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Microelectronics Circuit Analysis and Design 4th chapter 2 probl....pdf
REVIEW QUEsTl0Ns
Chapter2 Diode CircuⅡ s
111
lVhat characteristic of a diodc is used in the dcsign of diode signal proccssing
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二 Describc a simp1c ha1f-、 vavc diode rectincr circuit and skctch tllc Output vo1tage
Ⅴersus t1mc.
∶ Describc a siInp1e fuI1-wave diode rcctiner circuit and sketch the output Voltage
Ⅴersus timc.
三 VVhat is the advantage of connecting an RC nlter t。 tllc Output of a diode rccti-
ner circuit?
J Denncr匀 pple vo1tage。 How can the magnitude ofthc ripple voltage be rcduced?
- Describe a simple Zencr diode vo1tage reference circuit。
ˉ
、、厂hat Cfect does the Zcner diodc rcsistance haⅤ e on the voltage referencc circuit
opcration?lDe丘 ne load regulation.
3~ lVhat are the genera1characteristics of diode c1ippcr circuits?
; Dcscribe a sirnpIe diodc c1ippcr circuit that lirnits the negativc portion of a sinu-
soida1input vo1tage to a specincd va1ue。
∶0 V1厂hat are the general charactcristics of diodc c1amper circuits?
l二 lVhat one circuit c1cment, bcsides a diodc, iβ present in a11 diode clamper
cicu⒒ s?
12 Dcscribe the proccdure uscd in the anralysis of a circuit containing t、 vo diodcs
How many initia1assumptions conccming tlle state of the circuit are possib1c?
13 Dcscribe a diode OR1ogic circuit,Compare a1ogic1value at the output com-
pared to a1ogic1va1ue atthe input.Arethey the samc va1ue?
l亠 DcscⅡ be a dodc AND log妃 c虹 cuit Colllpare a log⒗ O value atthe output com-
pared to a logic0Ⅴa1ue atthe input Are they thc samc value?
15 Dcscribe a siFnp1c circuitthat can bc used to tum an IjED on or of、 vith a high
or lo、 v input vo1tage。
PRO:LEMs
「
:Xote∶ In thc№ llowing problems,assume勺 =0unless othcrwis9speoned,]
section2.1 Redfier CIrcu⒒ s
21 Considcr tlle circ缸t shown血 Figu￡ P2.1Let R=1kΩ ,Vr/=0,6Y and
I'=20Ω (a)Plot山 e voltage“ ansfer characteⅡ stks vε vesus v'ovcr tllC
△mge-10兰 vr兰 IO V(b)Assumc vF=10sin
ω r(V)。 (i)SkCtch v。
ve1sus time for tllc蚯 nus⒍dd i11ptIt (io Find tlle average valuc of v@.
(iii)DCtC∏nine thc peak diode cuⅡ cnt (iv)What is the PIV of thc diode?
22 For thc circu⒒ shown in Figure P2,1,show that for vJ≡ 0,thC Output volt-
agc is approximatc1y givcn by
Figure P2,1
V@=叻 -
n巧
丶
/
f
l
\
l
J
/竺
w
2.3 A ha1iwave rcctiner such as shown in Figure2,2(a)has a2kΩ load,
Thc inputis a120V(mns),60Hz signal and山 c订ansforlllcr`a10∶ 1step-
down“ aIlsfom⒃⒈Tlle dode h凼 a cut~in volta罗 of‰ =0· 7V(勺 =0),
(a)Whatis the peak output voltagc?(b)Detennnine the pe岔 k diode current。
1日 2
Part1 semicondudor Devices and Basic App"cati° ns
2,4
2.5
(c)What is曲 e frac吐 on ω∝cent)of a cyc1e that vε >0。 (Φ Determine the
aVerage output vo1tagc (c)Find the average cuⅡ entin the1oad。
Con蚯 der the brdtteγ char鲈 ng c△ c"t shown血 F熄ure2.4(a)Assume that
yB=9V ys=15Y and
ω =2万 (60).(⑻ DCteⅡ mne the Value of R such
thatthe aVerage battery charging currentis氵 D=0.8A.(b)Find the fraction
of dme thatthe cⅡ ode is conducting。
Figure P2.5 sho、 vs a siInp1e fu11-wave battery charging circuit。 Assume
‰ =9、 ‰ =⒍ 7、 and vs=15sin[2万 (60)r]α),(a)Detemine R sLlch
th肫 the peak b·dttcry charging cLlrrent“ ⒈2A.O)Detemine the ax/erage
ba钆 ery ch盯ging cuⅡent.o)Deterlllhe the fracdon oftime that each dode
is conducting.
