188
1 Topic 4 Bipolar Junction Transistors
1
Topic 4
Bipolar Junction Transistors
2
Introduction to BJTa three-terminal component – Emitter, Collector and Base.
(a)npn transistor which has an n – type emitter and collector and a p – type base, and
(b) pnp transistor which consist of p – type emitter and collector and an n – type base.
3
The arrow on the schematic symbol represents:
(1) the emitter terminal. The terminal opposite the emitter is the collector and the center terminal is the base.
(2) the direction of the arrow is always point towards the n – type of the material.
(3) the direction of the emitter current (conventional).
4
Power Transistor Water Analogy
For transistor working at the linear region, The current gain must be obey.
For transistor working as switches, this formula must be abandon because Ic(sat) is almost a constant value. The formula now is,
B
C
I
I
B
satcforce I
I )(
5
Transistor Currents• IE, IC and IB – emitter, collector and base
currents.IE > IC >> IB
(mA) (A)
IE IC
• Current directions of the npn transistor are opposite those for the pnp transistor.
• The direction of IE is always opposite that of IB and IC.
6
BJT - Current-controlled device• The values of the collector and emitter currents are
determined primarily by the value of the base current.
• The small increase in base current (from 10 A to 20 A) produces a larger increase in IC and in IE (from 2 mA to 4 mA).
= 200
7
BJT - Current-controlled device
• The value of IC is normally some multiple of the value of IB.
- forward current gain, of the device.
BC II
8
Example:
Assume that a transistor has values of IB = 50 A and = 120. The collector current for the transistor is
Solution:
IC = 6 mA)50)(120(C
BC
AI
II
9
ExampleDetermine the values of collector current for the values of the base current shown below.
10
Solution
The base current has an initial value of 20 A. The beta rating of the component is 300. The initial value of the collector current is found as
mA
When IB increases to 50 A, the collector current increases tomA
Thus, a 30 A change in base current causes a 9 mA change in collector current.
6 AII BC 20300
15 AII BC 50300
11
Transistor Voltages
o VCC and VBB are dc voltage sources that are used to bias the transistor.
o Some circuits also contain a dc biasing source in the emitter circuit labeled VEE.
12
Transistor Voltages
o VC, VB and VE are transistor terminal voltages.
o Each voltage is measured from the identified transistor terminal to ground.
13
Transistor Voltages
o VCE, VCB, and VBE are measured across the identified transistor terminals.
14
Transistor Construction• made up of three separate semiconductor materials
joined together so that they form two pn junctions.
the base – emitter junction. the collector – base junction.
15
Transistor Operation
Base-emitter junction
Collector-base junction
Operating region
BJT Characteristics
Reverse biased Reverse biased Cutoff IB = IC = IE = 0, VCE = VCC
Forward biased
Reverse biased Active IC = DCIB
Forward biased
Forward biased
Saturation
IC = IC(sat) = C
CC
R
V
*Breakdown
16
Zero Bias• no biasing potential applied.
• depletion layers due to recombination of free carriers produced by thermal energy.
• both junctions are in reverse biased at room temperature. Note: depletion layers extend farther into the base region due to its lower doping level.
17
Cutoff• both transistor junctions are
reverse biased.
• the depletion layers extend well into the emitter, base and collector regions.
• only an extremely small amount of reverse current passes from the emitter to the collector.
• the transistor is said to be in cutoff.
18
Example
A 2N3904 transistor with a collector – emitter voltage (VCE) of 40 V and a reverse base-emitter (VBE) as low as 3 V will allow only 50 nA of collector current IC.
This is extremely small compared to the 200 mA value of IC that the component is capable of handling when the base – emitter junction is forward biased.
19
Saturation • the opposite of cutoff.
• further increases in IB do not result in further increases in IC – it has reached its maximum possible value.
• In an ideal situation (VCE = 0 V), IC will depend completely on the values of VCC, RC, and RE
20
Saturation
• As IB increases from 0 A, IC increases according to IC = IB
until it reaches its maximum value.
• At this point, IC cannot increase any further – additional increases in IB do not increase the value of IC. The relationship
IC = IB no longer holds true.
• Both of the transistor junctions become forward biased.
21
Saturation
VCE for the transistor is shown to be approximately 0.3 V (typical for a saturated transistor).
With the 0.7 V value of VBE, the collector-base junction is biased to the difference between the two, which is 0.4 V.
Note that this voltage indicates that the collector-base junction of the transistor is forward biased (even though it is not fully).
22
Active Operation• Current is generated in the emitter
and base region when VBE > B.
• Base region is very lightly doped, its resistance is greater than the resistance of the reverse-biased collector-base junction.
• A vast majority of the emitter current continues through the reverse-biased collector-base junction to the collector circuit. Recall Zener diode.
• The collector-base junction is designed to allow a reverse current without damaging the junction.
23
Transistor Currents and Voltages • The transistor is a current-controlled device. • In many applications, the base current is varied to produce
variations in IC and IE
i.e. a small change in IB results in a large change in the other terminal currents.
24
Example
Determine the values of collector current for the values of the base current shown below.
Solution
The initial value of the collector current
When IB increases to 50 A, the collector current is
IB = 30 A,
IC = 9 mA,
mA
AII BC
20300
mA
AII BC
50300
25
Transistor Currents and Voltages The Relationship Among IE, IC and IB
• Kirchhoff’s current law: the current leaving a component must be equal to the current entering the component.
IE = IB + IC
• IB is normally much less that IC, IC IE
• The current relationships shown above hold true for both the npn and pnp transistors.
26
Transistor Currents and Voltages
• IB = 20 A, IC = 6 mA
IE = 6mA + 20 A
= 6.02 mA IC
• IB = 50 A, IC = 15 mA
IE = 15.05 mA
27
Transistor Currents & Voltages • The BE junction is forward biased,
VBE ≈ 0.7 V. • Emitter is at ground. The voltage across
RB isVR(B) = VBB − VBE --(1)
• Also, from Ohm’s Law,VR(B) = IBRB --(2)
• Let (1) = (2)IBRB = VBB − VBE
• Solving for IB:
B
BEBBB R
VVI
28
Transistor Currents & Voltages• The voltage at the collector with respect
to grounded emitter isVCE = VCC – VR(C)
• As the potential drop across RC is VR(C) = ICRC
The voltage at the collectorVCE = VCC − ICRC
where IC = DCIB.
• The voltage across the reverse biased collector-base junction is
VCB = VCE − VBE
29
Example
Determine IB, IC, IE, VBE, VCE, and VCB in the following circuit. The transistor has DC = 150.
RC
100 W
VCC
10 V
VBB
5 V
RB
10 kW
30
Solution• Known VBE = 0.7 V (forward biased),
IB = (VBB - VBE) / RB = 430 A
• The collector current,IC = DCIB = 64.5 mA
• Kirchhoff's current law:IE = IC + IB = 64.9 mA
• Solving for VCE and VCB:VCE = VCC – ICRC = 3.55 VVCB = VCE – VBE = 2.85 V
• The collector is at higher potential than the base (VCB > 0 VC > VB), the collector-base junction is reverse-biased.
31
Current gain, DC Beta
• The dc beta (DC) rating – ratio of dc collector current to dc base current.
• It is a ratio of current values, thus it has no unit of measure.
• Typical beta ratings can be as high as 300 i.e. the IC can be up to 300 times the value of IB.
• It can be used to define other terminal currents:IC = IB
IE = IC + IB = IB + IB
IE = IB( + 1)
B
C
I
IDC
32
DC Beta
• DC is also designated by an equivalent hybrid (h) parameter:
hFE = DC
• Transistor data sheets do not provide DC but hFE.
• DC is not truly constant. It varies slightly with IC and temperature.
• Transistors have both dc beta ratings and ac beta ratings (which will be discussed later)
33
Example
34
Example Determine the values of IC and IE for the circuit shown.
