Ch04 - electric circuit design solution PDF

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Summary

4     In the circuit of Fig. 4, IS 1  = IS 2  =  5   10 A. (a) Calculate the value of I X if VB = 0 V.(b) Find the value of IS to get IY= 3 mA.Figure 4.Solution(a) Let Is 1  = Is 2  = Is, then/ 2115 0 / 0.21e5102 mA.4 mA.VVBE T CQ CQ SXCQ CQIIIeII I...

Description

Chapter4(ProblemsandSolutions) 3 Dueto somefabricationerrors,thecross‐sectionalareaofemitterhasdoubled.Howdoesthe collectorcurrentchange? Solution IC 

AE qDnni2 NB WB

(eVBE / VT  1).



I C  AE .

SoifAEincreasesbyafactoroftwo,thenICincreasesbythesamefactor.                                  

4In thecircuitofFig.4.48,IS1=IS2=510–5A. (a)CalculatethevalueofIXifVB=0.7V. (b)FindthevalueofIStogetIY=3.5mA.

 Figure4.48 Solution (a) LetIs1=Is2=Is,then I CQ 2  ICQ1  I S eVBE /VT  5  1015 e0.7/ 0.026  2.46 mA.



I X  I CQ2  I CQ1  4.92 mA.

(b)  I CQ3  IY  I S3 eVBE / VT  I S 3 e 0.7/ 0.026  3.5 mA 3

3.5 10 e0.7 /0.026  7.10 1015 A.

I S3 

        



5

In thecircuitofFig4.49,itisrequiredthatthecollectorcurrentofQ2istobetwicethatofQ1ifVBE1 –VBE2=0.Determinetheratioofbasewidthsoftwotransistors,ifotherdeviceparametersare identical.



 Figure4.49 Solution 2

I C1 

AE1qDn ni VBE 1 / VT  1). (e N BWB1 2



A qD n V / V I C2  E2 n i (e BE 2 T  1). NB WB 2 Since for VBE1  VBE2  0, I C 2  2IC 1  2

2

AE 2qDn ni VBE / VT A qD n  1)  2  E 1 n i ( eV BE / VT 1) (e N BWB2 N BWB1 1 2  W B 2 W B1 W B1  2W B2 .

                



ϲ Considerthe circuitofFig.4.50. (a)IfVB=0.6VandIX=3.5mA,determineIS1andIS2suchthatIS1=2IS2. (b)FindthevalueofRCwhichplacesthetransistorsattheedgeoftheactivemode. 

 Solution I C 1  I S 1eV BE/V T I C 2  I S 2e VBE /VT I X  3.5mA, IS 1  2 IS 2 I X  IC1  IC 2 IC1 IC 2



IS1e VBE/ VT I S 2 eVBE /VT



2 IS 2 eV BE/V T I S 2 eVBE /VT

2

I C 1  2I C 2 I X  2 IC 2  IC 2  3.5 mA I C 2  1.166 mA and I C1  2.334 mA 2.334 mA  2.217 10 13 A e0.6/0.026 1.166 mA I S 1  0.6/0.026  1.107  10 13A. e IS 1 

To place the transistors at the edge of active region,VB  V X ,

V B  V X  VCC  I X RC 0.6  3  3.5  10 3  RC  RC  685.7 .

     



ϳ 

Consider thecircuitofFig.4.51.CalculateVXifIS=610–15A.

 Solution VBE  VCC  IC  1k  IE  1k  VCC  IC  2k

( IE  IC )

I  VT ln C   VCC  IC  2 k  IS   IC   2  IC  2 k 0.026  ln  15   6 10   IC  0.026  ln   I C  2 k  2  15   6 10  I C  0.67 mA V X  I C  1k  0.67V.

                   





8: The required circuit is shown in figure 4.52:

Figure 4.52

In figure 1 transistor Q 1 is at the edge of the active region. That is, VC  V B …… (1) Here, Base voltage is VB , Collector voltage is VC  Apply Kirchhoff’s voltage law in the circuit shown in figure 1: VCC  RC I C  VC

 RC I S e

 VB     VT 

 VC

…… (2)

Substitute equation (1) in equation (2) VCC  RC I S e

 VB     VT 

 VB

…… (3)

Here, Supply voltage VCC  is 2 V Thermal voltage VT  is 26 mV Saturation current  I S  is 5 10 16 A

Collector resistance is  RC 

Substitute corresponding values in equation (3) 2  RC  5  10  

VB

  

16

 VB     26 10 3 

e

 VB

2 5 10 16 15  4 10 Use numerical methods or calculate by hit and trial VB  760 mV 3

RCe  2610

V B 

RC  500 

ϵ Consider thecircuitofFig4.53.CalculatethevalueofVCCthatplacesQ1attheedgeoftheactive region.AssumeIS=510–16A.

