Chapter 3:BJT Amplifiers
Amplifier Operations
Introduction
Amplifiers are some of the most widely used circuits that you will encounter. They are used extensively in audio, video, telecommunications systems, digital systems, biomedical systems and so on. The most obvious example is when you turn up the volume of a stereo. What actually happens is that you are taking a relatively weak signal and making it stronger i.e. increasing its power level. This is a form of amplification. Amplifiers are circuits that provide amplification. Some examples are shown in figure 4.1.
Figure 4.1
BJTs are introduced here 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.
The biasing of a transistor is purely a dc operation. The purpose of biasing is to establish a Q-point about which variations in current and voltage can occur in response to an ac signal. In applications where small signal voltages must be amplified – such as from a microphone or an antenna – variations about the Q-point are relative small. Hence, small-signal amplifiers are the one of the simplest BJT amplifiers to consider.
As we shall see, there are three types of BJT amplifiers: common-emitter, common-collector and common-base. The most commonly used one is the common-emitter amplifier.
Linear Amplifier Operation
In an amplifier, there are two sources of currents and voltages: dc and ac.
(1)The dc currents and voltages are used to bias the transistor, and
(2)the ac currents and voltages come from the small signal that we wish to amplify.
The dc and ac currents and voltages are mixed together in the circuit. Hence, when we analyze the operation of an amplifier, we need to separate the dc component from the ac component.
Q: How does a capacitor separate ac voltages from dc voltages?
A: Recall that the capacitor’s reactance is XC = 1 / 2 f C where f is the frequency of the voltage that passes through the capacitor. For dc voltages, f = 0 and hence XC. For dc voltages, the capacitor acts as an open. For ac components with f that is large enough, XC 0.
Figure 4.2 below shows a voltage-divider biased transistor with a sinusoidal ac source capacitively coupled to the base through C1 and a load capacitively coupled to the collector through C2.
Figure 4.2
The coupling capacitors block dc and prevent the internal source resistance RS and the load resistance RL from changing the dc bias voltages at the base and collector. The sinusoidal source voltage causes the base voltage to vary sinusoidally above and below its dc bias level. The resulting variation in base current produces a larger variation in the collector current because of the current gain of the transistor.
Note that the voltage created at the collector (VCE) is out-of-phase with the input voltage. This is peculiar to the type of amplifier shown above, which is known as a common-emitter amplifier. The other two types of BJT amplifiers do not exhibit this behaviour.
Q: In a common-emitter amplifier, why is the output voltage out-of-phase with the input voltage?
A: It can be explained as follows: As IC increases, the collector voltage decreasesand vice versa. This is caused by the relationship VCE = VCC – IC(RC + RE). Note the negative sign on the right-hand-side of the equation. This sign cause VCE and IC to be out-of-phase.
Transistor AC Equivalent Circuits
To visualize the operation of a transistor in an amplifier circuit, it is often useful to represent the device by an ac equivalent circuit. The circuit uses internal transistor parameters to represent the transistor’s operation. There are two types of equivalent circuits:
i)Based on resistance (r) parameters.
ii)Based on hybrid (h) parameters.
r - parameters
We will concentrate on the first case (r – parameters). The r – parametersare:
i)ac : ac alpha ()
ii)βac : ac beta ()
iii): ac emitter resistance
iv): ac base resistance
v) : ac collector resistance
An r – parameterequivalent circuit for a BJT (i.e., neglecting all the external resistances) is shown below in figure 4.3(a). All resistances are internal to the transistor. For most general analysis work, we do not need to work with figure 4.3(a) because of complexity. Figure 4.3(a) is the complete equivalent circuit. Figure 4.3(b) is a simplified r – parameter equivalent circuit and is most useful to us. To go from figure 4.3(a) to figure 4.3 (b), we assume that:
i) is small enough to be neglected replace r’b with a short.
ii)is very large (hundreds of k) replace r’c with an open.
Figure 4.3
With figure 4.3(b), we interpret the simplified equivalent circuit, figure 4.4, as follows:
i)is the resistance seen looking into the emitter of a forward-biased transistor. It appears between the base and the emitter terminals.
ii)The collector current Ic is equal to αac Ie. Also, Ic = βac Ib.
Figure 4.4
The most important parameter in the simplified equivalent circuit is . To determine it, we use the following formula:
=.
This formula is just an approximation. We show its derivation below.