D1
Ⅳ
1
屿
彐
Figure P2.5
2.6
2,7
28
2.9
2,10
The fu1⒈ waVe recjner c廿 cu⒒ shown h F瑭 ure2.5(⒆ in the text k to delher
0.2A and12V(pe·dk v缸ues)to a1oad.The⒒pp1e vo1tage is to bc1i血 ted to
0.25V Theinputs圯 nal。 120Ⅴ 【l△n9扯 60Hz,Assume dode cut-in voltages
of0.7Vo)Detαmine the req证 red mnsI血 o of曲 e transfom∝ lb)Find the
requ△ ed Value of仇 e capac姒觅 O)Ⅵ△rat o the PIV mung。f仇e山 ode?
The input signa1voltage to the fuⅡ wave recdner circu⒒ in Figure2,6(a)in
the text^vr=160sh⒓ 万“Ol莎 ]Ⅴ .As阢 me‰ =0· 7Ⅴ for each dode。
DeteⅡ nine the required tums rauo of the transfomcr to produce a pe犭 k
output voltage of(ω 25V and(b)100V(c)VVhat must bc thc dodc PIV
rating for each casc?
The output rcsistancc ofthc fu11-wave recunerin F坨 urC26(a)in thC tCxt is
R=150Ω 。An1tcr capac⒒ or is connected in para11c1with R.Assume
‰ =0· 7v ThC pcak output vo1tage`to be12V and山 c ripplc vo1tagc`
to bc no more than0.3V The input frequency is60Hz。 (a)Deter而 nc伍 e
rcquicd rms va1ue of Vs,(b)DeteⅡ 匝ne thc required n1ter capaotance
Va1ue.(c)DeterInine the peak cLlrrent through each diode,
Repeat ProbIem2.8forthe haliwaVe recdnerin F屯 ure2.∝ a)
Consider d1e ha1iwaVe rectiner circuit shown in Figure2.8(a)in the text.
Assume vs=10sin[2万 ∞ Q)扌 ](V),‰ =⒍ 7、 alld R=sO0Ω 。(a)Whatis
the peak output Voltage?(b)Deterrnine the value of capacitance required
such that the ripple voltage is no more that砗 =0,5V。 o)Ⅵ△at is the PIV
rating of the diode?
2.11 The parameters of the ha1⒈ wave rccti丘er circuit in Figure2.8(a)in the
text are R=1kΩ , C=350u只 and ‰ =0·7v Assume vs(莎 )=
V愆 ⒍nlzJT(6⑴ 莎](V)whCre ys is in the range of11兰 ys兰 13Vo)What
o the range in output volt· age?lb)DeteⅡmne由 e range in npple voltage。 rc)
If thc五pple voltagc is to be Ⅱ臼血ed to⒕ =0.4V deteⅡ Ⅲne the血 nimum
va1uc of capacitancc rcquircd.
△e thc
c that
such
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cragc
diodc
·hvcr
ed to
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d tllc
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odc,
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xt is
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1um
Chapter2 Diode Circu"s
113
l i 丁he血 l1-wave recti丘 cr circuit shown in Figurc P2,12has an input signa1
示hosc frequency is60IIz,Thc Itns va1ue of vs=8,5V Assume each diode
:ut-in vdtage is‰ =0· 7v.(⒆ Wh·at is山e maximum Ⅶ1uc of%?(bl If
R= 10Ω ,determinc山 e va1uc of(· such that the ripple voltage is no largcr
△m025V(c)What must be the PIV rating of each modc?
%
c一
一
ˉ—
R
+乞
Vs
i
+
丶
一
D
2
门
Figure P2.12
Considcr the fu11-waⅤ e rectincr circuit in Figure27of thc text The output
rcsistancc is Rz=125Ω ,each dodc cut-in voltagc`‰ =07v,and tllc
frcquency ofthe input signa1is60Hz An1ter capacitor is connccted in par-
a11c1witll RL,The n1agnitude oftlle pcak output vo1tage isto be15V and the
Hpple voltage is to bc no more tllan035V(al Det∝ mine the rms valuc of
1· s and(b)the requircd value of the capacitor.
