Appendix
,
Additional Solved
Problems
1. A 230 V, 60 Hz voltage is applied to the primary of a 5:1 step down,
center tapped transformer used in a full wave rectifier having a load of 900
W. If the diode resistance and the secondary coil resistance together has a
resistance of 100 W, determine
(i) dc voltage across the load
(ii) dc current flowing through the load
(iii) dc power delivered to the load
(iv) PIV across each diode
(v) Ripple voltage and its frequency
Solution:
The voltage across the two ends of secondary =
Voltage from center tapping to one end,Vrms =
(a) d.c. voltage across the load, Vd.c. =
230
= 46 V
5
46
= 23 V
2
2 V m 2 × 23 × 2
=
= 20. 7 V
p
p
(b) d.c. current flowing through the load, Id.c.
=
Vd.c.
= 20. 7 = 20. 7 mA
( rs + r f + R L ) 1000
(c) d.c. power delivered to the load,
Pd.c. = (Id.c.)2 ¥ RL= (20.7 ¥ 10–3)2 ¥ 900 = 0.386 W
(d) PIV across each diode = 2 Vm = 2 ¥ 23 ¥
2 = 65 V
D.2
Appendix
(e) Ripple voltage,Vr, rms = (Vrms ) 2 − (Vd.c. ) 2
= ( 23) 2 − ( 20. 7 ) 2 = 10. 05 V
Frequency of ripple voltage = 2 ¥ 60 = 120 Hz
2. Find the capacitor C and hfe for the transistor to provide a resonating
frequency of 10 kHz of a transistorized phase shift oscillator. Assume
R1 = 25 kW, R2 = 60 kW, Rc = 40 kW, R = 7.1 kW and hie = 1.8 kW.
Solution:
For a phase shift oscillator,
fr = 10 kHz
R1 = 25 kW
R2 = 60 kW
R c = 40 kW
R = 7.1 kW
hie = 1.8 kW
(i) To find capacitance, C:
Frequency of oscillation is
fr =
1
2 p RC
6 + 4K
1
C=
2 p fr R 6 +
4 Rc
R
1
=
2 p ¥ 10 ¥ 103 ¥ 7.1 ¥ 103
(ii) To find hfe:
We know that hfe
≥ 23 + 29
≥ 23 + 29
Therefore,
hfe
6+
4 ¥ 40 ¥ 103
= 0.41 nF
7.1 ¥ 103
Rc
R
+4
Rc
R
7.1 ¥ 103
40 ¥ 103
+4¥
40 ¥ 103
7.1 ¥ 103
≥ 23 + 5.1475 + 22.53 = 50.67
≥ 50.67
3. A HWR has a load of 3.5 kW. If the diode resistance and secondary coil
resistance together have a resistance of 800 W and the input voltage has a
signal voltage of peak value 240 V. Calculate
Appendix
(i)
(ii)
(iii)
(iv)
D.3
Peak, average and rms value of current flowing.
dc power output
ac power input
efficiency of the rectifier
Solution:
Load resistance in a HWR, RL = 3.5 kW
Diode and secondary coil resistance, rf + rs = 800 W
Peak value of input voltage = 240 V
Vm
240
=
= 55.81 mA
(i) Peak value of current,
Im =
4300
rs + r f + RL
Average value of current,Idc =
RMS value of current, Irms =
Im
p
Im
2
=
=
55.81 ¥ 10 -3
p
55.81 ¥ 10 -3
2
= 17.77 mA
= 27.905 mA
(ii) DC power output is
Pdc = (Idc)2 RL = (17.77 ¥ 10–3)2 ¥ 3500 = 1.105 W
(iii) AC power input is
Pac = (Irms)2 ¥ (rf + RL) = (27.905 ¥ 10–3)2 ¥ 4300 = 3.348 W
= (0.15)2 ¥ 800 W = 18 W
(iv) Efficiency of the rectifier is
h=
Pdc
Pac
=
1.105
¥ 100 = 33%
3.348
4. Calculate the values of IC and IE for a transistor with adc = 0.99 and
ICBO = 5 µA. IB is measured as 20 µA.
