4. Bipolar Junction Transistors
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4.1 Basic Operation of the npn Bipolar Junction Transistor
npn BJT consists of thin p-type layer between
two n-type layers;
Layers: emitter, base, collector;
Two interacting pn junctions: emitter-base and
base-collector;
Emitter region is doped very heavily, compared
with the base region
Figure 4.1 The npn BJT.
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Basic Operation in the Active region
Figure 4.2 An npn transistor with variable biasing sources
(common-emitter configuration).
Shokley equation for the emitter current
  vBE  
 − 1
iE = I ES exp
  VT  
IES = 10-12 .. 10-17A – saturation current
VT = 26mV – thermal voltage
4. Bipolar Junction Transistors
(4.1)
Figure 4.3 Current flow for an npn BJT in the active region.
Most of the current is due to electrons moving from the emitter
through the base to the collector. Base current consists of holes
crossing from the base into the emitter and of holes that
recombine with electrons in the base.
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First - Order Common - Emitter Characteristics
Figure 4.4 Common-emitter characteristics of a typical npn BJT.
Amplification by the BJT
Factors Affecting the Current Gain
In Figure 4.4: If iB = 30µA, iC = 3mA – 100 times
more
i
β= C
(4.2)
iB
β - common-emitter current gain.
Typically β = 10 .. 1000
4. Bipolar Junction Transistors
• Doping of the emitter area compared with the
base area
• Base region should be thin;
• Geometry of the device
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Device Equations
iE = iC + iB
(4.3)
i
α= C
(4.4)
iE
α - common-base current gain.
Typically α = 0.9 .. 0.999
 v  
iC = αI ES exp BE  − 1
  VT  
I s = αI ES
iB = (1 − α )iE
 v  
iB = (1 − α )I ES exp BE  − 1
  VT  
(4.9)
iC = β iB
(4.11)
α
i
β= C=
iB 1 − α
(4.10)
(4.5)
(4.6)
Is – scale current
 vBE 

iC ≅ I s exp
 VT 
4. Bipolar Junction Transistors
(4.8)
(4.7)
TLT-8016 Basic Analog Circuits
α=
β
1+ β
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Example 4.1 Using Device Curves to Determine
α and β
Determine the values of α and β for the transistor
with the characteristics shown in Figure 4.4.
Solution:
For example:
at vCE=4 V and iB=30 µA; iC=3 mA;
i
3 mA
= 100
β= C=
iB 30 µA
α=
β
β +1
= 0.99
Figure 4.4 Common-emitter characteristics of a typical npn BJT.
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Secondary Effects
Base - Width Modulation
Base-width modulation: the dependence of the
base width from vCE.
Base-width modulation affects iB and iC.
VA – Early voltage
Collector Breakdown
Avalanche breakdown in the depletion region
of the collector-base junction
Punch-through
Leakage Current
ICO – reverse leakage current. Flows from
collector to the base.
Figure 4.5 Common-emitter characteristics displaying
exaggerated secondary effects.
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4.2 Load - Line Analysis of a Common - Emitter Amplifier
Analysis of the Input Current
VBB + vin ( t ) = RBiB ( t ) + vBE ( t )
(4.13)
Analysis of the Output Circuit
VCC = RC iC + vCE
Figure 4.10 Common-emitter amplifier.
(4.14)
Figure 4.11 Load-line analysis of the amplifier of Figure 4.10.
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Example 4.2 Graphical Determination of Q-point and Peak Signal Swings
Suppose that the current of Figure 4.10 has VCC=10 V, VBB=1.6 V, RB=40 kΩ and RC=2 kΩ. The
input signal is a 0.4 V peak, 1 kHz sinusoid given by vin(t)=0.4sin(2000πt). The common-emitter
characteristics for the transistor are shown in Figure 4.12a and b. Find the maximum, minimum and Qpoint values for vCE.
Solution
Figure 4.12 Load-line analysis for Example 4.2.
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Figure 4.13 Voltage waveforms for the amplifier of Figure 4.10. See Example 4.2.
Gain in Example 4.2:
Amplitude at the input: 0.4V
Amplitude at the output
7 – 5 = 2V
Av =
4. Bipolar Junction Transistors
2
=5
0.4
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Distortion
Figure 4.14 Output of the amplifier of Example 4.2 for
vin (t) = 1.2 sin(2000πt) showing gross distortion.
Cutoff: vBE < 0.6V and iB ≈ 0
Saturation region: iB is large and iC is not
proportional to it.
