analysis of differential amplifiers by muhammad irfan yousuf [peon of holy prophet (p.b.u.h)]

7/23/2019 Analysis of Differential Amplifiers by Muhammad Irfan Yousuf [Peon of Holy Prophet (P.B.U.H)]

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Muhammad Irfan Yousuf [Peon of Holy Prophet (P.B.U.H)] 2000-E-41 1Cell: 0300-8454295; Tel: 042-5421893

Analysis ofDifferential Amplifier


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).


+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


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


+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


+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)


+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.


+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


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)


+VCC

+

RC RC IC


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)


-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:


+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


B C C B

IC Q1 Q2IC

0V 0VE 0V E

tail current

7.5 k

-15 VFigure 17.6 (c)


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



0 (-VEE)IC +IC =

RE

VEE2IC =

RE

VEE2IE =

RE

VEEIE =

2RE

IT 2 mAIE = 1 mA

2 2




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)


+VCC

RC RC


B C C B

IC Q1 Q2IC

E E

RE

-VEEFigure 17.7 (a)


0.7 VReverse biased

0.7 VForward biased

Figure 17.7 (b)


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+VCC

+RC RC IC


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)


-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



-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


+12 V

+

3 k IC


B C C B

IC Q1 Q2IC

0V 0VE 0V E

tail current

5 k



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



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0 (-VEE)IC +IC =

RE

VEE2IC =

RE

VEE2IE =

RE

VEEIE =

2RE

IT 2.4 mAIE = 1.2 mA

2 2




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



+12

+

3 k IC


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


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


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


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

M h dI f Y f [P f H l P h (PB U H)] 2000E 41 35 M h dI f Y f [P f H l P h (PB U H)] 2000E 41 36


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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


behaves when an ac signal is present. When analyzing a transistor amplifier, we canreplace each transistor by aT model.

ic

ib

re

ie


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ic

ib

re

RC RC- +

VC1 VC2Vout

ic

ib

V1 reie ie

RE

17.9 (c)



re

RE

17.9 (d)

RE re

RE

+re

RE >>re

RE re

RE

=re


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RC RCVC1 - + VC2

Vout

ic ic

V1 reie

ie

re


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:


Vout 2icRC

V1 2iere

ie =ib +icHere


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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ic

ib

re

ie


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ic

ib

re

RC R

C

- +VC1 VC2

Vout

ic

ib

reie ie V2

RE

18.9 (c)


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re

RE

18.9 (d)

RE re

RE +re

RE >>re

RE re

RE

=re


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RC RCVC1 - + VC2

Vout

ic ic

ib

re V2

ie

ie

re


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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ic

ib

re

ie


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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


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)


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)


Cell: 0300-8454295; Tel: 042-5421893

re

RE

19.9 (d)

RE re

RE +re

RE >>re

RE re

RE

=re


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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)





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ic

ib

re

ie


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)





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ic

ib

re

ic

ib

re

ie


RC- +

VC1 VC2Vout

ic

ib

V1 reie ie

RE

21.9 (c)


re

RE

21.9 (d)

RE re

RE +re

RE >>re

RE re

RE

=re





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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)





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ic

ib

re

ie


ic

ib

re

RCVout

ic

ib

V1 reie ie

RE

22.8 (c)





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22.8 (d)

RE re

RE +re

RE >>re

RE re

RE

=re


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:




35/133

+15 V

5 k 5 k


B C C B

IC Q1 Q2IC

E E

7.5 k

-15 VFigure 17.11 (b)

+15 V

+

5 k 5 k IC


B C C B

IC Q1 Q2IC

0V 0V

E 0V E

tail current

7.5 k

-15 VFigure 17.11 (c)


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


0 ( V )


V 200(V )


36/133

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


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


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


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)


-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


-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


Vout RC 5 k190.6

V1 re 26.233


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)


DC Analysis


+15V


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DC Analysis

Ideal Approximation:+15 V

5 k 5 k


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


B C C B

IC Q1 Q2IC

0V 0V

E 0V E

tail current

7.5 k



0 (-VEE)

IT =RE

VEE 15 VIT = 2 mA

RE 7.5 k



0 (-VEE)


Vout =200 (-V2)


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IC +IC =RE

VEE2IC =

RE

VEE2IE =

RE

VEEIE =

2RE

ITIE = 1 mA

2


25 mVre =

IE

25 mVre =

1 mA

re =25


Vout RC 5 k200

V2 re 25


Vout =200 (-1 mV)

Vout =-200 mV

Input impedance of inverting input and differential output

Zin=2re

Zin=2(300) (25)

Zin=15 k


+15 V

5 k 5 k


B C C B

IC Q1 Q2IC

E E

7.5 k

-15 VFigure 17.11 (d)


+15 V


-0.7 (-VEE)I I


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+

5 k 5 k IC


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)


-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


IC +IC =RE

VEE 0.72IC =

RE

VEE 0.72IE =

RE

VEE 0.7IE =

2RE

IT 1.907 mAIE = 0.953 mA

2 2


25 mV

re =IE

25 mVre =

0.953 mA

re =26.233


Vout RC 5 k

190.6V1 re 26.233


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)


DC Analysis


+15 V


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Ideal Approximation:+15 V

1 M


B C C B

IC Q1 Q2

IC

E E

1 M

-15 VFigure 17.5 (b)