SolutionIB = 125 A
IC = IB
IC = (200)(125 A)IC = 25 mA
IE is found asIE = IC + IB IE = 25.125 mA
35
Example Determine the values of IC and IB for the circuit shown.
SolutionIB can be determine from
Now, IC can be found as IC = IE − IB= 14.9 mA
orIC = IB = 14.9 mA
A
6.741
E
BI
I
36
Example Determine the values of IB and IE for the circuit shown.
Solution…
The emitter current IE = IC + IB = 50.125 mA
37
DC Alpha • dc alpha () rating – the ratio of collector current to emitter
current.
• Typical values of DC range from 0.95 to 0.99 or greater. • Like beta, it has no units.
• It can be used to define other terminal currents, as follows:
IB = IE − IC = IE (1 − )
1E
C
I
I
C
E
ECI
I
II
38
The Relationship between Alpha and Beta
• The spec. sheet for a given transistor lists the value of beta for the device, but not the value of alpha as beta is used far more commonly than alpha.
• Alpha can be determined using the value of beta with the following equation:
1
39
ExampleDetermine the value of alpha for the transistor shown. Then, determine the value of IC using both the alpha and the beta ratings of the transistor.
Solution:
IC = IE = 29.9 mA
IC = IB = 30.0 mA
9967.01
40
Maximum Current Ratings
• Most transistor spec sheets list maximum collector current ratings for both saturation and cutoff.
• When the transistor is saturated, the collector current can go as high as several hundred mA. High-power transistors typically have current ratings as high as several amperes.
• The maximum allowable base current for a given transistor can be found by dividing its maximum IC value by its maximum dc rating.
max
(max)(max)
CB
II
41
ExampleThe transistor shown has the following ratings:
IC(max) = 500 mA and max = 300.Determine the maximum allowable value of IB for the device.
Solution:
If IB > IB(max) then IC > IC(max)
The transistor will probably bedestroyed.
42
Maximum Cutoff Current Ratings
• These ratings are usually in the low nanoampere (nA) range and are specified for exact values of VCE and reverse VBE.
• The 2N3904 has a maximum cutoff current rating of 50 nA when the reverse value of VBE is 3 V and the value of VCE is 40 V.
43
44
Transistor Voltage Ratings • It indicates the maximum amount of reverse bias that can be
applied to the collector-base junction (reverse biased for active region operation) without damaging the transistor.
• The value of VCB is equal to the difference between the other two voltages: 39.25 V. If this voltage exceeds the VCB rating of the transistor, the component will probable be destroyed.
45
Transistor Voltage Ratings
• Every transistor has three breakdown voltage ratings. • These ratings indicate the maximum reverse voltages that
the transistor can withstand. • For the 2N3904, these voltage ratings are as follows:
Rating Value (Vdc)
VCBO60
VCEO40
VEBO6
46
Transistor Characteristic Curves
• The three curves are (i) The collector curves, (ii) The base curves and (iii) The beta curves.
• The emitter curve is not part of the discussion as its current characteristics are the same as those of the collector.
47
Collector Curves• The collector characteristics
curve illustrates the relationship among IC, IB and VCE.
• Each collector curve is derived for a specified value of IB.
• Note that the IB = 0 A line represents the operation of the transistor when it is in cutoff.
• The collector curve is divided into three parts – saturation, active region, and breakdown.
48
Saturation Region Let VBB > 0, then IB ≠ 0. Set VCC = 0V VCE = 0
• For this condition, both the BE junction and CB junction are forward biased because the base is approximately 0.7V while the emitter and the collector are at 0V.
VCE = VCC – ICRC = 0 – 0(RC) = 0
• This represents the origin of the characteristic curve and is independent of IB.
O V
O.7 V
O V
49
Saturation Region
VCE < VK, VCE ≠ 0
• As VCC increases, VCE ↑ gradually, thus IC ↑.
• However, VCE remains less than 0.7V asVCE = VCC – ICRC
• This represents the portion of the graph where VCE < VK.
50
ExampleDetermine whether or not the transistor in the circuit below is in saturation. Assume VCE(sat) = 0.2 V.
gain = 50
RB
10kW
RC
1kW
VBB
3V
VCC
10V
51
SolutionFirst determine IC(sat):
IC(sat) = (VCC – VCE(sat)) / RC
IC(sat) = (10 V – 0.2V) / 1kW = 9.8 mA
Now let’s determine whether IB is large enough to produce IC(sat):IB = (VBB − VBE) / RB
= (3 V – 0.7 V) / 10kW = 0.23 mAIC =DCIB = (50) (0.23 mA) = 11.5 mA
• This shows that with the specified DC, this base current is capable of producing an IC greater than IC(sat).
• Thus, the transistor is saturated, and the collector current value of 11.5 mA is never reached. The collector current remains at its saturation value.
52
Active Region
• There is little change in the value IC when VCE increases from VK to VBR.
• IC is not controlled by the value of VCE when a transistor is operated in its active region.
Let VCC = constant,• IC increases with IB
• VCE decreases Until it reaches VCE(sat) somewhere below VK and it is usually only a few tenths of a volt for a silicon diode.
• IC cannot increase further(IC = DCIB is no longer valid).
53
Active Region
• Consider value held constant at 100, and IB is increased to 150 A: IC increases proportionately with increase in IB, and it is still relatively independent of changes in VCE.
• Alternatively, when VCE > 0.7V, the CB junction becomes reverse biased (transistor in active region)
• IC remains essentially constant for a given value of IB as VCE continues to increase.
• A slight increase in IC due to the widening of the CB depletion region. In the active region, the relation IC = DCIB holds.
54
Breakdown Region
• It happens when the value of VCE exceeds the breakdown voltage rating of the transistor.
• IC increases dramatically
• Note how VCE affects the CB junction. (1) If VCE < 0.7V, the CB junction is forward-biased. (2) If VCE > 0.7V, the CB junction becomes reverse-biased.
55
Cutoff• There is no current flowing in
the BE junction as VB = VE = 0. Hence, IB = IE = 0.
• The CB junction is reverse biased and hence, IC = 0.
• The two junctions are reverse-biased and ideally, no current should flow through the BE and CB junctions.
CB junctionreverse - biased
BE junction reverse - biased
RC
VCC
IB = 0
56
Cutoff
• Similar cutoff state can also be obtained by introducing a negative bias to the base.
• The negative bias is provided by VBB.
• The resistor RB is added to prevent the transistor from damage.
VC > 0
VE = 0RB
RC
VBB
VCC
VB < 0
57
Base Curves
• The base curve of a transistor plots IB as a function of VBE.
• This curve closely resembles the forward operating curve of a typical pn junction diode.
58
Beta Curves
59
Notations• Amplifier circuits have both dc and ac quantities for current,
voltage and resistance.
• Italic capital letters are used for both dc and ac currents (I) and voltages (V). Lowercase i and v for ac current and voltage are reserved for instantaneous values.
• DC quantities always carry an UPPERCASE Roman, nonitalic subscript. Example: IC, IB
• AC and all time-varying quantities always carry a lowercase italic subscript. Example: Ic, Ib
• Internal transistor resistances are designated by lowercase r’ with an appropriate subscript e.g. r’e refers to the internal ac emitter resistance.
60
Transistor as an Amplifier
• The figure above shows the basic transistor amplifier circuit. • Vin is superimposed on VBB. They are in series with the base resistor
RB. • VCC is connected to the collector through the collector resistor, RC.
r’e
RCRB
VCC
VBB
VinVcVb
61
VC
Vin
VBB VCC
Transistor as an Amplifier
• The ac input voltage produces an ac base current.• The collector current is related to the base current by
IC = DCIB
• This results in a much larger ac collector current.• The ac collector current produces an ac voltage across RC,
producing an amplified, but inverted, reproduction of the ac input voltage in the active region of operation.
