The basic arrangement of the common-base connection (or configuration) is shown in Fig. 4. The transistor has an alternating source of e.m.f. V_{s} volts r.m.s. and internal resistance R_{s} ohms connected to its input terminals. The alternating source is connected in series with the emitter-base voltage V_{EB} and it varies the forward bias applied to the emitter-base junction.
During negative half-cycles of the source e.m.f. the forward bias applied to the junction is increased, the potential barrier is lowered and an enhanced emitter current flows into the transistor. Conversely, during positive half-cycles the emitter current is decreased and in this way the collector current is caused to very in accordance with the waveform of the applied signal voltage. The collector-base bias battery V_{CB }has negligible internal resistance and so the collector-base voltage remains constant as the collector current varies . the collector circuit is said to be short-circuited so far as alternating currents are concerned .
In a common-base amplifier circuit an important parameter is the . short-circuit current gain of the transistor, symbol h_{fb}. The short-circuit current gain is defined as the ratio of a change in collector current to the change in emitter current producing it, with the collector-base voltage maintained constant, that is
h_{fb} = δI_{C}/δI_{E} = I_{c}/I_{e} when V_{CB} is constant (1.3)
The short-circuit current gain is specified since the current gain is a function of the value of any resistance connected in the collector circuit. For the common-base circuit, however, the difference between the short-circuit current gain and the current gain for any particular collector load resistance is very small for all resistance values used in practical circuits and it is usually neglected.
Example 1
In a certain transistor a change in emitter current of 1 mA produces a change n collector current of 0.99 mA. Determine the current gain of the transistor.
solution
Current gain h_{fb} = δI_{c}/δI_{E} = I_{c}/I_{e} = 0.99/1 = 0.99 (Ans.)
This is a typical value for the short-circuit current gain of a transistor connected in the common-base configuration. It should be evident that h_{fb}, must be less than unity, because the emitter current is the sum of the base and collector currents. Clearly, then, a common-base connected transistor must have a current gain of less than unity but, if a resistor is connected in the collector circuit, as shown in Fig, 3.5, both voltage and power gains are possible.
The output voltage is developed across the collector load resistor and, since the internal resistance of the collector supply is negligible, the top end of me resistor is effectively at earth potential so far as alternating currents are concerned. Thus the output signal voltage is taken from between the collector terminal and earth.
The source of the output power is the collector-base bias battery, the transistor effectively acting as a device for the conversion of d.c. power from the battery into the a.c. power supplied to the load.
In the common-base amplifier the input signal voltage and the output signal voltage are in phase with each other, as shown by the waveforms of Fig. 5. Consider the input signal voltage to be passing through zero and increasing in the negative direction. The forward bias of the base-emitter junction is then increased and this results in an increase in the emitter current. The collector current is increased and the voltage drop across the collector load resistor R_{L} increases also and this makes the collector-base potential less positive. Thus a negative increment in the input signal voltage produces a negative increment in the output signal voltage.
The common-base connected transistor is rarely, if ever, used at audio frequencies because of its low current gain and its inconvenient values of input and output impedance.
In practice, transistors are most often used in the common-emitter configuration shown in Fig. 6.
The emitter-base junction is forward-biased by the battery V_{BE} and the collector-base junction is reverse-biased by a potential equal to (v_{ce} — v_{be}). However, since the voltage of the collector-emitter bias battery V_{CE} is usually much larger than the emitter-base bias voltage V_{BE}, the reverse-bias voltage may often be taken as merely equal to V_{CE} volts.
When a transistor is connected in the common-emitter configuration, the input current is the base current. The operation of the transistor is unchanged from that previously described but the d.c. current gain h_{fE} is the ratio
h_{fE} = i_{c}/i_{b} (1.4)
During the positive half-cycles of the input signal voltage Vs, the forward bias of the emitter-base junction is increased, and so the emitter current i_{e} is increased by an amount δI_{E}. The collector current is also increased, by an amount δI_{C} = h_{fb}δI_{E}, and so is the base (input) current, by an amount
δI_{B} = δI_{E} – δI_{C} = δI_{E}(1 – h_{fb})
Conversely, during negative half-cycles of the input signal voltage all the three currents are reduced in magnitude.
The short-circuit current gain of a common-emitter connected transistor, symbol h_{fe}, is defined as the ratio of a change in collector current δI_{c} to the change in base current δI_{B} producing it, the collector-emitter voltage being maintained constant, that is
h_{fe} = δI_{C}/δI_{B}= I_{C}/I_{b} when V_{CE} is constant (1.5)
h_{fe} = h_{fb}I_{e}/(I_{e} - h_{fb}I_{e})
= h_{fb}I_{e}/I_{e}(1- h_{fb})
= h_{fb}/(1-h_{fb}) (1.6)
Typical values for die short-circuit current gain h_{fb}, of a common-base transistor are in the neighbourhood of unity and thus the common-emitter connection can give a considerable current gain.
The a.c. and me d.c. current gains of a transistor are not usually of the same value, their difference being dependent on the d.c. collector current flowing. Usually, little error is introduced by assuming that h_{fe} . is equal to h_{fe}. In any case, the value of h_{FE} (or h_{fe}) can vary considerably between two transistors of the same type and hence of the same nominal current gain.
Example 2
A transistor exhibits a change of 0.995 mA in its collector of 1 mA in its emitter current. Calculate (a) its common-base short-circuit current gain .
solution
(a) Common-base short-circuit current gain
h_{fb} = I_{c}/I_{e} = 0.995/1 = 0.995 (Ans.)
