Download CLASS A AMPLIFIERS

CLASS A AMPLIFIERS
When a common-emitter common-collector or common-base amplifier that it operates in the
linear region for the full 3600 of the input cycle, it amplifier. In this mode of operation, the
amplifier does not go into cut-off or therefore, the output voltage waveform has the same
shape as the input waveform. A class A amplifier can be either inverting or noninverting. In
this section, you will learn about large-signal class amplifiers.
After completing this section, you should be able to
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Explain and analyse the operation of large-signal class A amplifiers F Explain class A
operation
Discuss how the location of the Q-point on the ac load line affects the c the amplifier
Analyse ac load line operation
Explain how to centre the Q-point on the load line
Determine large-signal voltage gain
Discuss distortion
Calculate power gain and quiescent power
Calculate ac output power for several Q-point conditions
Determine the efficiency of a class A amplifier
Determine the maximum load power
The operation of a class A large-signal amplifier is illustrated in Figure 1, output waveform is
an amplified replica of the input and may be either in phase out of phase with the input
(inverted)
Figure 1 Basic class A amplifier operation. Output is shown 1800 out of phase with the input
(inverted).
Q-Point is centred for Maximum Output Signal
When the Q-point is at the centre of the ac load line (midway between saturation off), a
maximum class A signal can be obtained. This is graphically illustrated load line in Figure
2(a). Ideally, the collector current can vary from its Q point Value ICQ, up to its saturation
value IC(sat), and down to its cut-off value of zero. This operation indicated in Figure 2(b).
As you can see in Figure 2(b), the collector current swings from ICQ the its peak value equals
ICQ. Also, the collector-to-emitter voltage swings from zero so its peak value equals VCEQ.
This is the largest signal achievable from an amplifier. When the input signal is too large, the
amplifier is driven into cut-off and saturation which results in clipping of the peaks, as
illustrated in Figure 3..
FIGURE 3
Waveforms are clipped at cut-off and saturation because the amplifier is overdriven (too
large an input signal).
A Noncentered Q-Point Limits the Output
Q-Point Closer to Cut-off
If the Q-point is not centred on the ac load line, VCE is limited to less than the possible
maximum amplitude. Figure 4(a) shows an ac load line with the Q-point moved away from
centre toward cut-off. Vce is limited by cut-off in this case. The collector current can swing
only down to near zero and an equal amount ICQ The collector-to-emitter voltage can swing
only up to its cut-off value and amount below VCEQ. If the amplifier is driven any further than
this by an increase in the input signal, it will go into cut-off, as shown in Figure 4(b), and the
waveform clipped off on one peak.
Q-Point Closer to Saturation Figure 5(a) shows an ac load line with Q point moved away
from centre toward saturation. In this case, VCE is limited by saturation. The collector current
can swing only up to near saturation and an equal amount below ICQ. The collector-to-emitter
voltage can swing only down to near its saturation value equal amount above VCEQ. If the
amplifier is driven any further than this by an input signal, it will go into saturation, as shown
in Figure 5(b), and the' will be clipped off on one peak.
Large-Signal Load Line Operation
Recall that an amplifier such as that shown in Figure 6 can be represented either its dc or its
ac equivalent.
DC Load Line Using the dc equivalent circuit in Figure 7(a), you can de dc load line as
follows: k(sat) occurs when VCE  0, so
I C(sat) 
VCC
RC  RE
VCE(cut-off) occurs when Ic  0, so
VCE(cut-off) = VCC
The dc load line is shown in Figure 7(b).
AC Load Line From the ac viewpoint, the circuit in Figure 6 looks different than it does
from the dc viewpoint. The collector resistance is different because RL is in parallel with Rc
due to the coupling capacitor C3, and the emitter resistance is zero due to the bypass capacitor
C2; therefore, the ac load line is different from the dc load line. How much collector current
can there be under ac conditions before saturation occurs? To answer this question, refer to
the ac equivalent circuit and ac load line in Figure 7. Remember that a lowercase italic
subscript indicates an ac quantity and an uppercase non-italic subscript indicates a dc
quantity. For example, RC is the ac collector resistance and Rc is the dc collector resistance.
ICQ and VCEQ are the dc Q-point values. Going from the Q-point to the saturation point, the
collector-to-emitter voltage swings from VCEQ to near 0; that is, VVCE = VCEQ. The change in
collector current going from the Q point to saturation is therefore
I C 
VCEQ
RC
where RC = RC // RL is the ac collector resistance. The ac collector current at saturation is
I C  I CQ  I C
Thus,
I C  I CQ 
VCEQ
RC
Going from the Q-point to the cut-off point, the collector current swings from ICQ to near 0;
that is, Ic = ICQ. The change in collector-to-emitter voltage going from Q-point to cut-off is
therefore
VCE  (I C ) RC  I CQ RC
The cut-off value of ac collector-to-emitter voltage is
VCE (cot off )  VCEQ  I CQ RC
These results are shown on the ac load line of Figure 9. The corresponding dc load line is
shown for comparison.
FIGURE 9
Solution the Q-point values for this amplifier are determined as follows (neglecting
RIN(base)):
 R2 
 4.7k 
VCC  
VBQ  
10V  3.2V
R