Solution:
Given,
adc = 0.99, ICBO = 5 µA and IB = 20 µA
IC =
Therefore,
a dc I B
1 - a dc
+
0.99 ¥ 20 ¥ 10-6 5 ¥ 10 –6
I CBO
=
+
= 2.48 mA
1 - a dc
1 - 0.99
1 – 0.99
IE = IB + IC = 20 ¥ 10–6 + 2.48 ¥ 10–3 = 2.5 mA
5. An npn transistor if b = 50 is used in common emitter circuit with Vcc =
10 V and Rc = 2 kW. The bias is obtained by connecting 100 kW resistor
from collector to base. Find the quiescent point and stability factor.
Solution:
Given,
Vcc = 10 V, Rc = 2 kW,
b = 50 and collector to base resistor RB = 100 kW
To determine the quiescent point:
We know that for the collector to base bias transistor circuit
Vcc = b IB R c + IBRB + VBE
D.4
Appendix
Therefore,
IB =
=
Vcc - VBE
RB + b ◊ RC
10 - 0.7
100 ¥ 103 + 50 ¥ 2 ¥ 10 +3
= 46.5 µA
Ic = b ◊ IB = 50 ¥ 46.5 ¥ 10–6 = 2.325 mA
VCE = Vcc – IcRc = 10 – 2.325 ¥ 10–3 ¥ 2 ¥ 103 = 5.35 V
Therefore, the co-ordinates of the new operating point are
VCEQ = 5.35 V and ICQ = 2.325 mA
To find the stability factors S:
Hence,
S=
=
1+ b
È Rc
˘
1+ b Í
˙
Î Rc + RB ˚
1 + 50
È
˘
2 ¥ 103
1 + 50 Í
˙
3
3
ÎÍ 2 ¥ 10 + 100 ¥ 10 ˙˚
= 25.75
6. Calculate the gain input impedance, output impedance of voltage series
feed back amplifier having A = 300, Ri = 1.5kW, Rc = 50 kW and b = 1/20.
Solution:
Given
A = 300, R = 1.5 kW, Rc = 50 kW and b = 1/20 = 0.05
Voltage gain Af =
Input Resistance,
300
A
= 18.75
=
1 + Ab 1 + 300 ¥ 0.05
Rif = (1 + Ab )Ri
= (1 + 300 ¥ 0.05) ¥ 1.5 ¥ 103 = 24 kW
Output Resistance, Rof =
Rof
1 + Ab
=
50 ¥ 103
1 + 300 ¥ 0.05
= 3.125 kW
7. A phase shift oscillator is to be designed with FET having gm = 5000 µS,
rd = 40 kW, while the resistance in the feedback circuit is 9.7 kW. Select the
proper value of C and RD to have the frequency of oscillation as 5 kHz.
Solution:
Given,
gm = 5000 µS, rd = 40 kW, R = 9.7 kW, fr = 5 kHz
We know that fr =
5 ¥ 103 =
1
2 p RC
6
1
2 p ¥ C ¥ ( 9.7 ¥ 103 ) 6
Appendix
Therefore, C =
D.5
1
3
3
2 p ¥ 5 ¥ 10 ¥ 9.7 ¥ 10 ¥
6
=
1
304.58 ¥ 106 ¥
6
= 1.34 nF
Let us assume that the voltage gain, AV, of the FET amplifier is 40.
We know thatAV = gmr L
Therefore
40 = 5000 ¥ 10–6 ¥ r L
rL = 8 kW
We know that rL =
8 kW =
Therefore,
rd ◊ RD
rd + RD
40 kW ◊ RD
40 kW + RD
R D = 10 kW
8. In an electrostatic deflecting CRT, the length of the deflection plate is 2
cm and spacing between deflecting plates is 0.5 cm. The distance from the
centre of the deflecting plate to the screen is 20 cm and the deflecting
voltage is 25 V. Find the deflection sensitivity, the angle of deflection and
velocity of the beam. Assume final anode potential is 1000 V.
Solution:
Given, in an electro static deflecting CRT,
Length of deflection plate, l = 2 cm
Spacing between deflecting plates d = 0.5 cm
Distance from center of deflecting plate to screen, L = 20 cm
Deflecting potential, Vd = 25 V
Anode potential, Va = 1000 V
(i) Velocity of beam v
=
2 qVa
m
2 ¥ 1.6 ¥ 10-19 ¥ 1000
=
9.1 ¥ 10-31
= 18.75 ¥ 106 ms–1
(ii) Deflection sensitivity =
where D =
ILVd
2 Va d
=
D
Vd
2 ¥ 10-2 ¥ 20 ¥ 10-2 ¥ 25
2 ¥ 1000 ¥ 0.5 ¥ 10
-2
= 10–2 cm
-2
Therefore, the deflection sensitivity = 10
= 0.0004 cm/V
(iii) To find the angle of deflection, q :
tan q =
10-2
10 -2
D
=
=
( L - l ) 20 - 2
18
Therefore, q = tan–1 Ê 1 ˆ = 0.0318°
Ë 1800 ¯
25
D.6
Appendix
9. Determine the forward resistance of a p-n junction diode, when the forward current is 5 mA at T = 300º K. Assume Silicon diode.
