Download Electronic Electronic devices-9 e-solutions and more Essays (high school) Electronics in PDF only on Docsity!
BJT A mplifiers
CHAPTER OUTLINE
6–1 Amplifier Operation
6–2 Transistor AC Models
6–3 The Common-Emitter Amplifier
6–4 The Common-Collector Amplifier
6–5 The Common-Base Amplifier
6–6 Multistage Amplifiers
6–7 The Differential Amplifier
6–8 Troubleshooting
Device Application
CHAPTER OBJECTIVES
◆ ◆ (^) Describe amplifier operation
◆ ◆ (^) Discuss transistor models
◆ ◆ (^) Describe and analyze the operation of common-emitter amplifiers
◆ ◆ (^) Describe and analyze the operation of common-collector amplifiers
◆ ◆ (^) Describe and analyze the operation of common-base amplifiers
◆ ◆ (^) Describe and analyze the operation of multistage amplifiers
◆ ◆ (^) Discuss the differential amplifier and its operation
◆ ◆ (^) Troubleshoot amplifier circuits
KEY TERMS
VISIT THE WEBSITE Study aids, Multisim, and LT Spice files for this chapter are available at https://www.pearsonhighered.com /careersresources/
INTROdUCTION The things you learned about biasing a transistor in Chapter 5 are now applied in this chapter where bipolar junction tran- sistor (BJT) circuits are used as small-signal amplifiers. The term small-signal refers to the use of signals that take up a relatively small percentage of an amplifier’s operational range. Additionally, you will learn how to reduce an ampli- fier to an equivalent dc and ac circuit for easier analysis, and you will learn about multistage amplifiers. The differential amplifier is also covered.
dEVICE APPLICATION PREVIEW The Device Application in this chapter involves a preampli- fier circuit for a public address system. The complete system includes the preamplifier, a power amplifier, and a dc power supply. You will focus on the preamplifier in this chapter and then on the power amplifier in Chapter 7.
◆ ◆ r parameter
◆ ◆ Common-emitter
◆ ◆ ac ground
◆ ◆ Input resistance
◆ ◆ Output resistance
◆ ◆ Attenuation
◆ ◆ Bypass capacitor
◆ ◆ Common-collector
◆ ◆ Emitter-follower ◆ ◆ Common-base ◆ ◆ Decibel ◆ ◆ Differential amplifier ◆ ◆ Common mode ◆ ◆ CMRR (Common-mode rejection ratio)
256 ◆^ BJT Amplifiers
6–1 A mplifier O perATiOn
The biasing of a transistor is purely a dc operation. The purpose of biasing is to es- tablish a Q-point about which variations in current and voltage can occur in response to an ac input signal. In applications where small signal voltages must be amplified— such as from an antenna or a microphone—variations about the Q-point are relatively small. Amplifiers designed to handle these small ac signals are often referred to as small-signal amplifiers. After completing this section, you should be able to ❑ (^) Describe amplifier operation ❑ (^) Identify ac quantities ◆ (^) Distinguish ac quantities from dc quantities ❑ (^) Discuss the operation of a linear amplifier ◆ (^) Define phase inversion ◆ (^) Graphically illustrate amplifier operation ◆ (^) Analyze ac load line operation
AC Quantities
In the previous chapters, dc quantities were identified by nonitalic uppercase (capital) subscripts such as I C, I E, V C, and V CE. Lowercase italic subscripts are used to indicate ac quantities of rms, peak, and peak-to-peak currents and voltages: for example, I (^) c , I (^) e , I (^) b , Vc , and V (^) ce (rms values are assumed unless otherwise stated). Instantaneous quantities are represented by both lowercase letters and subscripts such as i (^) c , i (^) e , i (^) b , and v (^) ce. Figure 6– illustrates these quantities for a specific voltage waveform.
Vce V (^) ce
V (^) ce
v (^) ce
V CE
0 0
t
V (^) ce
rm
peak s avg
▶ (^) FIGURE 6–1 V Vce can represent rms, average, peak, or peak-to-peak, but rms will be assumed unless stated otherwise. vce can be any instantaneous value on the curve.
In addition to currents and voltages, resistances often have different values when a cir- cuit is analyzed from an ac viewpoint as opposed to a dc viewpoint. Lowercase subscripts are used to identify ac resistance values. For example, Rc is the ac collector resistance, and R C is the dc collector resistance. You will see the need for this distinction later. Resistance values internal to the transistor use a lowercase r 9 to show it is an ac resistance. An exam- ple is the internal ac emitter resistance, r (^9) e.