R
14
.
7
k



2 
 1
Then,
 VBQ  0.7   3.2  0.7 

I EQ  
  470   5.3mA
 RE

Therefore, since I C  I E
ICQ = 5.3 mA The collector voltage is
VCQ= VCC-ICQRC= 1OV- (5.3 mA)(l k) = 4.7V Therefore, the collector-to-emitter voltage is
VCEQ = VCQ - IEQRE = 4.7 V - 2.5 V = 2.2 V
The point of saturation under ac conditions is determined as follows:
VCE(sat) = 0 V
Ic(sat) = ICQ +
VCEQ
RC
The ac collector resistance is
R C  R C || R L 
R CR L
1k x 1.5k

 600 
R C  R L 1k  1.5k
Thus, Ic(sat) = 5.3 mA +
2.2V
 5.3 mA  3.67 mA  8.97 mA
600 
The point of cut-off under ac conditions is determined as follows:
IC(cut-off) = 0 A
VCE(cut-off)) = VCEQ + ICQRC = 2.2 V + (5.3 mA)(600 ) = 5.38 V
These results show that the collector current can swing up to almost 8.97 mA the collector-toemitter voltage can swing up to almost 5.38 V without the peaks being clipped due to
saturation or cut-off.
Related Exercise Determine IC and VCE at the points of saturation and cut-off in Figure 10 for
the following circuit values: VCC = 15 V and ac = 150.
As you learned earlier, a centred Q-point allows a maximum undistorted o. swing. A closer
look at Example 1 shows that the Q-point is not centred, and therefore, the unclipped output
swing is somewhat less than it could be if the Q-point centred on the ac load line. This is
examined further in Example 2.
Figure 11 shows the ac load line for the circuit in Example 1. notice mar me Q-point is not
centred. The maximum output swings are
I C  I c(sat) - I CQ  8.97 InA - 5.3 mA  3.67 mA and
VCE  VCEQ  2.2V
Determine the maximum output swings for collector current and collector-to-emitter voltage
when the Q-point is centred, assuming the same load line is maintained.
Solution The maximum output swings for a centred Q-point are
I C  I CQ 
I c(sat)
2

8.97
 4.49mA
2
VCE (max)  VCEQ 
VCE (cut  off ) 5.38

 2.69V
2
2
Related Exercise Explain what happens to the output voltage if the Q-point is shifted to VCE
= 3 V.
Centring the Q-Point on the AC Load Line
To have a centred Q-point on the ac load line in Figure 7-9, VCEO must be midway between
zero and the lower (cut-off) end of the ac load line, expressed as
VCEQ 
VCEQ  I CQ R C
2
must be midway between zero and the upper saturation end of the ac load line, expressed as
I CQ 
I CQ  VCEQ / R C
2
The condition for a centered Q-point is developed as follows using the preti equation for
VCEQ:
2VCEQ  VCEO  I CQ R C
2VCEQ - VCEQ  I CQ R C
VCEQ  I CQ R C
The Q-point can be moved to an approximate center position on the load changing ICQ until
both sides of Equation (7-3) are approximately equal.
To move the Q-point toward cut-off on the ac load line without affecting line itself, ICQ
should be decreased by increasing RE. To move the Q-point toward saturation on the load
line, ICQ should be increased by decreasing RE.
EXAMPLE 3 As you saw in Example 2, the circuit in Figure 10 does not have a centred Qpoint. Select a value for RE that will produce a Q-point that is approximately centred on the
ac load line.
Solution You know from Example 2 that the collector current for a centre Q-point should be
4.49 mA. RC = 600 from Example 1.
VCEQ = ICQRC = (4.49 mA)(600 ) = 2.69 V
Determine RE by the following steps:
VCEQ = VCC - ICQ(RC + RE)
= VCC - ICQRC - ICQRE
Transpose terms:
VCEQ - VCC + ICQRC = ICQRE
Then,
ICQRE = VCC - VCEQ - RC
10 V -2.69 V -(4.49 mA)(l k) = 628
4.49 mA
Choose the nearest standard value:
RE = 620 
This will produce an approximately centred Q-point.
Related Exercise If RC in Figure10 is increased to 1.2 k, what should you RE to keep a
centred Q-point?
Voltage Gain
The voltage gain of a class A large-signal amplifier is determined in the same way smallsignal amplifier with the exception that the formula re' = 25 mV/IE is not the large-signal
amplifier. Because the signal swings over a large portion of the transconductance curve, re'
varies significantly. Since re' = VBE/IC, the value is much larger at the lower end of the
transconductance curve than it is at the higher end, as illustrated in Figure 12. As you can see,
for the same VBE, the value of IC is greater the higher on the curve you go.
As you know, the voltage-gain formula for a common-emitter amplifier with RE completely
bypassed is
R
Av  C
r e
In large-signal amplifiers, however, the gain varies significantly as the signal swings over the
transconductance curve because of the variation in re'. Therefore, only an average gain can
be determined using an average value of re'
Find the average large-signal voltage gain of the amplifier in Figure 13. Assume that re' has
been found to have an average value of 5  from graphical data.
Solution
R C  R C || R L 
Av 
R CR L
1k x 1.5k