62
Transistor as an Amplifier• The forward-biased base-emitter junction presents a
very low resistance to the ac signal. This internal ac emitter resistance is designated r’e. The ac emitter current is
Ie Ic =
• The ac collector voltage, Vc equals the ac voltage drop across RC.
Vc = IcRC IeRC
• The ac voltage gain, Av, of the transistor circuit
'e
b
r
V
''e
C
ee
Ce
b
cv
r
R
rI
RI
V
VA
Vb = Vin IbRB
63
Example
• Determine the voltage gain and the ac output voltage in the circuit on the right if r’e = 50 W.
SolutionThe voltage gain is
Av =
Av =
Therefore the ac output voltage is Vout = AvVb
= (20)(100 mV) = 2 Vrms
RC
1.0 kRB
VCC
VBB
VinVout
100 mV
'C
er
R
20WW
50
0.1 k
64
Transistor as a Switch• In part (a) the transistor is in the cutoff region because the base-
emitter junction is not forward-biased. There is ideally an open between the collector and emitter.
• Neglecting leakage current, all of the currents are zero, and VCE is equal to VCC.
VCE(cutoff) = VCC
65
Transistor as a Switch
• In part (b), the transistor is in the saturation region (base emitter and base collector junctions are forward-biased) and the base current is made large enough to cause the collector current to reach its saturation value.
• There is a short between collector and emitter. Actually, a voltage drop of up to a few tenths of a volt normally occurs, which is saturation voltage, VCE(sat).
• The maximum collector current,
• The minimum value of base current needed to produce saturation is
•
C
)(CECC)(C R
VVI sat
sat
DC
)(C(min)B
satII
66
A Simple Application of a Transistor Switch
• The transistor is used as a switch to turn the LED on and off.
• A square wave with a period of 2 s is applied to the input.
• When the square wave is at 0 V, the transistor is in cutoff and, there is no collector current, the LED does NOT emit light.
• Square wave goes to its high level, the transistor saturates. It forward-biases the LED. The resulting collector current through the LED causes it to emit light.
RB
RC
VCC
ON ON
OFF1 s
Result: A blinking LED that is on for 1 s and off for 1 s.
67
Example
The LED shown requires 30 mA to emit sufficient level of light. Therefore, the collector current should be approximately 30 mA.
For the following circuit values: VCC = 9 V, VCE(sat) = 0.3 V,
RC = 270 W, RB = 3.3 kW, and DC = 50.
IC(sat) =
IB(min) =
mAVV
R
VV2.32
270
3.09
C
CE(sat)CC W
AmAI
64450
2.32
DC
C(sat)
68
To ensure saturation, use twice the value of IB(min) which gives 1.29 mA. From the circuit,
IB =
Rearrange the expression above to obtain the voltage amplitude of the square wave input Vin.
Vin − 0.7 V = 2IB(min)RB = (1.29 mA)(3.3 kW)
Vin = 4.96 V
W
k
VV
R
VV
R
VininRB
3.3
7.0
B
BE
B
69
Example For the circuit shown, What is VCE when VIN = 0 V?
Sol: When VIN = 0 V, the transistor is in cutoff (it acts like an open switch) and
VCE = VCC = 10 VRB
RC
1.0kW
VCC
+10 V
VIN
70
Example
What minimum value of IB is required to saturate
this transistor if DC is 200? Neglect VCE(sat).
Sol: Since VCE(sat) is neglected (assumed to be zero),
This is the value of IB necessary to drive the transistor to the point of saturation. Any further increase in IB will drive the transistor deeper into saturation but will not increase IC.
RB
RC
1.0kW
VCC
+10 V
VIN
mAk
V
R
VI sat 10
0.1
10
C
CC)(C
W
A50
200
10
DC
)(C(min)B
mAII sat
71
Example (c) Calculate the maximum value of RB
when VIN = 5 V.
Sol: When the transistor is on, VBE = 0.7 V. The voltage across RB is
VR(B) = VIN – VBE
= 5 V − 0.7 V = 4.3 V
The maximum value of RB needed to allow a minimum IB of 50 A by Ohm’s law as follows:
RB
RC
1.0kW
VCC
+10 V
VIN
W 86kA
V
I
VR
B
RB
B
50
3.4
(min)(max)
72
Chapter 2 BJT – DC Biasing Circuits
DC Load Line Q-point Base Bias, Emitter Bias, Voltage Divider Bias, and Collector Feedback Bias
73
Introduction
• The purpose of dc biasing is to set the initial values of IB, IC and VCE. The ac operation of a transistor amplifier depends on its initial dc values of IB, IC, and VCE.
• As IB varied from an initial value, IC and VCE are varied from their initial values.
74
DC Operating Point
• DC operating point/Q-point (quiescent point)• It is a point on the collector characteristic curve:
(IC, VCE) with IB = constant. It is set by properly biasing the transistor amplifier.
• Amplifiers are the most common linear devices. A linear amplifier preserves its input/output waveform.
75
DC Operating Point • If an amplifier is not properly biased, it will go
either into cutoff or saturation and the amplified signal will be distorted – nonlinear amplifier.
• A nonlinear amplifier does not preserve its input/output waveform.
• Example of a nonlinear amplifier – a device that turns your voice into a chipmunk (cartoon) voice. Novel purpose but in daily phone conversation?
76
The DC Load Line • Notice that when IB
increases, IC increases and VCE decreases.
• When IB decreases, IC decreases and VCE
increases.
• As VBB is adjusted up or down, the dc operating point of the transistor moves along a sloping straight line, called the dc load line.
77
The DC Load Line • The dc load line is a graph that represents all
the possible combinations of IC and VCE for a given amplifier.
• The dc load line represents all the IC-VCE combinations for the circuit.
78
The DC Load Line• The IC(sat) point represents the ideal value of
saturation current for the circuit. • Consider a saturated transistor as being short-
circuited from emitter to collector, then VCE is zero, thus
VCC = ICRC (saturation)• Rearranging the expression for IC, we get• Recall:
C
CCsatC R
VI )(
79
The DC Load Line • When the transistor is in cutoff, it looks like an
open circuit from emitter to collector. Collector current, IC = 0
VCE(off) = VCC
• These equations are used to plot the end points of the dc load line.
Recall:
80
ExamplePlot the dc load line for the
circuit shown.
SolutionFrom the circuit values shown,
VCE(off) = VCC = 12 V
and
The load line therefore has end points of VCE(off) = 12 V and IC(sat) = 6 mA
6mAW
kR
VI
C
CCsatC 2
12)(
81
Example(a) Plot the dc load line for the circuit
shown below. (b) Then, verify the load line VCE values for IC
= 1 mA, IC = 2 mA and IC = 5 mA.
82
Solution From the circuit values shown,VCE(off) = VCC = 10 V
and
Using IC(sat) and VCE(off) as the end points, the dc load line for the circuit is plotted as shown.
IC (mA)
1 2 5
VCE (V) 9 8 5
C
CCsatC R
VI )(
As shown in the figure, we would expect to obtain:
83
Solution These values are verified by the following equation
VCE = VCC − ICRC
For IC = 1 mA,VCE = VCC − ICRC =1V − (1 mA)(1 kW) = 9V
For IC = 2 mA, VCE = = 8V
For IC = 5 mA,VCE = = 5V
These calculations verify the values obtained from the dc load line.
84
The Q-Point • When a transistor does not have an input signal,
its output rests at specific dc values of IC and VCE. i.e. the Q-point on the dc load line.
• The Q-point can be easily determined if the dc load line is superimposed onto the collector curve for the transistor.
85
The Q-Point (Linear Amp)
• For linear operation of an amplifier, it is desirable to have the Q-point centered on the load line.
• VCE is half the value of VCC, and IC is half the value of IC(sat).