Solution:
Given, for a silicon diode, the forward current, I = 5 mA, T = 300°K
hVT
Forward resistance of a p-n junction diode, rf =
I
h = 2 for silicon
2¥
Therefore, rf =
T
11, 600
5 ¥ 10-3
=
2 ¥ 300
11, 600 ¥ 5 ¥ 10-3
where VT =
T
and
11, 600
= 10.34 W
10. In a Bridge rectifier, the transformer is connected to 200 V, 60 Hz mains
and the turns ratio of the step down transformer is 11:1. Assuming the diode
is ideal, find
(i) Idc
(ii) Voltage across the load
(iii) PIV
Solution:
Given, in a bridge rectifier, input voltage = 200 V, 60 Hz and turns ratio = 11:1
(i) To find the voltage across load, Vdc:
Vdc =
Where
2 Vm
p
V m = Vrms
Vrms (secondary) =
2
Vrms(primary)
Turns Ratio
Therefore
V m = 18.18 ¥
Hence,
Vdc =
2 ¥ 25.7
p
=
200 V
11
= 18.18 V
2 = 25.7
=16.36 V
(ii) To find Idc:
Assuming that , RL = 600 W, then
Idc =
Vdc
RL
=
16.36
= 27.26 mA
600
(iii) To find PIV:
PIV = Vm = 25.7 V
11. The reverse leakage current of the transistor when connected in CB
configuration is 0.2 µA and it is 18 µA when the same transistor is connected in CE configuration. Calculate adc and bdc of the transistor.
Appendix
D.7
Solution:
The leakage current ICBO = 0.2 µA
ICEO = 18 µA
Assume that
IB = 30 mA
IE = IB + Ic
Ic = IE – IB = bIB + (1 + b )ICBO
We know that ICEO =
b=
I CEO
1-a
I CEO
I CBO
= (1 + b )ICBO
-1=
18
- 1 = 89
0.2
Ic = b IB + (1 + b ) ICBO
= 89 (30 ¥ 10–6) + (1 + 89)(0.2 ¥ 10–6)
= 2688 ¥ 10–6 = 2.688 ¥ 10–3 A
adc = 1 = 1bdc =
=
I CBO
I CEO
0.2 ¥ 10-6
18 ¥ 10-6
= 0.988
IC ¥ I CBO
I B - I CEO
2.688 ¥ 10 -3 - 0.2 ¥ 10-6
3 ¥ 10-3 - 18 ¥ 10-6
= 89.539
12. If adc = 0.99 and ICBO = 50 µA, find emitter current.
Solution:
Given
adc = 0.99
Assume that IB = 1mA
Ic
=
=
a dc I B
1 - a dc
and
+
ICBO = 1µA,
0.99 (1 ¥ 10-3 ) 5 ¥ 10-6
I CBO
=
+
1 - a dc
1 - 0.99
1 - 0.99
0.99 ¥ 10 -3
50 ¥ 10-6
= 99 mA + 5 mA = 104 mA
0.01
0.01
IE = Ic + IB = 104 mA + 1 mA = 105 mA
+
13. For the given circuit, find Av, Rif and also the resistance seen by Vs,
Assume Rc = 4 kW, R¢ = 40 kW, RS = 10 kW and the transistor parameters are
hie =1.1 kW, hfe = 50, hre = 0 and 1/hoe = 40 kW. (Assume Av >> 1)
D.8
Appendix
R¢
IO
Ir
Rs
RC
VO
Ii
+
Vi
Vs
–
Rif
Rof
Fig. 13
Solution:
Rc = 4 kW
R1 = 40 kW
RS = 10 kW
hie = 1.1 kW, hfe = 50, hre = 0, 1/hoe = 40 kW.