 600 
R C  R L 1k  1.5k
R c 600

 120
r e
5
Related Exercise Find the average large-signal voltage gain of the amplifier in Figure 13 if
the supply voltage is changed to 9 V, RL is changed to 1 k, and r: has average value of 10.
Distortion
When the collector current swings over a large portion of the transconductance curve
distortion can occur on the negative half-cycle because of the greater nonlinearity a lower end
of the curve and because of lower gain due to a larger re' as shown in Figure14.Distortion can
be sufficiently reduced by keeping the collector current on the more linear part of the curve
(at higher values of ICQ and VBEQ). Increasing the base bias volt-age will result in more
collector current and an increase in VBE due to more voltage drop across re'. Also, distortion
can be reduced by using a swamping resistor in the emitter to minimise the effect of the
variation in re'.
If there is a swamping resistor, the voltage gain becomes A p 
swamping resistor.
Rc
R E 1  re 
where RE1 is the
Power Gain
The main purpose of a large-signal amplifier is to achieve power gain. If we assume that the
large-signal current gain, Ai, is approximately equal to DC then the power gain, AP, for a
common-emitter amplifier is
A p  A i A v   dc A v
A p   dc
Rc
r e
Quiescent Power
The power dissipation of a transistor with no signal input is the product of its Q-point current
and voltage.
PDQ  I CQ VCEQ (1-6)
The quiescent power is the maximum power that the class A transistor must handle;
therefore, its power rating should exceed this value.
Output Power
In general, for any Q-point location on the ac load line, the output power of a commonemitter amplifier is the product of the rms collector current and the rms collector-to-emitter
voltage.
Pout  VCE I C
Let's now consider the output power for three cases of Q-point location.
Q-Point Closer to Saturation When the Q-point is closer to saturation, the maximum
collector-to-emitter voltage swing is VCEQ, and the maximum collector current swing is
VCEO/R~, as shown in Figure 15(a). The ac output power is, therefore,
Pout
V
 0.707VCEQ 
0.707VCEQ  0.5 CEQ
 