Derivation of =
The equation for the current in a pn junction, in this case the base −emitter pn junction, is:
IE = IR (eVQ / kT– 1)
where
IE = total forward current across the base-emitter junction
IR= reverse saturation current
V = voltage across the depletion layer
Q = the charge on an electron
k = Boltzmann’s constant
T = absolute temperature
At ambient temperature, Q / kT ≈ 40, so
IE = IR (e40V– 1)
Differentiating the above equation yields
= 40 IRe40V
But IR e40V = IE + IR (from IE = IR (e40V– 1)), so
= 40 (IE + IR)
Assume IRIE, then
≈ 40IE
Note, however, that the ac resistance of the base – emitterjunction can be expressed as the change in emitter current with respect to the voltage across the depletion region. That is = . Thus
= = =
hParameters
Data sheets do not provide r – parameters. Rather, they provide hparameters (hybrid parameters). These parameters are listed next. In parenthesis are the conditions required to get these parameters.
i)hi:Input impedance (Output shorted)
ii)hr:Voltage feedback ratio (Input open)
iii)hf:Forward current gain (Output shorted)
iv)ho:Output admittance (Input open)
Each of these parameters appears in the three possible configurations for an amplifier: common-emitter, common-base, and common-collector. A second subscript is added to indicate which configuration we are working with.
For example, the hparameters for the common-emitter (usually the one given in data sheets) configuration are: hie, hre, hfe, and hoe. The conversion equations between r and h parameters are:
1
ac =
ac=
=
=
=
1
Common-Emitter (CE) Amplifier
Figure 4.5 below shows a common-emitter BJT amplifier with voltage – dividerbias and coupling capacitors C1 and C3 on the input and output and a bypass capacitor C2 from the emitter to ground. The circuit has a combination of dc and ac operation, both of which must be considered. The input ac signal Vin is capacitively coupled into the base and the output signal Vout is capacitively coupled from the collector.
Figure 4.5
CE amplifiers exhibit high voltage and current gains and are the most common type of BJT amplifiers. The input is applied to base of transistor and the output is taken from collector.
It is called common-emitter because:
i)The emitter is common to both input and output circuits.
ii)The emitter terminal of the transistor is normally set to 0 Vac and thus is at ac ground. The ac ground is provided by the “bypass capacitor” (C2 in circuit above) connected to the emitter terminal of the transistor.
If what is mentioned in (i) and (ii) does not make much sense to you, do not worry as they will be explained below. After going through the notes below come back and read it again.
DC Analysis
Remember that dc voltages in an amplifier circuit are used to bias the transistor. Hence, let us determine the dc bias values. To do this, a dc equivalent circuit is developed by replacing the coupling and bypass capacitors with opens (remember, a capacitor is open to dc), as shown in figure 4.6:
Figure 4.6
As seen previously, the input resistance is
RIN(base) = βDC RE = (150) (560 Ω) = 84 kΩ
which is 10 times more than R2 for this particular case. Thus, we can ignore it. This yields the following base voltage:
And the emitter voltage is
VE = VB – VBE = 2.83 V – 0.7 V = 2.13 V
Therefore, the emitter current is
IE = VE / RE = 2.13 V / 560 Ω = 3.80 mA
Since IC ≈ IE, we get the collector current
VC = VCC – ICRC= 12 V – (3.80 mA) (1.0 kΩ) = 8.20 V
Finally, the common-emitter voltage is
VCE = VC – VE = 8.20 V – 2.13 V = 6.07 V
AC equivalent circuit
To analyze the ac operation of the amplifier, we develop an ac equivalent circuit by using the following rules:
1)Short all capacitors because we assume that their reactance XC 0 at the signal frequency.
2)Replace all dc sources with a ground symbol.
These rules apply to all amplifier circuits, not just common-emitter amplifiers.
For the CE circuit above, the capacitors C1, C2 and C3 are replaced with shorts. The dc sources are replaced with ground (assuming the internal resistance is very small). Therefore, the VCC terminal is at zero – voltac potential, i.e., at ac ground. AC and dc grounds are both assumed to be at the same potential (0 V).
The emitter bypass capacitor, C2 in figure 4.5, provides an effective short to the ac signal around the emitter resistor, thus keeping the emitter at ac ground. This allows the amplifier to have maximum gain (RC /). It must be large enough so that its reactance over the frequency range is very small compared to RE. A good rule – of – thumb is
10XCRE
The ac equivalent circuit for the CE amplifier is shown below in figure 4.7(a). Notice that both RC and R1 have one end connected to ac ground (where in the original circuit, they are connected to VCC).
Figure 4.7
If the internal resistance, Rsof the ac source is 0 , then all of the source voltage appears at the base terminal.