- h fe
-50 ¥ 4 ¥ 103
(i) Av =
= –181.81
=
hie
1.1 ¥ 103
(ii) Rif = hie + hre AI RL
where AI = –hfe = –50
Therefore, Rif = 1.1 ¥ 103 + 0(–50) = 1.1 kW
Ri ¢ =Ri ||R¢
= 1.1 ¥ 103 || 40 ¥ 103 =
1.1 ¥ 40
1.1 + 40
(iii) Input resistance as seen by the source is
kW = 1.07 kW
Rs + R¢ = 10 kW + 1.07 kW = 11.07 kW
14. The gain of an amplifier is decreased to 1000 with –ve feedback from its
gain of 5000. Calculate the feedback factor and the amount of negative
feedback in dB.
Solution:
The gain of the amplifier is decreased from 5000 to 1000 with negative feedback.
Voltage gain of the amplifier with negative feedback is
A n¢ =
1000 =
An
1 + b ◊ An
5000
1 + b ◊ 5000
Appendix
D.9
5000
=5
1000
5000 b = 4
(1 + 5000 b ) =
b=
4
= 0.8 ¥ 10–3
5000
= 20 log 0.8 ¥ 10–3 = –62 dB
15. Show that the gain of Wien bridge oscillator using BJT amplifier must
be at least 3 for the oscillations to occur.
Solution:
Let us assume that R1= R2 = R and C1 =
C2 = C.
Then the feedback circuit is as shown
in Fig. Q 15.
Therefore,
Vf (s)
R 1
sC
Vf (s) = Vo(s)
1
+R 1
R+
sC
sC
Vf (s) = Vo(s)
= Vo(s)
R 1/sC
V(f)
R
V O (s )
1/sC
Fig. Q 15
R
1 + sRC
R+
1
R
+
sC 1 + sRC
sRC
s R C + 3 sRC + 1
2
2
2
Hence, the feedback factor is
b=
Vf ( s )
Vo ( s )
=
sRC
s 2 R 2 C 2 + 3 sRC + 1
We know that Ab = 1
Therefore, the gain of the amplifier,
s 2 R 2 C 2 + 3 sRC + 1
A= 1 =
b
sRC
Substituting s = jwr , where the frequency of oscillation fr =
1
, i.e. wr =
2 p RC
1 , in the above equation and simplifying, we get A = 3. Hence the gain of
RC
the Wien bridge oscillator using BJT amplifier is at least equal to 3 for oscillations to occur.
D.10
Appendix
16. Determine the velocity and kinetic energy of an electron accelerated
through potential of 3 kV.
Solution:
The velocity of the electron
v=
=
2 qV
m
2 ¥ 1.6 ¥ 10-19 ¥ 3 ¥ 103
9.1 ¥ 10-31
= 3.247 ¥ 107 m/s
The kinetic energy = q ¥ V
= 1.6 ¥ 10–19 ¥ 3000 = 4.8 ¥ 10–17 joules = 3000 eV
17. For the circuit shown below hfe = 100, hie = 1 kW and other two parameters are negligible if Re = 1 kW. Find the value of Av =
AVS =
Vo
Vs
,
Vo
Vt
Rif = Rof
+VCC
10 kW
100 kW
1 kW
Vi
+
–
VO
VS
Re
Rif
Rof
Fig. Q 18
Solution:
We know that AV =
Rc
Re
=
10 ¥ 103
1 ¥ 103
= –10
Rif = RB || hfe Re = 50 kW
AVS =
Vo
Vs
= AV ◊
Rif
RS + Rif
= ( -10 ) ◊
50 ¥ 103
51 ¥ 103
= –9.8
Appendix
D.11
18. A quartz crystal has the following constants. L = 50 mH, Cs =0.02 pF,
R = 500 W and Cp = 12 pF. Find the values of series and parallel resonant
frequencies. If the external capacitance across the crystal changes from 5 pF
to 6 pF, find the change in frequency of oscillations.
Solution:
Given, for a crystal oscillator,
L = 50 mH
Cs = 0.02 PF
R = 500 W
Cp = 12 pF
External capacitance changes from 5 pF to 6 pF
(i) Series resonant frequency is,
fs =
1
2p
LCs
=
1
2p
50 ¥ 10-3 ¥ 0.02 ¥ 10-12
= 5033 kHz
(ii) Parallel resonant frequency is
fp =
1
2p
=
(iii) When
When
Cs + C p
LC s C p
0.02 ¥ 10-12 + 12 ¥ 10-12
1
2p
( 50 ¥ 10-3 ¥ 0.02 ¥ 10-12 ¥ 12 ¥ 10-12 )
Cp = 5 pF,
Cp = 6 pF,
= 5037 kHz
fp = 5043 kHz
fp = 5041 kHz
19. A bridge rectifier uses four identical diodes having forward resistance of
5 W and the secondary voltage is 30 V (rms). Determine the dc output
voltage for Idc = 200 mA and value of the output ripple voltage.