The circuit in Figure P2.14is a comp1cmentaγ output recti丘 cr.If vs==26
sh⒓ 万6Olr]V,sktc· ll tlle outpLlt Wavcforms v产 al△ dv歹 vCrsLls time,a⒌
suΠⅡng‰ =⒍ 6V for cach山 odc
Figure P2.14
215 A fu11-、 vaⅤ e rectiner is to be dcsigncd using the centcr-tappcd transformcr
con丘 guration The peak output Vo1tage is to be 12V thc noΠ Ⅱnal load
cuⅡ ent is to bc05卢、 and tllc ripp1c Ⅴoltagc is to bc1imitcd to3percent.
Asmmc‰ =0.8Ⅴ alld let tf=120√Ξs血 [2万 ∞o)r]V ra)w№ t is tllc
饣aljsformer turns ratio?rb)、 汛严h·dt诋 the minimum value of C required?
(c)Ⅸ′atis the pc迂 酞dode currcnt?(d)DCteⅡ 匝nc山 e avcrage diode cuⅡ ent,
(Cl Ⅵ%atis the PIV rating oftllc dodcs
216 A fuⅡ -wave rectincris to be designed using the bridge circuit connguration。
The pcak output vo1tage is to bc9ˇ ∶thC nomina11oad cuⅡ entis to bc100
mA,and tlle rⅡ ple voltagc is to bc1imitcd to 怖 =02V Assumc
‰ =08v al.d1ct vr=120√ Ξsin[2万 (60)r](V).Q)What k thC trans-
folIller tLlms mto?rb)what。 the minimum valuc of C req诬 ∞d?o)What
114
Port1 sen】 icondudor Devices and Basic AppⅡ cations
o the peak diode cuⅡent?ld)DeteⅡ m⒚e the average diode cuⅡ ent.(e)、 、冂Lat
is the PIV ra住 ng ofthe diodes。
米2.17 sketch v@versus住 me for the circuit in Figure P2.17with the input shown。
A“ume V‰ =0.
Figure P2.17
Figure P2.18
米2.18 (a)Sketch v@versus time for the circu⒒ in Figurc P2.18。 The inputis a sinc
w⒏汜 gi妮n by叻 =10“ nω 莎V As叩 mc‰ =0。 O)Detem血 e the rms
va1ue ofthe output vo1tage
section2.2 zener E)iode(:″ cuits
2.19 Consider曲 c circuit shown in Figure P2.19.The zener diode vo1tage is
yz=3,9V andthe Zener dod0incremental re蛀 stance is△ =0,(a)Detcr-
血ne Jz,Jz,and the po、 ver dsslpated in the山 ode。 (1,)Repeat part o)if the
4kΩ 1oad resistor is increased to10kΩ
12⒑
120Ω
Figure P2.19
Figure P2.20
2.20 Consider me zcner diode circuitshGvn in Figure P2.20.Assume yz=12V
and饯 =0。 (a)Calcu1ate the zcn∝ dode current and the power dssipated
in the zener dode for RL=∞ .(b)W腕 t is the value of RL such that the
current in the zener diode is one-tcnth of the cuⅡ ent supp⒒ ed by the40V
source?O)DeteⅡmne the poWer do蚯 pated in由 e Zener dode for the
conditions of part(b)。
221 Consider the zen∝ 山ode ckcⅡt shown h Figure P2,21.Let yr=60V
Rj=150Ω ,and yz@=15.4V Assume饯 =0。 The power raung of the
dode is4W andthe mh⒒ mum dode cuⅡ entisto be15mA。 o)Det∝mine
the range of diode currents。 (b)Detemine the range of1oad resistance,
Figure P2.21
Chapter2 Diode Circuits
115
米222 In thc voltage rcgu1ator circu⒒ in Figure P2,21, yr:=2oˇ 1 Vz=10ˇ ∶
ˉ 0)DCt∝ mhe JI,Jz,alld rr,if
R氵 =222Ω ,and Pz(ma0=400汛
RL=380Ω 。(b)Determine thc valuc of RL that wi11cstab1ish Pz(max)in
tlle m。 de o)RCpc扯 p盯 tlb)if R氵 =175Ω ·
223 A Zencr diode is connectcd in a vo1tage regulator circuit as shown in Fig-
ure P2.21.The zcner vo1tage is yz==1o V and the Zener res^tance is
assumed to be rz=O ra)Determine tlle value of R氵 such that tlle Zen∝
diode rcmains in brcakdown if thc 1oad cuⅡ cnt varies from rL==5o to
500mA and if tl△ c inpLlt volnge v盯 ⒗s from yr=15to20V Assume
rz(m血 )=o.1rz(maxl,rb)Det∝ mine tlle power rat血 g∞ q“∞d for the
zener diode and the1oad resistor.
