analysis of differential amplifiers by muhammad irfan yousuf [peon of holy prophet (p.b.u.h)]
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Analysis ofDifferential Amplifier
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Differential amplifier
A differential amplifier is a type of electronic amplifier that multiplies the differencebetween two inputs by some constant factor. Given two inputsVin
+andVin-, a practical
differential amplifier gives an outputVout:
Vout =Ad (Vin+- Vin-) +0.5 Ac(Vin++Vin-)
Where Ad is the differential mode gain and Ac is the common mode gain.Note: Differential amplifier is a more general form of amplifier than one with a singleinput; by grounding one input of a differential amplifier, a single ended amplifier results.
Applications: Differential amplifiers are found in many systems that utilize negativefeedback, where one input is used for the input signal, the other for the feedback signal.A common application is for the control of motors or servos, as well as for signalamplification applications. In discrete electronics, a common arrangement forimplementing a differential amplifier is the long-tailed pair.
Long-tailed pair: A common design in electronics for implementing a differentialamplifier. It consists of two BJTs, connected so that the BJT emitters are connectedtogether. The common electrodes are then connected to a large voltage source through alarge resistor, forming the long tail of the name, the long tail providing an approximateconstant current source. In a long tailed pair formed using BJTs, the emitters areconnected together, and then through the current source to ground or to a negative supply.In this form, one of the transistors can be thought of as an amplifier operating in commonemitter configuration, and the other as an emitter follower, feeding the other input signalinto the emitter of the first stage.
Differential Input and Output
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+VCC
RC1 RC2
- + Vout
Q1 Q2 one stage
V1 V2RE
-VEEFigure 17.1 (a)
Figure 17.1 shows a differential amplifier. It is two CE stages in parallel with a commonemitter resistor. The ac output voltage Vout is defined as the voltage between thecollectors with the polarity shown in figure 17.1 (a).Vout =VC2 VC1
When V1 is greater than V2, the output voltage has the polarity shown in figure 17.1 (a).When V2 is greater than V1, the output voltage is inverted and has the polarity shown infigure 17.1 (b).
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+VCC
RC1 RC2
+ -Vout
Q1 Q2 one stage
V1 V2RE
-VEEFigure 17.1 (b)
When both the noninverting and inverting input voltages are present, the total input iscalled a differential input because the output voltage equals the voltage gain times thedifference of the two input voltages. The equation for the output voltage is:Vout =A (V1 V2)
A =Voltage gain
Single ended output
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+VCC
RC2
Vout
Q1 Q2 one stage
V1 V2RE
-VEEFigure 17.2 (a)
As you can see, the ac output signal is taken from the collector on the right side. Thecollector resistor on the left has been removed because it serves no useful purpose.Because the input is differential, the ac output voltage is still given by A (V1 V2).However, the voltage gain is half as much as with a differential output. Because the
output is coming from only one of the collectors.
Block Diagram Symbol for a Differential Input Single Ended Output
V1
Vout
V2
Noninverting Input Configurations
Often, only one of the inputs is active and the other is grounded as shown in figure 17.3(a). This configuration has a non-inverting input and a differential output. Since V 2 =0
A
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Vout =A (V1 0)Vout =A V1
+VCC
RC1 RC2
- + Vout
Q1 Q2 one stage
V1RE
-VEEFigure 17.3 (a)
Figure 17.3 (b) shows another configuration for the differential amplifier. This one has a
non-inverting input and a single ended output. Since Vout is the ac output voltage, but thevoltage gain A will be half as much because the output is taken from only one side of thedifferential amplifier.Vout=A V1
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+VCC
RC2
Vout
Q1 Q2 one stage
V1
RE
-VEEFigure 17.3 (b)
Inverting Input Configurations
In some applications, V2 is the active input and V1 is the grounded input, as shown in
figure 17.4 (a). In this caseVout =A (0 V2)Vout =-A V2The minus sign in equation indicates phase inversion.Figure 17.4 (b) shows the final configuration. Here we are using the inverting input witha single ended output. In this case, the ac output voltage is given byVout =A (0 V2)Vout =-A V2
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+VCC
RC1 RC2
- + Vout
Q1 Q2 one stage
V2
RE
-VEEFigure 17.4 (a)
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+VCC
RC2
Vout
Q1 Q2 one stage
V2RE
-VEEFigure 17.4 (b)
DC Analysis of a Differential Amplifier
Figure 17.5 (a) shows the dc equivalent circuit for a differential amplifier. Throughoutthis discussion, we will assume identical transistors and equal collector resistors. Also,
both bases are grounded in this preliminary (introductory) analysis.
Using Ideal Approximation
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+VCC
RC1 RC2
C C
B BQ1 Q2
E EIE IE
RE
2IE-VEE
Figure 17.5 (a)
Figure 17.5 (b) shows the ideal approximation of a transistor. We visualize the emitterdiode as an ideal diode. In this case, VBE =0. This allows us to calculate base currentquickly and easily. This equivalent circuit is often useful for troubleshooting when all weneed is a rough approximation of base current. The collector side of the transistor acts
like a current source that pumps a collector current ofdcIB through the collector resistor.
B C
Ideal dcIB
EFigure 17.5 (b)
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+VCC
RC RC
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2IC
E E
RE
-VEEFigure 17.5 (c)
Ideal diode
Figure 17.5 (d) shows an ideal diode. Therefore, an ideal diode acts like a switch thatcloses when forward biased and opens when reverse biased. We just saidzeroresistancewhen forward biased and infiniteresistance when reverse biased.
Ideal
Reverse biased
Forward biased
Figure 17.5 (d)
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+VCC
+
RC RC IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0V 0VE 0V E
tail current
RE
-VEEFigure 17.5 (e)
According to ohms Law
0 (-VEE)IT =
RE
VEEIT =
RE
IT =tail currentApplying KCL at node labeled 0 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all the
currents leaving that junction
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0 (-VEE)IC +IC =
RE
VEE2IC =
RE
VEE2IE =
RE
VEEIE =
2RE
ITIE =
2
Hint:VBE =VB - VEHereVB =VE =0Therefore,VBE =0 VIE =IB +ICHere
IB =0A (VBE =0 V)IE =IC
The dc collector voltage on either side is
VC=VCC - ICRC
Using Second ApproximationWe can improve the dc analysis by including the VBE drop across each emitter diode. InFigure 17.5 (f), the voltage at the top of the emitter resistor is one V BE drop belowground.
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+VCC
RC1 RC2
C C
B BQ1 Q2
E EIE IE
RE
2IE-VEE
Figure 17.5 (f)
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+VCC
RC RC
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2IC
E E
RE
-VEEFigure 17.5 (g)
Equivalent circuit for second approximation
0.7 VReverse biased
0.7 VForward biased
Figure 17.5 (h)
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+VCC
+
RC RC IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0.7 V 0.7 V
0V 0V
E -0.7V E
tail current
RE
-VEEFigure 17.5 (i)
According to ohms Law
-0.7 (-VEE)IT =
RE
VEE 0.7IT =
RE
IT =tail currentApplying KCL at node labeled -0.7 VAccording to KCL
The algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
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-0.7 (-VEE)IC +IC =
RE
VEE 0.72IC =
RE
VEE 0.72IE =
RE
VEE 0.7IE =
2RE
ITIE =
2
Hint:VBE =VB - VEHereVB =0 VVE =-0.7 VTherefore,VBE =0 V (-0.7 V)VBE =0.7 VIE =IB +ICHere
IB =0A or negligibleIE =ICThe dc collector voltage on either side is
VC=VCC - ICRC
Example 17.1What are the ideal currents and voltages in figure 17.6 (a)?Solution:
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+15 V
5 k 5 k
C C
B BQ1 Q2
E EIE IE
7.5 k
2IE-15 V
Figure 17.6 (a)Ideal diode
Figure 17.5 (d) shows an ideal diode. Therefore, an ideal diode acts like a switch thatcloses when forward biased and opens when reverse biased. We just saidzeroresistancewhen forward biased and infiniteresistance when reverse biased.