258 ◆^ BJT Amplifiers
intersects the horizontal axis ( V CE) at the ac value of the collector-to-emitter cutoff voltage Vce(cutoff). These values are determined as follows: Ic ( sat ) = V CEQ > Rc + I CQ Vce ( cutoff ) = V CEQ + I CQ Rc Where Rc is the parallel combination of R C and RL. Lines projected from the peaks of the base current, across to the I C axis, and down to the V CE axis, indicate the peak-to-peak variations of the collector current and collector- to-emitter voltage, as shown. The ac load line differs from the dc load line because the capacitors C 1 and C 2 effectively change the resistance seen by the ac signal. In the circuit in Figure 6–2, notice that the ac collector resistance is RL in parallel with R C, which is less than the dc collector resistance R C alone. This difference between the dc and the ac load lines is covered further in Chapter 7 in relation to power amplifiers.
Given the Q-point value of I CQ = 4 mA, V CEQ = 2 V, R C = 1 kV, and RL = 10 kV for a certain amplifier, determine the ac load line values of Ic(sat) and Vce(cutoff).
Solution The ac load line values of Ic(sat) and Vce(cutoff) are
Rc = R C 7 RL = 1 kV 7 10 kV = 909 V
Ic ( sat ) = V CEQ > Rc + I CQ = 2 V> 909 V + 4 mA = 6.2 mA Vce ( cutoff ) = V CEQ + I CQ Rc = 2 V + 4 mA (909 V) = 5.64 V
Related Problem* If the Q-point is changed to 3 V and 6 mA, what is the intersection values of the ac load line on the two axes?
EXAMPLE 6–
The ac load line operation of a certain amplifier extends 10 mA above and below the Q-point base current value of 50 mA, as shown in Figure 6–4. Determine the resulting peak-to-peak values of collector current and collector-to-emitter voltage from the graph.
EXAMPLE 6–
1
2
3
0 1 2 3 4
V CE (V)
10 A
20 A
30 A
40 A
50 A
70 A
I C (mA)
7
8
6 5 4
60 A I (^) c
V (^) ce
Ib
Q
m
m
m
m
m
m
60 mA
50 mA
40 mA
m
▶ (^) FIGURE 6–
*** Answers can be found at www.pearsonhighered.com/floyd.**
Transistor AC Models ◆^259
Solution Projections on the graph of Figure 6–4 show the collector current varying from 6 mA to 4 mA for a peak-to-peak value of 2 mA and the collector-to-emitter voltage varying from 1 V to 2 V for a peak-to-peak value of 1 V.
Related Problem What are the Q-point values of I C and V CE in Figure 6–4?
- When Ib is at its positive peak, Ic is at its _____ peak, and Vce is at its _____ peak.
- What is the difference between V CE and Vce?
- What is the difference between Re and re?
- Why is the ac resistance seen by the collector different from the dc resistance?
SECTION 6– CHECKUP Answers can be found at www .pearsonhighered.com/floyd.
6–2 T rAnsisTOr AC mOdels
To visualize the operation of a transistor in an amplifier circuit, it is often useful to represent the device by a model circuit. A transistor model circuit uses various internal transistor parameters to represent its operation. Transistor models are described in this section based on resistance or r parameters. Another system of parameters, called hybrid or h parameters, is briefly described. After completing this section, you should be able to ❑ (^) Discuss transistor models ❑ (^) List and define the r parameters ❑ (^) Describe the r- parameter transistor model ❑ (^) Determine r (^9) e using a formula ❑ (^) Compare ac beta and dc beta ❑ (^) List and define the h parameters
r Parameters
The r parameters that are commonly used for BJTs are given in Table 6–1. Strictly speaking, a ac and b ac are current ratios, not r parameters, but they are used with the resistance param- eters to model basic transistor circuits. The italic lowercase letter r with a prime denotes resistances internal to the transistor.