2,24 Consider伍 e zener chode crcuitin Figure219in the text Assume paralnct∝
valtles of yz@=5.6V (mode v。 ⒒agc when rz≡ o), △ =3Ω , and
R氵 =50Ω ,DcteⅡ mne VI,Jz,r.,and tl.c pow∝ dssipated h the山 ode for
(Θ yPs=10Y Rz=∞
RL=oo;alld ld)yPs=12V RL=200Ω .
;(b)yPs=10Y RL=200Ω
;rc)vPs=12Y
D2,25 Design a vo⒒ agc regu1ator crcuit such as sho、vn in Figure P221 sO tllat
Vz=7.5V mlc zen∝ diode vo1tage is yz=7.5V狨 rz=1oIllA The in-
crclnenta1 山odc resistance is rz==12Ω Tlle no洫 nal supp1y Vo1tage is
⒕ =12v alld tlle nO11nlldl load￡ sistaIlce k RL=1kΩ .0)DC℃ Ⅱ匝ne R卜
lb)Ⅱ yr vancs by± 10p∝cellt,calCtll狨e tlle sourcc regula伍on.氓
`at is
tlle vaJiation in output Ⅴoltagc? (c) If RL v缸 Cs ov∝ the rangc of
1kζ2兰 RL兰 oo,what^the variation in output vo1tagc?Detcrl1unc tlle1ordd
regu1ation.
2.26 The perccnt regu1ation ofthe Zencr diode regu1ator shown in Figure2,16is
5percent The zcncr vo1tagc is Vlz@=6V and the zener re蛀 stance is
r之 ==3‘’ Also, the load rcsistancc Ⅴ舶 cs between 500 and 1000Ω , thc
input resistance is R氵 =280Ω ,and the∏1inimum powcr supply vo1tage is
yPs(min)=15V DeteⅡ吐ncthe maximum powcr supp1y vo1tage a11oWed.
扌227 A Vo1tagc rcgu1ator is to havc a nominal output voltage of10、 l Thc speci-
ned zener diodc has a ra伍 ng of1※、has a10V drop at Jz=25111A,and
has a Zencr resistance of rt==5Ω .The input power supply has a non1ina1
va1uc ofˇPs=20V and can vary by=± 25perccnt The output1oad cuⅡ ent
o to Ⅴary betwccn J‘ =0and20mA,o)r tllc minimum Zenσ cuⅡcnt`
to bc Jz=5mA,determinc the rcq“ ℃dRⅡ Φ)DeterIn血 ethc ma虹 mum
variation in output vo1忱 ge.o)Deterrmne tllc perccnt regu1ation
叮 28 ConsⅡ er tllc orcmtin Figurc P2,28Let V%=0ThC⒃ cond盯 y voltagc`
ghcn by vs=⒕ sin c,,莎 ,Wherc炼 =24V The zen∝ 山odc has p盯 amctcrs
yz=16V rat Ⅰz=40mA and rz=2Ω .Determinc R氵 such that the1oad
cm犯nt Can v盯 y ovcrthe nngc40兰 r.兰 40o mA witll Jz(min)=40IllA
and nnd c such thatthc ripp1e voltage is no1arger than1′ 11
‘
⒈
￡
Ⅳ
¨e
d
屺
w
屺
)、
thc
1nc
Figure P2.28
116
Part1 semicondudor Devices and Basic App"cations
米2,29 Thc secondaΓ
`vo1tage in thc circuitin Figure P2.28is vs==12sin c/,莎
、l The
zener mode has paraltleters yz=8V荻 rz=10o mA alld严 z=O.5Ω 。Let
匕'=0and R氵 =3Ω ,Detcrlll血 e tlle percellt regt11aton for load cuⅡ ellts
bctween JL=0,2and1A Find C such thatthe ripp1e vo1tage is no1arger
than0.8、 1
section2.3 CⅡ pper and Clamper(△ rcuits
2.30 The parameters in the c△ cot shown in F熄 urc P2· 30a∞ lo=07Y
yzl=⒉ 3、 alld VIz2=⒌ 6VP1ot vo versus vr over山 c range of
-10兰 v'兰 +10V
R=05kΩ
ˇz=3v
Figure P2.30
Figure P2.31
2,31 Consider the circuit in Figurc P2.31,Let Vn/=0(a)P1ot v@versus V∫ ovcr
thc range-10兰 vr兰 +10V.(b)P1ot f10verthc same input vo1tage range
as pa⒒ (a).