 Solution VCC  IC RC  VC  I C RC  VC  I C eVBE/ VT  RC  VC  16

 5 10

1.59 V.

                         

e

0.75/0.026

  500  0.75

–16 1ϭ Consider thecircuitofFig4.55,assuming =100andIS=610 A.IfRB=10k,determineVB suchthatIC=1mA.

  Solution VB  I B R1  VBE  IC RE 

IC



1  103  10 10 3  V BE  1  10 3  10 10 3. 100



R1  V BE  I CR E



But, I  VBE  VT ln  C   IS   1 10 3   0.026ln   0.7316 V. 16   6  10 

Therefore, V BE  0.1  0.7316 1   1.8316 V.

      



1Ϯ

culatethecollectorcurrent. InthecircƵitofFig4.55,ifVB=2VandRB10k,caů

Solution VB  I B R1  VBE  IC RE  2

IC

 IC

R1 V BE  I CR E

  10  103  0.7316  IC  1 103

100 I C  1.153 mA.

                                



1ϯ In thecircuitdepictedinFig4.56,ifIS1=2IS2=510–16A, 1= 2=100andR1=10k,computeVB suchthatIX=1mAandIY=2mA.

  Figure4.56 Solution I X  IS 1eVBE 1 VT /

I Y  I S 2 eVBE 2/ VT VB E1  V BE 2  V BE  I 1 IX  S1  . IY IS 2 3

Also, I X  1 IB1 and IY  2 IB2 ; 1  2  100 I B1 I X 1   IB2 IY 3 VB   I B1  I B 2  R1  VBE . I  V BE  VT ln  X   I s1   1  103   0.026  ln  16   5 10   0.736 V. I B1 

IX

 1  10   10 A. 100 IY 3

 I B2 



3 10 3  30  A. 100 VB   10  A  30  A  10 10 3  0.736 

 1.136 V.





1ϰ In thecircuitofFig4.56,ifIS1=310–16AandIS2=510–16A, 1=2=100andR1=5k,VB= 800mV,calculateIXandIY. Solution V BE1 V BE2 V BE V B   I B1  I B 2  R1 V BE 

R1



I X

IS 2  2 IS1 ; VB  

R1



I   I Y   VT ln  X   I s1  IY  2 IX  IX    Is1 

 3I X  VT ln

5  10 3 IX  3 I X  0.026  ln   16  100  3 10 

 900 mV I X 1 mA and IY  2 mA.

             



1ϱ The base‐emitterjunctionofatransistorisdrivenbyaconstantvoltage.Supposeavoltage sourceisappliedbetweenthebaseandcollector.Ifthedeviceoperatesintheforwardactiveregion, whatchangeswilltakesplaceinIBandIC? Solution NochangeinVBEmeansnochangeinbasecurrentIB,andhencetherewillbenochangeinIC.                     

1ϲ In abipolardeviceVBEchangesby10mV.Calculatethechangeingm,ifthedeviceisbiasedat IC=2mA. Solution gm 

I C I S eVBE/ VT  VT VT

g m 

I S e VBE /VT  VBE IC  VBE  VT 2 VT 2

g m 

gm V BE. VT

gm 

IC VT

2 10  0.026 1  13 

3



g m   g m 

I Se VBE / VT V BE V T2 I CV BE 2

VT

gm  V BE VT

1 (10 mV) 13 0.026 1  . 33.8 

        





ϭϳ A transistorgivesatransconductanceof1/13withbase‐emittervoltageof800mV.Calculate thevalueofISofthetransistor. Solution gm   IS 

IC VT I SeVBE /VT VT gm VT eVBE / VT 0.026 



 1 13

e 0.8/ 0.026   8.67  10 17 A.

               

2Ϯ The collectorvoltageofabipolartransistorvariesfrom1Vto4Vwhilethebase‐emittervoltage remainsconstant.WhatEarlyvoltageisnecessarytoensurethatthecollectorcurrentchangesbyless than2%? Solution  V  I C  I S eVBE /VT  1 CE  VA    VCE  I C  I S eVBE /VT VA I C  IC

I S eV BE /V T

 VCE VA

 V  I SeVBE /VT 1  CE  VA   VCE  V A  VCE

I C  0.02 IC  VCE   0.02 VA  VCE V CE  3V V CE (min)  1V 50V CE  V A  V CE (min) 150  V A  1 V A  149 V.

                



2ϯ In thecircuitofFig4.59,IS=610–15A.DeterminethevalueofcollectorcurrentandVXfor (a)VA=and (b)VA=4V.