◀ (^) TABLE 6– r parameters.
r PARAMETERS DESCRIPTION r (^9) e ac emitter resistance rb (^9) ac base resistance rc (^9) ac collector resistance a ac ac alpha ( Ic > Ie ) b ac ac beta ( Ic > Ib )
r-Parameter Transistor Model
An r -parameter model for a BJT is shown in Figure 6–5(a). For most general analysis work, it can be simplified as follows: The effect of the ac base resistance ( r (^9) b ) is usually
Transistor AC Models ◆^261
Comparison of the AC Beta ( B ac) to the dC Beta ( B dC)
For a typical transistor, a graph of I C versus I B is nonlinear, as shown in Figure 6–7(a). If you pick a Q-point on the curve and cause the base current to vary an amount D I B , then the collector current will vary an amount D I C as shown in part (b). At different points on the nonlinear curve, the ratio D I C >D I B will be different, and it may also differ from the I C > I B ratio at the Q-point. Since bDC = I C > I B and b ac = D I C >D I B , the values of these two quanti- ties can differ slightly.
Determine the r (^9) e of a transistor that is operating with a dc emitter current of 2 mA.
Solution r (^9) e >
25 mV I E
25 mV 2 mA
= 12.5 V
Related Problem What is I E if re (^9) = 8 V?
EXAMPLE 6–
0 I BQ
I CQ
I C
I B
Q
(a) bDC = I C/ I B at Q-point
D I C^ Q
I C
I B D I B
( I B, I C )
0 (b) b ac = D I C /D I B
◀ (^) FIGURE 6– I C-versus- I B curve illustrates the dif- ference between BDC 5 I C , I B and Bac 5 D I C , D I B.
h Parameters
A manufacturer’s datasheet typically specifies h (hybrid) parameters ( h (^) i , hr , h (^) f , and h (^) o ) be- cause they are relatively easy to measure. The four basic ac h parameters and their descriptions are given in Table 6–2. Each of the four h parameters carries a second subscript letter to designate the common-emitter ( e ), common-base ( b ), or common-collector ( c ) amplifier configuration, as listed in Table 6–3. The term common refers to one of the three terminals (E, B, or C) that is referenced to ac ground for both input and output signals. The characteristics of each of these three BJT amplifier configurations are covered later in this chapter.
◀ (^) TABLE 6– Basic ac h parameters.
◀ (^) TABLE 6– Subscripts of h parameters for each of the three amplifier configurations.
h PARAMETER DESCRIPTION CONDITION h (^) i Input impedance (resistance) Output shorted h (^) r Voltage feedback ratio Input open h (^) f Forward current gain Output shorted h (^) o Output admittance (conductance) Input open
CONFIGURATION h PARAMETERS Common-Emitter h (^) ie , hre , hfe , hoe Common-Base h (^) ib , hrb , hfb , hob Common-Collector h (^) ic , hrc , hfc , hoc
262 ◆^ BJT Amplifiers
Relationships of h Parameters and r Parameters
The ac current ratios, a ac and b ac , convert directly from h parameters as follows: a ac = hf b b ac = hfe Because datasheets often provide only common-emitter h parameters, the following formulas show how to convert them to r parameters. We will use r parameters throughout the text because they are easier to apply and more practical.
r (^9) e =
hre hoe
r (^9) c =
hre + 1 hoe
r (^9) b = hie -
hre hoe
(1 + hfe )
- Define each of the parameters: A ac , B ac , re , rb , rc.
- Which h parameter is equivalent to B ac?
- If I E 5 15 mA, what is the approximate value of re.
- What is the difference between B ac and B DC?