When transistor enters in cutoff or in saturation,
clipping of the output signal occurs.
4. Bipolar Junction Transistors
Figure 4.15 Amplification occurs in the active region. Clipping
occurs when the instantaneous operating point enters
saturation or cutoff. In saturation, vCE < 0.2 V.
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4.3 The pnp Bipolar Junction Transistor
pnp transistor: thin n type semiconductor layer
between two p type semiconductor layers
Basic charge carriers: holes
All relationships between the currents and
voltages in a pnp BJT are the same as in npn
BJT. There are two basic differences:
• The currents flow in opposite directions;
• The voltages have opposite polarities.
iC = α iE
(4.15)
iB = (1 − α )iE
(4.16)
iC = β iB
(4.17)
iE = iC + iB
(4.18)
 −v  
iE = I ES exp BE  − 1
  VT  
(4.19)
Figure 4.16 The pnp BJT.
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Figure 4.17 Common-emitter characteristics for a pnp BJT. Pay attention that the voltages are negative.
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4.4 Large - Signal DC Circuit Models
Figure 4.20 Regions of operation on the characteristics of an npn BJT.
Active region: IB > 0; VCE > 0.2V
Saturation region: IB > 0; βIB > IC > 0
Cutoff region: VBE < 0.5V; VBC < 0.5V;
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Active - Region Model
Figure 4.19a BJT large-signal models. (Note: Values shown are appropriate for
typical small-signal silicon devices at a temperature of 300K.)
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Saturation - Region Model
Figure 4.19b BJT large-signal models. (Note: Values shown are appropriate for typical small-signal
silicon devices at a temperature of 300K.)
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Cutoff - Region Model
Figure 4.19c BJT large-signal models. (Note: Values shown are appropriate for typical
small-signal silicon devices at a temperature of 300K.)
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Example 4.3 Determination of BJT Operation Region
A given transistor has β =100. Determine the region of operation if
(a) IB = 50µA and IC = 3mA; (b) IB = 50µA and VCE = 5V;
(c) VBE = -2V and VCE = -1V.
Solution:
(a) IB = 50µA > 0 – active or saturation region;
βIB = 100×50×10-6 = 5mA > IC – saturation region.
(b) IB = 50µA > 0 – active or saturation region;
VCE = 5V > 0.2V – active region.
(c) VBE = -2V < 0.5V – most probably cutoff;
VCE = -1V < 0.5V – this confirms cutoff region.
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4.5 Large - Signal DC Analysis of BJT Circuits
Step 1: Assume an operation region for the BJT and replace it by the
corresponding large signal equivalent circuit.
Step 2: Solve the circuit to find IC, IB, and VCE.
Step 3: Check to see if the values found in Step 2 are consistent with
the assumed operating state. If so the solution is complete;
otherwise return to Step 1.
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Example 4.4 The Fixed - Base Bias Circuit
The DC bias circuit shown in Figure 4.21a has
RB=200 kΩ, RC=1 kΩ and VCC=15 V. The
transistor has β = 100. Solve for IC and VCE.
Solution:
First assumption: Cutoff. The equivalent circuit is
in Figure 4.21(b).
IB = 0 – thus the voltage drop across RB is zero.
Thus VBE = VCC = 15V > 0.5V.
The assumption is not valid.
Figure 4.21 Bias circuit of Examples 4.4 and 4.5.
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Second assumption: Saturation. The equivalent
circuit is in Figure 4.21c.
VCC − 0.7 15 − 0.7
=
= 71.5 µA
RB
200 × 103
I C = β I B = 7.15mA
V − 0.2 15 − 0.2
I C = CC
=
= 14.8mA
3
RC
1×10
IB =
Third assumption: Active region. The equivalent
circuit is in Figure 4.21d.
IB =
VCC − 0.7 15 − 0.7
= 71.5 µA
=
3
RB
200 × 10
VCE = VCC − RC I C = 15 − 7.15 × 10 −3 × 1× 10 3 = 7.85V
βIB = 100×71.5×10-6 = 7.15mA < IC
The assumption is not valid.
IB > 0; VCE > 0.2V. The conditions are met.
Figure 4.21 Bias circuit of Examples 4.4 and 4.5.
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Example 4.5 The Fixed - Base Bias Circuit with
Higher Beta
Repeat example 4.4 with β=300.
Solution: First, we assume that the circuit
operating in the active region.
V − 0.7
I B = CC
= 71.5 µA
RB
I C = β I B = 21.45 mA
One of the requirements for the active region
VCE>0.2 V is not met.
