The comparison of various characteristics of the three connections is given below in the tabular
The following points are worth noting about transistor arrangements :
The input resistance (ri) of CB circuit is low because IE is high. The output resistance (ro ) is high because of reverse voltage at the collector. It has no current gain (α < 1) but voltage gain can be high. The CB circuit is seldom used. The only advantage of CB circuit is that it provides good stability against increase in temperature.
The input resistance (ri ) of a CE circuit is high because of small IB . Therefore, ri for a CE circuit is much higher than that of CB circuit. The output resistance (ro ) of CE circuit is smaller than that of CB circuit. The current gain of CE circuit is large because IC is much larger than IB . The voltage gain of CE circuit is larger than that of CB circuit. The CE circuit is generally used because it has the best combination of voltage gain and current gain. The disadvantage of CE circuit is that the leakage current is amplified in the circuit, but bias stabilisation methods can be used.
The input resistance (ri ) and output resistance (ro ) of CC circuit are respectively high and low as compared to other circuits. There is no voltage gain (Av < 1) in a CC circuit. This circuit is often used for impedance matching.
Out of the three transistor connections, the common emitter circuit is the most efficient. It is used in about 90 to 95 percent of all transistor applications. The main reasons for the wide spread use of this circuit arrangement are :
In a common emitter connection, IC is the output current and IB is the input current. In this circuit arrangement, collector current is given by :
IC= β IB\ICEO
As the value of β is very large, therefore, the output current IC is much more than the input current IB . Hence, the current gain in CE arrangement is very high. It may range from 20 to 500.
Due to high current gain, the common emitter circuit has the highest voltage and power gain of three transistor connections. This is the major reason for using the transistor in this circuit arrangement.
(iii) Moderate output to input impedance ratio.
In a common emitter circuit, the ratio of output impedance to input impedance is small (about 50). This makes this circuit arrangement an ideal one for coupling between various transistor stages. However, in other connections, the ratio of output impedance to input impedance is very large and hence coupling becomes highly inefficient due to gross mismatching.
Fig. 8.33 shows the common emitter npn amplifier circuit. Note that a battery VBB is connected in the input circuit in addition to the signal voltage. This d.c. voltage is known as bias voltage and its
magnitude is such that it always keeps the emitter-base junction forward *biased regardless of the
polarity of the signal source.
During the positive half-cycle of the **signal, the forward bias across the emitter-base junction is increased. Therefore, more electrons flow from the emitter to the collector via the base. This causes an increase in collector current. The increased collector current produces a greater voltage drop across the collector load resistance RC. However, during the negative half-cycle of the signal, the forward bias across emitter-base junction is decreased. Therefore, collector current decreases. This results in the decreased output voltage (in the opposite direction). Hence, an amplified output is obtained across the load.
When no signal is applied, the input circuit is forward biased by the battery VBB. Therefore, a d.c. collector current IC flows in the collector circuit. This is called zero signal collector current. When the signal voltage is applied, the forward bias on the emitterbase junction increases or decreases depending upon whether the signal is positive or negative.
During the positive half-cycle of the signal, the forward bias on emitter-base junction is increased,
causing total collector current iC to increase. Reverse will happen for the negative half-cycle of the
signal. Fig. 8.34 shows the graph of total collector current iC versus time. From the graph, it is clear that total collector current consists of two components, namely ;
(i) The d.c. collector current IC (zero signal collector current) due to bias battery VBB. This is the current that flows in the collector in the absence of signal.
(ii) The a.c. collector current ic due to signal.
∴ Total collector current, iC= ic + IC
The useful output is the voltage drop across collector load RC due to the a.c. component Ic. The purpose of zero signal collector current is to ensure that the emitter-base junction is forward biased at all times. The table below gives the symbols usually employed for currents and voltages in transistor applications.