Rc
Rc


2
Q-Point Closer to Cutoff When the Q-point is closer to cutoff, the maximum collector
current swing is 1C0' and the collector-to-emitter voltage swing is ICQRC as shown in
Figure15(b). The ac output power is, therefore,
Pout  (0.707I CQ )(0.707I CQ R c )  0.5I CQ 2 R C
Q-Point Centered When the Q-point is centered, the maximum collector current swing is ICQ,
and the maximum collector-to-emitter voltage swing is VCEO, as shown in Fi 7-15(c). The
ac output power is, therefore,
P0Ut = (O.7O7VCEO)(O.7071CO)
')OUt = O.5VCEOl'CQ
(7-
This is the maximum ac output power from a class A amplifier under signal conditio Notice
that it is one-half the quiescent power dissipation.
Efficiency
Efficiency ~) of an amplifier is the ratio of ac output power to dc input power. The input
power is the dc supply voltage times the current drawn from the supply.
')DC VCCICC
The average supply current ~CC equals 1CO and the supply voltage VCC is tW VCEO when
the Q-point is centered. The maximum efficiency is therefore
POUt O.5VCEQICO _ O.5VCEOICO _ 0.5
2
~DC - VCCICC
VCEOICO 2
=0.25 (7
Thus, 0.25 or 25 percent is the highest possible efficiency available from a class amplifier
and is approached only when the Q-point is at the center of the ac load line.
EXAMPLE 7-5
Determine the following values for the amplifier in Figure 7-16 when
operated with
the
maximum possible output signal:
(a)
Minimum transistor power rating
(b) AC output power (c) Efficiency
FIGURE 7-16 tcC
+24 V
Rc
Rt
33011 C3
4.7 k11
(o~I
10~~
RL
'3DC=150
011
10 ~F
R2
~k11 RE
C2
~0011 10 ~F
Solution First determine the dc values. Neglect RtN(base).
(RlR;R2) VCC=(517~)24V=4.2V
Then
VE = VE - VBF = 4.2 V -0.7 V = 3.5 V
VE
3.5V
=35mA
TOOQ
Therefore, the collector current at the Q-point is
~CQ=~3S EuA
and
VC = VCC - 1CQRC = 24 V - (35 mA)(330 Q) = 12.5 V VCEQ = VC - VE = 12.5 V -3.5 V =
9.0 V
(a)
The transistor power rating must be greater than
~D(min) = PDQ = VCEQICQ = (9.0 V)(35 mA) =315 mW
(b)
To calculate the ac output power under a maximum signET condition, you must know
the location of the Q-point relative to the center. This will tell you whether 1CQ or VCEQ is
the limiting factor if the Q~point is not centered. The ac Toad Tine values are as follows:
RC=RCiIRL=330Q11330Q= 165Q VC -35 mA 9.OV
1
C(sat) = 1CQ +
_
= 89.5 mA
REQ + 165 Q
and
Vce(culoft) = VCEQ + 1C~c = 9.0 V + (35 mA)(165 Q) = 14.8 V A centered Q-point is at
89.5
mA 44.8 mA
1
CQ 2
and
14.8 V
VCEQ=
2
=7.37V
These Q-point values are shown on the ac load line in Figure 7-17. The actual Q-point for
this amplifier is closer to cutoff, as shown in the figure. There-fore, the maximum collector
current swing is 1CQ and the ac output power is
POW = O.S~2cQRc = 0.5(35 mA)2(165 Q) 101 mW
FIGURE 7-17 1c
89~5 mA
Q (centered)
(c)
The efficiency is
~
~OW ~Ou'
= ~oc VccIcC
101 mW
= 0.12
VccIcQ = (24 V)(35 mA)
This efficiency of 12 percent is considerably less than the maximum possible efficiency (25
percent) because the actual Q-point is not centered.
Related Exercise Suggest ways to increase the efficiency of the amplifier in Figure 7-16
i Open file FG07- 1 6.CA4 on your circuit disk. Measure 1CQ and VCEQ and compare with
the calculated values.
Maximum Load Power
The maximum power to the load in a class A amplifier occurs when the Q-point is ce tered,
as you can see in Figure 7-15(c). The maximum peak load voltage equals VCE~ assuming a
negligible voltage drop across the output coupling capacitor.
2
VL _ (O.7O7VCEQ)2
~L(niax) - RL RL
-
O.SVc2EQ
~inax) -
RL
(7-9
Often the load power is defined as the output power.
F 7-6 Determine the maximum load power for the amplifier in Example 7-5.
Solution When the Q-point is centered, VCEQ = 7.37 V.
O.SVC2EQ - 0.5(7.37 V)2 - 82.3 mW
RL
330Q
CLASS A PUSH-PULL AMPLIFIERS
=
When an amphfier is biased at cutoff such that it operates in the linear region for 1800 of the
input cycle and is iii cutoff for 180; it is a class B amplifier. Class AB amplifiers are biased
to conduct for slightly more than 1800. The primary advantage of a class B or class AB
amplifier over a class A amphfier is that they are more efficient; you can get more output
power for a given amount of input power. A disadvantage of class B or class AB is that it is
more difficult to implement the circuit in order to get a liiiear reproduction of the iaput
waveform. As you will see in this section, the term push-pull refers to a common type of class
B or class AB amplifier circuit in which the hiput wave shape is reproduced at the output.
After completing this section, you should be able to

Explain and analyze the operation of class B and class AB amplifiers
F Explain class B operation
F Discuss Q-point location for class B amplifiers
F Describe class B push-pull operation
F Explain crossover distortion and its cause
F Explain class AB operation
F Analyze class AB push-pull amplifiers
F Calculate maximum output power
F Calculate dc input power
F Determine class B maximum efficiency
F Find the input resistance
F Analyze a darlington push-pull amplifier
F Discuss methods of driving a push-pull amplifier