SECTION 6– CHECKUP
6–3 T he COmmOn - emiT Ter A mplifier
As you have learned, a BJT can be represented in an ac model circuit. Three amplifier configurations are the common-emitter, the common-base, and the common-collector. The common-emitter (CE) configuration has the emitter as the common terminal, or ground, to an ac signal. CE amplifiers exhibit high voltage gain and high current gain. The common-collector and common-base configurations are covered in the Sections 6–4 and 6–5. After completing this section, you should be able to ❑ (^) Describe and analyze the operation of common-emitter amplifiers ❑ (^) Discuss a common-emitter amplifier with voltage-divider bias ◆ (^) Show input and output signals ◆ (^) Discuss phase inversion ❑ (^) Perform a dc analysis ◆ (^) Represent the amplifier by its dc equivalent circuit ❑ (^) Perform an ac analysis ◆ (^) Represent the amplifier by its ac equivalent circuit ◆ (^) Define ac ground ◆ (^) Discuss the voltage at the base ◆ (^) Discuss the input resistance at the base and the output resistance ❑ (^) Analyze the amplifier for voltage gain ◆ (^) Define attenuation ◆ (^) Define bypass capacitor ◆ (^) Describe the effect of an emitter bypass capacitor on voltage gain ◆^ Discuss voltage gain without a bypass capacitor ◆^ Explain the effect of a load on voltage gain ❑ (^) Discuss the stability of the voltage gain ◆ (^) Define stability ◆ (^) Explain the purpose of swamping r (^9) e and the effect on input resistance ❑ (^) Determine current gain and power gain
264 ◆^ BJT Amplifiers
I E =
V TH - V BE
R E + R TH >bDC
2.83 V - 0.7 V
560 V + 34.6 V
= 3.58 mA
I C > I E = 3.58 mA V E = I E R E = (3.58 mA)(560 V) = 2.00 V V B = V E + 0.7 V = 2.70 V V C = V CC - I C R C = 12 V - (3.58 mA)(1.0 kV) = 8.42 V V CE = V C - V E = 8.42 V - 2.00 V = 6.42 V
AC Analysis
To analyze the ac signal operation of an amplifier, an ac equivalent circuit is developed as follows:
1. The capacitors C 1 , C 2 , and C 3 are replaced by effective shorts because their values are selected so that XC is negligible at the signal frequency and can be considered to be 0 V. 2. The dc source is replaced by ground. A dc voltage source has an internal resistance of near 0 V because it holds a constant volt- age independent of the load (within limits); no ac voltage can be developed across it so it appears as an ac short. This is why a dc source is called an ac ground. The ac equivalent circuit for the common-emitter amplifier in Figure 6–8 is shown in Figure 6–10(a). Notice that both R C and R 1 have one end connected to ac ground (red) be- cause, in the actual circuit, they are connected to V CC which is, in effect, ac ground.
R C 1.0 kV
R 1 22 kV
R 2 6.8 kV
R C 1.0 kV
R 1 22 kV
R 2 6.8 kV
R (^) s
V (^) s
(a) Without an input signal voltage (^) (b) With an input signal voltage (AC ground is shown in red.)
▶ (^) FIGURE 6–10 ac source
AC equivalent circuit for the ampli- fier in Figure 6–8.
In ac analysis, the ac ground and the actual ground are treated as the same point electri- cally. The amplifier in Figure 6–8 is called a common-emitter amplifier because the bypass capacitor C 2 keeps the emitter at ac ground. Ground is the common point in the circuit. Signal (AC) Voltage at the Base An ac voltage source, Vs , is shown connected to the input in Figure 6–10(b). If the internal resistance of the ac source is 0 V, then all of the source voltage appears at the base terminal. If, however, the ac source has a nonzero inter- nal resistance, then three factors must be taken into account in determining the actual signal
voltage at the base. These are the source resistance ( R s ), the bias resistance ( R 1 7 R 2 ), and
the ac input resistance at the base of the transistor ( Rin ( base ) ). This is illustrated in Figure 6–11(a) and is simplified by combining R 1 , R 2 , and Rin ( base ) in parallel to get the total input resistance , Rin ( tot ) , which is the resistance “seen” by an ac source connected to the input, as shown in Figure 6–11(b). A high value of input resistance is desirable so that the amplifier will not excessively load the signal source. This is opposite to the requirement for a stable
The Common-Emitter Amplifier ◆^265
Q-point, which requires smaller resistors. The conflicting requirement for high input re- sistance and stable biasing is but one of the many trade-offs that must be considered when choosing components for a circuit. The total input resistance is expressed by the following formula:
Rin ( tot ) 5 R 1^7 R 2^7 Rin ( base ) Equation 6–
As you can see in the figure, the source voltage, V (^) s , is divided down by R (^) s (source resist- ance) and Rin ( tot ) so that the signal voltage at the base of the transistor is found by the voltage-divider formula as follows:
Vb = a
Rin ( to t ) Rs + Rin ( to t )
b Vs
If Rs 66 Rin ( tot ) , then Vb > Vs where Vb is the input voltage, Vin , to the amplifier.
Input Resistance at the Base To develop an expression for the ac input resistance look- ing in at the base, use the simplified r -parameter model of the transistor. Figure 6–12 shows the transistor model connected to the external collector resistor, R C. The input resistance looking in at the base is
Rin ( b as e ) =
Vin Iin
Vb Ib
The base voltage is
Vb = Ie r (^9) e
and since Ie > Ic ,
Ib >
Ie b ac
Substituting for Vb and Ib ,
Rin ( b as e ) =
Vb Ib
Ie re 9 Ie >b ac
Cancelling Ie ,
Rin ( base ) 5 B acr (^9) e
R 1
Rs
R 2 R (^) in ( base )
Base
V (^) s
(a)
Rs
R (^) in ( tot ) = R 1 || R 2 || R (^) in ( base )
Base
V (^) s
(b)
V (^) in
◀ (^) FIGURE 6– AC equivalent of the base circuit.