Ideal
Reverse biased
Forward biased
Figure 17.6 (b)
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+15 V
+
5 k 5 k IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0V 0VE 0V E
tail current
7.5 k
-15 VFigure 17.6 (c)
According to ohms Law
0 (-15)IT =
7.5 k
15IT =
7.5 k
IT =2 mAApplying KCL at node labeled 0 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all the
currents leaving that junction
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0 (-VEE)IC +IC =
RE
VEE2IC =
RE
VEE2IE =
RE
VEEIE =
2RE
IT 2 mAIE = 1 mA
2 2
Hint:VBE =VB - VEHereVB =VE =0Therefore,VBE =0 VIE =IB +ICHere
IB =0A (VBE =0 V)IE =IC
The dc collector voltage on either side is
VC=VCC - ICRC
VC=15 (1 mA)(5 k)VC=15 (1 10
-3)(5 10+3)VC=15 5 10
-3+3
VC=15 5 100
VC=15 5 1
VC=10 Volts
Example 17.2:Recalculate the currents and voltages for figure 17.6 (a) using the secondapproximation.
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+15 V
5 k 5 k
C C
B BQ1 Q2
E EIE IE
7.5 k
2IE-15 V
Figure 17.6 (a)
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+VCC
RC RC
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2IC
E E
RE
-VEEFigure 17.7 (a)
Equivalent circuit for second approximation
0.7 VReverse biased
0.7 VForward biased
Figure 17.7 (b)
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+VCC
+RC RC IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0.7 V 0.7 V
0V 0V
E -0.7V E
tail current
RE
-VEEFigure 17.7 (c)
According to ohms Law
-0.7 (-VEE)IT =
RE
VEE 0.7 15 V 0.7 VIT = 1.907 mA
RE 7.5 k
Applying KCL at node labeled -0.7 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all the
currents leaving that junction
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-0.7 (-VEE)IC +IC =
RE
VEE 0.72IC =
RE
VEE 0.72IE =
RE
VEE 0.7IE =
2RE
ITIE =
2
IE =0.954 mAHint:VBE =VB - VEHereVB =0 VVE =-0.7 VTherefore,VBE =0 V (-0.7 V)VBE =0.7 VIE =IB +IC
HereIB =0A or negligibleIE =ICThe dc collector voltage on either side is
VC=VCC - ICRC
VC=15 (0.954 mA)(5 k)VC=15 (0.954 10
-3)(5 10+3)VC=15 4.77 10
-3+3
VC=15 4.77 100
VC=15 4.77 1
VC=10.23 Volts
Example 17.3:
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What are the currents and voltages in the single ended output circuit of figure 17.7 (a)?Solution:
+12
3 k
Q1 Q2 one stage
5 k
-12Figure 17.7 (a)
Using Ideal Approximation
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+12 V
+
3 k IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0V 0VE 0V E
tail current
5 k
-12 VFigure 17.6 (c)
According to ohms Law
0 (-12)IT =
5 k
12IT =
5 k
IT =2.4 mAApplying KCL at node labeled 0 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all the
currents leaving that junction
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0 (-VEE)IC +IC =
RE
VEE2IC =
RE
VEE2IE =
RE
VEEIE =
2RE
IT 2.4 mAIE = 1.2 mA
2 2
Hint:VBE =VB - VEHereVB =VE =0Therefore,VBE =0 VIE =IB +ICHere
IB =0A (VBE =0 V)IE =IC
The dc collector voltage on either side is
VC=VCC - ICRC
VC=12 (1.2 mA)(3 k)VC=12 (1.2 10
-3)(3 10+3)VC=12 3.6 10
-3+3
VC=12 3.6 100
VC=12 3.6 1
VC=8.4 Volts
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Equivalent circuit for second approximation
+12
+
3 k IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2
IC0.7 V 0.7 V0V 0V
E -0.7V E
tail current
5 k
-12
Figure 17.7 (c)According to ohms Law
-0.7 (-VEE)IT =
RE
VEE 0.7 12 V 0.7 VIT = 2.26 mA
RE 5 k
Applying KCL at node labeled -0.7 V
According to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
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-0.7 (-VEE)IC +IC =
RE
VEE 0.72IC =
RE
VEE 0.72IE =
RE
VEE 0.7IE =
2RE
ITIE =
2
IE =1.13 mAHint:VBE =VB - VEHereVB =0 VVE =-0.7 VTherefore,VBE =0 V (-0.7 V)VBE =0.7 VIE =IB +IC
HereIB =0A or negligibleIE =ICThe dc collector voltage on either side is
VC=VCC - ICRC
VC=12 (1.13 mA)(3 k)VC=12 (1.13 10
-3)(3 10+3)VC=12 3.39 10
-3+3
VC=12 3.39 100
VC=12 3.39 1
VC=8.61 VoltsAC Analysis of a Differential Amplifier
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Voltage gain of Noninverting input and single ended output
Theory of operation:
Figure 17.8 (a) shows a non-inverting input and single ended output. The left transistorQ1 acts like an emitter follower that produces an ac voltage across the emitter resistor. On
the positive half cycle of input voltage, the Q1 emitter current increases, the Q2 emittercurrent decreases, and the Q2 collector voltage increases. Similarly, on the negative halfcycle of input voltage, the Q1 emitter current decreases, the Q2 emitter current increases,and the Q2 collector voltage decreases. This is why the amplified output sine wave is inphase with the non inverting input.
+VCC
RC
Vout
Q1 Q2 one stage
V1RE
-VEEFigure 17.8 (a)
To analyze the ac operation of a transistor amplifier, we need an ac equivalent circuit fora transistor. In other words, we need a model for the transistor that simulates how itbehaves when an ac signal is present. When analyzing a transistor amplifier, we canreplace each transistor by aT model.
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ic
ib
re
ie
17.8 (b)AC Equivalent circuit
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ic
ib
re
RCVout
ic
ib
V1 reie ie
RE
17.8 (c)
The biasing resistor RE is in parallel with the re of the right transistor. In any practicaldesign, RE is much greater than re. Because of this, we can ignore RE in an introductoryanalysis.
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17.8 (d)
RE re
RE +re
RE >>re
RE re
RE
=re
Simplified Equivalent Circuit
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RCVout
ic ic
V1 reie
ie
re
17.8 (e)AC Input VoltageAccording to KVLSum of all the voltage rise =sum of all the voltage dropV1 =iere +iere
V1 =2iere
AC Output VoltageAccording to ohms Law:Vout =icRCDividing Vout by V1 gives voltage gain:
Vout icRC
V1 2iere
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ie =ib +icHere
ib =0A or negligibleie =ic
Vout icRC
V1 2icre
Vout RC
V1 2re
Voltage gain of Noninverting input and differential output
+VCC
RC RC
VC1 - + VC2Vout
Q1 Q2 one stage
V1RE
-VEEFigure 17.9 (a)
To analyze the ac operation of a transistor amplifier, we need an ac equivalent circuit fora transistor. In other words, we need a model for the transistor that simulates how it
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behaves when an ac signal is present. When analyzing a transistor amplifier, we canreplace each transistor by aT model.
ic
ib
re
ie
17.9 (b)AC Equivalent circuit
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ic
ib
re
RC RC- +
VC1 VC2Vout
ic
ib
V1 reie ie
RE
17.9 (c)
The biasing resistor RE is in parallel with the re of the right transistor. In any practicaldesign, RE is much greater than re. Because of this, we can ignore RE in an introductoryanalysis.