If the ac source internal resistance Rs is non – zero, we need to take it into account, as shown in figure 4.7(b). In this case, three factors must be taken into account in determining the actual signal voltage at the base. They are
(i)the source resistance (Rs),
(ii)the bias resistance (R1 || R2), and
(iii)the input resistance (Rin(base)).
Input Resistance, Rin(base)
Note that Rin(base)is an ac quantity (in contrast to RIN(base) which is a dc quantity) and it isalso known as the input impedance. This is shown in figure 4.8(a) below. The simplified circuit is shown in figure 4.8(b):
Figure 4.8
As you can see, the source voltage Vs is divided down by Rs and Rin(tot) so that the signal voltage at the base of the transistor Vb is
If RsRin(tot), then Vb ≈ Vs.
To develop an expression for the input resistance Rin(base)as seen by an ac source looking in at the base, we will use the simplified r – parametermodel shown in figure 4.9:
Figure 4.9
The figure above shows the transistor connected with the external collector resistor RC.The input resistance looking in at the base is
Rin(base)= Vin / Iin = Vb / Ib
The base voltage is
Vb = Ie
and the base current
Ib≈ Ic/ βac.
Substituting for Vband Ib in the first equation above and since Ie ≈ Ic, we get
Rin(base) = βac
Hence, the total input resistance seen by the source Vbis R1|| R2 || Rin(base)
Rin(tot) = R1|| R2 || Rin(base)
Example
Determine the signal voltage at the base of the transistor in the circuit shown below. This circuit is the ac equivalent of the amplifier (also shown next). Assume a 10 mV, 300 Ω signal source. IE was already found to be 3.80 mA.
Solution
We need to determine the ac emitter resistance.
= = 25 mV / 3.80 mA = 6.58 Ω
Then,
Rin(base) = βac = 160 (6.58 Ω) = 1.05 kΩ
Next determine the total input resistance viewed from the source:
Rin(tot) = R1|| R2|| Rin(base)
=(1 / 22kΩ + 1 / 6.8 kΩ + 1.05 kΩ)−1
= 873 Ω
The source voltage is divided down by Rsand Rin(tot), so the signal voltage at the base is the voltage across Rin(tot).
=
So there is attenuation of the source voltage due to the source resistance and amplifier’s input resistance acting as a voltage divider. So, instead of getting the full 10 mV at the base, we get only 7.44 mV.
Voltage Gain of the CE Amplifier
Voltage gain (Av) of the CE amplifier is the ratio between the ac output voltage (Vc)and input voltage (Vb). We use the simplified r – parameterequivalent circuit to find the voltage gain as shown in figure 4.10:
Figure 4.10
From the figure above, we see that
Vb = Ie
Vc = acIeRC ≈ IeRC
since αac 1. The voltage gain is then
The gain Av is the voltage gain from the base to collector.
To get the overall gain of the amplifier from the source voltage to collector voltage, the attenuation of the input circuit must be included. Attenuation means that the signal voltage is reduced as it passes through the circuit.
This signal reduction is due to the internal source resistance Rs. The attenuation from source to base multiplied by the gain from base to collector is called the overall amplifier gain. The overall gain is illustrated in the figure 4.11:
Figure 4.11
The attenuation in the example we had before is equal to 7.44 mV / 10 mV = 0.744. Thus, the overall amplifier gain, , would be = (0.744)Av.
Remember to always put a bypass capacitor in the emitter.
Example
Calculate the base – to – collectorvoltage gain of the amplifier shown next, with and without an emitter bypass capacitor. There is no load resistor.
Solution
From a previous example we know that = 6.58 Ω. Without C2, the gain is
Av = RC / ( + RE) = 1 kΩ / 566.58 Ω = 1.76
With C2 included, the gain is
Av = RC / = 1 kΩ / 6.58 Ω = 152
The bypass capacitor makes a big difference!
A few more notes:
i)The effect of the load in the gain of an amplifier can be analyzed as follows. If a load of resistance RL is connected across the output of the amplifier, the voltage gain may or may not be affected. When a load is added, RL is connected parallel to RC. The total resistance is then Rtot= RCRL / (RC + RL). The voltage gain can then be written as
Av = Rtot /
If the load resistance RLRC, then Rtot RC and there is no change in the gain.But if RLRC, then Rtot = RL. The voltage gain is reduced. Ideally, we require RL to be as large as possible.
ii)As we have seen in the example above, bypassing RE produces the maximum voltage gain. But it also produces a stability problem. Since the ac voltage gain is dependent on (from Av = RC / ) and depends on IE and on temperature, the gain will be unstable over temperature changes.
