Chapter 3:
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 decreases and 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 – parameters are:
i) αac : ac alpha ([pic])
ii) βac : ac beta ([pic])
iii) [pic] : ac emitter resistance
iv) [pic] : ac base resistance
v) [pic] : ac collector resistance
An r – parameter equivalent 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) [pic] is small enough to be neglected ( replace r’b with a short.
ii) [pic] 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) [pic] 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 [pic]. To determine it, we use the following formula:
[pic] = [pic].
This formula is just an approximation. We show its derivation below.
Derivation of [pic] = [pic]
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
[pic] = 40 IR e40V
But IR e40V = IE + IR (from IE = IR (e40V – 1)), so
[pic] = 40 (IE + IR)
Assume IR RC, then Rtot ( RC and there is no change in the gain. But if RL > 10[pic]), then the effect of [pic] 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 [pic].
[pic] = 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 ([pic] + 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 Av can 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
Min Vc(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 – divider biased. 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 [pic]. From the ac equivalent model shown in figure 4.14, we see that
Vout = Ie Re
and
Vin = Ie ([pic] + Re)
Figure 4.14
Thus, the gain is
[pic]
Note that
Re = RE || RL
and
Re = RE
when there is no load.
In practice, Re >> [pic], hence
Av ≈ 1
that is, the voltage gain of a CC amplifier is approximately unity.
Input Resistance Rin(base)
The CC amplifier has high input resistance. This is what makes it a useful circuit. Because of this high input resistance, it can be used to minimize loading effects when a circuit is driving a low resistance load. Note that the emitter resistance is never bypassed (as was the case in the common-emitter amplifier). The derivation of the CC amplifier input resistance is similar to that of the CE amplifier:
Rin(base) = Vin / Iin = Vb / Ib = [pic]
Since Ie ≈ Ic = βac Ib
Rin(base) = [pic]
= βac ([pic] + Re)
If Re >> [pic], the input resistance becomes
Rin(base) = βac Re
From the CC amplifier circuit, we can see that the bias resistors appear in parallel with Rin(base), looking from the input source. Thus, the total input resistance becomes
Rin(tot) = R1 || R2 || Rin(base)
Output Resistance
The output resistance is very low. It is given by the following expression:
Rout [pic]
Where Rs is the resistance of the input source.
FYI
[pic] [pic]
Derivation (Floyd 920):
The output of the transistor is taken at node E, thus, across RE. The emitter current is[pic]. With Vs = 0 and with Ib produced by Vout, and neglecting the base-to-emitter voltage drop (and therefore [pic]),
Ib = Ve / (R1 || R2 || Rs)
Assuming R1 >> R2 >> Rs,
Ib ≈ Ve / Rs
Iout = Ie = βac Ve / Rs
Vout / Iout = Ve / Ie = Ve / (βac Ve / Rs) = Rs / βac
Looking into the emitter, RE appears in parallel with Rs / βac. Therefore
Rout = (Rs / βac) || RE
FYI
Current Gain
The current gain is [pic]. The input current is given by Iin = Vin / Rin(tot). If [pic], then most of input current goes into the base. Thus, current gain of amplifier is almost equal to current gain of the transistor, βac, that is
Ai ( βac = [pic].
This is because very little signal is diverted to the bias resistors.
Otherwise,
[pic]
βac is the maximum achievable current gain in both CC and CE amplifiers.
Recall Ie ( βacIb and Iin = Ib.
Power Gain
The CC power gain is
[pic].
Since Av ≈ 1, the overall power gain is
Ap ≈ Ai.
Example
Determine the total input resistance of the emitter follower shown below. Also find the voltage gain, current gain, and power gain in terms of power delivered to the load, RL. Assume βac = 175. And that the capacitive reactances are negligible at the frequency of operation.
Solution
The ac emitter resistance external to the transistor, Re, is:
Re = RE || RL = 1 kΩ || 1kΩ = 500 Ω
The approximate resistance, looking in at the base, is:
Rin(base) ≈ βac Re = (175) (500 Ω) = 87.5 kΩ
The total input resistance is
Rin(tot) = R1 || R2 || Rin(base) = 18kΩ || 18kΩ || 87.56kΩ = 8.16 kΩ
The voltage gain is almost unity (since it is a CC amplifier). By using [pic] we can determine a more precise value for Av:
VE = R2 / (R1 + R2) VCC – VBE = (0.5) (10 V) – 0.7 V = 4.3 V
IE = VE / RE = 4.3 V / 1.0 kΩ = 4.3 mA
[pic] = 25 mV / IE = 25 mV / 4.3 mA = 5.8 Ω
So
Av = Re / ([pic] + Re) = 500 Ω / 505.8 Ω = 0.989
The small difference in Av as a result of considering [pic] is insignificant in most cases.