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re
RE
17.9 (d)
RE re
RE
+re
RE >>re
RE re
RE
=re
Simplified Equivalent Circuit
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RC RCVC1 - + VC2
Vout
ic ic
V1 reie
ie
re
17.9 (e)AC Input VoltageAccording to KVLSum of all the voltage rise =sum of all the voltage dropV1 =iere +iere
V1 =2iere
AC Output VoltageAccording to ohms Law:Vout =VC2 VC1VC2 =icRCVC1 =-icRC Hint: minus sign appears because the VC1signal is 180
0 out of phaseVout =icRC (-icRC)Vout =2icRCDividing Vout by V1 gives voltage gain:
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Vout 2icRC
V1 2iere
ie =ib +icHere
ib =0A or negligibleie =ic
Vout 2icRC
V1 2icre
Vout RC
V1 re
Voltage gain of inverting input and differential output
+VCC
RC RC
VC1 - + VC2Vout
Q1 Q2 one stage
V2RE
-VEEFigure 18.9 (a)
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To analyze the ac operation of a transistor amplifier, we need an ac equivalent circuit fora transistor. In other words, we need a model for the transistor that simulates how itbehaves when an ac signal is present. When analyzing a transistor amplifier, we canreplace each transistor by aT model.
ic
ib
re
ie
18.9 (b)AC Equivalent circuit
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ic
ib
re
RC R
C
- +VC1 VC2
Vout
ic
ib
reie ie V2
RE
18.9 (c)
The biasing resistor RE is in parallel with the re of the right transistor. In any practicaldesign, RE is much greater than re. Because of this, we can ignore RE in an introductoryanalysis.
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[ y p ( )]Cell: 0300-8454295; Tel: 042-5421893
re
RE
18.9 (d)
RE re
RE +re
RE >>re
RE re
RE
=re
Simplified Equivalent Circuit
[ y p ( )]Cell: 0300-8454295; Tel: 042-5421893
RC RCVC1 - + VC2
Vout
ic ic
ib
re V2
ie
ie
re
18.9 (e)AC Input VoltageAccording to KVLSum of all the voltage rise =sum of all the voltage dropV2 =iere +iere
V2 =2iere
AC Output VoltageAccording to ohms Law:Vout =VC2 VC1VC2=icRCVC1=-icRC Hint: minus sign appears because the VC1 signal is 180
0 out of phaseVout =icRC (-icRC)Vout =2icRCDividing Vout by V1 gives voltage gain:
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Vout 2icRC
V2 2iere
ie =ib +icHere
ib=0A or negligible
ie =ic
Vout 2icRC
V2 2icre
Vout RC
V2 re
Voltage gain of differential input and differential output
+VCC
RC1 RC2
- + Vout
Q1 Q2 one stage
V1 V2RE
-VEEFigure 19.1 (a)
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To analyze the ac operation of a transistor amplifier, we need an ac equivalent circuit fora transistor. In other words, we need a model for the transistor that simulates how itbehaves when an ac signal is present. When analyzing a transistor amplifier, we canreplace each transistor by aT model.
ic
ib
re
ie
18.9 (b)AC Equivalent circuit
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ic
ib
re
RC RC- +
VC1 VC2Vout
ic
ib
reV1 ie ie V2
RE
18.9 (c)AC Input Voltage
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re reV1 V2
ie ie
REI1 I2
2ie
18.9 (d)Using Loop Analysis
Loop I1:According to KVLSum of all the voltage rise =sum of all the voltage drop
V1 =iere +2ieRE (i)
Loop I2:According to KVLSum of all the voltage rise =sum of all the voltage drop0 =iere +2ieRE +V2
V2 =-iere - 2ieRE (ii)
Subtracting equation (ii) from (i)V1 V2 =2iere +4ieREV1 V2 =2iere ignoring higher order term
Hint: re is in ohms whereas RE is in kilo-ohms
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AC Output VoltageAccording to ohms Law:Vout =VC2 VC1VC2 =icRCVC1 =-icRC Hint: minus sign appears because the VC1signal is 180
0 out of phaseVout =icRC (-icRC)Vout =2icRCDividing Vout by V1 gives voltage gain:Vout 2icRC
V1 - V2 2iere
ie =ib +icHere
ib =0A or negligibleie =ic
Vout 2icRC
V1 - V2 2icre
Vout RC
V1 - V2 re
Input impedance of a differential amplifier
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+VCC
RC
VC1 - + VC2Vout
Q1 Q2 one stage
V2RE
-VEEFigure 19.9 (a)
To analyze the ac operation of a transistor amplifier, we need an ac equivalent circuit fora transistor. In other words, we need a model for the transistor that simulates how itbehaves when an ac signal is present. When analyzing a transistor amplifier, we canreplace each transistor by aT model.
ic
ib
re
ie
19.9 (b)
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ic
ib
re
AC Equivalent circuit
RC- +
VC1 VC2Vout
ic
ib
reie ie V2
RE
19.9 (c)
The biasing resistor RE is in parallel with the re of the right transistor. In any practicaldesign, RE is much greater than re. Because of this, we can ignore RE in an introductoryanalysis.
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re
RE
19.9 (d)
RE re
RE +re
RE >>re
RE re
RE
=re
Simplified Equivalent Circuit
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RCVC1 - + VC2
Vout
ic ic
ib
re V2
ie
ie
re
19.9 (e)
AC Input VoltageAccording to KVLSum of all the voltage rise =sum of all the voltage dropV2 =iere +iere
V2 =2iere
ie =ib +icHere
ib =0A or negligibleie =icHere
V2 =VinVin =2icre
ic =ib
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Vin =2ibre
Vin
Zin 2reib
NOTE
In a differential amplifier, the input impedance of either base is twice as high. Equation isvalid for all configurations.
Input Impedance of differential input and differential output
+VCC
RC1 RC2
- + Vout
Q1 Q2 one stage
V1 V2RE
-VEEFigure 20.9 (a)
To analyze the ac operation of a transistor amplifier, we need an ac equivalent circuit fora transistor. In other words, we need a model for the transistor that simulates how itbehaves when an ac signal is present. When analyzing a transistor amplifier, we canreplace each transistor by aT model.
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ic
ib
re
ie
20.9 (b)AC Equivalent circuit
ic
ib
re
RC RC
- +VC1 VC2
Vout
ic
ib
re
V1 ie ie V2
RE
20.9 (c)AC Input Voltage
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re reV1 V2
ie ie
REI1 I2
2ie
20.9 (d)Using Loop Analysis
Loop I1:According to KVLSum of all the voltage rise =sum of all the voltage drop
V1 =iere +2ieRE (i)
Loop I2:According to KVLSum of all the voltage rise =sum of all the voltage drop0 =iere +2ieRE +V2
V2 =-iere - 2ieRE (ii)
Subtracting equation (ii) from (i)V1 V2 =2iere +4ieREV1 V2 =2iere ignoring higher order term
Hint: re is in ohms whereas RE is in kilo-ohmsie =ib +icHere
ib =0A or negligibleie =icHereV1 - V2 =VinVin =2icre
ic =ibVin =2ibre
Vin
Zin 2reib
Input Impedance of Noninverting input and differential output
+VCC
RC
VC1 - + VC2Vout
Q1 Q2 one stage
V1RE
-VEEFigure 21.9 (a)
To analyze the ac operation of a transistor amplifier, we need an ac equivalent circuit fora transistor. In other words, we need a model for the transistor that simulates how itbehaves when an ac signal is present. When analyzing a transistor amplifier, we canreplace each transistor by aT model.