Solution:
Given
Transformer secondary resistance = 5 W
Secondary voltage Vrms = 30 V, Idc = 200 mA
Since only two diodes of the bridge rectifier circuit will conduct during
positive or negative half cycle of the input signal, the diode forward resistance
rf = 2 ¥ 5 W = 10 W
We know that,
Vdc =
Therefore,
Vdc =
2 Vm
π
2×
– Idc (rf + rs) where Vm =
2 Vrms =
2 × 30
– 200 ¥ 10–3 (10 + 5) = 24 V
π
2 ¥ 30 V
D.12
Appendix
Ripple factor =
Therefore,
0.48 =
rms value of ripple at the output
Average value of output voltage
rms value of ripple at the output
24
Hence, rms value of ripple at the output = 0.48 ¥ 24 = 11.52 V
20. If h = 0.8, VBB = 15 V, and VD = 0.7 V, find the value of VP.
Solution:
We know that
Therefore,
VP = hVBB
VP = 0.8 ¥ 15 = 12 V
21. In a full wave rectifier, the required dc voltage is 9 V and the diode drop
is 0.8 V. Calculate ac rms input voltage required in case of bridge rectifier
circuit and centre tapped full wave rectifier circuit.
Solution:
The dc voltage across the load of the full wave rectifier circuit,
Vdc =
where Vrms
Vrms =
− 0.8 =
2 2 × Vrms
= − 0.8
π
π
is the rms input voltage from centre tapping to one end. That is,
9.8 =
Therefore,
2Vm
2 2 Vrms
π
9.8 π
= 10.885 V.
2 2
Hence, the voltage across the two ends of the secondary = 2 ¥ 10.885 =
21.77 V
In the bridge rectifier, Vdc = 9 =
2 2 Vrms
π
– 2 ¥ 0.8
Therefore, the voltage across two ends of secondary, Vrms =
10.6 π
2 2
= 11.77 V.
22. For an N-Channel silicon FET with a = 3 ¥ 10–4 cm and ND = 1015
electrons/cm3, find the pinch-off voltage.
Solution:
We know that the pinch-off voltage is
|VP | =
where
qN D
2ε
a2
e = dielectric constant of channel material (Si) = e r e0 = 12e0,
q = magnitude of electronic change = 1.602 ¥ 10–19 C
Appendix
D.13
a is in metres and ND is in electrons/m3
Therefore, |VP | =
1.602 × 10−19 × 1021
2 × 12 × 8.854 × 10−12
¥ (3 ¥ 10–6)2 = 6.8 V
23. Determine the quiescent current and collector to emitter voltage for a
germanium transistor with b = 50 in self biasing arrangement. Draw the
circuit with a given component value with VCC = 20 V, RC = 2 kW, RE = 100
W, R1 = 100 kW and R2 = 5 kW. Also find the stability factor.
Solution:
For a germanium transistor, VBE = 0.3 V and b = 50
To find the coordinates of the operating point:
Thevenin’s voltage, VT =
=
R2
R1 + R2
VCC
5 × 103
105 × 103
R1 R2
Thevenin’s resistance, RB =
R1 + R2
¥ 20 = 0.95 V
=
100 × 103 × 5 × 103
= 4.76 kW
105 × 103
The loop equation around the base circuit is
VT = IB RB + VBE + (IB + IC)RE
=
0.95 =
IC
 IC

RB + VBE + 
+ I C  RE
β
β


IC
50
¥ 4.76 ¥ 103 + 0.3 + IC ¥
0.65 = 197.2IC
51
¥ 100
50
0.65
= 3.296 mA
197.2
Since IB is very small, IC = ª IE = 3.296 mA
Therefore,
VCE = VCC – IC RC – IE RE
= VCC – IC (RC + RE)
= 20 – 3.296 ¥ 10–3 ¥ 2.01 ¥ 103 = 13.375 V
Therefore, the coordinates of the operating point are IC = 3.296 mA and
VCE = 13.375 V.