• When a circuit is designed to have a centered Q-point, the amplifier is said to be midpoint biased.
86
The Q-Point (Midpoint biased)
• It allows optimum ac operation of the amplifier.
• When an ac signal is applied to the base of the transistor, IC and VCE both vary around their Q-point values.
• When the Q-point is centered, IC and VCE can both make the maximum possible transitions above and below their initial dc values.
87
ExampleThe circuit on the right shows the linear operation of a transistor. Assume a sinusoidal voltage Vin is superimposed on VBB, causing the base current to vary sinusoidally 100 A above and below its Q-point value of 300 A. Consider its operation as a linear amplifier.
88
• The collector current varies 10 mA above and below its Q-point value of 30 mA.
• The collector-to-emitter voltage varies 2.2 V above and below its Q-point value of 3.4 V.
mAAII bDCC 20)200)(100(
mAAII bDCC 40)400)(100(
VmAVRIVV
mAAII
Ak
VV
R
VVI
CCQCCCEQ
BQDCCQ
B
BBBQ
4.3)220)(30(10
30)300)(100(
30010
7.07.37.0
W
W
Q-point values
VmAVRIVV CCCCCE 2.1)220)(40(10 W
VmAVRIVV CCCCCE 6.5)220)(20(10 W
89
Point A on the load line corresponds to the positive peak of the sinusoidal input voltage while point B corresponds to the negative peak as shown.
90
• When the Q-point is above center of the dc load line, the input may cause the transistor to saturate. When this happens, part of the output sine wave is clipped off.
• If the Q-point is below midpoint on the load line, the input may cause the transistor to go into cutoff. This can also cause a portion of the output sine wave to be clipped.
91
Transistor is driven into saturation and cutoff because the input signal is too large.
IBQ
ICQ
VCEQ
92
ExampleDetermine the Q-point for the circuit below and find the maximum peak value of base current for linear operation. Assume DC= 200.
93
Solution
The Q-point is defined by ICQ and VCEQ.
IBQ =
= 198 μA
ICQ = βDCIB = (200)(198 μA) = 39.6 mA
VCEQ = VCC – ICRC = 20 V – 13.07V = 6.93 V
Thus the Q-point is at ICQ = 39.6 mA and VCEQ = 6.93 V.
B
BB
R
VV 7.0
94
Solution Since IC(cutoff) = 0, calculate IC(sat) to determine how much variation in collector current can occur and still maintain linear operation.
IC(sat) = 60.6 mA
Thus, IC can increase, ideally, by
IC(sat) – ICQ = 21 mA
And it can decrease by 39.6 mA before cutoff (IC = 0) is reached.The limiting value is then, 21 mA.The maximum peak variation of the base current is:
Ib(peak) = Ic(peak) / βDC = 105 μA
95
Bias Methods(A)Base Bias / fixed bias• Simplest form of biasing
96
Base Bias – circuit analysis• The primary goal of biasing circuit analysis is to
determine the Q-point values (IC, VCE) to determine whether the circuit is midpoint biased.
Step 1: Determining the base current, IB (Ohm’s law)
The voltage across RB equals the difference between
VCC and VBE. By formula,
VBB = VCC − VBE
Upon substitution, the base current
B
BECCB R
VVI
B
BBB R
VI
97
Base Bias – circuit analysis
Step 2: The collector current is then calculated from
Step 3: VCE can be found as
VCE = VCC − ICRC
The last two equations give the Q-point values of IC and VCE.
BC IhI FE
98
Example(a) Determine the Q-
point values of IC and VCE for the circuit shown below.
(b) Construct the dc load line and plot the Q-point from the values in part (a). Determine whether the circuit is midpoint biased.
(c) Determine whether the circuit is midpoint biased without drawing a dc load line for the circuit.
99
Solution
First, IB is found as
Next, IC is found as
Finally, VCE is found as
VCE = VCC − ICRC = 3.94 V
A2820 .360
7.08
W
k
VV
R
VVI
B
BECCB
028mA2. BFEC IhI
100
Solution (b) The end points of the dc load line are found as
IC(sat) =
and VCE(off) = VCC = 8 V
4mAW
k
V
R
V
C
CC
2
8
The dc load line and the Q – point is as plotted. The amplifier is very nearly midpoint biased.
101
Solution
By definition, a circuit is midpoint biased when the Q-point value of VCE is ½ VCC.
From part (a), VCE = 3.94 V which is approximately one-half of VCC.
The circuit is midpoint biased.
102
Base Bias – Q-point Shift• Base bias circuits are easy to build and analyze
but they are extremely susceptible to a problem called Q-point shift (change in the Q-point values).
• From circuit analysis performed earlier :VCE = VCC − ICRC
(1) VCE will change if IC changes.
(2) IC is dependent on either IB or hFE changes.
• This circuit is referred to as a beta – dependent circuit.
• It is primarily used in switching applications where it mainly operates in saturation and cutoff regions.
BC IhI FE
103
Base Bias – Q-point Shift• The dc gain varies with temperature.
• If temperature increases, hFE also increases, thus, IC increases. This, in turn, causes a decrease in VCE. When this occurs, the circuit will no longer be midpoint biased.
104
ExampleThe transistor below has values of hFE = 100 when T = 25 oC and hFE = 150 when T = 100 oC. Determine the Q-point values of IC and VCE at both of these temperatures.
105
Solution• This is the same circuit as in the previous
example. • At T = 25 oC, hFE = 100 (IB = 20.28 A):
IC = 2.028 mA and VCE = 3.94 V
• When T = 100 oC, hFE = 150 (assume IB does not change) :
IC = hFEIB = (150)(20.28 A) = 3.04 mA
andVCE = VCC − ICRC = 1.92 V
• The percentage change in IC:
• The percentage change in VCE:
%50%100028.2
028.204.3%
mA
mAmAIC
%51%10094.3
94.392.1%
V
VVVCE
106
Voltage-Divider Bias• It is the most commonly used biasing scheme, similar in form
to base bias, with the following exceptions:(1) A resistor (R2) has been added between the base terminal of the transistor and ground.(2) A resistor (RE) has been added to the emitter circuit.
• These modifications results in a biasing circuit with values of ICQ and VCEQ that are
relatively stable against variations in hFE.
107
Circuit Operation • Resistors R1 and R2 form a simple voltage divider that sets
the value of base voltage, VB as follows:
• The value of VE can be found asVE = VB − 0.7 V
• Then the value of IE is determined from Ohm’s law:
• Assume that ICQ IE, thus, VCEQ can be found asVCEQ = VCC − ICQRC − ICQRE
VCEQ = VCC − ICQ(RC + RE)
CC21
2B V
RR
RV
E
EE R
VI
108
Circuit Operation
• The value of IB can be found from
• Recall: hFE can be found from the spec sheet of the transistor.
• A transistor spec sheet will list any combination of the following values:(a) a maximum value of hFE
(b) a minimum value of hFE
(c) a typical value of hFE
1E
B
FEh
II
109
Circuit Operation • If only one value of hFE is listed, it can be used in any
circuit analysis.
• When two or more values are listed, identify the typical value and use the typical value in circuit analysis.
• If only a minimum value and a maximum value are shown, use the geometrical average of these two values. The geometrical average of hFE is found as
(max)(min))( FEFEaveFE hhh
110
Example
Determine the value of ICQ and VCEQ for the circuit shown below.
SolutionThe base voltage is determined from the voltage divider
VE is found as VE = VB − 0.7 V = = 1.37 V
Because ICQ IE,
Finally,VCEQ = VCC − ICQ(RC + RE)
= 4.87 V
VVRR
RV CCB 07.2
21
2
25mA1.1.1
37.1
W
k
V
R
VI
E
ECQ
111
Base input resistance, RIN(base)
• From the previous example,
and I2 =
• As IB << I2, it can be ignored.
