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Unformatted text preview: 282 Chapter 4 BipolarJunction Transistors operation in the active region. Repeat for the saturation region.
Repeat for the cutoff region. 4.27. Repeat Problem 4.26, for a pnp transistor. . 4.28. Determine the region of operation for a room—
temperature silicon npn transistor that has 18 = 100 if
(a) V5}; = 10V and 13 = ZOMA; [C = [B = 0;
(c) VCE = 3V and V3,; = 0.4V; (11) Ic = 1mA and
13 = ﬂA. 4.29. Determine the region of operation for a room temperature silicon pnp transistor that has ﬂ = 100 if
(a) VCE = —5V and VBE = —0.3V; [C = 10mA
and 13 = 1mA; (c) 13 = 0.05 mA and VCE = —5 V. Section 4.5: LargeSignal DC Analysis
of BJT Circuits 4.30. Brieﬂy discuss the procedure for performing a dc analysis
of a BJT circuit using the largesignal circuit models. 4.31. Draw the ﬁxedbase bias circuit. What is the principal
reason that this circuit is unsuitable for the mass production of
ampliﬁer circuits? 4.32. Draw the fourresistor bias circuit for the BJT. Give the
rule—ofthumb design guidelines for this circuit. 4.33. Use the large—signal models for the transistors illustrated
in Figure 4.19 to ﬁnd IC and VCE for the circuits of Figure P4.33.
Assume that [3 = 100. Repeat for 13 = 300 and compare the
results for both values. +20V +20V +15V +15V
1 M9 4.7 m 470 m 6.8 m
1 MD J
(a) =— 15 V (b) '
+15 v +15 v +15 v
1 M9 10““
470 kQ 220 k9.
1 m
—15 V = : =
(c) (d) Figure P4.33 4.34. Find I and V in the circuits shown in Figure P434. F0
all transistors, assume that 13 = 100 and IV”! : 0.7 in both
the active and saturation regions. Repeat for ,3 = 300. 1 +10V +10V
2.7 kg
ll
v
390 m l1 ~
22 k9 10 m i
(a) (M
+15 V +15V
+5 v .L, —15 v 15 M9 92
1m 1m QI l,
+
:1 1m
(6) (d) Figure P4.34 4.35. Consider the circuit displayed in Figure P435. A Q—point
value for IC between a minimum of 4 mA and a maximum of 5 mA is required. Assume constant resistor values. and suppose that [3 ranges from 100 to 300. It is desired that R 3 have the largest possible value while meeting the other constraints. Find _7
the values of RB and R E. The resistors in this problem are not ’ required to be nominal values. +15 V 1 k9
Rs VBEQ = 0.7 v wt:— Figure P4.35 4.36. Consider the four—resistor bias network of Figure 4.283 VCC = R1 = R2 = RC = awvmm‘\«uanrwr>l~3*ﬂ'/Mrkumr«:gv,~wfW’MWWV.W,_‘NA < ’ t r H ‘ ’ V ’ ’ w... ...,«..m. u . M; 7m”. . m. P434~ F0 2R5 = 4.7 kSZ. Suppose that ,8 ranges from 50 to 200,
—— 0.7 in both 0.7 V, and the resistors have a tolerance of i5%. Find
= 300. aximum and minimum values of Ic.
HOV ,‘ Consider the circuit shown in Figure P437. Find R, and
a bias point of V55 2 5V and IC : 2mA is required.
'7 m ‘t are the closest 5%tolerance nominal values for R1 and Figure P4.37
Find 1c and VCE in the circuit of Figure P438. +15 V 35. A Qpo I
maximum of
s, and suppmé , ‘
RB have the _ nstraints. Find]
oblem are not j Figure P438 . Fourresistor bias circuit design. Suppose that VCC = "V, RC = 1 k9, and a Qpoint of ICQ E 5 mA is desired. The
Wtor has ﬂ ranging from 50 to 150. Design a fourresistor
!_ circuit. Use standard 5%tolerance resistor values. Many third of the supply voltage is dropped across RC, onethird across
transistor (VCE), and onethird across R E. 'on 4.6: SmallSignal Equivalent 7V A certain npn silicon transistor at room temperature
[3 = 100. Find the corresponding values of gm and r,r if Chapter 4 Problems 283 ICQ = lmA, 0.1mA, and luA. Assume that the device is
operating in the active region. Section 4.7: The CommonEmitter
Ampliﬁer 4.43. Why are coupling capacitors often used to connect the
signal source and the load to discrete ampliﬁer circuits? Should
coupling capacitors be used if it is necessary to amplify dc
signals? Explain. 4.44. Draw the circuit diagram of a commonemitter ampliﬁer circuit that uses the four—resistor biasing network. Include a
signal source and a load resistance. 4.45. Consider the commonemitter ampliﬁer of Figure P445.