V (^) b
I (^) b B
E
r (^) e 9 I (^) e
b ac Ib = I (^) c
C
+V (^) CC
R C
▲ (^) FIGURE 6– r -parameter transistor model (inside shaded block) connected to external circuit.
Equation 6–
Output Resistance The output resistance of the common-emitter amplifier is the re- sistance looking in at the collector and is approximately equal to the collector resistor.
Rout @ R C Equation 6–
Actually, Rout = R C 7 r 9 c , but since the internal ac collector resistance of the transistor, r 9 c , is
typically much larger than R C, the approximation is usually valid.
The Common-Emitter Amplifier ◆^267
The Ie terms cancel, so
Av 5
R C
r (^9) e
Equation 6–
Equation 6–5 is the voltage gain from base to collector. To get the overall gain of the amplifier from the source voltage to collector, the attenuation of the input circuit must be included. Attenuation is the reduction in signal voltage as it passes through a circuit and cor- responds to a gain of less than 1. For example, if the signal amplitude is reduced by half, the attenuation is 2, which can be expressed as a gain of 0.5 because gain is the reciprocal of attenuation. Suppose a source produces a 10 mV input signal and the source resistance combined with the load resistance results in a 2 mV output signal. In this case, the attenu- ation is 10 mV>2 mV = 5. That is, the input signal is reduced by a factor of 5. This can be expressed in terms of gain as 1> 5 = 0.2. Assume that the amplifier in Figure 6–15 has a voltage gain from base to collector of Av and the attenuation from the source to the base is V (^) s > Vb. This attenuation is produced by the source resistance and total input resistance of the amplifier acting as a voltage divider and can be expressed as
Attenuation =
Vs Vb
Rs + Rin ( to t ) Rin ( to t )
The overall voltage gain of the amplifier, A (^9) v , is the voltage gain from base to collector, Vc > Vb , times the reciprocal of the attenuation, Vb > Vs.
A (^9) v = a
Vc Vb
ba
Vb Vs
b =
Vc Vs
R 1 || R 2
R C
Attenuation Vs / V (^) b
R (^) s
V (^) s
V (^) b
Voltage gain base-to- collector V (^) c / V (^) b
V (^) out
Overall voltage gain V (^) c / V (^) s
V (^) c
▲ (^) FIGURE 6–
Base circuit attenuation and overall voltage gain.
Effect of the Emitter Bypass Capacitor on Voltage Gain The emitter bypass capaci- tor , which is C 2 in Figure 6–8, provides an effective short to the ac signal around the emitter resistor, thus keeping the emitter at ac ground, as you have seen. With the bypass capacitor, the gain of a given amplifier is maximum and equal to R C > r (^9) e. The value of the bypass capacitor must be large enough so that its reactance over the frequency range of the amplifier is very small (ideally 0 V) compared to R E. A good rule of thumb is that the capacitive reactance, XC , of the bypass capacitor should be at least 10 times smaller than R E at the minimum frequency for which the amplifier must operate.
10 XC … R E
268 ◆^ BJT Amplifiers
Select a minimum value for the emitter bypass capacitor, C 2 , in Figure 6–16 if the am- plifier must operate over a frequency range from 200 Hz to 10 kHz.
EXAMPLE 6–
V CC +12 V
R C 1.0 kV
R 2 6.8 kV
R E 560 V
R 1 22 kV C 1
C 2
C 3
2N
Vout
V (^) in
Solution The XC of the bypass capacitor, C 2 , should be at least ten times less than R E.
XC 2 =
R E
560 V
= 56 V
Determine the capacitance value at the minimum frequency of 200 Hz as follows:
C 2 =
2 p f XC 2
2 p(200 Hz)(56 V)
= 14.2 M F
This is the minimum value for the bypass capacitor for this circuit. You can always use a larger value, although cost and physical size may impose limitations.
Related Problem If the minimum frequency is reduced to 100 Hz, what value of bypass capacitor must you use?