+

1 M IC


B C C B

IC Q1 Q2IC

0V 0V

E 0V E

tail current

1 M



0 (-VEE)

IT =RE

VEE 15 V

IT = 15A

RE 1 M



0 (-VEE)IC +IC =


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


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)


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


+15V

5 k 5 k


B C C B

IC Q1 Q2IC

E E

7.5 k

-15VFigure 17.11 (d)


+15 V


-0.7 (-VEE)IC +IC =


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+

5 k 5 k IC


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)


-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


C C

RE

VEE 0.72IC =

RE

VEE 0.72IE =

RE

VEE 0.7IE =

2RE

IT 14.3A

IE = 7.15A2 2


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)


The ac output voltage is:Vout =143.02 (V1)


+15 V


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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)


V3error =1 mV


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


+V


+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


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).


+VCC


RE


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+VCC

RC

Vout


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


2RE 2RE

-VEE Figure 17.19 (f)

VOLTAGE GAINa.c. equivalent circuit



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:


+15 V



51/133

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:


Vout ieRC

Vin(CM) ie(re +2RE)


+15 V


52/133

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.


+VCC


Potential difference =VCC - VBEHereVBE =0 V


53/133

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


IC =IBSuppose=50

IC =50 (15A)


+VCC

R


54/133

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


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


For drawing the dc equivalent circuit,

VCCIB IC


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)


IDEAL APPROXIMATION

According to KVLSum of all the voltage rise =sum of all the voltage drop


ic


56/133

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.


IDEAL APPROXIMATION 2nd APPROXIMATI ON

VCCIB

VCC - VBEIB


10 109rL

1010000

10 109rL

1010000


57/133

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


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)


10 k10 V

IL

106+10000


+VCC


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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


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)



+VCC


60/133

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




61/133

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)



+15 V


62/133

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


+15 V


+15 V


63/133

7.5 k 7.5 k


B C RL C B

IC Q1 Q2IC

E E

7.5 k

-15 VFigure 17.26 (b)

7.5 k 7.5 k


B C RL C B

IC Q1 Q2IC

E 0 V E

7.5 k

-15 VFigure 17.26 (c)


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


0 (-VEE)IC +IC =

RE

VEE2I


RTH =15 k

3V R 15k


64/133

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:


+15 V


+15 V


65/133

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


B C C B

IC Q1 Q2IC

E E

7.5 k

-15 VFigure 17.27 (b)


+15 V


0 (-VEE)IC +IC =

RE

VEE2IC =


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7.5 k 7.5 kA


B C C B

IC Q1 Q2IC

E 0 V E

7.5 k

-15 VFigure 17.27 (c)


0 (-15)

IT =7.5 k

15IT =

7.5 k


2IC =RE

VEE2IE =

RE

VEEIE =

2RE

IT 2 mAIE = 1 mA

2 2


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


RTH =15 k

A3V


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:


+12 V


+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)


+12 V


0 (-VEE)IC +IC =

RE

VEE2IC =


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12 k 12 k


B C C B

IC Q1 Q2IC

E 0 V E

6 k

-12 VFigure 17.29 (c)


0 (-12)

IT =6 k

12IT =

6 k


RE

VEE2IE =

RE

VEEIE =

2RE

IT 2 mAIE = 1 mA

2 2


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:


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


+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)


+12 V


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)


0 (-12)

IT =6 k

12IT =

6 k


RE

VEE2IE =

RE

VEEIE =

2RE

IT 2 mAIE = 1 mA

2 2


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:


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


+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)


+12 V


0 (-VEE)IC +IC =

RE

VEE2IC =

RE


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12 k 12 k


B C C B

IC Q1 Q2IC

E 0 V E

6 k

-12 VFigure 17.30 (c)


0 (-12)IT =

6 k

12IT =

6 k

IT =2 mAApplying KCL at node labeled 0 VAccording to KCL


RE

VEE2IE =

RE

VEEIE =

2RE

IT 2 mAIE = 1 mA

2 2


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 17.17:

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


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)?


Solution:IDEAL APPROXIMATION

+12 V

180k 180k


+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


B C C B

IC Q1 Q2IC

E E

270 k

-12 VFigure 17.34 (b)


+12 V

+


0 (-VEE)IC +IC =

RE

VEE2IC =

RE


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180 k 180 k IC


B C C B

IC Q1 Q2IC

0V 0VE 0V E

tail current

270 k

-12 VFigure 17.34 (c)


0 (-VEE)IT =

RE

VEE 12 V

IT = 44.445A

RE 270 k


VEE2IE =

RE

VEEIE =

2RE

IT 44.445AIE = 0.022 mA

2 2




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:


+12 V

180 k 180 k


+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


B C C B

IC Q1 Q2IC

E E

270 k

-12 VFigure 17.35 (b)


+12 V

+


-0.7 (-VEE)IC +IC =

RE

VEE 0.72IC =

RE


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180 k 180 k IC


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)


-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


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:


IDEAL APPROXIMATION

+15 V

200 k


+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)


+15 V

+

Muhammad Irfan Yousuf [Peon of Holy Prophet (P.B.U.H)] 2000-E-41 160

analysis of differential amplifiers by muhammad irfan yousuf [peon of holy prophet (p.b.u.h)]

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