Next we assume that the transistor is in
saturation.
V − 0.2
I C = CC
= 14.8 mA
RC
V − 0.7
I B = CC
= 71.5 µA
RB
VCE = VCC − RC I C = −6.45V < 0.2V
βI B = 300 × 71.5 ×10 −6 = 21.45mA > 14.8mA
Figure 4.21 Bias circuit of Examples 4.4 and 4.5.
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Analysis for the Four - Resistor Bias Circuit
RB =
1
= R1 || R2
1 / R1 + 1 / R2
VB =
R2
VCC
R1 + R2
VB = RB I B + VBE + RE I E
(4.21)
(4.22)
(4.23)
I E = (β + 1)I B
IB =
VB − VBE
RB + (β + 1)RE
VCE = VCC − RC I C − RE I E
(4.24)
(4.25)
Figure 4.28 Four-resistor bias circuit.
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Discrete Bias - Circuit Design
The principal goal of bias circuit design is to
achieve nearly identical operating point for the
BJTs, even though the BJT parameters may vary
significantly from unit to unit.
In the design usually are given the supply voltage
VCC, the collector current IC in the quiescent point
and often VCE in the quiescent point.
The design steps are:
1. Choice of VCE, if it is not specified. A good
choice is
V
VCE = CC
3
2. Determining of the voltage drop VE across RE
and the voltage VCC – VC, which is across RC. A
good choice is
V − VCE
VE = VCC − VC = CC
2
3.
5. Since IE = (β+1)IB ≈ IC
RE = VE I E ≈ VE I C
6. VB = VE + VBE = VE + 0.7
7. R2 = VB I 2
R1 = (VCC − VB ) (I B + I 2 )
RC = (VCC − VC ) I C
I2+IB
R1
RC
IB
VB
VC +
R2
+
VCC
VCE
+
- - V
E
VBE
I2
IC
RE
-
IE
I B = IC β
4. Choice of the current I2 to be I2 = (10..20)IB.
4. Bipolar Junction Transistors
TLT-8016 Basic Analog Circuits
Four-resistor bias circuit.
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How the circuit stabilize the quiescent point?
Assume that the emitter current IC is increased, due to
some reason. Then
• Emitter current IE is increased also, since IE ≈ IC.
• VE increases since VE = IERE
• I2 is at least 10 times more than IB. It is the basic part
of the current through R1. Thus the both currents are
stable and depend very weak on the variation of the
currents in the BJT.
• The stable currents through R1 and R2 define a stable
voltage VB.
• VBE = VB - VE and the increasing of VE decreases VBE.
• Smaller VBE means smaller base current IB (see the
input characteristic in Figure 4.4).
I2+IB
R1
RC
IB
VB
VC +
R2
+
VCC
VCE
+
- - V
E
VBE
I2
IC
RE
-
IE
Four-resistor bias circuit.
• IC = βIB and the smaller base current returns the
collector current to its initial value.
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Problem D4.39. Four-resistor bias circuit
design.
Suppose that VCC = 20V, RC = 1kΩ, and a
quiescent point of ICQ = 5mA is desired. The
transistor has β ranging from 50 to 150. Design a
four resistor bias circuit. Use standard 5%tolerance resistors.
Solution:
Since RC is specified we can determine the voltage
drop across it and the voltage VC
VCC − VC = I CQ RC = 5 × 10 −3 × 1×103 = 5V
VC = 20 − 5 = 15V
R2 = VB I 2
(
The 5%-tolerance standard values for the resistors
are RE = 1.5kΩ, R1 = 8.2kΩ, R2 = 11kΩ.
I2+IB
RC
IB
I2
+
VCC
VCE
+
R2
IC
VC +
- - V
E
VBE
I B = I C β = 5 × 10 −3 50 = 100µA
4. Bipolar Junction Transistors
R1
VB
To determine IB, we take the smallest value of β.
In this way we will determine the largest value of
IB and the largest value of I2. If β is higher, the
condition for I2 will be also satisfied.
)
= (20 − 8.2 ) 1× 10 −3 + 100 × 10 −6 = 10.7 kΩ
VB = VE + VBE = 7.5 + 0.7 = 8.2V
I 2 = 10 I B = 10 ×100 × 10
−3
R1 = (VCC − VB ) (I 2 + I B )
VE = VCE = VC 2 = 15 2 = 7.5V
−6
(
)
= 8.2 (1× 10 ) = 8.2kΩ
RE = VE I E = 7.5 5 × 10 −3 = 1500Ω
RE
-
IE
Four-resistor bias circuit.