Draw the dc circuit and ﬁnd ICQ. Find the value of r”. Then
calculate values for A”, Aw, Z", A;, G, and Zn. Assume that
the circuit is operating in the midband region for which the
coupling and bypass capacitors are short circuits. +15 V +15 V s=oo
o.7v VBEQ = Figure [MAS 4.46. Repeat Problem 4.45 if all resistance values, including
R. and RL, are increased in value by a factor of 100. If you have also worked Problem 4.45, prepare a table comparing
the results for the lowimpedance ampliﬁer with those for the highimpedance ampliﬁer. (Comment: When we consider the
highfrequency response of these circuits, we will ﬁnd that the
gain of the highimpedance circuit falls off at lower frequencies
than the gain of the lowimpedance circuit does. Thus, if we
want constant gain to extend to very high frequencies, we should
use the lowimpedance circuit.) Section 4.8: The Emitter Follower 4.47. Draw the circuit diagram of a discrete emitter follower. 4.48. For a smallsignal midband analysis of an ampliﬁer,
with what do we replace the coupling capacitors? Dc voltage
sources? Dc current sources? Very large inductors? 284 Chapter 4 Bipolarjunction Transistors
follower ampliﬁer of Figure P451.
d ICQ. Find the value of 1,,_ Then Aim, Zin, Aia Giand Z“. 4.51. Consider the emitter 4.49. Draw the smallsignal equi
illustrated in Figure P4.49. Draw the dc circuit and tin
' calculate midband values for A” , t
+15 V +15 V valent circuits for the circuits emitter ampliﬁer with unbypassed emitter
Figure P4.51 4.52. Repeat Problem 4.51 if all resistance values, including RJ , A
and R L , are increased in value by a factor of 100. Prepare a C2 i table comparing the results for the low—impedance ampliﬁer, * with those for the high—impedance ampliﬁer.
+ 4.53. Draw the smallsignal equivalent circuit for the ampliﬁer “ shown in Figure P453. Derive expressions for the voltage gain ,5;
and input impedance in terms of the resistor values, r,,, and ﬁ. 3"
"' Assume that the capacitors are short circuits for the Signals (a) Common
resistor emitter follower using a dc current (b) Variation of the
source for biasing +Vcc Figure P453
A”, and Zn for the circuit 5 I , v, “in
— 4.54. Find the values of ICQ , rn ,
_ j: V of Problem 4.53 if Vcc = 15V, [8 = 100, VBEQ = 0.7, 
‘ ' ‘ CC RB = 270m, RC = 1m, R5 = 100:2, and RL = 1kg.
(c) Variation of the common—base ampliﬁer [assume that Repeat for R E = 0 and prepare a table comparing the resulis
the mi?'freque"cy wake (RFC) is mops“ circuit for 4.55. Find an expression for the output impedance of the
the 3‘: s‘gnals] ampliﬁer displayed in Figure P453.
Figure P4.49 Ampliﬁer circuits 4.56. Draw the smallsignal equivalent circuit of the Circui‘ I
shown in Figure P456, and derive expressmns for the “19m”
in. Assume that the capacitors are _ impedance and voltage ga edure for
short circuits for the signals describe the small—signal analysis proc 4.50. Brieﬂy
of an ampliﬁer. ﬁnding the output resistance Chapter 4 Problems 2 85 +VCC +VCC 4.59. Consider the commonemitter ampliﬁer illustrated in
Figure P459. The ac source in series with the dc supply
represents power~supply hum. (a) Assume that a large bypass
capacitance is connected between points A and E. Draw the
smallsignal equivalent circuit, including the hum source. Solve
for Alum, = va/vhum, assuming that v, = 0. (b) Now consider connecting the emitter bypass capacitor
from point E to ground, and again ﬁnd Ahm. Notice that if
vhum = 0, the two equivalent circuits are identical, so they
perform equally as far as the source signal v; is concerned.