• However, if IB is NOT small enough to be neglected compared to I2, then the dc input resistance RIN(base) must be taken into consideration.
AkR
VB 4407.4
07.2
2
W
AmA
h
II
FE
5.2451
25.1
1E
B
112
Derivation of RIN(base)
• The input resistance of the transistor isRIN(base) =
• Applying KVL and assuming VBE << IERE VIN = VBE + IERE IERE
• Since IE ≈ IC = βDCIB, VIN becomesVIN ≈ βDCIBRE
• The input current is the base current, IIN = IB.
IN
IN
I
V
Note: Arrows indicated the direction of conventional current flow.
o Substituting VIN and IIN into the RIN(base):
The input resistance of the transistor is the gain βDC times the emitter resistance RE.
EDCbaseIN
B
EBDCbaseIN
RRI
RIR
)(
)(
113
Example
Determine the dc input resistance looking in at the base of the transistor in the circuit shown in previous slide if RE = 1.0 kW and DC = 125.
SolutionRIN(base) = βDC RE
RIN(base) = (125)(1.0 kW) = 125 kW
114
Analysis of a Voltage-Divider Bias Circuit
When the base current, IB is NOT negligible (??), then the base input resistance, RIN(base) must be considered in the determination of the value of base voltage.Note that the RIN(base) is parallel to the resistor R2 of the voltage-divider.
115
Analysis of a Voltage-Divider Bias Circuit
• The total resistance from base to ground is R2 || RIN(base) = R2 || DCRE
and it is in series with resistor R1. Applying the voltage divider formula yields:
• From this point onwards, the analysis follows that the procedure as in “circuit operation” section. Note that if DCRE > 10R2 then the expression above will revert to
CCE21
E2B )||(
||V
RRR
RRV
DC
DC
CC21
2B V
RR
RV
116
ExampleDetermine VCE and IC in the voltage-divider biased
transistor circuit on the right if DC = 100.SolutionDetermine the dc input resistance at the base:RIN(base) = βDCRE =(100)(560) = 56 kΩ
Since RIN(base) = 10R2, we may neglect RIN(base).
Thus, = 3.59 V
So, VE = VB – VBE = 2.89 V and from Ohm’s law,IE = 5.16 mAIC ≈ 5.16 mA
and VCE ≈ VCC – IC(RC + RE) = 1.95 VSince VCE > 0, the transistor is NOT in saturation.
CC21
2B V
RR
RV
117
Bias Stability
The Thevenin voltage and resistance are:
RTH = R1R2 / (R1 + R2)
Apply KVL around the equivalent base-emitter loop:VTH – VR(th) – VBE – VR(E) = 0
VTH = IBRTH + VBE + IERE
But IB = IE / βDC,VTH = IE (RE + RTH/βDC) + VBE
Solve for IE:IE = (VTH – VBE) / (RE + RTH/βDC)
As usual, assume RE >> RTH / βDC, then
IE = (VTH – VBE)/RE
CCTh VRR
RV
21
2
118
Bias Stability
Note that the expression for IE is independent of βDC. Varying βDC will not affect IE and IC is also unaffected by βDC.
Remember that for this type of biasing (voltage-divider) we must make sure RE is at least ten times RTH/βDC. This is not an unreasonable assumption, since βDC is generally large.
119
Emitter Bias
An emitter-bias circuit consists of several resistors and a dual-polarity power supply, a positive and a negative supply voltage
Current action:
(i) the emitter current originates at the emitter power supply (− VEE).
(ii) A small portion of the emitter current leaves the transistor through the base terminal. Base current passes through RB to ground.
(iii) the majority of the emitter current continues to the collector. Collector current passes through RC to VCC.
120
Circuit Currents and Voltage • Emitter-bias circuits are always
designed so that VB 0 V.
• Assume IB = 100 A, then the base voltage
VB = IBRB = (100 A)(100 W) = 10 mV
• The circuit can be represented by figure (a).
• Figure (b) shows an equivalent circuit using a diode to represent the base-emitter junction of the transistor. As this circuit illustrates,
VE = 0 V − VBE = − VBE
VE
121
Circuit Currents and Voltage • According to KVL, VR(E) must equal the difference between VE and
VEE and since, VE = − VBE,
VR(E) = VEE + VBE
• The value of the emitter current,
• Assuming that ICQ IE and VBE = 0.7 V,
• Note: 1) The above absolute value is used to obtain a positive value of ICQ.
2) hFE is not involved in the equation - the emitter bias provides output values that are highly stable against variations in beta.
E
BEEEE R
VVI
E
EECQ R
VVI
7.0
122
Circuit Currents and Voltage • The approximate value of VCEQ can be determine using
VCEQ = VCC − ICRC
• This equation is based on the fact that VE = − 0.7 V. If we assume this potential to be correct, then KVL tells us that
− 0.7 V + VCEQ + ICQRC = VCC
orVCEQ = VCC − ICQRC + 0.7 V
• When ICQRC >> 0.7V, which is normally the case, we can ignore − VE, leaving us with the
VCEQ = VCC − ICQRC
123
ExampleDetermine the values of ICQ and VCEQ for the amplifier shown in figure below.
SolutionFirst, the value of ICQ is approximated as
The value of VCEQ is found asVCEQ = VCC − ICQRC
VCEQ = 12V−(7.53mA)(750W) = 6.35 V
mAk
VV
R
VVI
E
EECQ 53.7
5.1
7.0127.0
W
124
Collector – Feedback Bias• The term feedback is used to
describe a circuit connection that “feeds” a portion of the output voltage or current back to the input to control the circuit’s operating characteristics.
• In this case, the circuit is constructed so that the collector voltage, VC has a direct effect on the base voltage, VB.
• These feedback connections reduce the effects that variations in hFE have on Q-point values of each circuit.
125
Collector – Feedback Bias
• If IC tries to increases, it drops more voltage across RC, thereby causing VC to decrease.
• When VC decreases, there is a decrease in voltage across RB, which then decreases IB.
• The decrease in IB produces less IC which, in turn, drops less voltage across RC and thus offsets the decrease in VC.
126
Circuit Analysis• By Ohm’s law, the base current can be expressed as
• Substitute VC = VCC − ICRC and
into the first equation:
• Rearranging the above equation and solving for IC:
• Since the emitter is grounded, VCE = VC:VCE = VCC − ICRC
B
BECB R
VVI
DC
CB
II
B
BECCCC
DC
C
R
VRIVI
DCBC
BECCC RR
VVI
127
Circuit Analysis
• The last 2nd equation shows a dependency on DC and VBE. This can be minimized by making
RC >> and VCC >> VBE.
• This circuit essentially eliminates DC and VBE dependency even if the stated conditions are met.
• This circuit is able to maintain a relatively stable Q-point values as temperature changes, either increases or decreases.
128
ExampleCalculate the Q-point values for the circuit below.
Solution
VCE = VCC − ICRC
VCE = 10 V − (2.82mA)(1.5 kW) = 5.77 V
DCBC
BECCC RR
VVI
82mA2.1001805.1
7.010
WW
kk
VVIC
129
Summary 1. The purpose of biasing a circuit is to establish a proper,
stable dc-operating point (Q-point) i.e. ICQ and VCEQ.
1. Graphically, the Q-point is the intersection of the dc load line and the transistor’s collector curves
1. A dc load line intersects the vertical axis at approximately IC(sat) and the horizontal axis at VCE(cutoff).
1. The linear (active) region of a transistor lies along the load line, below saturation, and above cutoff.
1. The dc input resistance at the base of a BJT is approximately βDCRE.
130
Summary Voltage-divider provides good Q-point stability with single polarity supply voltage. It is the most popular bias circuit.
Base bias circuit arrangement has poor stability because its Q-point varies widely with βDC.
Emitter bias generally provides good Q-point stability, but requires both positive and negative supply voltages.