= 1mA
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4.6 Small - Signal Equivalent Circuits
Small - Signal Current - Voltage Relationship
 v  
iB = (1 − α )I ES exp BE  − 1
  VT  
(4.28)
 VBEQ + vbe ( t ) 

I BQ + ib ( t ) = (1 − α )I ES exp
VT


 v (t ) 
I BQ + ib (t ) ≅ I BQ 1 + be 
VT 

(4.34)
v (t )
ib (t ) = be
rπ
(4.35)
(4.29)
 VBEQ   vbe ( t ) 
 exp
 (4.30)
I BQ + ib ( t ) = (1 − α )I ES exp
 VT   VT 
 VBEQ 

(
)
I BQ = 1 − α I ES exp
 VT 
(4.31)
 v (t ) 
I BQ + ib ( t ) = I BQ exp be 
 VT 
(4.32)
exp( x ) ≅ 1 + x
(4.33)
4. Bipolar Junction Transistors
TLT-8016 Basic Analog Circuits
rπ =
VT
I BQ
rπ =
(4.36)
βVT
I CQ
iC (t ) = β iB (t )
I CQ + ic (t ) = β I BQ + β ib (t )
ic (t ) = β ib (t )
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(4.37)
(4.38)
(4.39)
(4.40)
27
Small - Signal Equivalent Circuit for the
BJT
ic (t ) =
β
vbe (t )
rπ
β
gm =
rπ
gm =
I CQ
VT
(4.41)
(4.42)
vbe (t ) = rπ ib (t ) and ic (t ) = g m vbe (t )
Exercise 4.19 At room temperature, a certain
Figure 4.33 Small-signal equivalent circuits for the BJT.
transistor has β=100. Compute the values of gm and
rπ for ICQ=10 mA. Repeat for ICQ=1 mA
Solution:
βV 100 × 0.026
β 100
At I C = 10mA : rπ = T =
=
260
Ω
;
=
=
≈ 385mS
g
m
10 ×10 −3
I CQ
rπ 260
At I C = 1mA : rπ =
βVT
4. Bipolar Junction Transistors
I CQ
=
100 × 0.026
β 100
=
2600
Ω
;
=
=
≈ 38.5mS
g
m
1× 10 −3
rπ 2600
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4.7 The Common - Emitter Amplifier
R1, R2, RC and RE1+RE2 form the four
resistor biasing circuit;
C1 and C2 are coupling capacitors;
CE is bypass capacitor.
For ac signal the supply voltage is short
circuit.
Figure 4.34 Common-emitter amplifier.
4. Bipolar Junction Transistors
If RE1 = 0 the input signal is between the
base and the emitter (ground); the output
signal is between collector and the emitter
(+VCC, which is ground for ac signal) –
common-emitter amplifier.
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The Small - Signal Equivalent Circuit
Rs
+
+
-
vs
vin
C2
C1
R1
R2
RE1
RE2
-
RC
CE
Rs
+
+
+
vo
RL
+
-
-
The first step in creating the small-signal equivalent circuit: the
dc voltage sources are replaced by short circuits
vs
vin
R1
R2
RE1
RC
-
vo
RL
-
The second step in creating the small-signal equivalent circuit
for mid-band region: the coupling capacitors and the bypass
capacitors are replaced by short circuits
Figure 4.34 Common-emitter amplifier. (b) Final small-signal mid-band equivalent circuit.
4. Bipolar Junction Transistors
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Voltage Gain
vin = vbe + ie RE1
= rπ ib + RE1 (β + 1)ib
vo = − R'L β ib
(4.46)
vo
βR'L
Av =
=−
vin
rπ + (β + 1)RE1
If RE1 = 0:
Figure 4.34 Common-emitter amplifier.
R'L = RL || RC =
4. Bipolar Junction Transistors
(4.47)
βRL'
rπ
βRL'
Simplifications:
RB = R1 || R2 =
Av = −
(4.45)
1
1 / R1 + 1 / R2
(4.43)
1
1 / RL + 1 / RC
(4.44)
'
R
L
If (β+1)RE1 >> rπ: Av ≅ −
≈−
(β + 1)RE1 RE1
and the voltage gain doesn’t depend on
BJT parameters.