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ic
ib
re
ic
ib
re
ie
21.9 (b)AC Equivalent circuit
RC- +
VC1 VC2Vout
ic
ib
V1 reie ie
RE
21.9 (c)
The biasing resistor RE is in parallel with the re of the right transistor. In any practicaldesign, RE is much greater than re. Because of this, we can ignore RE in an introductoryanalysis.
re
RE
21.9 (d)
RE re
RE +re
RE >>re
RE re
RE
=re
Simplified Equivalent Circuit
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RC
VC1 - + VC2
Vout
ic ic
V1 re
ie
ie
re
21.9 (e)AC Input Voltage
According to KVLSum of all the voltage rise =sum of all the voltage dropV1 =iere +iere
V1 =2iere
ie =ib +icHere
ib =0A or negligibleie =icHereV1 =Vin
Vin =2icreic =ibVin =2ibre
Vin
Zin 2reib
Input Impedance of Noninverting input and single ended output
+VCC
RC
Vout
Q1 Q2 one stage
V1RE
-VEEFigure 22.8 (a)
To analyze the ac operation of a transistor amplifier, we need an ac equivalent circuit fora transistor. In other words, we need a model for the transistor that simulates how itbehaves when an ac signal is present. When analyzing a transistor amplifier, we canreplace each transistor by aT model.
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ic
ib
re
ie
22.8 (b)AC Equivalent circuit
ic
ib
re
RCVout
ic
ib
V1 reie ie
RE
22.8 (c)
The biasing resistor RE is in parallel with the re of the right transistor. In any practicaldesign, RE is much greater than re. Because of this, we can ignore RE in an introductoryanalysis.
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22.8 (d)
RE re
RE +re
RE >>re
RE re
RE
=re
Simplified Equivalent Circuit
RCVout
ic ic
V1 reie
ie
re
22.8 (e)
AC Input VoltageAccording to KVLSum of all the voltage rise =sum of all the voltage dropV1 =iere +iere
V1 =2iere
ie =ib +icHere
ib =0A or negligibleie =icHere
V1 =VinVin =2icre
ic =ib
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Vin =2ibre
Vin
Zin 2reib
ie =ib +ic
Hereib =0A or negligibleie =icHereV1 =VinVin =2icre
ic =ibVin =2ibre
Vin
Zin 2reib
Example 17.4:In Fig. 17.11 (a), what is the ac output voltage? If =300, what is the input impedanceof the differential amplifier?Solution:
+15 V
5 k 5 k
VC1 - + VC2Vout
Q1 Q2 one stage
1 mV7.5 k
-15 VFigure 17.11 (a)
DC Analysis
Ideal Approximation:
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+15 V
5 k 5 k
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2IC
E E
7.5 k
-15 VFigure 17.11 (b)
+15 V
+
5 k 5 k IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0V 0V
E 0V E
tail current
7.5 k
-15 VFigure 17.11 (c)
According to ohms Law
0 (-VEE)
IT =RE
VEE 15 VIT = 2 mA
RE 7.5 k
IT =tail currentApplying KCL at node labeled 0 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
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0 ( V )
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V 200(V )
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0 (-VEE)IC +IC =
RE
VEE2IC =
RE
VEE2IE =
RE
VEEIE =
2RE
ITIE = 1 mA
2
Hint:VBE =VB - VEHereVB =VE =0Therefore,VBE =0 VNow, we can calculate the ac emitter resistance
25 mVre =
IE
25 mVre =
1 mA
re =25
Voltage gain of Noninverting input and differential output
Vout RC 5 k200
V1 re 25
The ac output voltage is:
Vout =200 (V1)Vout =200 (1 mV)
Vout =200 mV
Input impedance of non inverting input and differential output
Zin=2re
Zin=2(300) (25)
Zin=15 k
Using Second Approximation
+15V
5 k 5 k
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2IC
E E
7.5 k
-15VFigure 17.11 (d)
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+15 V
+
5 k 5 k IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0.7 V 0.7 V
0V 0V
E -0.7V E
tail current
7.5 k
-15 VFigure 17.11 (e)
According to ohms Law
-0.7 (-VEE)IT =
RE
VEE 0.7 15 0.7IT = 1.907 mA
RE 7.5 k
IT =tail currentApplying KCL at node labeled -0.7 VAccording to KCL
The algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
-0.7 (-VEE)IC +IC =
RE
VEE 0.72IC =
RE
VEE 0.72IE =
RE
VEE 0.7IE =
2RE
IT 1.907 mAIE = 0.953 mA
2 2
Hint:VBE =VB - VEHereVB =0 VVE =-0.7 VTherefore,VBE =0 V (-0.7 V)VBE =0.7 VNow, we can calculate the ac emitter resistance
25 mVre =
IE
25 mVre =
0.953 mA
re =26.233
Voltage gain of Noninverting input and differential output
Vout RC 5 k190.6
V1 re 26.233
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Theacoutputvoltageis:
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The ac output voltage is:Vout =190.6 (V1)Vout =190.6 (1 mV)
Vout =190.6 mV
Input impedance of non inverting input and differential output
Zin=2reZin=2(300) (26.233)
Zin=15.74 k
Comparison:
IDEAL SECONDIT 2 mA IT 1.907 mA
IE 1 mA IE 0.953 mA
re 25 re 26.233
A 200 A 190.6Vout 200 mV Vout 190.6 mV
Zin 15 k Zin 15.74 k
Example 17.5:Repeat Example 17.4 for V2 =1 mV and V1 =0.Solution:
+15 V
5 k 5 k
VC1 - + VC2Vout
Q1 Q2 one stage
V2=1 mV7.5 k
-15 VFigure 17.5 (a)
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DC Analysis
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+15V
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DC Analysis
Ideal Approximation:+15 V
5 k 5 k
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2
IC
E E
7.5 k
-15 VFigure 17.5 (b)
+15 V
+
5 k 5 k IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0V 0V
E 0V E
tail current
7.5 k
-15 VFigure 17.5 (c)
According to ohms Law
0 (-VEE)
IT =RE
VEE 15 VIT = 2 mA
RE 7.5 k
IT =tail currentApplying KCL at node labeled 0 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
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0 (-VEE)
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Vout =200 (-V2)
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IC +IC =RE
VEE2IC =
RE
VEE2IE =
RE
VEEIE =
2RE
ITIE = 1 mA
2
Hint:VBE =VB - VEHereVB =VE =0Therefore,VBE =0 VNow, we can calculate the ac emitter resistance
25 mVre =
IE
25 mVre =
1 mA
re =25
Voltage gain of inverting input and differential output
Vout RC 5 k200
V2 re 25
The ac output voltage is:
Vout =200 (-1 mV)
Vout =-200 mV
Input impedance of inverting input and differential output
Zin=2re
Zin=2(300) (25)
Zin=15 k
Using Second Approximation
+15 V
5 k 5 k
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2IC
E E
7.5 k
-15 VFigure 17.11 (d)
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+15 V
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-0.7 (-VEE)I I
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+
5 k 5 k IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0.7 V 0.7 V0V 0V
E -0.7V E
tail current
7.5 k
-15 VFigure 17.11 (e)
According to ohms Law
-0.7 (-VEE)IT =
RE
VEE 0.7 15 0.7IT = 1.907 mA
RE 7.5 k
IT =tail currentApplying KCL at node labeled -0.7 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all the
currents leaving that junction
IC +IC =RE
VEE 0.72IC =
RE
VEE 0.72IE =
RE
VEE 0.7IE =
2RE
IT 1.907 mAIE = 0.953 mA
2 2
Hint:VBE =VB - VEHereVB =0 VVE =-0.7 VTherefore,VBE =0 V (-0.7 V)VBE =0.7 VNow, we can calculate the ac emitter resistance
25 mV
re =IE
25 mVre =
0.953 mA
re =26.233
Voltage gain of inverting input and differential output
Vout RC 5 k
190.6V1 re 26.233
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The ac output voltage is:V 1906 ( V )
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Vout =190.6 (-V2)Vout =190.6 (-1 mV)