Collector-feedback bias provides good stability using negative feedback from collector to base.
131
Amplifier Operations Introduction
Amplifiers are some of the most widely used circuits encounter e.g. in audio, video, telecommunications systems, digital systems, biomedical systems etc. etc.
For example: turning up the volume of a stereo. What actually happens is that you are taking a relatively weak signal and making it stronger i.e. increasing its power level. This is a form of amplification.
Amplifiers are circuits that provide amplification.
132
BJT Amplifiers
• They are known as small-signal amplifiers - it uses signals that take up a relatively small percentage of an amplifier’s operational range.
• The biasing of a transistor is purely a dc operation (C2). It establishes a Q-point about which variations in current and voltage can occur in response to an ac signal.
• Variations about the Q-point are relative small. • There are three types of BJT amplifiers:
- common-emitter (most commonly used), - common-collector and - common-base.
133
Linear Amplifier OperationIn an amplifier, there are two sources of currents and voltages: dc and ac. They are mixed together in the circuit. When we analyze the operation of an amplifier, we need to separate the dc component from the ac component.
134
• The coupling capacitors block dc and prevent the internal source resistance RS and the load resistance RL from changing the dc bias voltages at the base and collector.
• The sinusoidal source voltage causes Vb to vary sinusoidally above and below its VB. The resulting variation in Ib produces a larger variation in the Ic because of the current gain of the transistor.
135
• Note that the voltage created at the collector (VCE) is out-
of-phase with the input voltage (common-emitter amplifier).
(CB and CC amplifiers do not exhibit this behaviour.)
• As IC increases, the collector voltage decreases and vice versa. • This is caused by the relationship
VCE = VCC – IC(RC + RE).
• Note the negative sign on the right-hand-side of the equation. This sign cause VCE and IC to be out-of-phase.
136
Transistor AC Equivalent Circuits
• To visualize the operation of a transistor in an amplifier circuit, it is often useful to represent the device by an ac equivalent circuit.
• The elements forming the equivalent circuits relate the changes in voltages and currents about the operating point.
• The hybrid- equivalent circuit is as shown:
hie
hfev2 1 hoe
b
e
c
e
hrei1
137
h-Parameters (hybrid parameters)
• Data sheets only provide h parameters. In parenthesis are the conditions required to get these parameters.
1. hi : Input impedance (Output shorted)2. hr : Voltage feedback ratio (Input open)3. hf : Forward current gain (Output shorted)4. ho : Output admittance (Input open)
• A second subscript is added to indicate the configuration (common-emitter, common-base, and common-collector) for each of these parameters.
• E.g.: the h-parameters for the common-emitter (usually the one given in data sheets) configuration are: hie, hre, hfe, and hoe.
138
r - parameters
The r-parameters are:1. ac : ac alpha ( )
1. βac : ac beta ( )
1. : ac emitter resistance
1. : ac base resistance
1. : ac collector resistance
e
c
I
I
b
c
I
I
'er
'br
'cr
Assume that: is small enough to be neglected replace r’b with a short. is very large (hundreds of kW) replace r’c with an open.
'br'
cr
139
The conversion equations between r and h parameters:
ac =
ac =
h fb
feh
'er
oe
re
h
h
'cr
oe
re
h
h 1
'br fe
oe
reie h
h
hh 1
140
Interpretation of simplified equivalent circuit:
• is the resistance seen looking into the emitter of a forward-biased transistor. It appears between the base and the emitter terminals.
• The collector current Ic is equal to αac Ie. Also, Ic = βacIb.
The most important parameter in the simplified equivalent circuit is
'er
'er
EI
mV25
141
Common-Emitter (CE) Amplifier
Common-emitter BJT amplifier with voltage – divider biasing
142
Common-Emitter (CE) Amplifier
• C1 and C3 are the bias and coupling capacitors on the input and output, while C2 acts as bypass capacitor between the emitter to ground.
• The circuit has a combination of dc and ac operation, both of which must be considered.
• The input ac signal Vin is capacitively coupled into the base and the output signal Vout is capacitively coupled from the collector.
• It exhibit high voltage and current gains.• The input is applied to base of transistor and the output is
taken from collector.
143
DC Analysis
A dc equivalent circuit is developed by replacing the coupling and bypass capacitors with opens.
Recall: a capacitor is open to dc.
144
• As seen previously, the input resistance is RIN(base) = βDCRE = (150)(560) = 84 kΩ
which is 10 times more than R2 for this particular case. Thus, we can ignore it.
• This yields the following base voltage:
• And the emitter voltage isVE = VB – VBE = (2.83) – (0.7) = 2.13 V
• Therefore, the emitter current isIE = VE / RE = (2.13) / (569) = 3.80 mA
• Since IC ≈ IE, we get the collector currentVC = VCC – ICRC = (12) – (2.80m)(1.0kΩ) = 8.20 V
• Finally, the common-emitter voltage isVCE = VC – VE = (8.20) – (2.13)=6.07 V
VVk
kV
RR
RV 83.212
8.28
8.6CC
21
2B
WW
145
AC equivalent circuit
• For ac operation of the amplifier, develop an ac equivalent circuit by using the following rules:
1. Short all capacitors because we assume that their reactance XC 0 at the signal frequency.
2. Replace all dc sources with a ground symbol.
• These rules apply to all amplifier circuits, not just common-emitter amplifiers.
146
AC equivalent circuit
Capacitors C1, C2 and C3 are replaced with shorts. The dc sources are replaced with ground (assuming the internal resistance is very small). VCC terminal is at ac ground. AC and dc grounds are both assumed to be at the same potential (0 V).
147
AC equivalent circuit
• The emitter bypass capacitor, C2 provides an effective short to the ac signal around the emitter resistor, thus keeping the emitter at ac ground.
• This allows the amplifier to have maximum gain (RC/ ). It must be large enough so that its reactance over the frequency range is very small compared to RE.
• A good rule-of-thumb is 10XC < RE
'er
148
AC equivalent circuit• If the internal resistance, Rs of the ac source is 0 W, then
all of the source voltage appears at the base terminal. • If the ac source internal resistance Rs is non-zero, three
factors must be taken into account in determining the actual signal voltage at the base.
They are: 1. the source resistance (Rs), 2. the bias resistance (R1||R2), and 3. the input resistance (Rin(base)).
149
Input Resistance, Rin(base)• Rin(base) is an ac quantity (in contrast to RIN(base) which is a dc
quantity) and it is also known as the input impedance.
• The source voltage Vs is divided down by Rs and Rin(tot) so that the signal voltage at the base of the transistor Vb is
• If Rs << Rin(tot) , then Vb ≈ Vs ， s)(
)( VRR
RV
totins
totinb
150
Derivation for the input resistance Rin(base)
• The transistor is connected with the external collector resistor RC. The input resistance looking in at the base is
Rin(base) = Vin / Iin = Vb / Ib
• The base voltage isVb = Ie
• and the base currentIb ≈ Ic / βac.
• Substituting for Vb and Ib in the first equation above and since Ie ≈ Ic, we get
Rin(base) = βac
'er
'er
151
ExampleDetermine the signal voltage at the base of the
transistor in the circuit shown below. This circuit is the ac equivalent of the amplifier. Assume a 10 mV, 300 Ω signal source. IE was found to be 3.80 mA.
r ' e=25m
3.80m=6.58W
SolutionDetermine the ac emitter resistance,
Then, Rin(base) = βac = 1.05 kW
Next determine the total input resistance viewed from the source:Rin(tot) = R1 || R2 || Rin(base) = 873 Ω
152
SolutionThe source voltage is divided down by Rs and Rin(tot), so the signal voltage at the base is the voltage across Rin(tot).
There is attenuation of the source voltage due to the source resistance and amplifier’s input resistance acting as a voltage divider.
