This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework Solution (Chapter 1) Electronics Devices (332:361) Fall 2010 EXERCISE 1.10 Voltage: gain = 20 log 100 = 40 :13 1.14 Avg =40 :13 = 100 ‘WV
Currant gain = 20 Log 1000 = 60 (15 U: RL 2
Power gain = 10 log AP= 10105 (AVAE) PL = E = (A”HU'RL+R9) RL
=10: 105:50d3 1
0g =v§x£100x;) 10003251}?
1 + 1
10 5 ? v?“
1.12 va=l>< 6 #10 v=1ouv P£=ﬂ= *
10 + 10 R; 10,000
~15 2 P 2.5:)?
1951:3an = ﬁlo—xi.) =10'”w v Apsf= A 2=2.5><10‘wa
10 i 10 1:;
IO log AP: 44 (13
With the buffer ampliﬁer: 119 R:
._ 1 X R‘ x A x R!“
Ila "" Rl+ I w RL+Rﬂ 1+1 10+10
2 2
pL=Eazﬂ=6zsmw
EL 10
. v 0.25 V
= —° = —— = 0.25 VN
Voltage gam v: 1 v
=—12dB
ﬂ Power gain (AP) E P_ where PL = 6.25 111W and P; = viz}, 1V
_ 2: =05 A
“"0511 and “ 1MQ+1MQ ’" Thugr
Pi = 0.5 X 0+5 = and, _3
6.25 x10 = 25 X“): 0.25 x 10'“
10103 AP = 44 dB AP: vi, = ibr#+(ﬂ+ l)i&Rf = ib[rn.+ (ﬁ—I 1)R,] But 11;, = v; and t}, = ix. thus RinE$=ﬂ=rx+(B+l)Rp I: ll, PROBLEM . ' ode ‘elds .
1.16 (a) NOde equatm“ at the 39mm n 3" Now, I... II. and 13 can be easﬂy found as __iSV_lS—2.375= 2.
Using the fact that the sum oi: the voitage drops across ’1 " 10 " 10 13625 m L06 ma
RI and R3 equals 15 V, we wnte I _ ID— V = = L125 m =l 13 mA
l5=IIR1+13R3 1" 5 S .
=1011+Ut+19><2 13=1=ZEE=L1375111A==L19M
= 121.1212 R31 2 Method (b) is much preferred; faSIer. more insightful
and less prone to errors. In general. one attempts to
identify the least possible number of variables and write
the corresponding minimum number of equations. That is.
Similarly. the voltage drops across R2 and R3 add up to
10 V, thus
10 = Isz + [3R3
=5I2+UI +12)1<2
which yields Dividins {1) by (1) Rim
2!. + 7n= 10 (2) _' “Rs” 10} ... 3
_ . 1+ 1111;111:011
Equattons (l) and (2) can be solved together by multt =3 R “136m
1 ‘ ' ‘
p ymg (2) by 6‘ Substituting in {2) gives I
1211 + 4211: 60 (3) 11,: 3315 my Now. subtracting (1) from (3) yields
40!; = 45
:21 I; = 1.125 m
Substituting in (i) giVes
31:107 x1.125 mA
= I, = 1.0625 mm I3=II+I2 11250 v" R‘ XA x R"
311:1. v: “a V=I333 (a) 3.? _ xA x 10R...
= 1.1375 x 2 = 13,750 V v, 10R, + R, "“ 103., + R”
T0 Summarize: = 39 x 10 x 19 = 3.26 wv
1.=1.06mA Iz=l13mﬁ 11 11
I =1.19mA V=233V . =
(b) A node :quation at the common node can he wnttﬁﬁ or! 20 mg 8.26 18.3 dB R
in [611115 0f V35 b E =' R: a
15_v+10_v=1 U5 R5+RIXAv”xRo+Ra
R1 RI R3 = 0.5 11101:: 0.5 = 2.5 wv
Thus ‘
15_v+1o_v=g or.2010g2.5=3tlB
19 5 2 (C) E, __ R,/10 x A x 110/10
=‘ 03“: 35 v, (R,/10)+ R, "a (110/111) + R...
11 =—><le——=0.033va 1.44 ZOIth Av”: 40 (113 a: Av“: 100 VA! +
1MB 1:,
A =93
I! 100
" llick)":me+10
=90.9wv or. 10103 90.9 = 39.1 dB 2
AP: m2 = Aﬁx 10“ = 3.3 x107 WIW
iii/1 MQ
or 1010g(8.3 x 10"): 79.1 dB.
For a peak output sinewave current of 100 Q, the peak output voltage will be 100 mA at 100 f1 = 10 V.
Correspondineg 1:. will be a sine wave with a peak
value of 10 V/Au = 10/ 90.9 or an rms value of
10/“(903 x Ji) = 0.03 v. 2
Corresponding output power = { 10/215) /100 £1
= 0.5 W 1.56 Current ampliﬁer. To limit the change in i9 resulting from R, varying over
the range 1 to 10 1:9, to 10% we select RE sufficiently.r low so that. R, ERIN“HO .rhUS. R; = 9
To limit the change in t"ﬂ as RL changes from 1 to
10 REP... 10 10% we select Ra sufficiently' large; Ru 2 IORLM
111115. R... = 100 1(1)
NOW for f, = 10 _. R ~ R
_ 5 3mm AF :1
tam“ Rsmin +Ri JR0+RLm
Thus.