(b) Common-emitter short-circuit current gain
h_{fb} = I_{c}/I_{e} = h_{fb}/(1- h_{fb}) = 0.995/(1 - 0.995 ) = 199 (Ans.)
When a load resistor R_{L} is connected in the collector circuit (Fig.7) the current gain of the transistor is no longer equal to the short-circuit value but is somewhat less. The actual value of the current gain is dependent on the value of the collector load resistor R_{L}, decreasing with increase in R_{L}. The circuit now has both a voltage gain and a power gain.
Fig. 8. shows a voltage source of e.m.f. V_{s} volts (r.m.s.) and internal resistance R_{s} ohms connected to the input terminals of a transistor whose input resistance is R_{in}.
The a.c. input current to the transisior, I_{h}, is
I_{b} = V_{S}/(R_{S} + R_{in})
and the voltage V_{in} appearing across the transistor input terminals is
V_{in} = I_{b}R_{in} or I_{b} = V_{in}/R_{in}
The output or collector current I_{c} is
I_{c} = h_{fe}V_{in}/R_{in}
This current flows through the collector load resistance R_{L} and develops the output voltage V_{out} across it, therefore
V_{out = }h_{fe}V_{in} R_{L}/R_{in}
and the voltage gain A_{v} is .
A_{v} = V_{out}/V_{in} = h_{fe}R_{L}/Rin (1.7)
Since the short-circuit current gain h_{fe} is greater than unity and the collector load resistance R_{L} is usually greater than the input resistance R_{IH} of the transistor, the circuit provides a voltage gain.
The power gain of a common-emitter transistor is the ratio of the power delivered to the load to the power delivered to the transistor.
The input power P_{in} to the transistor is (see Fig. 8)
P_{in} = (I_{b})^{2}R_{in}
and the output power P_{out} is
P_{out} = (h_{fe}I_{b})^{2}R_{L }
Therefore the power gain A_{p} is
A_{p} = P_{out}/P_{in} =(h_{fe})^{2}(I_{b})^{2}R_{L}/(I_{b})^{2}R_{in}
A_{p} = (h_{fe})^{2}R_{L}/R_{in} = A_{v}A_{i }(1.8)
Again , since R_{L} is greater than R_{in} the circuit provides a power gain .
Example 3
A transistor is connected with common-emitter in a circuit a collector load resistance of 2000 Ω. the short-circuit gain of the transistor is 100 and its input resistance is 1000 Ω. Calculate the voltage and power gains of the transistor .
Solution
From equation
Voltage gain = h_{fe}R_{L}/R_{in} = (100 X 2000)/1000 = 200 (Ans.)
From equation
Power gain = (h_{fe})^{2}R_{L}/R_{in} = 200 X 100 = 20 000 (Ans.)
Example 4
A source of e.m.f. 50 mV and internal resistance 1000Ω is applied to the circuit of example 3 calculate (a) the input voltage and (b) the output voltage of the circuit .
Solution
(a) V_{in}_{ }= (50 X 1000)/(1000 + 1000) = 25mV (Ans.)
(b) V_{out} = A_{v}V_{in} = 200 X 25 10-3 = 5 v (Ans.)
The third way in which a transistor may be connected is shown in in Fig. 9. The collector terminal is now common to both input and output circuits and the load resistor is connected in the emitter circuit. With this configuration the base current is the input current and the emitter current is the output current. The short-circuit current gain is defined as
Short-circuit current gain
h_{fe} = δI_{E}/δI_{c} = I_{c}/I_{b} when V_{CE} is constant
= I_{e}/(I_{e} – I_{e})
= I_{e}/I_{e} (1-h_{fb})
= 1/(1-h_{fb}) (1.10)
= h_{fe} + 1 (1.11)
But since h_{fe} is a variable it is usual to take h_{fe}. as being equal to h_{fe}. The short-circuit current gain of a transistor connected in the common-collector configuration is approximately equal to the short-circuit gain of the same transistor connected with common-emitter. The current gain when a load is connected in the emitter circuit is not equal to the short-circuit current gain but is reduced by an amount that is dependent on the value of the emitter load.
Expressions (7) and (8) may be used to determine the voltage and power gains of a common-collector circuit.
The input resistance of the common-collector circuit is equal to h_{fe}R_{L} + R_{in} the voltage gain is
A_{v} = A_{i}R_{L}/R_{in} = h_{fe}R_{L}(/h_{fe}R_{L} + R_{in}) (1.12)
Example 5
A transistor is connected in a common-collector circuit with an emitter load resistance of 2000 Ω. The current gain h_{fe} of the transistor is 100 and its input resistance is 1000 Ω. Calculate (a) the voltage gain and (b) the power gain of the circuit.
Solution
(a) Av = (100 x 2000)/(100 X 2000 + 1000) = 0.995 (Ans.)
(b) A_{p} = 100 + 0.995 = 99.5 (Ans.)
The main use of the common-collector circuit or the emitter-follower as it is usually called — is as a buffer which is connected between a high impedance source and a low impedance load.
A comparison between the main characteristics of the three transistor configurations is given in Table.1.
Parameter | Common-base | Common-emitter | Common-collector |
Short-circuit | |||
current gain | h_{fb} | h_{fe} = h_{fe}/(1 - h_{fe}) | H_{fe} = 1/(1 –h_{fb}) |
Voltage gain | Good | Better than common-base | Unity or less |
Input | Low 30 to | Medium 800 to | High 5000 to |
resistance | 100 Ω | 5000 Ω | 500 000 Ω |
Output | High 10^{5} to | Medium 10 000 | Low 50 to |
resistance | 10^{6} Ω | to 50 000 Ω | 1000 Ω |