The current gain is Ai = Ie / Iin:
Ie = Ve / Re = Av Vb / Re ≈ 1 V / 500 Ω = 2 mA
Iin = Vin / Rin(tot) = 1 V / 8.16 kΩ = 123 μA
Ai =Ie / Iin = 2 mA / 123 μA = 16.3
The overall power gain is
Ap ≈ Ai = 16.3
Since RL = RE, one half of the total power is dissipated in RL. So, in terms of power to the load, the power gain is one half of the overall power gain.
Ap(load) = Ap / 2 = 16.3 / 2 = 8.15
The Darlington Pair
As we have seen, βac is a major factor in determining the input resistance of an amplifier. The βac of a transistor limits the maximum achievable input resistance you can get from a given CC circuit. One way to boost the input resistance is to use a darlington pair as shown in figure 4.15.
In a darlington pair, the collectors of the two transistors are connected and the emitter of the first drives the base of the second. This configuration achieves βac multiplication, i.e.,
Ie2 = βac2 βac1 Ib1.
Thus, the effective current gain in the darlington pair is
(ac = βac2 βac1
Neglecting r’e by assuming that it is much smaller than RE, the input resistance is
Rin = βac2 βac1RE
Figure 4.15
As an application, CC is often used as an interface between a circuit with a high output resistance and a low resistance load. In such an application, the CC circuit is called a buffer.
For example, suppose a common-emitter amplifier with a 1.0 kΩ collector resistance (output resistance) must drive a low-resistance load such as an 8 Ω low-power speaker. If the speaker is capacitively coupled to the output of amplifier, the 8 Ω load (appears to the ac signal) in parallel with the 1 kΩ collector resistor. This results in an ac collector resistance of
Rc = RC || RL = 1kΩ || 8Ω = 7.94 Ω.
Obviously this is not acceptable, since most of the voltage gain is lost (Av = Rc / [pic]). For example, if [pic]= 5 Ω, the voltage gain is reduced from
Av = RC / [pic] = 1 kΩ / 5 Ω = 200
to
Av = Rc / [pic] = 7.94 Ω / 5 Ω = 1.59
We can add a darlington pair to interface the amplifier and the speaker as shown in the circuit in figure 4.16 below.
Figure 4.16
Example
The circuit below shows a common-emitter stage driving a darlington pair connected as an emitter follower. The β values for the silicon transistors are β1 = 200, β2 = 100, and β3 = 100. Q2 and Q3 are the first and second stage of the darlington pair, respectively.
a) Find gain of the whole circuit, Av = vL / vS.
b) Find the gain Av = vL / vS if the darlington pair is removed and the 100 Ω load is capacitor coupled to the collector of Q1.
Solution
a) The bias voltage at the base of Q1 results from the voltage divider bias:
Vth ≈ 10 kΩ / (47 kΩ + 10 kΩ) (15 V) = 2.6 V
RB = 10 kΩ || 47 kΩ = 8.25 kΩ
IC1 ≈ IE1 ≈ (2.6 – 0.7) / (68 Ω + 910 Ω + 8.25kΩ / 200)
= 1.9 mA
[pic]1 ≈ 25 mV / IE1 = 25 mV / 1.9 mA = 13.7 Ω
rin = RB || β1([pic]1 + Re) = (8.25 kΩ) || (13.7 Ω + 68 Ω)
= 5.48 kΩ
Notice the collector of Q1 is direct-coupled to the base of Q2 in the darlington pair. So,
VB2 = VC1 − IC1RC1 = 15 V – (1.9 mA) (3.3 kΩ) = 8.7 V
The dc emitter voltage of Q3 is about 1.4 V less that the base voltage of Q2, since there are two forward biased base-emitter junctions between those two points:
VE3 = VB2 – 1.4 V = 8.7 V – 1.4 V = 7.3 V
IE3 = VE3 / RE3 = 7.3 V / 47 Ω = 155 mA
[pic]3 = 25 mV / IE3 = 25 mV / 155 mA = 0.17 Ω
The internal emitter resistance of the darlington pair is twice that of the [pic]3
[pic](dp) = 2[pic]3 = 2 (0.17 Ω) = 0.34 Ω
The gain of the darlington pair is βdp = β1 β2 = 10000. The total internal resistance, looking into the darlington pair, is:
ri = βdp ([pic](dp) + 47 || 100)
= 10000 (0.34 Ω + 47 Ω || 100 Ω)
= 323 kΩ
The input resistance is large enough in comparison to the 3.3 kΩ collector resistance of Q1. No loading will occur. Thus, the gain Av1 becomes
Av1 ≈ Rc1 / ([pic]1 + 68 Ω) = 3.3 kΩ / 81.7 Ω = 40.4
The voltage gain of the emitter-follower stage is
Av(dp) ≈ (RL || RE3) / ((RL || RE3) + [pic](dp))
= (47 Ω || 100 Ω) / (0.34 Ω + (47 Ω || 100 Ω) = 0.99
= 32 Ω / (0.34 Ω + 32 Ω) = 0.99
Taking into account the gain drop caused by the source resistance, RS = 5 kΩ, the overall gain is
vL / vS = [rin / (rS + rin)] Av1 Av(dp)
= 5.48 kΩ / (5 kΩ + 5.48 kΩ) (40.4) (0.99)
= 20.9
b) With the 100 Ω load connected to the collector of Q1, the ac load on Q1 is
rL1 = (3.3 kΩ) || (100 Ω) ≈ 97 Ω.