—3 —5 1000 100 1.61 Apply kirehhoff’s current law in the mesh on the left 111 471+ n.1,. + RE(Bi,, + i.) = 0
5:51qu +113 + DRE] Uh : New vi. = —3 >< ihXRL
Ur? : _ BRL
vb rat + + 1.63 Using the voltage divider rule =10 A
gm 0 gmvi R1L = Skﬂ
. 7 R .
L iio 1 .65
+ +
R R
V: C V0 Vi C V0
{3) (b) 1.67 Using tHe voltagedivider rule. l HSC J
f V = V —
m (a) ” ’(1 are + R
E 2 1
VI, 1 + SCR
where k=1 Cu?! + R1)
1 which is from Table 1.2ie of the hi ghpase type with
mg 3 from table 1.2 it is low pass.
R1+ R2 “ one1 + RE)
ﬁx ([3) v“ = v! R As afurther veriﬁcation thetthie is ahigh—pase
R + i network and HS) is a hi ghpeee transfer function,
SC weessumeasLrzbﬂ,T(s)=}D;andaeS——}OG+
ﬂ} = ﬂ T(5) = REEL»?l + R2).Also, from the circuit
V! i + SCH
observe as s—em, (.1 MC) —}U and
E3 = S VOIVI. = ream?L +R1).Now,for
V, S + 1 "
E R1 = 10 kﬂﬁg = 40 kﬂ,and C = 0.] nF.
where k=l f 2 E i 1
r; —E 3
ten = éfrornteble 1.2itie high pass. 27] 211' X 01 X 10 (10 + 40} X [0
= 31.8 Hz
K 40 l
= — = m— = 05’?va IV DESIGN EXERCISE 1.24 i xn
K; = Rt ___l_._ R: 5 Overall power gain a v 5 x If m
I R3+Ri$+; _ ﬁrm) i }'
SC C (R, + RE) 2
which is a HP STC function. / 1000
1 2
= —— S 100 Hz = ———————
f3” 2eC(R,+ R.) [10x10'3)x(ﬂ.1x10'6)
C a ————L3——— = 0.16 as Ji Ji
2::(l+9)10 x100 =9xlotwm
PROBLEM 1'49 .
Required overall voltage gain. = 2 V/ 10 mV = 200 (T1115 takes into acct. the power dissipated in the internal
V/V. Each stage is capable of providing a maxi resistance of the spurce.) mum voltage gain of 10 (the open—circuit gain
value). For :1 stages in cascade the maximum (unattainable) voltage gain in 10". We thus see that (c) If (Aware) has itspeak value limited to 5 v. the larg
est value of R0 is found from  . . h l R '
we need at. least stages For 3 stages t e overal . 5 X L = 3 = R0 = gRL = 667 Q
voltage gam obtamed IS RL + .Rﬂ
U_o = 10 X 10 X '1 Q x 10 X _ 10 (If Ra were greater than this value, the output voltage
I’3 10 + 10 1 + 10 I + 10 across RL would be less than 3 V.)
X 10 X ﬂ {(1) For Rf = 90 kn. and R, = 667 E1. the required value
of A c h f
2 2066 WV 1,” an e ound frog) 1
Thus. three stages sufﬁce and provide a gain 300 VJV = 90 10 x A". 3: my
slightly larger than required. The output voltage + + '
actually obtained is 10 mV X 206.6 = 2.07 V. = A1,“ = 555.7 VN (e) ai=100 m (13:105 a}
aﬂ=1oom1x1ozm 3on=_.1_99_xgu xﬂ
100+10 ' 10004100 =1: Aug = 363 VN 1.68 Using the voltage divider rule, (:1) Required voltage gain 5 i’ b : ——RL——
5
= = V
0.01 v 300 N RL 3
{b} {The smallest R. allowed is obtained from _ RL + Res J. _1.___
C(RL + RS)
0,1 M = ‘0 "1V __., Riv. Rf = 100 m which is of the highpass STC type (see Table 1.2)
R, + Rl' Thus R, = 90 m. ' RL 1
For Ri=90 ko, 5,.=0.1 ,uApealt. and K 2 RL + R3 03’” = C(RL + as)
/R  ‘
Overall current gain = v" * L For f E 10 H2;
III a If)
"I
3 mA 4 ————— i: 10
= = 3 x
0.1 M 10 MA 217C(RL + RS)
=> C E 1 2e ><10(20 + 5)><103
Thus. the smallest value of C that will do the job
is C = 0.64 uF. ...
View
Full
Document
This note was uploaded on 10/23/2011 for the course EE 101 taught by Professor Hoa during the Spring '11 term at American InterContinental University.
 Spring '11
 Hoa

Click to edit the document details