Introduction of Single Stage Transistor Amplifier
it was discussed that a properly biased transistor raises the strength of a weak signal and thus acts as an amplifier. Almost all electronic equipment must include means for amplifying electrical signals. For instance, radio receivers amplify very weak signals—sometimes a few millionths of a volt at antenna–until they are strong enough to fill a room with sound. The transducers used in the medical and scientific investigations generate signals in the microvolt (µV) and millivolt (mV) range. These signals must be amplified thousands and millions of times before they will be strong enough to operate indicating instruments. Therefore, electronic amplifiers are a constant and important ingredient of electronic systems. Our purpose here will be to discuss single stage transistor amplifier. By a stage, we mean a single transistor with its bias and auxiliary equipment. It may be emphasized here that a practical amplifier is always a multistage amplifier i.e. it has a number of stages of amplification. However, it is profitable to consider the multistage amplifier in terms of single stages that are connected together. In this article, we shall confine our attention to single stage transistor amplifiers.
What is Single Stage Transistor Amplifier
When only one transistor with associated circuitry is used for amplifying a weak signal, the circuit is known as single stage transistor amplifier. A single stage transistor amplifier has one transistor, bias circuit and other auxiliary components. Although a practical amplifier consists of a number of stages, yet such a complex circuit can be conveniently split up into separate single stages. By analysing carefully only a single stage and using this single stage analysis repeatedly, we can effectively analyse the complex circuit. It follows, therefore, that single stage amplifier analysis is of great value in understanding the practical amplifier circuits.
How Transistor Amplifies ?
Fig. 10.1 shows a single stage transistor amplifier. When a weak a.c. signal is given to the base of transistor, a small base current (which is a.c.) starts flowing. Due to transistor action, a much larger (β times the base current) a.c. current flows through the collector load RC. As the value of RC is quite high (usually 4-10 kΩ), therefore, a large voltage appears across RC. Thus, a weak signal applied in the base circuit appears in amplified form in the collector circuit. It is in this way that a transistor acts as an amplifier. The action of transistor amplifier can be beautifully explained by referring to Fig. 10.1. Suppose a change of 0.1V in signal voltage produces a change of 2 mA in the collector current. Obviously, a signal of only 0.1V applied to the base will give an output voltage = 2 mA × 5 kΩ = 10V. Thus, the transistor has been able to raise the voltage level of the signal from 0.1V to 10V i.e. voltage amplification or stage gain is 100.
Graphical Demonstration of Transistor Amplifier
The function of transistor as an amplifier can also be explained graphically. Fig. 10.2 shows the output characteristics of a transistor in CE configuration. Suppose the zero signal base current is 10 µA i.e. this is the base current for which the transistor is biased by the biasing network. When an a.c. the signal is applied to the base, it makes the base, say positive in the first half-cycle and negative in the second half cycle. Therefore, the base and collector currents will increase in the first half-cycle when base-emitter junction is more forward-biased. However, they will decrease in the second half-cycle when the base-emitter junction is less forward biased.
For example, consider a sinusoidal signal which increases or decreases the base current by 5 µA in the two half-cycles of the signal. Referring to Fig. 10.2, it is clear that in the absence of signal, the base current is 10μA and the collector current is 1 mA. However, when the signal is applied in the base circuit, the base current and hence collector current change continuously. In the first half-cycle peak of the signal, the base current increases to 15 µA and the corresponding collector current is 1.5 mA. In the second half-cycle peak, the base current is reduced to 5 µA and the corresponding collector current is 0.5 mA. For other values of the signal, the collector current is in between these values i.e. 1.5 mA and 0.5 mA. It is clear from Fig. 10.2 that 10 µA base current variation results in 1mA (1,000 µA) collector current variation i.e. by a factor of 100. This large change in collector current flows through collector resistance RC. The result is that output signal is much larger than the input signal. Thus, the transistor has done amplification.