 Solution (a) VA   I C  I S e VBE /VT  6 1017  e0.8/ 0.026  1.38 mA V X  2.5  I C RC



 2.5 1.38 10 3 1 103  1.11 V.

(b) VA  5 V V  V /V  IC  IS e BE T 1  CE  VA   VX  VCC  IC RC IC 

VCC  VX RC

 V  VCC VX  I S eV BE /V T 1  CE  RC VA   V  V /V  V CC  V X  R CI Se BE T 1  CE   VA  V 2.5  VX  1 103  6 10 17  e0.8/ 0.026  1  X  5   VX  0.874 V and IC  1.625 mA.





Ϯϲ Abipolar currentsourceistobedesignedforaspecificoutputcurrent.IfVA=2Vandoutput resistanceisgreaterthan10k,findtheoutputcurrent. Solution  V  I C  I Se VBE /VT  1  CE  VA   dI 1  C r0 dVCE 1 1  I S eVBE /VT   IC  VA VA  V r0  A IC VA  10 k IC  IC 

                          

2V  0.2 mA. 10 k 

ϮϵForthe VA=.   

circuitdepictedinFig.4.64,calculatethevalueofcollector‐basebias,ifIS=710–16Aand

 VCC  IC RC  VCE IC 

V CC  V CE RC

 IS eVBE /VT V BE  V CE  RC I Se VBE / VT V BE  0.2  1 10  6  10 VBE  VCC 3

 0.7144 V IC  0.5144 mA.

                     

 16

e

VBE /0.026

3ϯ If IS1=3IS2=610–16A,calculatethevalueofVBinFig.4.68,requiredforgettingIX=10mA. 

 Solution I C 1  I S 1e VBE 1/VT  6  10 16e VBE 1/0.026 I C2  I S2 e VBE 2 / VT  2  10 16 e0.82/ 0.026  9.95 mA I X  I C1  I C 2 10 mA  I C1  9.95 mA I C 1  0.05mA V EB 1  VT e 0.05/0.026  0.026 e 0.05/ 0.026  0.654 V VB  2  0.654  1.346 V.

               



–16 3ϰ In thecircuitofFig.4.69,calculatethevalueofIC,if =100andIS=610 A. 

 Solution VCC  VEB  RB I B  VEB  RB

IC



1.5  VEB  23  103 

IS eVBE / VT



  16

1.5  VEB  23  103  VEB  0.762 V I C  3.20 mA.

                      

6 10

e 100

V BE / 0.026

–1ϳ

ϯϴCalculatethecollectorcurrntofQϭinFiŐ4.73,ifIS=310 A.



 Solution VEB  2.5V  1.8V  0.7 V I C  I S eVBE /VT 17

 3  10

 e

0.7/0.026

 1.477  105 A.

              

4Ϯ A pnpcurrentsourcemustprovideanoutputcurrentof5mAwithanEarlyvoltageof2V.What istheoutputimpedance? Solution VA IC 20 V   5 mA

r0 

 4 k

                                 

ϰ5 In thecircuitinFig4.79,=100,andVA=. (a)Determinethevalueofcollector‐baseforwardbiasifIS=510–16AandVBEof0.8V. (b)Calculatethetransconductanceofthedevice. 

 Solution VC  VB  VCB RC IC  IB RB  VCB V EB  V CC  I B R B  2.5  I B  360 k 0.7  2.5  I B  360 k I B  5 A I C  0.5 mA V CB  R C I C  I B R B  4 k  0.5 mA  5  A  360 k  0.2 V.

                   



ϰϳConsider thecircuitofFig4.81whereIS1=IS2=510–16A,IC2=0.5mA, 1=100, 2=50,VA=,and RC=500.IfVin=1.45V,whatisthevalueofVBE1andVBE2?

 Solution Vout  IC 2RC  0.5 mA 500  0.25 V. I  V EB2  VT ln  C2   I S2   0.5 mA    0.026 ln 16   (5/3)  10   0.745 V. Vin  VBE1  VBE2 . V BE 1  V in  V BE 2  1.45  0.745  0.705 V.

48 In order to find out VBE1 and VBE 2 , we have to use the formula, I  VBE 2  VT ln  C 2   IS 2  and Vin V BE1  V BE2

In accordance to question, Vin  1.45 V

I S1  I S2  5 106 A IC1  0.5mA

1  100 2  50 VA   RC  500 

VBE is deified as,

I  VBE2  VT ln  C 2   IS 2 

To obtainVBE 2 , substitute 0.5 103 A for I C 2 and 5 10 16 A for IS 2 in the equation I  VBE2  VT ln  C 2  , we get,  IS 2 ...