Vout =-190.6 mV
Input impedance of inverting input and differential output
Zin=2reZin=2(300) (26.233)
Zin=15.74 k
Comparison:
IDEAL SECONDIT 2 mA IT 1.907 mA
IE 1 mA IE 0.953 mA
re 25 re 26.233
A 200 A 190.6
Vout -200 mV Vout -190.6 mVZin 15 k Zin 15.74 k
Example 17.6:What is the ac output voltage in Fig. 17.12 if =300, what is the input impedance of thedifferential amplifier?Solution:
+15 V
1 M
Vout
Q1 Q2 one stage
7 mV1 M
-15 VFigure 17.6 (a)
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DC Analysis
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Ideal Approximation:+15 V
1 M
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2
IC
E E
1 M
-15 VFigure 17.5 (b)
+
1 M IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0V 0V
E 0V E
tail current
1 M
-15 VFigure 17.5 (c)
According to ohms Law
0 (-VEE)
IT =RE
VEE 15 V
IT = 15A
RE 1 M
IT =tail currentApplying KCL at node labeled 0 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
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0 (-VEE)IC +IC =
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Vout =149.971 (V1)Vout =149971(7mV)
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IC +ICRE
VEE2IC =
RE
VEE2IE =
RE
VEEIE =
2RE
IT 15A
IE = 7.5A2 2
Hint:VBE =VB - VEHereVB =VE =0Therefore,VBE =0 VNow, we can calculate the ac emitter resistance
25 mVre =
IE
25 mVre =
7.5A
re =3.334 k
Voltage gain of non inverting input and single ended output
Vout RC 1 M149.971
V1 2re 2(3.334 k)
The ac output voltage is:
Vout 149.971 (7 mV)
Vout =1.05 V
Input impedance of non inverting input and single ended output
Zin=2re
Zin=2(300) (3.334 k)
Zin=2 M
Using Second Approximation
+15V
5 k 5 k
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2IC
E E
7.5 k
-15VFigure 17.11 (d)
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+15 V
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-0.7 (-VEE)IC +IC =
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+
5 k 5 k IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0.7 V 0.7 V0V 0V
E -0.7V E
tail current
7.5 k
-15 VFigure 17.11 (e)
According to ohms Law
-0.7 (-VEE)IT =
RE
VEE 0.7 15 0.7
IT = 14.3ARE 1 M
IT =tail currentApplying KCL at node labeled -0.7 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all the
currents leaving that junction
C C
RE
VEE 0.72IC =
RE
VEE 0.72IE =
RE
VEE 0.7IE =
2RE
IT 14.3A
IE = 7.15A2 2
Hint:VBE =VB - VEHereVB =0 VVE =-0.7 VTherefore,VBE =0 V (-0.7 V)VBE =0.7 VNow, we can calculate the ac emitter resistance
25 mV
re =IE
25 mVre =
7.15A
re =3.496 k
Voltage gain of non inverting input and single ended output
Vout RC 1 M
143.02V1 2re 2(3.496 k)
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The ac output voltage is:Vout =143.02 (V1)
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Vout =143.02 (7 mV)
Vout =1.002 V
Input impedance of non inverting input and single ended output
Zin=2reZin=2(300) (3.496 k)
Zin=2.098 M
Comparison:
IDEAL SECOND
IT 15A IT 14.3A
IE 7.5A IE 7.15A
re 3.334 k re 3.496 kA 149.971 A 143.02
Vout 1.05 V Vout 1.002 VZin 2.098 M Zin 2.098 M
Example 17.7:The differential amplifier of Figure 17.17 has A =200, Iin(bias) =3A, Iin(off) =0.5A,Vin(off) =1 mV. What is the output error voltage?Solution:
5 k 5 k
VC1 - + VC2Vout
1 k
Q1 Q2 one stage
10 mV7.5 k
-15 VFigure 17.17 (a)
V1error =(RB1 RB2) Iin(bias)Here
RB1 =1 k
V1error =(1 k 0)(3A)V1error =(1 10
+3)(3 10-6)
V1error =3 103-6V1error =3 10
-3
V1error =3 mV
V2error =0.5 (RB1 +RB2)Iin(off)V2error =0.5 (1 k +0)(0.5A)V2error =0.5 (1 10
+3)(0.5 10-6)V2error =0.25 10
3-6
V2error =0.25 10-3
V2error =0.25 mV
V3error =Vin (off)
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V3error =1 mV
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V1error =0 V
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The output error voltage is
Verror =200 (V1error +V2error +V3error)Verror =200 (3 mV +0.25 mV +1 mV)
Verror =200 (4.25 mV)
Verror =850 mV
Example 17.8:The differential amplifier of Figure 17.18 has A =300, Iin(bias) =80 nA, Iin(off) =20 nA,and Vin(off) =5 mV. What is the output error voltage?Solution:
+15 V
1 M 1 M
VC1 - + VC2Vout
10 k
Q1 Q2 one stage
10 k
10 mV1 M
-15 VFigure 17.18 (a)
V1error =(RB1 RB2) Iin(bias)Here
RB1 =RB2 =10 kV1error =(10 k 10 k)(80 nA)V1error =0 V
V2error =0.5 (RB1 +RB2)Iin(off)V2error =0.5 (10 k+10 k)(20 nA)V2error =0.5 (20 10
+3)(20 10-9)V2error =200 10
3-9
V2error =200 10-6
V2error =0.2 mV
V3error =Vin (off)
V3error =5 mV
The output error voltage is
Verror =300 (V1error +V2error +V3error)Verror =300 (0 V +0.2 mV +5 mV)
Verror =300 (5.2 mV)
Verror =1.56 V
Common-Mode Gain
Figure 17-19(a) shows a differential input and single ended output. The same inputvoltage, Vin(cm) is being applied to each base. This voltage is called a common modesignal. I f the differential amplifier is perfectly symmetrical, there is no ac output voltagewith a common mode input signal because V1 =V2. When a differential amplifier is notperfectly symmetrical, there will be a small ac output voltage.In Fig. 17.19 (a), equal voltages are applied to the non-inverting and inverting inputs
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+V
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+V
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+VCC
RC
Vout
Q1 Q2 one stageVin (CM) Vin (CM)
RE
-VEEFigure 17.19 (a)
Here is how a common mode signal appears: The connecting wires on the input bases actlike small antennas. If the differential amplifier is operating in an environment with a lotof electromagnetic interference, each base acts like a small antenna that picks up anunwanted signal voltage.
+VCC
RC
Vout
Q1 Q2 one stageVin (CM) Vin (CM)
RE
-VEEFigure 17.19 (b)
Here is an easy way to find the voltage gain for a common mode signal: we can redrawthe circuit, as shown in Fig. 17-19 (c), since equal voltages Vin(CM) drive both inputssimultaneously, there is almost no current through the wire between the emitters.Therefore, we can remove the connecting wire, as shown in Fig. 17-19 (f).
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+VCC
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RE
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+VCC
RC
Vout
Q1 Q2 one stageVin (CM) Vin (CM)
2RE 2RE
-VEE Figure 17.19 (c)
2RE 2RE
Figure 17.19 (d)2RE 2RE
2RE +2RE
4RE RE
4RE
0 V 0 V
Figure 17.19 (e)
+VCC
RC
Vout
Q1 Q2 one stageVin (CM) Vin (CM)
2RE 2RE
-VEE Figure 17.19 (f)
VOLTAGE GAINa.c. equivalent circuit
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Vout ieRC
Vin(CM) ie(re +2RE)
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ic
ib
re
RC
Vout
ic
ib
Vin(CM) re
ie ie Vin(CM)
2RE 2RE
17.19 (g)AC Input Voltage
According to KVLSum of all the voltage rise =sum of all the voltage dropVin(CM) =iere +2ieRE
Vin(CM) =ie(re +2RE)
AC Output VoltageAccording to ohms Law:Vout =icRCie =ib +icHere
ib =0A or negligible
ie =icVout =ieRCDividing Vout by Vin(CM) gives voltage gain:
Vin(CM) ie(re +2RE)
Vout RC
Vin(CM) re +2RE
Vout RC
Vin(CM) 2RE
Common-Mode Rejection Ratio
The common mode rejection ratio is defined as the voltage gain (differential input singleended output) divided by common mode voltage gain.