In the transistor circuit analysis, it is generally required to determine the collector current for various collector-emitter voltages. One of the methods can be used to plot the output characteristics and determine the collector current at any desired collector-emitter voltage. However, a more convenient method, known as load line method can be used to solve such problems. As explained later in this section, this method is quite easy and is frequently used in the analysis of transistor applications. d.c. load line. Consider a common emitter NPN transistor circuit shown in Fig. 8.35 (i) where no signal is applied. Therefore, d.c. conditions prevail in the circuit. The output characteristics of this circuit are shown in Fig. 8.35 (ii). The value of collector-emitter voltage VCE at any time is given by ;
As VCC and RC are fixed values, therefore, it is a first degree equation and can be represented by
a straight line on the output characteristics. This is known as d.c. load line and determines the locus
of VCE − IC points for any given value of RC. To add load line, we need two end points of the straight line. These two points can be located as under :
(i) When the collector current IC = 0, then collector-emitter voltage is maximum and is equal to
This gives the second point A (OA= VCC /RC) on the collector current axis as shown in Fig. 8.35 (ii). By joining these two points, d.c. *load line AB is
Importance. The current (IC) and voltage (VCE) conditions in the transistor circuit are represented by some point on the output characteristics. The same information can be obtained from the load line. Thus when IC is maximum (= VCC /RC), then VCE = 0 as shown in Fig. 8.36. If IC = 0, then VCE is maximum and is equal to VCC. For any other value of collector current say OC, the collector-emitter voltage VCE = OD. It follows, therefore, that load line gives a far more convenient and direct solution to the problem.
Note. If we plot the load line on the output characteristic of the transistor, we can investigate the behaviour of the transistor amplifier. It is because we have the transistor output current and voltage specified in the form of load line equation and the transistor behaviour itself specified implicitly by the output characteristics.
The zero signal values of IC and VCE are known as the operating point. It is called operating point because the variations of IC and VCE take place about this point when signal is applied. It is also called quiescent (silent) point or Q-point because it is the point on IC−VCE characteristic when the transistor is silent i.e.in the absence of the signal. Suppose in the absence of signal, the base current is 5µA. Then IC and VCE conditions in the circuit must be represented by some point on IB = 5 µA characteristic. But IC and VCE conditions in the circuit should also be represented by some point on the d.c. load line AB. The point Q where the
load line and the characteristic intersect is the only point which satisfies both these conditions. Therefore, the point Q describes the actual state of affairs in the circuit in the zero signal conditions and is called the operating point. Referring to Fig. 8.37, for IB = 5 µA, the zero signal values are :
VCE = OC volts
IC = OD mA
It follows, therefore, that the zero signal values of IC and VCE (i.e. operating point) are determined by the point where d.c. load line intersects the proper base current curve.
Example 8.22. For the circuit shown in Fig. 8.38 (i), draw the d.c. load line.
Solution. The collector-emitter voltage VCE is given by ;
This locates the point A of the load line on the collector current axis. By joining these two points,
we get the d.c. load line AB as shown in Fig. 8.38 (ii).
Example 8.23. In the circuit diagram shown in Fig. 8.39 (i), if VCC = 12V and RC = 6 kΩ, draw
the d.c. load line. What will be the Q point if zero signal base current is 20µA and β = 50 ?
Solution. The collector-emitter voltage VCE is given by :
Example 8.24. In a transistor circuit, collector load is 4 kΩ whereas quiescent current (zero
signal collector current) is 1mA.
(i) What is the operating point if VCC = 10 V ?
(ii) What will be the operating point if RC = 5 kΩ ?
Example 8.25. Determine the Q point of the transistor circuit shown in Fig. 8.40. Also draw the
d.c. load line. Given β = 200 and VBE = 0.7V
Solution. The presence of resistor RB in the base circuit should not disturb you because we can
apply Kirchhoff’s voltage law to find the value of IB and hence IC (= βIB). Referring to Fig. 8.40 and
applying Kirchhoff’s voltage law to base-emitter loop, we have,
Therefore, the Q-point is IC = 39.6 mA and VCE = 6.93V.
In order to draw the d.c. load line, we need two end points.VCE = VCC – ICRC When IC = 0, VCE=VCC = 20V. This locates the point B of the load line on the collector-emitter voltage axis as shown in Fig. 8.41. When VCE = 0, IC= VCC/RC= 20V/330Ω = 60.6 mA. This locates the point A of the load line on the collector current axis. By joining these two points, d.c. load line AB
is constructed as shown in Fig. 8.41.
Example 8.26. Determine the Q point of the transistor circuit shown in *Fig. 8.42. Also draw
the d.c. load line. Given β = 100 and VBE = 0.7V
Solution. The transistor circuit shown in Fig. 8.42 may look complex but we can easily apply Kirchhoff’s voltage law to find the various voltages and currents in the * circuit. Applying Kirchhoff’s voltage law to the base-emitter loop, we have,
This locates the second point A (OA = 3.51 mA) of the load line on the collector current axis. By
joining points A and B, d.c. load line AB is constructed as shown in Fig. 8.43.