▶ (^) FIGURE 6–
Voltage Gain Without the Bypass Capacitor To see how the bypass capacitor affects ac voltage gain, let’s remove it from the circuit in Figure 6–16 and compare voltage gains. Without the bypass capacitor, the emitter is no longer at ac ground. Instead, R E is seen by the ac signal between the emitter and ground and effectively adds to r (^9) e in the voltage gain formula.
Av 5
R C
r (^9) e 1 R E
Equation 6–
The effect of R E is to decrease the ac voltage gain.
Calculate the base-to-collector voltage gain of the amplifier in Figure 6–16 both with- out and with an emitter bypass capacitor if there is no load resistor.
Solution From Example 6–4, r (^9) e = 6.98 V for this same amplifier. Without C 2 , the gain is
Av =
R C
r (^9) e + R E
1.0 kV 567 V
EXAMPLE 6–
270 ◆^ BJT Amplifiers
Stability of the Voltage Gain
Stability is a measure of how well an amplifier maintains its design values over changes in temperature or for a transistor with a different b. Although bypassing R E does produce the maximum voltage gain, there is a stability problem because the ac voltage gain is depend- ent on r (^9) e since Av = R C > r (^9) e. Also, r (^9) e depends on I E and on temperature. This causes the gain to be unstable over changes in temperature because when r (^9) e increases, the gain decreases and vice versa. With no bypass capacitor, the gain is decreased because R E is now in the ac circuit ( Av = R C >( r (^9) e + R E )). However, with R E unbypassed, the gain is much less dependent on re 9. If R E W r (^9) e , the gain is essentially independent of r (^9) e because
Av >
R C
R E
Swamping re to Stabilize the Voltage Gain Swamping is a method used to minimize the effect of r (^9) e without reducing the voltage gain to its minimum value. This method “swamps” out the effect of r (^9) e on the voltage gain. Swamping is, in effect, a compromise between hav- ing a bypass capacitor across R E and having no bypass capacitor at all. Whenever a bypass capacitor is used, its reactance should be small compared to the ac emitter resistance at the lowest frequency at which the amplifier will be used. In a swamped amplifier, R E is partially bypassed so that a reasonable gain can be achieved, and the effect of r (^9) e on the gain is greatly reduced or eliminated. The total ex- ternal emitter resistance, R E, is formed with two separate emitter resistors, R E1 and R E2, as indicated in Figure 6–18. One of the resistors, R E2, is bypassed and the other is not. Both resistors ( R E1 + R E2) affect the dc bias while only R E1 affects the ac voltage gain.
Av =
R C
re (^9) + R E
Therefore,
Av =
Rc r (^9) e
833 V
6.98 V
The unloaded gain was found to be 143 in Example 6–6.
Related Problem Determine the base-to-collector voltage gain in Figure 6–16 when a 10 kV load resist- ance is connected from collector to ground. Change the resistance values as follows: R C = 1.8 kV, R E = 1.0 kV, R 1 = 33 kV, and R 2 = 6.8 kV. The emitter resistor is effectively bypassed and r (^9) e = 18.5 V.
Vin
V (^) out
R 1
C 3
C 1
R C
R 2
R E
R E
C 2
▶ (^) FIGURE 6– A swamped amplifier uses a partially bypassed emitter resistance to mini- mize the effect of re on the gain in order to achieve gain stability.
9
The Common-Emitter Amplifier ◆^271
If R E1 is at least ten times larger than r (^9) e , then the effect of r (^9) e is minimized and the ap- proximate voltage gain for the swamped amplifier is
Av @
R C
R E
Equation 6–
Determine the voltage gain of the swamped amplifier in Figure 6–19. Assume that the bypass capacitor has a negligible reactance for the frequency at which the amplifier is operated. Assume re (^9) = 15 V.
EXAMPLE 6–
V CC +10 V
R 1 33 kV
R C 2.7 kV
R E 470 V
R E 220 V
C 3
C 1
R 2 10 kV
1 F
V (^) in
V (^) out
C 2 100 F
1 F
m
m
m
▶ (^) FIGURE 6–
Solution R E2 is bypassed by C 2. R E1 is more than ten times r (^9) e so the approximate voltage gain is
Av >
R C
R E
2.7 kV 220 V
Related Problem What would be the voltage gain without C 2? What would be the approximate voltage gain if R E1 and R E2 were exchanged?