RC
2reCMRR =
RC
2RE
RECMRR =
re
CMRRdB =20 Log CMRR
Example 17-9: In Figure 17-20 (a), what is the common mode voltage gain? The outputvoltage?Solution:
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+15 V
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1 M
Vout
Q1 Q2 one stageVin (CM)= Vin (CM) =1 mV 1 mV
1 M
-15 VFigure 17.20 (a)
VOLTAGE GAINa.c. equivalent circuit
ic
ib
re
1 MVout
ic
ib
1 mV re
ie ie 1 mV
2M 2M
17.20 (b)AC Input Voltage
According to KVLSum of all the voltage rise =sum of all the voltage dropVin(CM) =iere +2ieRE
Vin(CM) =ie(re +2RE)
AC Output VoltageAccording to ohms Law:Vout =icRCie =ib +icHere
ib =0A or negligible
ie =icVout =ieRCDividing Vout by Vin(CM) gives voltage gain:
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Vout ieRC
Vin(CM) ie(re +2RE)
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Vout RC
Vin(CM) re +2RE
Vout RC 1 M0.5
Vin(CM) 2RE 2 M
Vout =ACMVin(CM)Vout =0.5(1 mV)
Vout =0.5 mV
Example 17-10:
In figure 17.21 (a), A =150, ACM =0.5, and Vin =1 mV. If the base leads are picking upa common mode signal of 1 mV, what is the output voltage?Solution:Vout1 =A VinVout1 =150 (1 mV)
Vout1 =150 mV
Vout2 =ACM VinVout2 =0.5 (1 mV)
Vout2 =0.5 mV
Vout =Vout1 +Vout2Vout =150 mV +0.5 mV
Vout =150.5 mV
1 M
Vout
1 k
Q1 Q2 one stage
1 kVin
1 M
-15 VFigure 17.21 (a)
Thisexample shows why the differential amplifier is useful as the input stage of an op-amp. It attenuates the common mode signal. This is a distinct advantage over the ordinaryCE amplifier, which amplifies a stray pickup signal the same way it amplifies the desiredsignal.
The current mirror
With ICs, there is a way to increase the voltage gain and CMRR of a differentialamplifier. Figure 17.22 (a) shows a compensating diode in parallel with the emitter diodeof a transistor. The current through the resistor is given by:
VCC - VBEIR
R
If the compensating diode and the emitter diode have identical current-voltage curves, thecollector current will equal the current through the resistor:
IC =IRA circuit like figure 17.22 (a) is called a current mirror because the collector current is amirror image of the resistor current.
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+VCC
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Potential difference =VCC - VBEHereVBE =0 V
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R
Fig. 17.22 (a)
Equivalentcircuitusing ideal approximation
+VCC
C
R dcIB
B
E
VBE
=0 V
Fig. 17.22 (b)
According to ohms Law:IR R =VCC - VBE
VCC - VBEIR
R
VCC VBE
VCCIR
RA
R IC VCC
B Fig. 17.22 (c)VCC =IRR =VABAgain
VCC
IR
R
+15 V
1 M
Fig. 17.22 (d)
15 VIR
1 M
IR =15A =IB =0.000015 A
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IC =IBSuppose=50
IC =50 (15A)
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R
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IC =750A =0.00075 A
Hence
IR =IC
Currentmirror sources thetail current
With a single ended output, the voltage gain of a differential amplifier is RC/2re and thecommon mode voltage gain is RC/2RE. The ratio of the two gains gives:
Single ended voltage gain of a differential amplifierCMRR
Common mode voltage gain
RC
2reCMRR
RC
2RE
RECMRR
re
The larger we can make RE, the greater the CMRR.
One way to get a high equivalent RE is to use a current mirror to produce the tail current,as shown in figure 17.23 (a). The current through the compensating diode is:
VCC (-VEE) - VBEIR =
R
VCC +VEE - VBEIR =
R
Because of the current mirror, the tail current has the same value. Since Q4 acts like a
current source it has very high output impedance. As a result, the equivalent RE of thedifferential amplifier is in hundreds of mega ohms and the CMRR is dramaticallyimproved.
RC
Vout
V1 V2R Q1 Q2
C
BQ4
Q3 VBE E
-VEEFigure 17.23 (a)
Equivalent Circuit of a CE Amplifier
Consider the simple CE amplifier circuit of figure 17.23 (b) in which base bias has beenemployed.
VCCIB IC
RCC2
RB
C1
is ib RLVout
VS
Figure 17.23 (b)DC Equivalent Circuit
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For drawing the dc equivalent circuit,
VCCIB IC
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B CIB IC
IB
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IB ICRC
C2RB
C1
is ib RLVout
VS
Figure 17.23 (c)
VCCIB IC
RC
RB
Figure 17.23 (d)Figure 17.23 (e) shows the dc equivalent circuit of an NPN transistor when connected inthe CE configuration.
IB
E
Figure 17.23 (e)
VCCIB IC
RC
RB
B C
IB
E
Figure 17.23 (f)
VCCIB IC
RC
RB
B C
IB
E
Figure 17.23 (g)
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IDEAL APPROXIMATION
According to KVLSum of all the voltage rise =sum of all the voltage drop
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ic
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g g pVCC =IBRB
VCCIB
RB
IC =IB
IE =IB +ICIE =IB +IB
IE =(1 +) IB
2nd APPROXIMAT ION
VCCIB IC
RC
RB
B C
VBE
IB
E
Figure 17.23 (h)According to KVLSum of all the voltage rise =sum of all the voltage dropVCC =IBRB +VBE
VCC - VBEIB
RB
AC EQUIVAL ENT CIRCUIT
ib
re
ie
Let us now analyze the ac equivalent circuit given in figure 17.23 (i).
IB ICRC
RB
icRL
ibis
re
figure 17.23 (i)Example 17.11:If in the CE circuit of figure 17.23 (a), VCC =20 V, RC =10 k, RB =1 M, RL =1 M, =50, find rin, rL.
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IDEAL APPROXIMATION 2nd APPROXIMATI ON
VCCIB
VCC - VBEIB
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10 109rL
1010000
10 109rL
1010000
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RB
20 V
IB 20A1 M
IC =IB
IC =50 (20A) =1 mA
IE =(1 +) IBIE =(1 +50) (20A)IE =1.020 mA
25 mVre
IE
25 mV
re 24.511.020 mA
RB re 106 50 24.51
rinRB +re 10
6+50 24.51
1225500000rin
1001225.5
rin =1224
RL RCrL
RL +RC
106 10 103rL
106 +10 103
RB
20 V 0.7 V
IB 19.3A1 M
IC =IB
IC =50 (19.3A) =0.965 mA
IE =(1 +) IBIE =(1 +50) (19.3A)IE =0.984 mA
25 mVre
IE
25 mV
re 25.4060.984 mA
RB re 10650 25.406
rinRB +re 10
6 +50 25.406
1270300000rin
1001270.3
rin=1268.688
RL RCrL
RL +RC
106 10 103rL
106 +10 103
Current source
It produces a constant load current for different load resistances. An example of a dccurrent source is a battery with a large source resistance.