If we take the bypass off IE, the gain is decreased but RE overpowers in the gain calculation. This actually makes the circuit much less dependent on it. We make a compromise by using a method called swamping. In swamping, we only partially bypass REas shown in figure 4.12:
Figure 4.12
Note that the total external emitter resistance RE is formed by two separate emitter resistors RE1 and RE2. One of the resistors RE2 is bypassed and the other is not. Both the resistors (RE1 + RE2) affect the dc bias but only RE1 affects the ac voltage gain:
Av = RC / ( + RE1)
If we make RE1 at least ten times larger than (RE1 > 10), then the effect of is minimized and the approximate voltage gain for the swamped amplifier is
Av ≈ RC / RE1
Example
For the amplifier shown next, determine the total collector voltage and the total output voltage, both dc and ac. Draw the waveforms.
Solution
We need to solve this in three steps:
Step 1) DC Analysis
Determine the dc bias values. For this, we need a dc equivalent circuit for the amplifier above. That is, let each capacitor be open.
RIN(base) = βDC (RE1 + RE2) = 150 (940 Ω) = 141 kΩ
Since RIN(base) > 10R2, it can be neglected in the dc base voltage calculation.
VB ≈ R2 / (R1+R2)VCC =(10kΩ) / (47kΩ + 10kΩ) 10V = 1.75 V
VE = VB – 0.7 V = 1.75 V – 0.7 = 1.05 V
IE= VE / (RE1 + RE2) = 1.05 V / 940 Ω = 1.12 mA
VC = VCC − ICRC = 10 V – (1.12 mA) (4.7 kΩ) = 4.74 V
Step 2) AC Analysis
The ac analysis is based on the ac equivalent circuit shown next.
The first thing to do in an ac analysis is calculate .
= 25 mV / IE = 25 mV / 1.12 mA = 22 Ω
Next determine the attenuation in the base circuit. Looking from the 600 Ω source resistance, the total Rin is
Rin(tot) = R1|| R2|| Rin(base)
Rin(base) = βac ( + RE1) = 175 (492 Ω) = 86.1 kΩ
Rin(tot) = 47 kΩ || 10 kΩ || 86.1 kΩ = 7.53 kΩ
The attenuation from source to base is
Attenuation = Vb / Vs = Rin(tot) /(Rs + Rin(tot))
= 7.53 kΩ / (600 Ω + 7.53 kΩ) = 0.93
Before Avcan be determined, we need to know the ac collector resistance:
Rc = RCRL / (RL + RC) = (4.7 kΩ) (47 kΩ) / (4.7 kΩ + 47 kΩ)
= 4.27 kΩ
Now we are ready to calculate the gain from base to collector:
Av≈ Rc /RE1 = 4.27 kΩ / 470 Ω = 9.09
And the overall voltage gain is the attenuation times the amplifier voltage:
A’v = (Vb/ Vs) Av= (0.93) (9.09) = 8.45
Since the source produces 10 mVrms, the rms voltage at the collector will be
Vc = A’v Vin= (8.45) (10 mV) = 84.5 mV
Step 3) Plot waveforms
The total collector voltage is the signal voltage of 84.5 mVrms riding on a dc level of 4.74 V. This is shown in the next graph. The peaks are
Max Vc(p)= 4.74 + (84.5 mV) (1.414) = 4.86 V
MinVc(p)= 4.74 - (84.5 mV) (1.414) = 4.62 V
The coupling capacitor C3 keeps the dc level from getting to the output, so Vout is equal to the ac portion of the collector voltage (Vout(p)= 119 mV). Source voltage is shown to emphasize phase inversion.
Common-Collector (CC) Amplifier
The common-collector amplifier is also called the emitter-follower amplifier (EF). The input is applied to base through a coupling capacitor and the output is at the emitter. The voltage gain of a CC amplifier is approximately 1 (i.e. the voltage amplitude is roughly the same in both the input and the output). Its main advantage is its high input resistance and current gain.
Figure 4.13
The circuit in figure 4.13 shows the CC amplifier which is voltage – dividerbiased. Notice that the input signal is capacitively coupled to the base, the output signal is capacitively coupled to the emitter. There is no phase inversion at the output and the output is approximately the same amplitude as the input.
Q: In the common-collector amplifier, why is the output voltage in-phase with the input voltage?
A: Since the output voltage is at the emitter, it is in phase with the base voltage. So there is no inversion from input to output. Because there is no inversion and because the voltage gain is approximately 1, the output voltage closely follows the input voltage in phase and amplitude, thus the term emitter-follower amplifier.
Voltage Gain
The voltage gain is . From the ac equivalent model shown in figure 4.14, we see that
Vout = Ie Re
and
Vin = Ie (+ Re)