Instead of getting the full 10 mV at the base, we get only 7.44 mV.
V b=Rin( tot)
Rs +Rin(tot )V s= =7 . 44 mV
153
Voltage Gain of the CE Amplifier
• Voltage gain (Av) of the CE amplifier is the ratio between the ac output voltage (Vc) and input voltage (Vb).
• From the figure, Vb = Ie
Vc = acIeRC ≈ IeRC ???since αac 1.
• The voltage gain, from the base to collector , is then
'er
''e
C
ee
Ce
b
cv r
R
rI
RI
V
VA
154
Overall GainTo get the overall gain of the amplifier from the source voltage to collector voltage, the attenuation (due to the
internal source resistance Rs) of the input circuit must be included.
• The attenuation from source to base multiplied by the gain from base to collector is called the overall amplifier gain.
• The overall gain,= (attenuation) Av
'vA
155
ExampleCalculate the base-to-collector voltage gain of the amplifier shown next, with and without an
emitter bypass capacitor. There is no load resistor.Solution
From a previous example we know that = 6.58 Ω.
Without C2, the gain is
Av = RC / ( + RE) = 1.76
With C2 included, the gain is
Av = RC / = 152
The bypass capacitor makes a big difference!
'er
'er
'er
156
Further Notes
1. If a load of resistance RL is connected across the output of the amplifier, the total resistance is then
Rtot = RCRL / (RC + RL). The voltage gain is written as
Av = Rtot /
• If the load resistance RL >> RC, then Rtot RC and there is no change in the gain.
• If RL << RC, then Rtot = RL. The voltage gain is reduced.
• Ideally, we require RL to be as large as possible.
'er
157
Bypassing RE produces the maximum voltage
gain and stability problem. The ac voltage gain is dependent on re’, IE and on temperature, it will be unstable over temperature changes.
If we take the bypass off IE, the gain is decreased but RE overpowers in the gain calculation. This actually makes the circuit much less dependent on it.
We make a compromise by using a method called swamping. In swamping, we only partially bypass RE as shown next.
158
SwampingNote that the total external emitter resistance RE is
formed by two separate emitter resistors RE1 and RE2. • One of the resistors RE2 is bypassed
and the other is not. Both resistors affect the dc bias but only RE1 affects the ac voltage gain:
Av = RC / (re’ + RE1)
• If we make RE1 at least ten times larger than re’ (RE1 >> 10re’), then the effect of re’ is minimized and the approximate voltage gain for the swamped amplifier is
Av ≈ RC / RE1
159
ExampleFor the amplifier shown next, determine the total
collector voltage and the total output voltage, both dc and ac. Draw the waveforms.
160
SolutionStep (1): DC Analysis
RIN(base) = βDC(RE1 + RE2) = 150 (940 Ω) = 141 kΩ
Since RIN(base) > 10R2, it can be neglected in the dc base voltage calculation.
VB ≈ [R2 / (R1 + R2)]VCC = [(10kΩ) / (47kΩ + 10kΩ)] 10V = 1.75 V
VE = VB – 0.7 V = 1.75 V – 0.7 = 1.05 VIE = VE / (RE1 + RE2) = 1.05 V / 940 Ω = 1.12 mA
VC = VCC − ICRC = 10 V – (1.12 mA) (4.7 kΩ) = 4.74 V
161
Step(2) AC Analysis
• Calculate re’:re’ = 25 mV / IE = 25 mV / 1.12 mA = 22 Ω
• Determine the attenuation in the base circuit. Looking from the 600 Ω source resistance, the total Rin is
Rin(tot) = R1||R2||Rin(base)
Rin(base) = βac(re’ + RE1) = 175 (492 Ω) = 86.1 kΩRin(tot) = 47 kΩ||10 kΩ||86.1 kΩ = 7.53 kΩ
162
• The attenuation from source to base isAttenuation = Vb / Vs = Rin(tot) /(Rs + Rin(tot))
= 7.53 kΩ / (600 Ω + 7.53 kΩ) = 0.93
Before Av can be determined, we need to know the ac collector resistance:
Rc = RCRL/(RL + RC) = (4.7 kΩ) (47 kΩ) / (4.7 kΩ + 47 kΩ)= 4.27 kΩ
• Now we are ready to calculate the gain from base to collector:Av ≈ Rc /RE1 = 4.27 kΩ / 470 Ω = 9.09
• And the overall voltage gain is the attenuation times the amplifier voltage:
A’v = (Vb / Vs) Av = (0.93) (9.09) = 8.45• Since the source produces 10 mVrms , the rms voltage at the
collector will be Vc = A’v Vin = (8.45) (10 mV) = 84.5 mV
163
Step (3) Plot waveforms
The total collector voltage is the signal voltage of 84.5 mVrms riding on a dc level of 4.74 V. The peaks are
Max Vc(p) = 4.74 + (84.5 mV) (1.414) = 4.86 VMin Vc(p) = 4.74 - (84.5 mV) (1.414) = 4.62 V
164
• The coupling capacitor C3 keeps the dc level from getting to the output, so Vout is equal to the ac portion of the collector voltage (Vout(p) = 119 mV).
• Source voltage is shown to emphasize phase inversion.
165
Common-Collector (CC) Amplifier Emitter-follower amplifier (EF).
• The input is applied to base through a coupling capacitor and the output is at the emitter. The voltage gain of a CC amplifier is approximately 1.
• Its main advantage is its high input resistance and current gain.
166
Voltage Gain• From the ac equivalent circuit
Vout = IeRe
andVin = Ie(re’ + Re)
• The voltage gain is
• Note that Re = RE||RL and Re = RE
when there is no load.
ee
e
eee
ee
in
outv Rr
R
RrI
RI
V
VA
''
In practice, Re >> re’ , henceAv ≈ 1
that is, the voltage gain of a CC amplifier is approximately unity.
167
b
eee
I
RrI )( '
b
eebac
I
RrI )( ' 'er
Input Resistance Rin(base)
• CC amplifier has high input resistance - it can be used to minimize loading effects when a circuit is driving a low resistance load. Note: the emitter resistance is never bypassed.
• The derivation of the CC amplifier input resistance is similar to that of the CE amplifier:
Rin(base) = Vin / Iin = Vb / Ib
=
• Since Ie ≈ Ic = βacIb ,
Rin(base) = = βac ( + Re)
168
• If Re >> , the input resistance becomesRin(base) = βacRe
• From the CC amplifier circuit, we can see that the bias resistors appear in parallel with Rin(base), looking from the input source. Thus, the total input resistance becomes
Rin(tot) = R1 || R2 || Rin(base)
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Output Resistance
• The output resistance is very low. It is given by the following expression:
Rout
where Rs is the resistance of the input source.(FYI: Floyd/ pg. 937)
Eac
s RR
||
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Current Gain
The current gain is
• The input current, Iin = Vin / Rin(tot). If , then most of input current goes into the base. Thus, current gain of amplifier is almost equal to current gain of the transistor, βac, that is
Ai βac =
• This is because very little signal is diverted to the bias resistors. Otherwise,
• βac is the maximum achievable current gain in both CC and CE amplifiers. Recall Ie βacIb and Iin = Ib.
in
e
in
outi I
I
I
IA
)(21 || baseinRRR
b
c
I
I
in
ei I
IA
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Power Gain
• The CC power gain is
• Since Av ≈ 1, the overall power gain is
Ap ≈ Ai.
viinin
outout
in
outp AA
VI
VI
P
PA
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ExampleDetermine the total input resistance of the emitter follower shown below. Also find the voltage gain, current gain, and power gain in terms of power delivered to the load, RL. Assume βac = 175. And that the capacitive reactances are negligible at the frequency of operation.