2.32 Forthe c立 rcuit in Figurc P232,(a)p1ot vo versus vr for0兰 Vr 兰 15、 l As冖
阢 mC‰ =0· 7v Indicate沮 1b1· cakpoil△ ts tb)Plot rD Ovcr tl△ e saIne ral鸭 e
of订put voltagc.(Θ Compa⒑ thC rcsu1ts ofparts(al and(b)witl△ a computcr
sirnu1ation,
R=4kΩ
+15V tJ
R=4kΩ
(a)
Figure P2.32
Figure P2.33
2.33 Each diodc cut-in vo1tagc is0.7V for the circuits shown in Figurc P2.33。
(a)Plot vo versus vr over the rallge-5兰 vr兰 +5V for the⒍ rcuit h
Figure P2.33(ω for(i)Vl:=1· 8V and Ci)V1B=-18V rb)RepCat pa⒒
(a)for the circuit shown in Figure P2.33(b)
米2,34 The diode in the c△ cuit of Figure P2.34(a)has picce、 visc1inear parameters
‰ =⒍ 7v alld I尸 =10Ω ·o)Plot v@Ⅴ ersus vr for-30兰 tr兰 30V lb)If
曲e trhngu1盯 wave,shown in Figurc P2.34(b),is appⅡ cd,p1ot the output
versus tⅡ ne.
ChaPter2 Diode Circuits
R=100Ω
Fˉ gvre P2.34
2· 35
∶;:llⅧ∶莞∮1Ⅰ
peat part(ω br曲 eo∞证t in∏ gu⒑ P2,35(ω 。
忒C;臂i昱畲￡圭犭堑苜∴萝∶:甘罕T
(D vr:=5v rdlld C0yB=-5V lb)R⒍
(a)
Figure P2.35
'
0
P1ot Vε for cach circuit in Figure
⑶ ‰ =O alld(b)‰ =⒍ 6v
P2,36 for thc input shown AssumC
=Igure P2.36
237 Considcr the para11e1 c1】 ppcr circuit in Figurc 226 in thC tCxt Assumc
lz1=6、 ‰ 2=4V and‰ =⒍ 7v for rd11⒍ odes For vJ=10sin ωr,
skctch v@versus ti1ne ovcr two pcriods ofthC input signa1,
兴238 A car’ s radio may be sut刂 c∝Cd to voltagc spikes induced by coupling f1· om
the ignition systcm。 Pu1scs on tlle order of=± 250V and1asting for120‘ ‘s
may exist.Dcsign a c1ipper circuit using resistors,diodes,and Zener diodes
to1imittllc input voltage bet△ l,ecn+14V alld一 O.7V Spco卸 poW∝ mtings
of tlle componellts,
h
c
`
c
t
ι
t
s
c
r
Y
o
f
q
g
e
L
s
.
g
e
ι
e
r
;
3
.
血
血
刈
旧
哑
118
Part1 semiconductor Devices and Basic App"cations
2.39 Sketch the steady-state output vo1tage vo Versus犭Lnc
Figure P2.39 Ⅵ汶曲 the input vo1tage given in Figure
y/=0and assumethe RC ume∞ nstant o large.
for each circuk h
P2.39(0)Assume
Vr
vr o-H
ta)
(
vo 叻 0-ˉ 丬
〓‰
一
V
+
十 ⊥
‰=5v二
Figure P2.39
D2.40 De⒍ gn a dode c1amper to generate a steady-state output vo1tage v@from
the i叩ut vo1tage vr血 own h∏ gure P2.40if0)‰ =0and(b)‰ =0·7v
Vo
+27V
0
-173V
Vr
+10V
0
=10V
(a)
F:gure P2.40
D2.41 De蛀 gn a dodc c1aInper to gcnerate a steady-state output vo⒒ age vo from
the i叫”t vo1tage Vrin Figwe P2.41if‰ =0.
Vr
+20Ⅴ
0
-20V
Figure P2.41
Vo
+30Ⅴ
0
-10V
—
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巳
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