Open circuit voltage gain (when RL = ∞)
βRC
v
Avo = o = −
vin
rπ + (β + 1)RE1
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(4.48)
31
Input Impedance
Current Gain and Power Gain
i
Z
Ai = o = Av in
iin
RL
(4.51)
G = Ai Av
(4.52)
Figure 4.34 Common-emitter amplifier.
v
Zit = in = rπ + (β + 1)RE1
ib
(4.49)
v
1
Zin = in =
iin 1 / RB + 1 / Zit
(4.50)
4. Bipolar Junction Transistors
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Output Impedance
Figure 4.34 Common-emitter amplifier.
ib = 0 and βib = 0. Thus there is an open circuit between C and E.
Z o = RC
4. Bipolar Junction Transistors
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(4.53)
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Example 4.9 Calculation of Common - Emitter
Amplifier Performance
a) Find Av , Avo , Zin , Ai , G, and Zo for the
amplifier shown in Figure 4.35.
b) Repeat part (a) if the emitter resistor RE is
split in into RE1=100 Ω and RE2=900 Ω, with
bypass capacitor in the parallel with RE2.
Solution:
β of the BJT is 100 and from dc analysis is
rπ =
found that the quiescent point is ICQ = 4.12mA
and VCE = 6.72V.
βVT
I CQ
=
100 × 0.026
= 631Ω
4.12 ×10 −3
RB =
1
1
=
= 3.33 kΩ
1 / R1 + 1 / R2 1 10 ×103 + 1 5 × 103
R 'L =
1
1
=
= 667 Ω
1 / RL + 1 / RC 1 2 ×103 + 1 1× 103
(
) (
(
) (
)
)
Figure 4.35 Common-emitter amplifier of Example 4.9.
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b)
a)
Av =
=−
vo
100 × 667
βR
=−
=−
= −106
631
vin
rπ
'
L
Avo =
βRC
vo
100 ×1×103
=−
=−
= −158
Avo =
631
vin
rπ
=
vin
1
=
iin 1 / RB + 1 / Z it
1
= 531 Ω
1 3.33 × 103 + 1 631
Ai =
(
)
io
Z
531
= Av in = −106
= −28.1
3
iin
RL
2 × 10
G = Ai Av = (− 28.1)× (− 106 ) = 2980
vo
βRC
=−
vin
rπ + (β + 1)RE1
Z it =
vin
= rπ + (β + 1)RE1 = 631 + (100 + 1)×100 = 10.7 kΩ
ib
Z in =
1
1
=
= 2.54 kΩ
3
3
1 / RB + 1 / Z it 1 3.33 ×10 + 1 10.7 × 10
(
) (
)
io
Z in
2.54 × 103
= −6.21
= −7.89
Ai = = Av
2 ×103
iin
RL
G = Ai Av = (− 7.89 )× (− 6.21) = 49.0
Z o = RC = 1kΩ
4. Bipolar Junction Transistors
100 × 667
= −6.21
631 + (100 + 1)×100
100 ×1×103
=−
= −9.31
631 + (100 + 1)×100
Z it = rπ = 631Ω
Z in =
vo
βRL'
=−
Av =
vin
rπ + (β + 1)RE1
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Z o = RC = 1kΩ
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4.8 The Emitter Follower
R1, R2, and RE are from the four resistor biasing
circuit.
C1 and C2 are coupling capacitors.
The input signal is between the base and the
ground (the collector, since the supply voltage is
short circuit for the ac signal); the output signal is
between the emitter and the ground (the collector)
– common-collector amplifier.
vin = vbe + vo
Figure 4.36 Emitter follower.
4. Bipolar Junction Transistors
Thus vin > vo and Av < 1. Usually Av ≈ 1 – emitter
follower.
Typical application as a buffer amplifier, since it
has high input impedance and small output
impedance.
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Small - Signal Equivalent Circuit
Rs
+
+
-
vs
vin
C2
C1
R1
R2
-
RE
Rs
+
+
+
vo
RL
+
-
-
The first step in creating the small-signal equivalent circuit: the
dc voltage sources are replaced by short circuits
vs
vin
R1
R2
RE
-
vo
RL
-
The second step in creating the small-signal equivalent circuit
for mid-band region: the the coupling capacitors and the bypass
capacitors are replaced by short circuits
Figure 4.36 Emitter follower. (b) Final small-signal mid-band
equivalent circuit.
4. Bipolar Junction Transistors
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Voltage Gain
Figure 4.36 Emitter follower.
vo = (1 + β )R'LiB
Simplifications:
1
RB = R1 || R2 =
1 / R1 + 1 / R2
R'L = RL || RE =
4. Bipolar Junction Transistors
1
1 / RL + 1 / RE
(4.54)
(4.55)
(4.56)
vin = rπ ib + (1 + β ) iB R'L
(4.57)
(
1 + β )RL'
Av =
rπ + (1 + β )RL'
(4.58)
<1
Since usually (1+β)RL′ >> rπ
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38
Input Impedance
Figure 4.36 Emitter follower.