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Solution
• The ac emitter resistance external to the transistor, Re, is:Re = RE || RL = 1 kΩ || 1kΩ = 500 Ω
• The approximate resistance, looking in at the base, is:Rin(base) ≈ βac Re = (175) (500 Ω) = 87.5 kΩ
• The total input resistance isRin(tot) = R1||R2||Rin(base) = 18kΩ || 18kΩ || 87.56kΩ = 8.16 kΩ
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Solution • The voltage gain is almost unity (CC amplifier). By using
we can determine a more precise value for Av:VE = [R2/(R1 + R2)]VCC – VBE
= (0.5)(10 V) – 0.7 V = 4.3 VIE = VE / RE = 4.3 V / 1.0 kΩ = 4.3 mA = 25 mV/IE = 25 mV / 4.3 mA = 5.8 Ω
SoAv = Re / ( + Re) = 500 Ω / 505.8 Ω = 0.989
• The small difference in Av as a result of considering re’ is insignificant in most cases.
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Solution
• The current gain is Ai = Ie / Iin:Ie = Ve / Re = AvVb / Re ≈ 1 V / 500 Ω = 2 mA
Iin = Vin / Rin(tot) = 1 V / 8.16 kΩ = 123 μAAi =Ie / Iin = 2 mA / 123 μA = 16.3
• The overall power gain is Ap ≈ Ai = 16.3
Since RL = RE, one half of the total power is dissipated in RL. So, in terms of power to the load, the power gain is one half of the overall power gain.
Ap(load) = Ap / 2 = 16.3 / 2 = 8.15
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The Darlington Pair• The maximum achievable input
resistance you can get from a given CC circuit is limited by ac.One way to boost the input resistance is to use a darlington pair.
• The collectors of the two transistors are connected and the emitter of the first drives the base of the second.
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• This configuration achieves βac multiplication, i.e.,Ie2 = βac2 βac1 Ib1.
• Thus, the effective current gain in the darlington pair is ac = βac2 βac1
• Neglecting r’e by assuming that it is much smaller than RE, the input resistance is
Rin = βac2βac1RE
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• CC is often used as an interface between a circuit with a high output resistance and a low resistance load. In such an application, it is called a buffer.
• E.g.: suppose a common-emitter amplifier with a 1.0 kΩ collector resistance (output resistance) must drive a low-resistance load e.g. an 8 Ω low-power speaker that is capacitively coupled to the output of amplifier. The 8 Ω load (appears to the ac signal) in parallel with the 1 kΩ collector resistor. This results in an ac collector resistance of
Rc = RC || RL = 1kΩ || 8Ω = 7.94 Ω.• Obviously this is not acceptable, since most of the voltage gain is lost
Av = Rc / .For example, if = 5 Ω, the voltage gain is reduced from
Av = RC / = 1 kΩ / 5 Ω = 200to
Av = Rc / = 7.94 Ω / 5 Ω = 1.59• We can add a darlington pair to interface the amplifier and the speaker.
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For the CE amplifier : VCC = 12 V, RC = 1.0 kWand re’ = 5 W.For the Darlington EF : R1 = 10 kW R2 = 22 kWRE = 22 W RL = 8 W.DC = ac = 100 for both transistors.
(a) Determine the voltage gain of the CE amplifier.(b) Determine the voltage gain of the Darlington EF
amplifier.(c) Determine the overall voltage gain and compare to the
gain of the CE amplifier driving the speaker directly without the Darlington EF.
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SolutionThe total input resistance of the Darlington emitter-follower amplifier will act as a load to CE amplifier. In order to determine the voltage gain for the CE amplifier for this
circuit, it is necessary to calculate Rin(tot) EF. VB = (R2 / R2 + R1)VCC = (22/32)12V = 8.25 V
VE = 8.25 V – 1.4 V = 6.85 V
IE = 6.85 V/22 W = 0.311 A
re’ = (25 mV/IE) = 80.3 mW
Re = RE || RL = 22 W || 8 W =5.87 W
The input resistance of the Darlington EF, Rin(base) EF = ac1 ac2Re = (100)2(80.3 mW5.87 W) =
59.5 kW The total input resistance,
Rin(tot) EF = R1|| R2||Rin(base) EF = 6162.9 W
Thus, the ac output resistance of CE, Rc = RC||Rin(tot) EF = 860.4 W
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Solution cont.
Voltage gain for CE amplifier with Darlington EF, Av = Rc /re’ = 860.4/5 = 172
Voltage gain for Darlington amplifier, Av EF = 0.987
Overall voltage gain, Av’ = 170
Voltage gain for CE amplifier w/out Darlington EF, Av = Rc /re’ = 7.94/5 = 1.59
Using the same circuit determine Av’ is a single transistor is used in the EF amplifier instead of Darlington pair.
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Common-Base (CB) Amplifier•The CB amplifier provides high voltage gain with a maximum current gain of 1 and it has low input resistance. •CB amplifiers are most appropriate for certain applications where sources tend to have very low-resistance outputs.
•The base is the common terminal and is at ac ground because of the capacitor C2. •The input signal is capacitively coupled to the emitter. •The output is capacitively coupled from the collector to a load resistor.
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Voltage Gain
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Since the input voltage is Vin = Ve and the output voltage is Vout = Vc, the gain becomesAv =
e
c
in
out
V
V
V
V Eee
cc
RrI
RI
||' Eee
ce
RrI
RI
||'
Assuming that RE >> , we get Av ≈ ≈ (RC||RL) /
Note: The gain expression is the same as for the common emitter amplifier. However, there is no phase inversion from emitter to collector.
'e
c
r
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Input Resistance
• The resistance looking into the emitter isRin(emitter) =
• Assuming RE >> , thenRin(emitter) ≈
• Typically, RE is much greater than , so the assumption
RE >> used is usually valid.
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e
Eee
e
e
in
in
I
RrI
I
V
I
V )||( '
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Output Resistance, Current Gain, Power Gain
Output Resistance• Looking into the collector, the ac collector resistance, rc’,
appears in parallel with RC. As rc’ is much larger than RC,Rout ≈ RC
Current Gain• It is defined as Ai = Iout / Iin. Thus we have
Ai = Ic / Ie ≈ 1
Power Gain• Since the current gain is approximately unity, by Ap = Ai Av, we
haveAp ≈ Av
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ExampleFind the input resistance, voltage gain, current gain, and power gain for the amplifier shown below. Firstly, find IE so that re’ can be determined.
Then Rin ≈ re’. Since βDCRE >> R2, thenVB = R2 / (R1 + R2) VCC = 1.76 V
VE = VB – 0.7 V = 1.06 VIE = VE / RE = 1.06 mA
Therefore,Rin ≈ re’ = 25 mV / IE = 23.6 Ω
The voltage gain is found as follows.Rc = RC||RL = 1.8 kΩAv = Rc / re’ = 76.3
We also know Ai = 1 and Ap ≈ Av = 76.3.
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Amplifier ComparisonsCE CC (or EF) CB
Input Base Base Emitter
Output Collector Emitter Collector
Inversion Yes No No
Voltage gain High (RC/r’e) Low (≈ 1) High (RC/r’e)
Current gain High (ac) High (ac) Low (≈ 1)
Power gain Very High (AiAv)
High (≈ Ai) High (≈ Av)
Input Resistance Low (acr’e) High (acRE) Very low (r’e)
Output Resistance High RC
Very low (Rs/βac)||RE
HighRC
Frequency Range Medium Medium High
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Summary
• A small signal amplifier uses only a small portion of its dc load line.
• r-parameters are easily identifiable and applicable with a transistor’s circuit operation.
• A common emitter amplifier has good voltage, current, and power gains, but a relatively low input resistance.
• A common collector amplifier has high input resistance and good current gain, but its voltage gain is approximately 1.
• The common base amplifier has a good voltage gain, but it has a very low input resistance and its current gain is approximately 1.
• A darlington pair provides β multiplication for increased input resistance.