Practical Circuit of Single Stage Transistor Amplifier
It is important to note that a transistor can accomplish faithful amplification only if proper associated circuitry is used with it. Fig. 10.3 shows a practical single stage transistor amplifier. The various circuit elements and their functions are described below :
(i) Biasing circuit. The resistances R1, R2 and RE form the biasing and stabilisation circuit. The biasing circuit must establish a proper operating point otherwise a part of the negative half-cycle of the signal may be cut off in the output.
(ii) Input capacitor Cin. An electrolytic capacitor Cin (~10 μF ) is used to couple the signal to the base of the transistor. If it is not used, the signal source resistance will come across R2 and thus change the bias. The capacitor Cin allows only a.c. signal to flow but isolates the signal source from R2.*
(iii) Emitter bypass capacitor CE. An emitter bypass capacitor CE (~ 100µF ) is used in parallel with RE to provide a low reactance path to the amplified a.c. signal. If it is not used, then amplified a.c. signal flowing through RE will cause a voltage drop across it, thereby reducing the output voltage.
(iv) Coupling capacitor CC. The coupling capacitor CC (~10μF) couples one stage of ampli-fication to the next stage. If it is not used, the bias conditions of the next stage will be drastically changed due to the shunting effect of RC. This is because RC will come in parallel with the upper resistance R1 of the biasing network of the next stage, thereby altering the biasing conditions of the latter. In short, the coupling capacitor CC isolates the d.c. of one stage from the next stage, but allows the passage of a.c. signal.
Various circuit currents. It is useful to mention the various currents in the complete amplifier circuit. These are shown in the circuit of Fig. 10.3.
(i) Base current. When no signal is applied in the base circuit, d.c. base current IB flows due to biasing circuit. When a.c. signal is applied, a.c. base current i b also flows. Therefore, with the application of signal, total base current i B is given by:
iB = IB + ib
(ii) Collector current. When no signal is applied, a d.c. collector current IC flows due to biasing circuit. When a.c. signal is applied, a.c. collector current i c also flows. Therefore, the total collector current i C is given by:
iC = IC + ic
where IC = β IB = zero signal collector current
ic = β ib = collector current due to signal.
(iii) Emitter current. When no signal is applied, a d.c. emitter current IE flows. With the application of signal, total emitter current i E is given by:
iE = IE + ie
It is useful to keep in mind that :
IE = IB + IC
ie = ib + ic
Now base current is usually very small, therefore, as a reasonable approximation,
IE ~ IC and ie ~ ic
Example 10.1. What is the role of emitter bypass capacitor CE in CE amplifier circuit shown in Fig. 10.3 ? Illustrate with a numerical example.
Solution. The emitter bypass capacitor CE (See Fig. 10.3) connected in parallel with RE plays an important role in the circuit. If it is not used, the amplified a.c. signal flowing through RE will cause a voltage drop across it, thereby reducing the a.c. output voltage and hence the voltage gain of the amplifier.
Let us illustrate the effect of CE with a numerical example. Suppose RE = 1000Ω and capacitive reactance of CE at the signal frequency is 100Ω (i.e. Xce = 100Ω). Then 10/11 of a.c emitter current will flow through CE and only 1/11 through RE. The signal voltage developed across RE is, therefore, only 1/11 of the voltage which would have been developed if CE were not present. In practical circuits, the value of CE is so selected that it almost entirely bypasses the a.c. signal (the name for CE is obvious). For all practical purposes, we consider CE to be a short for a.c. signals.
Example 10.2. Select a suitable value for the emitter bypass capacitor in Fig. 10.4 if the amplifier is to operate over a frequency range from 2 kHz to 10 kHz.
