ELECTRONIC CIRCUITS
Contents
1-Power Amplifiers.
2-Amplifier Frequency Response.
3-Operational Amplifiers and their
Applications.
4-Oscillators and PLL.
5-Programmable Analog Arrays (PAA)
1
References
1- T. L. Floyd, "Electronic Devices", Prentice Hall,
1996,2000,2005,…
2- R. L. Boylested and L. Nahlesky, "Electronic Devices and
Circuit Theory", Prentice Hall, 2002.
3- D. L. Shilling and C. Belove, "Electronic Circuits, Discrete and
Integrated", Mc Graw Hill, 1968.
4- B. Grob, "Electronic Circuits and Applications", Mc Graw Hill,
1982.
5- J. Millman & C. Halkais, "Integrated Electronics", Mc Graw
Hill, 1971.
2
Chapter1
Power Amplifiers
Introduction
Power amplifier are large signal amplifier. This generally
means that a much larger portion of the load line is used
during signal operation than in a small signal amplifier.
In this chapter we are going to study four classes of power
amplifier.
Class A power amplifier.
Class B power amplifier.
Class AB power amplifier.
Class C power amplifier.
Theses amplifier classifications are based on the percentage of
the input cycle for which the amplifier operates in its linear
region.
3
Class A Amplifier
• It is an amplifier is biased so that it operates in
the linear region for the full 360o of the input
cycle.
• The output voltage has the same shape as the
input voltage
• Class A amplifier can be either inverting or
non inverting.
4
FIG. 1 Basic class A amplifier operation.
Output is shown 180o out of phase
(inverted).
5
Centered Q-Point
Recall that the dc and ac load lines cross at the
Q-point. When the Q-point is at the center of
the ac load line, a maximum class A signal can
be obtained.
When the input signal is too large the amplifier
is driven into cut off and saturation.
6
Fig . 2a
Fig . 2a
7
Fig. 2b
Fig 2 Maximum class A output occurs when the
Q-point is centered on the ac load line
8
If the Q-point is not centered on the ac load line,
the output signal is limited. Figure 3 shows an
ac load line with the Q-point moved away
from center toward cutoff. The output
variation is limited by cutoff in this case as in
Fig3a.
If the amplifier is driven any further than
this, it will "clip" at cutoff, as shown in
Figure 3(b)
9
Fig. 3a Amplitude of VCE and IC limit by
10
cutoff
11
Fig 3-b Transistor driven into cutoff by a
further increase in input amplitude
Figure 4a shows an ac load line with the Q-point
moved away from center toward saturation.
In this case, the output variation is limited by
saturation.
If the amplifier is driven any further, it will "clip“
at saturation ,as shown in Figure 4(b).
12
Fig. 4-a Amplitude of VCE and IC limited
by saturation
13
14
Fig. 4-b Transistor driven into saturation by a
further increase in input amplitude
Large Signal load line operation
The amplifier shown in Fig. 6 can be represented
in terms its Ac or DC equivalent circuits.
VCC
V+
1R
Rs
RC
3C
1C
1Q
NPN
+
RL
Vs
-
2C
2R
RE
15
Fig.6
DC load line
Replacing the capacitors By Open circuit we can get the
DC equivalent circuit as shown in Fig. 7a.
VCC
+V
R1
RC
Q1
NPN
R2
RE
Fig .7a DC EQ. Circuit
16
We can determine the DC load line as follows:
VCC=VCE+IC(RC+RE)
When VCE=0
Ic(sat) = Vcc/(RC+RE)
When IC=0
VCE(cut off) = VCC.
The DC load line is shown in Fig. 7.7b.
17
E
18
We can get the AC Equivalent circuit by shorting
C1,C2, C3, and VCC as shown in Fig. 8a.
Rs
Q1
NPN
+
Vs
R1PR2
-
Fig. 8.a AC EQ Circuit
19
RCPRL
Reffering to AC equivalent circuit we notice that:
RE=0, Rc=RC//RL ,
Going from Q-point to saturation point
∆IC= VCEQ / Rc
and
IC(sat)=ICQ+VCEQ/Rc
Going from Q-point to cut off point
∆VCE= ICQRc
and
VCE(cutoff) = VCEQ+ICQRc
The AC load line is shown in Fig. 8b.
20
The AC and DC load line are shown together in Fig. 2a.
21
Example
Determine the collector current and the
collector to emitter voltage for the circuit
shown in Fig. 10 at the points of saturation
and cutoff with an ac input. Assume
Xc1=Xc2=Xc3=0 and bac=200.
22
VCC
10V
+V
R1
10k
Rs
RC
1k
C3
C1
Q1
NPN
+
Vs
-
R2
4.7k
Fig10
23
C2
RE
470
RL
1.5k
Solution
DC analysis
The Q-point values for the amplifier are determined as follows
(neglecting Rin(base))
VBQ=[R2/(R1+R2)]VCC =[ 4.7kW/14.7kW ] x10=3.2V
VEQ=VBQ-VBE=3.2- 0.7=2.5V
IEQ=VEQ /RE=2.5V/470W=5.3mA
ICQ=IEQ=5.3mA
VCQ=VCC-ICQRC=10V-(5.3mA)(1kW)=4.7V
VCEQ=VCQ-VEQ=4.7V-2.5V=2.2V
24
Solution cont.