Which option is best for the connection of the bypass capacitor? Why? r of Figure P451
Ialue of r,,, Thel;
\i , G, and 20. [5:100
VHE 20.7v Figure P4.56 m. Consider the circuit of Figure P456 with Vcc = 15 V,
x1: 10kg, R2 = 101:9, R3 : 100kS'Z, R5 = 10kg,
RL = 4.7 k9. Assume a transistor having [9 = 200 and = 0.7 V. Evaluate the expressions found in Problem 4.56 ’ input impedance and voltage gain. ‘ Consider the commonemitter ampliﬁer circuits displayed
gure P458. The ac sources shown in series with the dc y sources represent powersupply hum. Draw the small
equivalent circuits Be sure to include the burn source in
model. Notice that if mm, = 0, the two equivalent circuits
identical. Find an expression for the voltage gain va/vin
mm = 0. Then set via = 0 and solve for Ahum = va/vh.ml ‘ 'ach circuit. Evaluate the gain values for Vcc = 15V, I m = 0.7V, ﬂ = 100, R3 = IMO, and RC = 4.71:9.
“Which of these circuits is preferable? Why? alues, including R,
)f 100. Prepare a
edance ampliﬁer .it for the ampliﬁef
or the voltage gain I q
' values, rJ1 , andﬂ
'. for the signals Figure P4.59 4.60. Find the value of VCEQ for the circuit of Figure P460.
Draw the smallsignal equivalent circuit and ﬁnd an expression for the smallsignal output impedance Zn in terms of [3 , r,r , R1 ,
and R2. Evaluate 20 for the values shown in the ﬁgure. +15 V {in for the circuit VBEQ = 07’ VBEQ=O.7V
and RL ﬁ=100
paring the results pedance of the ‘ uit of the circuit
ans for the inpul
the capacitors ' Figure P458 Figure P4.60 ansistors 286 Chapter 4 Bipolarjunction Tr 4.61. Consider the voltage—reference circuits illustrated in Figure P461. The dynamic smallsignal resistance of each 100 S2. Find the dc output voltage of each Zener diode is rd =
nt circuit, and derive an circuit. Draw the smallsignal equivale
the output impedance of each circuit. Evaluate
ameters expression for
s for the resistances and transistor par the expression shown.
HSV +UV
R VBEQ = 0.7 v
R 10 k9 ﬂ = 200
Va
+ V,,
+ <e—4 55v A_ ‘9.
5.1 V_ 1 — RE 1 kg 1 Z”
Figure P4.61
Section 4.9: The BJT as a Digital Logic
Switch f an RTL inverter. In what 4.62. Draw the circuit diagram o
te if the input is high? region is the transistor intended to opera Low?
it diagram of an RTL NOR gate. 4.63. Draw the circu 4.64. Consider the transfer characteristic of the RTL inverter
shown in Figure 4.43. Sketch the transfer characteristic to
scale for the values shown in Figure P464. Sketch the output voltage v0 (t) to scale versus time if (a) vm(t) = 2.7 sin(20007r t);
(b) vin(t) = 2.7 + sin(20007rt); (c) vin(t) = 2.7 + 5 sin(20007tt). (d) For which of the previous parts does the RTL inverter
behave as a linear ampliﬁer for the ac signal? +12V RC 2.2 m
‘4;
R
,8 = 30 B
+ 22 k9
yin“) VBE = 0.7 V E
i
_ t
E _ Figure P464 4.65. Consider the
that we require the minimum 0 in the cutoff region to be V0”
number of driven gates (i.e., the fa
Assume that the driven gates
those of the RTL inverter. 4.66. If Vin = 6V for the circuit of
minimum value of [9 to ensure that the transis 4.67. If VOL (i.e., the output voltage in the low logic state)*~is
required to be less than 0.5 V for the circuit of Figure P464,
what is the maximum fanout allowed (i.e., the maximu
number of inputs that can be connected) if the driven gates . also RTL circuits? RTL inverter of Figure P464. Assuming _
utput voltage with the transistor = 6V, what is the maximum n—out) that can be connected? ,
have input circuits identical to"? _ ‘1 ...
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