Therefore,
Av1 ≈ rL1 / ([pic] + 68 Ω) = 97 / 81.7 = 1.2
The overall gain is, then
vL / vS = [rin / (rS + rin)] Av1
= 5.48 kΩ / (5 kΩ + 5.48 kΩ) (1.2) = 0.63
Note that the output wave will be shifted by 180o with respect to the input wave. This is because the first stage is a common emitter amplifier. Another way of indicating this is with a minus sign in the gain. For example, in the second case the overall gain would be vL / vS = –0.63.
Common-Base (CB) Amplifier
The CB amplifier provides high voltage gain with a maximum current gain of 1. Since it has low input resistance, CB amplifiers are most appropriate for certain applications where sources tend to have very low-resistance outputs.
A typical CB amplifier is shown in figure 4.17 below. The base is the common terminal and is at ac ground because of the capacitor C2. The input signal is capacitively coupled to the emitter. The output is capacitively coupled from the collector to a load resistor.
Figure 4.17
Voltage Gain
Since the input voltage is Vin = Ve and the output voltage is Vout = Vc, the gain becomes
Av = [pic]
= [pic]
≈ [pic]
Assuming that RE >> [pic], we get
Av ≈ [pic]
≈ (RC || RL) / [pic]
where Rc = RC || RL. Notice that the gain expression is the same as for the common emitter amplifier. However, there is no phase inversion from emitter to collector.
Input Resistance
The resistance looking into the emitter is
Rin(emitter) = [pic]
Assuming RE >> [pic], then
Rin(emitter) ≈ [pic]
Typically, RE is much greater than [pic], so the assumption RE >> [pic] used is usually valid.
Output Resistance
Looking into the collector, the ac collector resistance, [pic], appears in parallel with RC. Just like the CE amplifier, [pic] is much larger than RC. So a good approximation is
Rout ≈ RC
Current Gain
It is defined as Ai = Iout / Iin. Thus we have
Ai = Iout / Iin = Ic / Ie ≈ 1
Power Gain
Since the current gain is approximately unity, by Ap = Ai Av, we have
Ap ≈ Av
Example
Find the input resistance, voltage gain, current gain, and power gain for the amplifier shown next.
Solution
Firstly, find IE so that [pic] can be determined. Then Rin ≈ [pic]. Since βDCRE >> R2, then
VB = R2 / (R1 + R2) VCC = (12 kΩ) 68 kΩ) 10 V = 1.76 V
VE = VB – 0.7 V = 1.76 V – 0.7 = 1.06 V
IE = VE / RE = 1.06 V / 1.0 kΩ = 1.06 mA
Therefore,
Rin ≈ [pic] = 25 mV / IE = 25 mV / 1.06 mA = 23.6 Ω
The voltage gain is found as follows.
Rc = RC || RL = 2.2 kΩ || 10 kΩ = 1.8 kΩ
Av = Rc / [pic] = 1.8 kΩ / 23.6 Ω = 76.3
We also know Ai = 1 and Ap ≈ Av = 76.3.
Amplifier Comparisons
| |CE |CC (or EF) |CB |
|Input |Base |Base |Emitter |
|Output |Collector |Emitter |Collector |
|Inversion |Yes |No |No |
|Voltage gain |High |Low |High |
| |RC/r’e |≈ 1 |RC/r’e |
|Current gain |High |High |Low |
| |βac |βac |≈ 1 |
|Power gain |Very High |High |High |
| |AiAv |≈ Ai |≈ Av |
|Input Resistance |Low |High |Very low |
| |βacr’e |βacRE |r’e |
|Output Resistance |High |Very low |High |
| |RC |(Rs/βac)||RE |RC |
|Frequency Range |Medium |Medium |High |
Chapter Summary
1) A small signal amplifier uses only a small portion of its dc load line.
2) r-parameters are easily identifiable and applicable with a transistor’s circuit operation.
3) A common emitter amplifier has good voltage, current, and power gains, but a relatively low input resistance.
4) A common collector amplifier has high input resistance and good current gain, but its voltage gain is approximately 1.
5) The common base amplifier has a good voltage gain, but it has a very low input resistance and its current gain is approximately 1.
6) A darlington pair provides β multiplication for increased input resistance.
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[pic]
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