RS =1 M
VS =10 V IL RL =1
According to ohms Law
VSIL
RS +RL
RL =110 V
IL
106+1
IL =10A
10010 V
IL
106+100
IL =9.999A
rL =9.9 k (high output impedance) rL =9.9 k (high output impedance)
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10 k10 V
IL
106+10000
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IL =9.901A
1 M 10 VIL
106+1000000
IL =5A
Load current versus Load resistance
0
200000
400000
600000
800000
1000000
1200000
Load resistance
Load
current
Series1
Series2
Series3
Series4
Series110 1
Series2 9.999 100
Series3 9.901 10000
Series4 5 1000000
1 2
THE LOADED DIFFERENTIAL AMPLI FIER
When a load resistor is used, the analysis becomes much more complicated, especiallywith a differential output. Figure 17.24 (a) shows a differential output with a load resistorbetween the collectors.
RC RCRL
A B
Q1 Q2
V1 V2RE
-VEEFigure 17.24 (a)
Thevenin Equivalent Resistance
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ic
ib
re
RC RC
RTH
ic
ib
V1 reie ie V2
RE
17.24 (b)
ic
ib
re
RC RC
RTH
ic
ib
s.c.re
ie ie s.c.
RE
17.24 (c)
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RC RC
RTH
17.24 (d)RTH =RC +RC
RTH =2RCRTH =2RC
Vout RL
17.24 (e)
Here is how it is done: i f we open the load resistor in figure 17.24 (a), the theveninvoltage is the same as the Vout. Also, looking into the open AB terminals with all sourceszeroed, we see a thevenin resistance of 2RC.Note: Because the transistors are current sources, they become open when zeroed.differential output with a load resistor between the collectors.
SINGLE ENDED OUTPUT WITH A LOAD RESISTOR BETWEEN THE COLLECTORS
RCRL
A B
Q1 Q2
V1 V2RE
-VEE
Figure 17.25 (a)Thevenin Equivalent Resistance
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ic
ib
re
RC
RTH
ic
ib
V1 reie ie V2
RE
17.25 (b)
ic
ib
re
RC
RTH
ic
ib
s.c.re
ie ie s.c.
RE
17.25 (c)
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+15 V
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RC
RTH
17.25 (d)RTH =RC
RTH =RCRTH =RC
Vout RL
17.25 (e)Example 17.12:
What is the load voltage in figure 17.26 (a) when RL =15 k?Solution:
7.5 k 7.5 kRL
A B
Q1 Q2
10 mV
7.5 k
-15 VFigure 17.26 (a)
IDEAL APPROXIMATION
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+15 V
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+15 V
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7.5 k 7.5 k
EquivalentModel EquivalentModel
B C RL C B
IC Q1 Q2IC
E E
7.5 k
-15 VFigure 17.26 (b)
7.5 k 7.5 k
EquivalentModel EquivalentModel
B C RL C B
IC Q1 Q2IC
E 0 V E
7.5 k
-15 VFigure 17.26 (c)
According to ohms Law
0 (-15)
IT =7.5 k
15IT =
7.5 k
IT =2 mAApplying KCL at node labeled 0 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
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0 (-VEE)IC +IC =
RE
VEE2I
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RTH =15 k
3V R 15k
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2IC =RE
VEE2IE =
RE
VEEIE =
2RE
IT 2 mAIE = 1 mA
2 2
Hint:VBE =VB - VEHereVB =VE =0Therefore,VBE =0 V
25 mVre
IE
25 mV
re 251 mA
RC 7.5 kA 300 (unloaded voltage gain)
re 25
The thevenin or unloaded output voltage is:VTH =AV1VTH =300 (0.01) =3 VoltThe thevenin resistance is:
RTH =2RC =2 (7.5 k) =15 k
3 V RL =15 k
17.26 (d)According to voltage divider rule:
RLVL = 3 V
RL +RTH
15 kVL = 3 V
15 k +15 k
VL =1.5 Volt
Example 17.13:
An ammeter is used for the load resistance in figure 17.27 (a). What is the currentthrough the ammeter?Solution:
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+15 V
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+15 V
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7.5 k 7.5 kA
A B
Q1 Q2
10 mV
7.5 k
-15 VFigure 17.27 (a)
IDEAL APPROXIMATION
7.5 k 7.5 kA
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2IC
E E
7.5 k
-15 VFigure 17.27 (b)
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+15 V
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0 (-VEE)IC +IC =
RE
VEE2IC =
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7.5 k 7.5 kA
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2IC
E 0 V E
7.5 k
-15 VFigure 17.27 (c)
According to ohms Law
0 (-15)
IT =7.5 k
15IT =
7.5 k
IT =2 mAApplying KCL at node labeled 0 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
2IC =RE
VEE2IE =
RE
VEEIE =
2RE
IT 2 mAIE = 1 mA
2 2
Hint:VBE =VB - VEHereVB =VE =0Therefore,VBE =0 V
25 mVre
IE
25 mV
re 251 mA
RC 7.5 kA 300 (unloaded voltage gain)
re 25
The thevenin or unloaded output voltage is:VTH =AV1VTH =300 (0.01) =3 VoltThe thevenin resistance is:
RTH =2RC =2 (7.5 k) =15 k
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RTH =15 k
A3V
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and ground and also T2 and ground, (5) it can provide two separate outputs by means ofT3 and ground and T4 and ground, (6) it can provide a single output between T3 and T4that is, differential output.
ADVANTAGES
1 I f d d li b i All h i i
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3 V
17.27 (d)
VTHiL =
RTH
3 ViL =
15 k
iL =0.2 mA
DIFFERENTIAL AMPLIFIER+VCC
R2 R3
T3 T4
T1 T2
Q1 Q2
R1 R4
Figure 17.28 (a)
Figure 17.28 (a) shows the circuit of a differential amplifier or difference amplifier. As
seen (1) it contains two CE amplifiers, (2) it only uses resistors and transistors. (3) it is adirectly coupled emitter to emitter amplifier, (4) it can accept two inputs by means of T1
1. It uses no frequency-dependent coupling or bypass capacitors. All that it requiresis resistors and transistors both of which can be easily integrated on a chip. Hence,
it is extensively used in linear Integrated Circuits.2. It can compare any two signals and detect any difference. Thus, if two signals arefed into its inputs, identical in every respect except that one signal has beenslightly distorted, then only the difference between the two signals that is,distortion will be amplified.
3. It gives higher gain than two cascaded stages of ordinary direct coupling.4. It provides very uniform amplification of signal from dc up to very high
frequencies.5. It provides isolation between input and output circuits.6. It is almost a universal choice for amplifying dc.7. It finds a wide variety of applications such as amplification, mixing, signal
generation, amplitude modulation, frequency multiplication and temperature
compensation etc.
