mA ≈ 2.6KΩ
gm = IC/VT = .97mA/25mV ≈ 39mA/V
RL=2K

Prob.
b
c
e
+
vout
-
(c) find Rin1
Rin1 = RBB||Rpi
= 32K||2.6K
= 2.4KΩ
b
c
e
Rin1
(d) find Rin2
Rin2 = RBB||Rpi
= 2.4KΩ
Rin2
find vb1/vi
= Rin1/ [Rin1 + RS]
= 2.4K/[2.4K + 5K]
= 0.32
+
vb1
-
find vb2/vb1
vb2 = -gmvbe1[RC||RBB||Rpi]
vb2/vbe1 = -gm[RC||RBB||Rpi]
vb2/vb1 = -(39mA/V)[6.8||32K||2.6K]
= -69.1
+
vb2
-
RL=2K

Prob.
b
c
e
+
vout
-
(e) find vout/vb2
vout = -gmvbe2[RC||RL]
vout/vbe2 = -gm[RC||RL]
vb2/vb1 = -(39mA/V)[6.8K||2K]
= -60.3
b
c
e
(f) find overall voltage gain
vout/vi = (vb1/vi) (vb2/vb1) (vout/vb2)
vout/vi = (0.32) (-69.1) (-60.3)
vout/vi = 1332
+
vb1
-
+
vb2
-
RL=2K

Prob.
Find IE1, IE2, VB1, and VB2
IE2 = 2mA
IE1 = I20μA + IB2
IE1 = I20μA + IE2/β2
IE1 = 20μA + 10μA
IE1 = 30μA
IE1
IE2
Q1 has β1 = 20
Q2 has β2 = 200
Q1
Q2
IB2

Prob.
Find VB1, and VB2
Use Thevenin equivalent
VB1 = VBB1 - IB1(RBB2)
= 4.5 - (30μA/20)500K
= 3.8V
VB2 = VB1 - VBE
= 3.8V - 0.7
= 3.1V
Q1 has β1 = 20
Q2 has β2 = 200
IB2
+
vB1
-
+
vB2
-
IB1

Prob.
(b) find vout/vb2
vout = (ib2 + β2ib2)RL
vb2 = (ib2 + β2ib2)RL + ib2Rpi2
vout/vb2 = (1 + β2)RL/[(1 + β2)RL + Rpi2]
= (1 + 200)1K/[(1 + 200)1K + 2.5K]
≈ 0.988
b1
e1
c1
b2
e2
c2
+
vout
-
+
vB2
-
Rpi2 = VT/IB2
= VT β2/IE2
= 25mV(200)/2mA
= 2.5KΩ

Prob.
(b) find Rin2 = vb2/ib2
vb2 = (ib2 + β2ib2)RL + ib2Rpi2
Rin2 = vb2/ib2 = (1 + β2)RL + Rp
= (1 + 200)1K + 2.5K
≈ 204K
b1
e1
c1
b2
e2
c2
+
vB2
-
Rin2

Prob.
(c) find Rin1 = RBB1||(vb1/ib1)
= RBB1|| [ib1Rpi1 + (ib1 + β1ib1)Rin2]/ib1
= RBB1|| [Rpi1 + (1+ β1)Rin2],
where Rpi1 = VT β1/IE1 = 25mV(20)/30μA = 16.7K
= 500K||[16.7K + (1+ 20)204K]
≈ 500KΩ
b1
e1
c1
b2
e2
c2
+
vB1
-
Rin1
iB1

Prob.
(c) find ve1/vb1
ve1 = (ib1 + β1ib1)Rin2
vb1 = (ib1 + β1ib1)Rin2 + ib1Rpi1
ve1/vb1 = (1 + β1) Rin2 /[(1 + β1) Rin2 + Rpi1]
= (1 + 20)204K/[(1 + 20)204K + 16.7K]
≈ 0.996
b1
e1
c1
b2
e2
c2
+
ve1
-
iB1
+
vB1
-

Prob.
(d) find vb1/vi
vb1/vi = Rin1/[RS + Rin1]
= 0.82
b1
e1
c1
b2
e2
c2
+
vb1
-
(e) find overall voltage gain
vout/vi = (vb1/vi) (ve1/vb1) (vout/ve1)
vout/vi = (0.82) (0.99) (0.99)
vout/vi = 0.81

Voltage outputs at each stage
Output of
stage 2
Output of
stage 1
Input

Current
Input
current
Input to
stage 2 (ib2)

Current
output
current
Input to
stage 2 (ib2)

Current
output
current
Input to
stage 2 (ib2)
Input
current

Power and current gain
Input current = (Vi)/Rin = 1/500K = 2.0μA
output current= (Vout)/RL = (0.81V)/1K= 0.81mA
current gain = 0.81mA/ 2.0μA = 405
Input power = (Vi) (Vi)/Rin = 1 x 1/500K = 2.0μW
output power = (Vout) (Vout)/RL = (0.81V) (0.81V)/1K= 656μW
power gain = 656μW/2μW = 329

BJT Output Characteristics
Plot Ic vs. Vce for multiple values of Vce and Ib
From Analysis menu use DC Sweep
Use Nested sweep in DC Sweep section

Probe: BJT Output Characteristics
1 Result of probe
2 Add plot (plot menu) -> Add trace (trace menu) -> IC(Q1)
3 Delete plot (plot menu)

BJT Output Characteristics: current gain
Ib = 5μA
(Each plot 10μA difference)
β at Vce = 4V, and Ib = 45μA
β = 8mA/45μA = 178

BJT Output Characteristics: transistor output resistance
Ib = 5μA
(Each plot 10μA difference)
Ro = 1/slope
At Ib = 45μA,
1/slope = (8.0 - 1.6)V/(8.5 - 7.8)mA
Rout = 9.1KΩ

CE Amplifier: Measurements with Spice
Rin
Rout

Input Resistance Measurement Using SPICE
Replace source, Vs and Rs with Vin, measure Rin = Vin/Iin
Do not change DC problem: keep capacitive coupling
if present
Source (Vin) should be a high enough frequency so that
capacitors act as shorts: Rcap = |1/ωC|. For C = 100μF,
ω = 1KHz, Rcap = 1/2π(1K)(100E-6) ≅ 1.6Ω
Vin should have a small value so operating point does not change
Vin ≅ 1mV

Rin Measurement
Transient analysis

Probe results
I(C2)
Rin = 1mV/204nA
= 4.9KΩ

Output Resistance Measurement Using SPICE
Replace load, RL with Vin, measure Rin = Vin/Iin
Set Vs = 0
Do not change DC problem: keep capacitive coupling
if present
Source (Vin) should be a high enough frequency so that
capacitors act as shorts: Rcap = |1/ωC|. For C = 100μF,
ω = 1KHz, Rcap = 1/2π(1K)(100E-6) ≅ 1.6Ω
Vin should have a small value so operating point does not change
Vin ≅ 1mV

Rout Measurement
Transient analysis

Probe results
-I(C1)
Rout = 1mV/111nA
= 9KΩ
-I(C1) is current in Vin flowing out of + terminal

DC Power measurements
Power delivered by + 10 sources:
(10)(872μA) + (10)(877μA) = 8.72mW + 8.77mW = 17.4mW