Example 8.27. In the above example, find (i) emitter voltage w.r.t. ground (ii) base voltage w.r.t. ground (iii) collector voltage w.r.t. ground.
The common emitter circuits drawn so far can be shown in another convenient way. Fig. 8.45 shows the practical way of drawing CE circuit. In Fig. 8.45 (i), the practical way of drawing common emitter npn circuit is shown. Similarly, Fig. 8.45 (ii) shows the practical way of drawing common emitter pnp circuit. In our further discussion, we shall often use this scheme of presentation.
A transistor raises the strength of a weak signal and thus acts as an amplifier. Fig. 8.46 shows the common emitter amplifier. There are two ways of taking output from this transistor connection. The output can be taken either across RC or across terminals 1 and 2. In either case, the magnitude of output is the same. This is clear from the following discussion :
(i) First method. We can take the output directly by putting a load resistance RC in the collector circuit i.e. Output = voltage across RC = icRC …(i)
This method of taking output from collector load is used only in single stage of amplification.
(ii) Second method. The output can also be taken across terminals 1 and 2 i.e. from collector and emitter end of supply.
Output = Voltage across terminals 1 and 2
= VCC − ic RC
As VCC is a direct voltage and cannot pass through capacitor CC, therefore, only varying voltage
ic RC will appear across terminals 1 and 2.
∴ Output = − ic RC
From exps. (i) and (ii), it is clear that magnitude of output is the same whether we take output across collector load or terminals 1 and 2. The minus sign in exp. (ii) simply indicates the phase reversal. The second method of taking output is used in multistages of amplification.
The performance of a transistor amplifier depends upon input resistance, output resistance, effective collector load, current gain, voltage gain and power gain. As common emitter connection is universally adopted, therefore, we shall explain these terms with reference to this mode of connection.
(i) Input resistance. It is the ratio of small change in base-emitter voltage (ΔVBE) to the resulting change in base current (ΔIB) at constant collector-emitter voltage i.e.
The value of input resistance is quite small because the input circuit is always forward biased. It
ranges from 500 Ω for small low powered transistors to as low as 5 Ω for high powered transistors. In fact, input resistance is the opposition offered by the base-emitter junction to the signal flow. Fig. 8.47 shows the general form of an amplifier. The input voltage VBE causes an input current IB
Thus if the input resistance of an amplifier is 500 Ω and the signal voltage at any instant is 1 V, then, Base current, ib =1v/500Ω= 2 mA
(ii) Output resistance. It is the ratio of change in collectoremitter voltage (ΔVCE) to the resulting change in collector current (ΔIC) at constant base current i.e.
The output characteristics reveal that collector current changes very slightly with the change in collector-emitter voltage. Therefore, output resistance of a transistor amplifier is very high– of the order of several hundred kilo-ohms. The physical explanation of high output resistance is that collector-base junction is reverse biased.
(iii) Effective collector load. It is the total load as seen by the a.c. collector current. In case of single stage amplifiers, the effective collector load is a parallel combination of RC and RO as shown in Fig. 8.48 (i).
It follows, therefore, that for a single stage amplifier, effective load is equal to collector load RC. However, in a multistage amplifier (i.e. having more than one amplification stage), the input resistance Ri of the next stage also comes into picture as shown in Fig. 8.48 (ii). Therefore, effective
collector load becomes parallel combination of RC, RO and Ri i.e.
As input resistance Ri is quite small (25 Ω to 500 Ω), therefore, effective load is reduced.
(iv) Current gain. It is the ratio of change in collector current (ΔIC) to the change in base current (ΔIB) i.e.
The value of β ranges from 20 to 500. The current gain indicates that input current becomes β
times in the collector circuit.
Voltage gain. It is the ratio of change in output voltage (ΔVCE) to the change in input
voltage (ΔVBE) i.e.
Example 8.28. A change of 200 mV in base-emitter voltage causes a change of 100 µA in the
base current. Find the input resistance of the transistor.
Solution. Change in base-emitter voltage is
Example 8.29. If the collector current changes from 2 mA to 3mA in a transistor when collector-emitter voltage is increased from 2V to 10V, what is the output resistance?
Example 8.30. For a single stage transistor amplifier, the collector load is RC = 2kΩ and the
input resistance Ri = 1kΩ. If the current gain is 50, calculate the voltage gain of the amplifier.
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Reference: Principles Of Electronics By V K Mehta And Rohit Mehta
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