To find the stability factor S:
Therefore,
IC =
1+
S = (1 + b )
RB
RE
1+ β +
RB
RE
= (1 + 50)
1+
4.76 × 103
1 + 50 +
100
= 25.18
4.76 × 103
100
D.14
Appendix
24. An electron with a velocity of 3 ¥ 105 ms–1
enters an electric field of 910 V/m making an
angle of 60° with the positive direction. The direction of the electric field is in the positive y
direction. Calculate the time required to reach its
maximum height.
y
E
v0
q
x
z
Solution:
Given
v0 = 3 ¥ 105 m/sec
E = 910 V/m
q = 60°
The electron starts moving in the +y direction, but, since acceleration is along
the –y direction, its velocity is reduced to zero at time t = t ¢.
v0y = v0 cos q = 3 ¥ 105 ¥ cos 60° = 0.15 ¥ 106 m/sec
ay =
t¢ =
qE 1.602 × 10−19
¥ 910 = 1.5984 ¥ 1014 m/sec2
=
m 9.109 × 10−31
v0 y
=
ay
0.15 × 106
1.5984 × 1014
= 9.384 ¥ 10–10 sec = 0.938 nsec
25. Determine Av, AI, Ri, Ro for a CE amplifier using npn transistor with
hie = 1200 W, hre = 0, hfe = 36 and hve = 2 ¥ 10–6 mhos. RL = 2.5 kW, RS =
500 W (neglect the effect of biasing circuit).
Solution:
To find current gain (AI):
− h fe
−36
AI =
= –35.82
=
1 + hoe RL 1 + 2 × 10−6 × 2.5 × 103
To find input resistance (Ri):
Ri = hie –
h fe + hre
hoe +
1
RL
= 1200
To find voltage gain (AV):
AV = AI
RL
Ri
= –35.82 ¥
To find output resistance (Ro)
Yo = hoe –
h fe hre
hie + RS
2.5 × 103
1200
= –74.625
Appendix
= 2 ¥ 10–6 –
Therefore,
Ro =
36 × 0
1200 + 500
D.15
= 2 ¥ 10–6 mhos
1
1
=
= 500 kW
Yo 2 × 10−6
26. A FET phase shift oscillator has gm = 5 m mhos and rd = 50 kW. The
feedback resistance is 100 kW and the capacitor value is 64.97 pF. Calculate
the frequency of the oscillator and the value of RD.
Solution:
Given gm = 5 m mhos, rd = 50 kW, R = 100 kW, C = 64.97 pF.
We know that the frequency of the oscillation is fo =
Therefore,
fo =
1
2 π RC 6
1
2 π × 100 × 10 × 64.97 × 10−12
3
6
= 10 kHz
Let us assume that the voltage gain, AV, of the FET amplifier is 40.
We know that AV = gmr L.
Therefore
40 = 5 ¥ 10–3 ¥ rL
rL = 8 kW
We know that
rL =
8 ¥ 103 =
Therefore,
rd ⋅ RD
rd + RD
40 × 103 RD
40 × 103 + RD
R D = 10 kW.
27. The voltage across a silicon diode at room temperature (300° K) is 0.7
Volts when 2 mA current flows through it. If the voltage increases to 0.75
V, calculate the diode current (assume VT = 26 mV)
Solution:
Given data:
Room temperature = 300°K
Voltage across a silicon diode, VD1 = 0.7 V
Current through the diode ID1 = 2 mA
When the voltage increases to 0.75 V, (VD2), then
ID 2
ID 1
Therefore,
=
I o ( eVD 2 /VT η − 1)
VD 1 / VT η
Io ( e
− 1)
=
e 0.75/ 26 ×10
e 0.7 / 26 ×10
−3 × 2
−3
ID2 = 2.615 ¥ ID1
= 2.615 ¥ 2 ¥ 10–3 = 5.23 mA
×2
−1
−1
= 2.615
D.16
Appendix
28. Find C and hfe of a transistor to provide f0 of 50 kHz of a RC transistorized phase shift oscillator. Given R1 = 22 kW, R2 = 68 kW, R = 6.8 kW and
hie = 2 kW.