Solution. An amplifier usually handles more than one frequency. Therefore, the value of CE is so selected that it provides adequate bypassing for the lowest of all the frequencies. Then it will also be a good bypass (XC ∝ 1/f ) for all the higher frequencies. Suppose the minimum frequency to be handled by CE is f min. Then CE is considered a good bypass if at f min,
In common emitter connection, when the input signal voltage increases in the positive sense, the output voltage increases in the negative direction and vice-versa. In other words, there is a phase difference of 180° between the input and output voltage in CE connection. This is called phase reversal.* The phase difference of 180° between the signal voltage and output voltage in a common emitter amplifier is known as phase reversal. Consider a common emitter amplifier circuit shown in Fig. 10.5. The signal is fed at the input terminals (i.e. between base and emitter) and output is taken from collector and emitter end of supply. The total instantaneous output voltage
vCE is given by : **vCE = VCC − iC RC …(i)
When the signal voltage increases in the positive half-cycle, the base current also increases. The result is that collector current and hence voltage drop i C RC increases. As VCC is constant, therefore, output voltage vCE decreases. In other words, as the signal voltage is increasing in the positive halfcycle, the output voltage is increasing in the negative sense i.e. output is 180° out of phase with the input. It follows, therefore, that in a common emitter amplifier, the positive half-cycle of the signal appears as amplified negative half-cycle in the output and vice-versa. It may be noted that amplification is not affected by this phase reversal. The fact of phase reversal can be readily proved mathematically. Thus differentiating exp. (i), we get,
The negative sign shows that output voltage is 180° out of phase with the input signal voltage.
The fact of phase reversal in CE connection can be shown graphically with the help of output characteristics and load line (See Fig. 10.6). In Fig. 10.6, AB is the load line. The base current fluctuates between, say ± 5 µA with 10µA as the zero signal base current. From the figure, it is clear that when the base current is maximum in the positive direction, vCE becomes maximum in the negative direction (point G in Fig. 10.6). On the other hand, when the base current is maximum in the negative direction, vCE is maximum in the positive sense (point H in Fig. 10.6). Thus, the input and output voltages are in phase opposition or equivalently, the transistor is said to produce a 180º phase reversal of output voltage w.r.t. signal voltage. Note. No phase reversal of voltage occurs in common base and common collector amplifier. The a.c. output voltage is in phase with the a.c. input signal. For all three amplifier configurations; input and output currents are in phase.
Example 10.3. Illustrate the phenomenon of phase reversal in CE amplifier assuming typical circuit values.
Solution. In every type of amplifier, the input and output currents are in phase. However, common emitter amplifier has the unique property that input and output voltages are 180° out of phase, even though the input and output currents are in phase. This point is illustrated in Fig. 10.7. Here it is assumed that Q-point value of IB = 10 µA, ac signal peak value is 5 µA and β = 100. This means that input current varies by 5 µA both above and below a 10 µA dc level. At any instant, the output current will be 100 times the input current at that instant. Thus when the input current is 10 µA, output current is i C = 100 × 10 µA = 1 mA. However, when the input current is 15 µA, then output current is i C = 100 × 15 µA = 1.5 mA and so on. Note that input and output currents are in phase.
The output voltage, vC = VCC − iC RC
(i) When signal current is zero (i.e., in the absence of signal), iC = 1 mA.
∴ vC = VCC − i C RC = 10 V − 1 mA × 4 kΩ = 6V
(ii) When signal reaches positive peak value, iC = 1.5 mA.
∴ vC = VCC − i C RC = 10 V − 1.5 mA × 4 kΩ = 4 V
Note that as i C increases from 1mA to 1.5 mA, vC decreases from 6V to 4V. Clearly, output voltage is 180° out of phase from the input voltage as shown in Fig. 10.7.
(iii) When signal reaches negative peak, i C = 0.5 mA.
∴ vC = VCC − i C RC = 10 V − 0.5 mA × 4 kΩ = 8V
Note that as iC decreases from 1.5 mA to 0.5 mA, vC increases from 4 V to 8 V. Clearly, output voltage is 180° out of phase from the input voltage. The following points may be noted carefully about CE amplifier :
(a) The input voltage and input current are in phase.
(b) Since the input current and output current are in phase, input voltage and output current are in phase.
(c) Output current is 180° out of phase with the output voltage (vC). Therefore, input voltage and output voltage are 180° out of phase.
Input/Output Phase Relationships
The following points regarding the input / output phase relationships between currents and voltages for the various transistor configurations may be noted :
(i) For every amplifier type (CE, CB and CC), the input and output currents are in phase. When the input current decreases, the output current also decreases and vice-versa.