AC analysis
The point of saturation under AC condition is determined
as follows:
Rc=RC//RL=RCRL/(RC+RL)=(1KW)(1.5kW)/(1kW+1.5kW)=600W
Vce(sat)=0
Ic(sat)=ICQ+VCEQ /RC=5.3mA+2.2V/600W=8.97mA
The point of cut off under AC condition is determined as
follows:
Ic(cutoff)=0
Vce(cutoff)=VCEQ+ICQRC =2.2V+(5.3mA)(600 W )=5.38V
25
Centering the Q-point on the AC load line
A closer look to the above example shows that
the Q-point is not centered on the AC load
line.
The Q-point can be moved to an approximate
Center position on the AC load line until:
VceQ=IcQRc
Without affecting the load line itself.
26
This can be made by changing RE as follows:
VCC=VCEQ+ICQRC+ICQRE
RE=(VCC - VCEQ - ICQRC)/ICQ
Where ICQ and VCEQ are the centered Q point
Values.
ICQ(centered) =(ICQ+VCEQ /RC)/2
VCEQ(centered) =(VECQ+ICQRC)/2
27
Voltage Gain
AV=RC/r'e
and r'e=∆VBE/∆IC
Power Gain
AP = Av Ai =bDC (RC/r'e)
The power gain can also given by:
AP=PL/Pin =(V2L /RL)/(V2in/Rin) =(VL/Vin)2 (Rin/RL)
=Av2 (Rin/RL)
28
For a voltage-divider biased amplifier
and that for a CE or CC amplifier
29
DC Quiescent Power (PDQ)
The power dissipation of a transistor with no
signal input is the product of its Q-point current
and Q-point voltage.
The quiescent power, is the maximum power that a
class A amplifier must handle. The transistor's
power rating must exceed this value.
30
Output Power
Pout = Vce Ic
Where Ic and Vce are rms values.
We have three cases
1-When the Q-point is at the center of the
AC load line (Vmax=VCEQ and Imax=ICQ).
Pout = (0.707Imax)(0.707Vmax)
= 0.5 Imax Vmax = 0.5 ICQVCEQ
31
2- When the Q-point is close to saturation
(Vmax=VCEQ and Imax=VCQ /RC)
Pout = 0.5 V2CEQ /RC
3- When the Q-point is close to cutoff
(Vmax=ICQ RC and Imax=ICQ)
Pout = 0.5 I2CQRC
32
Efficiency (h)
h=Pout/PDC
PDC=VCCICC =(2VCEQ)(ICQ)
hmax=Poutmax / PDC =
=(0.5VCEQICQ)/(2VCEQICQ)
hmax = 0.25 = 25%
This value is approached only when the Q-point is
at the center of the AC load line.
33
Example
Determine the following values for the amplifier
shown in Fig.11 when operated With
maximum possible output signal:
a- Minimum transistor power rating.
b- AC output power
c- Efficiency
34
VCC
24V
+V
R1
4.7 kohm
RC
330 ohm
C3
10 uF
C1
10 uF
Q3
NPN
+
Vs
R2
1 kohm
-
Fig. 11
35
RE
100 ohm
C2
10uF
RL
330 ohm
Solution
a-First determine the DC values. Neglect Rin(base)
VB=(R2/(R1+R2))Vcc=(1kW/5.7kW)x24=4.4V
VE=VB-VBE=4.4V-0.7V=3.5V
IE=VE/RE=3.5V/100W=35mA
ICQ=35mA
VCQ=VCC - ICQRC
=24V-(35mA)(330W)=12.5V
VCEQ=VCQ-VE =12.5V-3.5V= 9V
36
b- To make a calculation of ac power under
maximum signal condition we must know the
location of the Q-point relative to the centre of
the AC load line.
The AC load line values are:
Rc=RC//RL=330//330=165W
Ic(sat)=ICQ+VCEQ /Rc
=35mA+9V/165W=89.5mA
Vce(cutoff)=VCEQ+ ICQRc =9V+35mAx165W=14.8V
37
A centered Q-point is at:
ICQ=89.5mA/2=44.8mA
VCEQ=14.8/2=7.4V
As shown in Fig.12 the actual
Q-point is closer to cutoff, Therefore
Pout=0.5 I2CQRc
=0.5(35mA)2(165W)=101mW
38
39
c-The efficiency is
h=Pout/PDC=Pout/(VCCICQ)
=101mW/(24Vx35mA)=0.12
h < 0.25 because the Q-point is not centered
40
Maximum load power
PL=V2L/RL
=(0.707VCEQ)2/RL
=0.5V2CEQ /RL
41