Solution:
We know that for a RC phase shift oscillator,
f0 =
where
K=
Therefore,
C=
=
1
=
2π 0 R
RC
R
=
1
6 + 4K
20 × 103
6.8 × 103
1
2 ππ 0 R
= 2.94
1
6 + 4K
1
⋅
2 π × 50 × 103 × 6.8 × 103
1
6 + 4 × 2.94
= 111.08 pF
To find hfe
hfe > 4K + 23 +
29
K
> 4 ¥ 2.94 + 23 +
> 44.62
29
2.94
29. Identify topology, with justification for the circuit shown in the figure
below. Transistors used are identical and have parameters hie = 2 k W , h fe = 50
and hre = hoe = 0. Determine Avf .
Solution:
In the given circuit, the feedback voltage (VF) across 3.3K is fed in series with
Vs. Hence the topology is identified as voltage series feedback.
Appendix
D.17
Given data:
hie = 2 kW; hfe = 50; hre = hoe = 0.
RC1 = 51 kW, R1 = 3.3 kW, R2 = 3.3 kW
The overall gain without feedback, Av = Av1 ◊ Av2
Av1 =
- h fe RL¢
h fe + (1 + h fe ) RE
R¢L = RC1||hie = 51 K||2 K = 1.92 kW
Therefore
Therefore
RE = R1||R2 = 1.65 kW
-50 ¥ 1.92 K
Av1 =
= 1.114
2 K + (1 + 50) 1.65 K
Av2 = –hfe
Here
Then,
Then,
R¢L2 =
R¢L2 =
R¢L2 =
Therefore,
Av2 =
Hence,
Av =
We know that,
Avf =
Here,
Therefore,
b=
Avf =
RL¢ 2
hie
R1||R, where R = R1 + R2 = 6.6 kW
3.3 K||6.6 K = 6.6 kW
3.3 K
-50 ¥ 6.6 K
= –165
2K
Av1 ◊ Av2 = (–1.114) (–165) = 183.81
Av
1 + b Av
R1
R1 + R2
Av
1 + b Av
=
3.3 K
= 0.5
3.3 K + 3.3 K
=
183.81
= 1.978
1 + ( 0.5 ¥ 183.81)
30. We have an amplifier of 60 dB gain. It has an output impendence
Z0 = 10 kW, it is required to modify its output impendence to 500 W by
applying negative feedback. Calculate the value of the feedback factor. Also
find the percentage change in the over all gain, for 10% change in the gain
of the internal amplifier.
Solution:
Since the output impedance of the amplifier with feedback is 500 W which is
less than that of without feedback, it is a voltage series feedback amplifier.
Ro
We know that, Rof =
1 + Ab
where Rof = output resistance of the amplifier with feedback
Ro = output resistance of the amplifier without feedback
D.18
Appendix
A = gain of the amplifier
b = feedback factor
Hence,
(1 = Ab) =
Ro
=
Rof
10 ¥ 103
500
= 20
Ab = 19
Given A = 60 dB; i.e. 20 log A = 60; log A = 3; therefore, A = 1000 b =
= 0.019
19
1000
dA
= 10
A
Hence the percentage change in gain of the amplifier with feedback is
Given the change in the gain of the internal amplifier = 10%; i.e.
dA f
Af
d
1
1
= A
= 10 ¥
= 0.5%
20
A (1 + Ab)
31. A 15-0-15 Volts (rms) ideal transformer is used with a full wave rectifier
circuit with diodes having forward drop of 1 volt. The load is a resistance of
100 ohm and a capacitor of 10,000 mF is used a s a filter across the load
resistance calculate the dc load current and voltage.
Solution:
Given transformer secondary voltage = 15-0-15 V (rms); Diode Amp = 1 V; RL
= 100 W; C = 10,000 mF
We know that, Vdc = Vm
Therefore, Vdc = Vm –
Vr , pp
2
Vdc
RL 4 fC
= Vm
,
I dc
4 fC
Vdc ˘
È
Í since I dc =
˙
RL ˚
Î
Simplifying, we get
È 4 fRL C ˘
Vdc = Í
˙ vm
Î 4 fRL C + 1 ˚
We know that Vm = Vrms ¥
2 = 15 ¥
2
È 4 ¥ 50 ¥ 100 ¥ 10000 ¥ 10 -6 ˘
Therefore, Vdc = Í
˙ ¥ 15 ¥
-6
ÎÍ 4 ¥ 50 ¥ 100 ¥ 10000 ¥ 10 + 1 ˚˙
Considering the given voltage drop of 1 volt due to diodes,
Vdc = 21.105 – 1 = 20.105 V
Idc =
Vdc
RL
=
20.105
= 0.20105 A
100
2 = 21.105 V