(ii) Remember that common emitter (CE) circuit is the only configuration that has input and output voltages 180° out of phase.
(iii) For both common base (CB) and common collector (CC) circuits, the input and output voltages are in phase. If the input voltage decreases, the output voltage also decreases and vice-versa.
D.C. And A.C. Equivalent Circuits
In a transistor amplifier, both d.c. and a.c. conditions prevail. The d.c. sources set up d.c. currents and voltages whereas the a.c. source (i.e. signal) produces fluctuations in the transistor currents and voltages. Therefore, a simple way to analyse the action of a transistor is to split the analysis into two parts viz. a d.c. analysis and an a.c. analysis. In the d.c. analysis, we consider all the d.c. sources at the same time and work out the d.c. currents and voltages in the circuit. On the other hand, for a.c. analysis, we consider all the a.c. sources at the same time and work out the a.c. currents and voltages. By adding the d.c. and a.c. currents and voltages, we get the total currents and voltages in the circuit. For example, consider the amplifier circuit shown in Fig. 10.8. This circuit can be easily analysed by splitting it into d.c. equivalent circuit and a.c equivalent circuit.
(i) D. C. equivalent circuit. In the d.c. equivalent circuit of a transistor amplifier, only d.c. conditions are to be considered i.e. it is presumed that no signal is applied. As direct current cannot flow through a capacitor, therefore, all the capacitors look like open circuits in the d.c. equivalent circuit. It follows, therefore, that in order to draw the equivalent d.c. circuit, the following two steps are applied to the transistor circuit :
(a) Reduce all a.c. sources to zero.
(b) Open all the capacitors.
Applying these two steps to the circuit shown in Fig. 10.8, we get the d.c. equivalent circuit shown in Fig. 10.9. We can easily calculate the d.c. currents and voltages from this circuit.
(ii) A.C. equivalent circuit. In the a.c. equivalent circuit of a transistor amplifier, only a.c. conditions are to be considered. Obviously, the d.c. voltage is not important for such a circuit and may be considered zero. The capacitors are generally used to couple or bypass the a.c. signal. The designer intentionally selects capacitors that are large enough to appear as short circuits to the a.c. signal. It follows, therefore, that in order to draw the a.c. equivalent circuit, the following two steps are applied to the transistor circuit :
(a) Reduce all d.c. sources to zero (i.e. VCC = 0).
(b) Short all the capacitors.
Applying these two steps to the circuit shown in Fig. 10.8, we get the a.c. *equivalent circuit shown in Fig. 10.10. We can easily calculate the a.c. currents and voltages from this circuit.
It may be seen that total current in any branch is the sum of d.c. and a.c. currents through that branch. Similarly, the total voltage across any branch is the sum of d.c. and a.c. voltages across that branch.
Example 10.4. For the transistor amplifier circuit shown in Fig. 10.8, determine :
(i) d.c. load and a.c. load
(ii) maximum collector-emitter voltage and collector current under d.c. conditions
(iii) maximum collector-emitter voltage and collector current when a.c. signal is applied
Solution. Refer back to the transistor amplifier circuit shown in Fig. 10.8.
(i) The d.c. load for the transistor is Thevenin’s equivalent resistance as seen by the collector and emitter terminals. Thus referring to the d.c. equivalent circuit shown in Fig. 10.9, Thevenin’s equivalent resistance can be found by shorting the voltage source (i.e. VCC) as shown in Fig. 10.11. Because a voltage source looks like a short, it will bypass all other resistances except RC and RE which will appear in series. Consequently, transistor amplifier will see a d.c. load of RC + RE i.e
(iii) When no signal is applied, VCE and IC are the collector-emitter voltage and collector current respectively. When a.c. signal is applied, it causes changes to take place above and below the operating point Q (i.e. VCE and IC).
Maximum collector current due to a.c. signal = *IC
∴ Maximum positive swing of a.c. collector-emitter voltage = IC × RAC
Total maximum collector-emitter voltage = VCE + IC RAC
Maximum positive swing of a.c. collector current = VCE/RAC
Total maximum collector current = IC + VCE/RAC