Example 17.14:
Calculate the approximate output voltage for the differential amplifier of figure 17.29 (a)which uses only a single ended non-inverting input of 1 mV. Take re=25 mV/IE andneglect VBE.Solution:
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+12 V
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+12 V
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12 k 12 k
- + Vout
Q1 Q2
1 mV6 k
-12 VFigure 17.29 (a)
IDEAL APPROXIMATION
12 k 12 k
EquivalentModel - Vout + EquivalentModel
B C C B
IC Q1 Q2IC
E E
6 k
-12 VFigure 17.29 (b)
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+12 V
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0 (-VEE)IC +IC =
RE
VEE2IC =
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12 k 12 k
EquivalentModel - Vout + EquivalentModel
B C C B
IC Q1 Q2IC
E 0 V E
6 k
-12 VFigure 17.29 (c)
According to ohms Law
0 (-12)
IT =6 k
12IT =
6 k
IT =2 mAApplying KCL at node labeled 0 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
RE
VEE2IE =
RE
VEEIE =
2RE
IT 2 mAIE = 1 mA
2 2
Hint:VBE =VB - VEHereVB =VE =0Therefore,VBE =0 V
25 mVre
IE
25 mV
re 251 mA
RC 12 kA 480
re 25
Vout =A (1 mV)Vout =480 (0.001 V)
Vout =0.48 Volt
Example 17.15:
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If a differential input signal of 1 mV is applied to the differential amplifier shown infigure 17.30 (a), calculate the output voltage. Neglect VBE and take re=25 mV/IE.Solution:
+12 V
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+12 V
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12 k 12 k
Vout
+Q1 Q2
1 mV
6 k
-
-12 VFigure 17.30 (a)
IDEAL APPROXIMATION
12 k 12 k
EquivalentModel Vout EquivalentModel
B C C B
IC Q1 Q2IC
E E
6 k
-12 VFigure 17.30 (b)
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+12 V
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0 (-VEE)IC +IC =
RE
VEE2IC =
R
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12 k 12 k
EquivalentModel Vout + EquivalentModel
B C C B
IC Q1 Q2IC
E 0 V E
6 k
-12 VFigure 17.30 (c)
According to ohms Law
0 (-12)
IT =6 k
12IT =
6 k
IT =2 mAApplying KCL at node labeled 0 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
RE
VEE2IE =
RE
VEEIE =
2RE
IT 2 mAIE = 1 mA
2 2
Hint:VBE =VB - VEHereVB =VE =0Therefore,VBE =0 V
25 mVre
IE
25 mV
re 251 mA
RC 12 kA 240
2re 50
Vout =A (V1 V2)Vout =A (1 mV)Vout =240 (0.001 V)
Vout =0.24 Volt
Example 17.16:
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A differential input signal of 1 mV is applied to the differential amplifier of figure 17.31(a) when used in double ended output modes. Calculate the approximate value of outputvoltage. Neglect VBE and take re=25 mV/IE.Solution:
+12 V
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+12 V
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12 k 12 k
- +Vout
+Q1 Q2
1 mV
6 k-
-12 VFigure 17.30 (a)
IDEAL APPROXIMATION
12 k 12 k
EquivalentModel Vout EquivalentModel
B C C B
IC Q1 Q2IC
E E
6 k
-12 VFigure 17.30 (b)
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+12 V
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0 (-VEE)IC +IC =
RE
VEE2IC =
RE
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12 k 12 k
EquivalentModel - Vout + EquivalentModel
B C C B
IC Q1 Q2IC
E 0 V E
6 k
-12 VFigure 17.30 (c)
According to ohms Law
0 (-12)IT =
6 k
12IT =
6 k
IT =2 mAApplying KCL at node labeled 0 VAccording to KCL
The algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
RE
VEE2IE =
RE
VEEIE =
2RE
IT 2 mAIE = 1 mA
2 2
Hint:VBE =VB - VEHereVB =VE =0Therefore,VBE =0 V
25 mVre
IE
25 mV
re 251 mA
RC 12 kA 480
re 25
Vout =A (1 mV)Vout =480 (0.001 V)
Vout =0.48 Volt
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Example 17.17:
Calculate the single ended and differential gain of the differential amplifier shown infigure 17.32 (a). Use re =25 mV/IE.Solution:DIFFERENTIAL AMPLIFIER
12V
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Example 17.18:
For the differential amplifier shown in figure 17.33 (a), voltage gain of each stage is 200,Vi1 =30 mV and Vi2 =20 mV. Find the voltages between (i) T3 and ground, (ii) T4 andground, (iii) T3 and T4.Solution:
+12 V
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+12 V
RC1=10 k RC2=10 k
T3 T4
T1 T2
Q1 Q2
0.5 mA 0.5 mA
1 mA
Figure 17.32 (a)Hence, single ended voltage gain is
25 mVre
IE
25 mV
re 500.5 mA
RC 10 kA 200 RC1 =RC2 =RC
re 50
Hence, each stage has a voltage gain of 200. I f we consider differential gain, its value istwice that is 2 200 =400.
IC1 IC2
RC1=10 k RC2=10 k
T3 T4
T1 T2
Q1 Q2
0.5 mA 0.5 mA
1 mA
Figure 17.33 (a)V0 (T3) =A1Vi1V0 (T3) =200 (30 mV) =6 V
V0 (T4) =A2Vi2V0 (T4) =200 (20 mV) =4 V
V0 (T3 T4) =A (Vi1 Vi2)V0 (T3 T4) =A (Vi1 Vi2)V0 (T3 T4) =200 (30 mV 20 mV)V0 (T3 T4) =2 V
Because Vi1 >Vi2; IC1RC1 >IC2RC2, hence, T4 will be positive with respect to T3.Hint: Vout =VCC - ICRCDifferential input differential output
PROBLEMS
DC ANALYSIS OF A DIFFERENTIAL AMPL IFI ER
Q#17.1: What are the ideal currents and voltages in figure 17.34 (a)?
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Solution:IDEAL APPROXIMATION
+12 V
180k 180k
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+12 V
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180 k 180 k
C C
B BQ1 Q2
E EIE IE
270 k
2IE-12 V
Figure 17.34 (a)
180 k 180 k
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2IC
E E
270 k
-12 VFigure 17.34 (b)
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+12 V
+
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0 (-VEE)IC +IC =
RE
VEE2IC =
RE
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180 k 180 k IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0V 0VE 0V E
tail current
270 k
-12 VFigure 17.34 (c)
According to ohms Law
0 (-VEE)IT =
RE
VEE 12 V
IT = 44.445A
RE 270 k
IT =tail currentApplying KCL at node labeled 0 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
VEE2IE =
RE
VEEIE =
2RE
IT 44.445AIE = 0.022 mA
2 2
Hint:VBE =VB - VEHereVB =VE =0Therefore,VBE =0 VIE =IB +ICHere
IB =0A (VBE =0 V)IE =IC
The dc collector voltage on either side is
VC=VCC - ICRC
VC=12 (0.022 mA)(180 k)VC=12 (0.022 10
-3)(180 10+3)VC=12 3.96 10
-3+3
VC=12 3.96 100
VC=12 3.96
VC=8.04 VoltsQ#17.2: Repeat problem 17.1 using the second approximation.Solution:
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+12 V
180 k 180 k
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+12 V
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C C
B BQ1 Q2
E EIE IE
270 k
2IE-12 V
Figure 17.35 (a)
180 k 180 k
EquivalentModel EquivalentModel
B C C B
IC Q1 Q2IC
E E
270 k
-12 VFigure 17.35 (b)
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+12 V
+
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-0.7 (-VEE)IC +IC =
RE
VEE 0.72IC =
RE
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180 k 180 k IC
EquivalentModel - EquivalentModelVC
B C C B
IC Q1 Q2IC
0.7 V 0.7 V0V 0V
E -0.7V E
tail current
270 k
-12 VFigure 17.35 (c)
According to ohms Law
-0.7 (-VEE)
IT =RE
VEE 0.7 12 V 0.7 V
IT = 41.852A
RE 270 k
Applying KCL at node labeled -0.7 VAccording to KCLThe algebraic sum of all the currents entering into the junction =algebraic sum of all thecurrents leaving that junction
VEE 0.72IE =RE
VEE 0.7IE =
2RE
IT 41.852AIE = 0.021 mA
2 2
Hint:VBE =VB - VEHereVB =0 VVE =-0.7 VTherefore,VBE =0 V (-0.7 V)VBE =0.7 VIE =IB +ICHere
IB =0A or negligibleIE =IC
The dc collector voltage on either side is
VC=VCC - ICRC
VC=12 (0.021 mA)(180 k)VC=12 (0.021 10
-3)(180 10+3)VC=12 3.78 10
-3+3
VC=12 3.78 100
VC=12 3.78 1
VC=8.22 Volts
Q#17.3: What are the ideal currents and voltages in figure 17.36 (a)?Solution:
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IDEAL APPROXIMATION
+15 V
200 k
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+15 V
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VoutC C
B BQ1 Q2
E EIE IE
200 k
2IE
-15 V Figure 17.36 (a)
200 k
EquivalentModel EquivalentModelVout
B C C B
IC Q1 Q2IC
E E
200 k
-15 VFigure 17.36 (b)
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+15 V
+
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