The Effect of Swamping on the Amplifier’s Input Resistance The ac input resist- ance, looking in at the base of a common-emitter amplifier with R E completely bypassed, is Rin = b acr (^9) e. When the emitter resistance is partially bypassed, the portion of the resistance that is unbypassed is seen by the ac signal and results in an increase in the ac input resist- ance by appearing in series with r (^9) e. The formula is
Rin ( base ) 5 B ac ( re (^9) 1 R E1) Equation 6–
For the amplifier in Figure 6–20, (a) Determine the dc collector voltage. (b) Determine the ac collector voltage. (c) Draw the total collector voltage waveform and the total output voltage waveform.
EXAMPLE 6–
The Common-Emitter Amplifier ◆^273
(a) The ac analysis is based on the ac equivalent circuit in Figure 6–22.
R E 470 V
R 1 || R 2 V (^) s 8.25 kV 10 mV
b ac = 175
V (^) b
R (^) c =
R C R (^) L R C + R (^) L
R (^) s 600 V
▶ (^) FIGURE 6–
AC equivalent for the circuit in Figure 6–20.
The first thing to do in the ac analysis is calculate re 9.
r (^9) e >
25 mV I E
25 mV 1.06 mA
= 23.6 V
Next, determine the attenuation in the base circuit. Looking from the 600 V source, the total R (^) in is
Rin ( tot ) = R 1 7 R 2 7 Rin ( base )
Rin ( base ) = b ac ( r (^9) e + R E1) = 175(494 V) = 86.5 kV Therefore,
Rin ( tot ) = 47 kV 7 10 kV 7 86.5 kV = 7.53 kV
The attenuation from source to base is
Attenuation =
Vs Vb
Rs + Rin ( tot ) Rin ( tot )
600 V + 7.53 kV 7.53 kV
Before Av can be determined, you must know the ac collector resistance Rc.
Rc =
R C RL
R C + RL
(4.7 kV)(47 kV) 4.7 kV + 47 kV
= 4.27 kV
The voltage gain from base to collector is
Av >
Rc R E
4.27 kV 470 V
The overall voltage gain is the reciprocal of the attenuation times the amplifier voltage gain.
A (^9) v = a
Vb Vs
b Av = (0.93)(9.09) = 8.
The source produces 10 mV rms, so the rms voltage at the collector is Vc = A (^9) vVs = (8.45)(10 mV) = 84.5 mV (b) The total collector voltage is the signal voltage of 84.5 mV rms riding on a dc level of 4.74 V, as shown in Figure 6–23(a), where approximate peak values are determined as follows: Max Vc ( p ) = V C + 1.414 Vc = 5.02 V + (84.5 mV)(1.414) = 5.14 V Min Vc ( p ) = V C - 1.414 Vc = 5.02 V - (84.5 mV)(1.414) = 4.90 V The coupling capacitor, C 3 , keeps the dc level from getting to the output. So, Vout is equal to the ac component of the collector voltage ( Vout ( p ) = (84.5 mV)(1.414) = 119 mV),
274 ◆^ BJT Amplifiers
as indicated in Figure 6–23(b). The source voltage, V (^) s , is shown to emphasize the phase inversion.
Related Problem What is A (^) v in Figure 6–20 with RL removed?
Open the Multisim file EXM06-09 or the LT Spice file EXS06-09 in the Examples folder on the website. Measure the dc and the ac values of the collector voltage and compare with the calculated values.
(a) Total collector voltage
+119 mV
0 V
–119 mV (b) Source and output ac voltages
V (^) out V (^) s
5.14 V
5.02 V
4.90 V
V (^) c
▶ (^) FIGURE 6– Voltages for Figure 6–20.
Current Gain
The focus in this section has been on voltage gain because that is the principal use for a CE amplifier. However, for completeness, we end this section with a discussion of current and power gain. The current gain from base to collector is Ic > Ib or b ac. However, the overall current gain of the common-emitter amplifier is
Ai 5
Ic Is
Equation 6–
Is is the total signal input current produced by the source, part of which ( Ib ) is base current
and part of which ( Ibias ) goes through the bias circuit ( R 1 7 R 2 ), as shown in Figure 6–24. The
source “sees” a total resistance of Rs + Rin ( tot ). The total current produced by the source is
Is =
Vs Rs + Rin ( tot )
R 1 || R 2
R (^) c I (^) bias
I (^) s
R (^) s
V (^) s
I (^) b Ic
R (^) in(tot) V (^) b
▶ (^) FIGURE 6–24 V (^) c Signal currents (directions shown are for the positive half-cycle of Vs ).