1
1 / RB + 1 / Zit
(4.59)
v
Zit = in = rπ + (1 + β )R'L
ib
(4.60)
Zi =
4. Bipolar Junction Transistors
For small power BJT RL′ is in the range of kOhms
and β ~ 100. Thus the range of Zit is hundreds of
kOhms.
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Output Impedance
Figure 4.36 Emitter follower.
v
Zo = x
ix
(4.61)
v
ib + β ib + ix = x
RE
1
Rs' =
1 / Rs + 1 / R1 + 1 / R2
v x + rπ ib + R i = 0
'
s b
4. Bipolar Junction Transistors
Zo =
(4.62)
vx
1
=
ix (1 + β ) / (R's + rπ ) + (1 / RE )
(4.65)
Rs' + rπ
Z ot =
(1 + β )
(4.66)
Z ot ≈
rπ
1
≈
(1 + β ) g m
(4.63)
If Rs′ ≈ 0
(4.64)
For small power BJT gm ~ 102..103mS. Thus the
range of Zot is Ohms or tens of Ohms.
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Example 4.10 Calculation of Emitter Follower
Performance. Compute the voltage gain, input
impedance, current gain, power gain, and output
impedance of the emitter-follower amplifier
displayed in Figure 4.37a.
The dc analysis gives the following quiescent
point: ICQ = 4.12mA and VCE = 11.7V.
R 'L =
βVT
200 × 0.026
= 1260Ω
I CQ
4.12 ×103
1
1
RB =
=
= 50kΩ
1 / R1 + 1 / R2 1 (100 × 103 ) + 1 (100 ×103 )
4. Bipolar Junction Transistors
=
)
(1 + 200)× 667 = 0.991
1260 + (1 + 200)× 667
1
1
=
= 36.5 kΩ
1 / RB + 1 / Z it 1 50 ×103 + 1 135 × 103
=
rπ =
=
) (
Z it = rπ + (1 + β )RL' = 1260 + (1 + 200 )× 667 = 135 kΩ
Rs' =
Solution
(
(
1 + β )RL'
Av =
rπ + (1 + β )RL'
Zi =
Figure 4.37 Emitter follower of Example 4.10.
1
1
=
= 667 Ω
3
3
1 / RL + 1 / RE 1 1×10 + 1 2 ×10
Zo =
=
(
) (
)
1
1 / Rs + 1 / R1 + 1 / R2
1
= 8.33 kΩ
1 10 ×103 + 1 100 × 103 + 1 100 ×103
(
) (
) (
)
1
(1 + β ) / (R's + rπ ) + (1 / RE )
1
= 46.6 Ω
(1 + 200) 8.33 ×103 + 1260 + 1 2 ×103
(
) (
)
Zi
36.5 ×103
Ai = Av
= 0.991
= 36.2; G = Av Ai = 38.8
RL
1× 103
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4.9 The BJT as a Digital Logic Switch
Figure 4.41 RTL inverter.
Figure 4.42b Load-line analysis of RTL inverter under noload conditions.
4. Bipolar Junction Transistors
The circuit in Fig. 4.41 is a modification of
common-emitter amplifier intended for operation
in saturation and in cut-off region only.
Cut-off state: Vin < 0.5 (logical zero). BJT is in
cut-off and IC = 0. There is no voltage drop across
RC and Vo = +VCC (logical one).
Saturation state: Vin is high (logical one), usually
Vin ≈ +VCC.
V − 0.7
iB = in
RB
In saturation VCE < 0.2V. This Vo = VCE < 0.2V
(logical zero).
V − 0.7
iC = CC
RC
To have saturation region, RB and RC must be
designed in such a way that βiB > iC.
The output has logically opposite voltage to the
input (zero-one; one-zero). The circuit is called
logical inverter.
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RTL NOR Gate
When at the three inputs are applied low voltages
(logical zeros) all BJT are in cutoff, no current
flows through RC and Vo = +VCC (logical one).
When a high voltage is presenting at one or more
of the inputs (logical ones at these inputs) the
corresponding BJTs are in saturation and Vo <
0.2V (logical zero).
Figure 4.45 Three-input RTL NOR gate.
4. Bipolar Junction Transistors
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