The ac or dynamic resistance of emitter-base junction diode of a transistor is called ac emitter resistance. It is defined as the change in base-emitter voltage divided by change in corresponding emitter current [See Fig. 10.23] i.e
For instance, suppose an ac base voltage change of 1 mV produces an ac emitter current change of 50 µA. Then emitter diode has an ac resistance of
It can be shown mathematically that the ac resistance of emitter diode is given by ;
Note the significance of this formula. It implies that ac emitter resistance can be found simply by substituting the quiescent value of emitter current into the equation. There is no need to have the characteristics available. It is important to keep in mind that this formula is accurate only for small signal operation. It is a usual practice to represent ac emitter resistance by re′.
The subscript e indicates emitter. The lower case r is used to indicate an ac resistance. The prime shows that it is internal resistance.
Example 10.13. Determine the ac emitter resistance for the transistor circuit shown in Fig. 10.24.
The voltage gain (Av) of an amplifier is equal to a.c. output voltage (vout) divided by a.c. input voltage (vin) i.e. Av = vout/vin. We have already seen that voltage gain of a CE amplifier is given by;
The above formula for Av can be used if we know the values of RC (or RAC), β and Rin. Generally, all these values are not known. In that case, we can find the value of Av in terms of total a.c. collector resistance and total a.c. emitter resistance. For the circuit shown in Fig. 10.25 (with CE connected across RE), it can be proved that the voltage gain is given by ;
Fig. 10.25 shows the circuit of unloaded CE amplifier (i.e. no load RL is connected to the circuit).Note that emitter bypass capacitor CE is connected in parallel with emitter resistance RE. The capacitor CE acts as a *short to the a.c. signal so that it bypasses the a.c. signal to the ground. Therefore, the emitter is effectively at a.c. ground. It is important to note that CE plays an important role in determining the voltage gain (Av) of the CE amplifier. If it is removed, the voltage gain of the amplifier is greatly reduced (soon you will see the reason for it).
Derivation. Fig. 10.25 shows the common emitter amplifier. The ac equivalent circuit of the amplifier is shown in Fig. 10.26. (i). Replacing the transistor by its *equivalent circuit, we get the circuit shown in Fig. 10.26 (ii). Note that current source is still connected between the collector and base terminals while the diode between the base and emitter terminals. Further, the input current is the base current (ib) while the output current is still ic. Note that input voltage (Vin) is applied across the diode and re′. Assuming the diode to be ideal (so that it can be replaced by a wire), the ac emitter current is given by :
Fig. 10.27 shows the simple a.c. model of CE amplifier with CE connected across RE. Note that CE behaves as a short so that RE is cut out from the emitter circuit for a.c. signal. Therefore, as for as a.c. signal is concerned, the total a.c. emitter resistance is re ′ . Voltage gain for loaded amplifier. Fig. 10.28 (i) shows a part of a.c. equivalent circuit of the 0v a.c.
CE amplifier. Note that load RL is connected to the circuit. Remember that for a.c. analysis, VCC = 0V i.e. at ground. Since both RC and RL are connected to the collector on one side and ground on the other, the two resistors are in *parallel as shown in Fig. 10.28 (ii).
When we remove the emitter bypass capacitor from the CE amplifier shown in Fig. 10.25, the voltage gain of the circuit is greatly reduced. The reason is simple. Without the emitter bypass capacitor CE, the emitter is no longer at the ac ground as shown in Fig. 10.29. Therefore, for the a.c. signal, both e r′ and RE are in series. As a result, the voltage gain of the amplifier becomes :
Example 10.14. For the amplifier circuit shown in Fig. 10.30, find the voltage gain of the amplifier with (i) CE connected in the circuit (ii) CE removed from the circuit.
Solution. We shall first find D.C. IE and hence re
What a difference the emitter bypass capacitor CE makes ! With CE connected, Av = 360 and when CE is removed, the voltage gain goes down to 5.38.
Example 10.15. If in the above example, a load of 6 kΩ is connected (with CE connected) to the collector terminal through a capacitor, what will be the voltage gain of the amplifier?
Solution. Amplifiers are used to provide ac power to the load. When load RL is connected to the collector terminal through a capacitor, the total ac resistance of collector changes to :′ .
Thus voltage gain of the amplifier is reduced from 360 to 120 when load is connected to the circuit. Comments. This example shows the fact that voltage gain of the amplifier is reduced when load is connected to it. Conversely, if the load is removed from an amplifier, the voltage gain will increase. If a load goes open circuit, the effect will be the same as removing the load entirely. Thus the primary symptom of an open load in an amplifier is an increase in the voltage gain of the circuit.
Example 10.16. For the circuit shown in Fig. 10.31, find (i) a.c. emitter resistance (ii) voltage gain (iii) d.c. voltage across both capacitors
Solution. (i) In order to find a.c. emitter resistance re ′ , we shall first find D.C. emitter current IE . To find IE, we proceed as under :
(iii) The d.c. voltage across input capacitor is equal to the d.c. voltage at the base of the transistor which is V2 = 1V. Therefore, d.c. voltage across Cin is 1V. Similarly, d.c. voltage across CE = d.c voltage at the emitter = VE = 0.3V.
Example 10.17. For the circuit shown in Fig. 10.32, find (i) the d.c. bias levels (ii) d.c. voltages across the capacitors (iii) a.c. emitter resistance (iv) voltage gain and (v) state of the transistor.
(i) D.C. bias levels. The d.c. bias levels mean various d.c. currents and d.c. voltages.
Example 10.18. An amplifier has a voltage gain of 132 and β = 200. Determine the power gain and output power of the amplifier if the input power is 60 μW
Power gain, Ap = current gain × voltage gain
= β × Av = 200 × 132 = 26400
Output power, Pout = Ap × Pin = (26400) (60 μW) = 1.584 W
Example 10.19. For the circuit shown in Fig. 10.33, determine (i) the current gain (ii) the voltage gain and (iii) the power gain. Neglect the a.c. emitter resistance for the transistor.
Solution. In most practical circuits, the value of a.c. emitter resistance re′ for the transistor is generally quite small as compared to RE and can be neglected in circuit calculations with reasonable accuracy
When one CE amplifier is being used to drive another, the input impedance of the second amplifier will serve as the load resistance of the first. Therefore, in order to calculate the voltage gain (Av) of the first amplifier stage correctly, we must calculate the input impedance of the second stage. The input impedance of an amplifier can be found by using the ac equivalent circuit of the amplifier as shown in Fig. 10.34.
Zin = R1 || R2 | | Zin (base)
where Zin = input impedance of the amplifier
Zin (base) = input impedance of transistor base
Now Zin (base) = *β re′
The input impedance [Zin] is always less than the input impedance of the base [Zin(base)].
Example 10.20. Determine the input impedance of the amplifier circuit shown in Fig. 10.35.
One important consideration for an amplifier is the stability of its voltage gain. An amplifier should have voltage gain values that are stable so that the output of the circuit is predictable under all normal conditions. In a standard CE amplifier, the entire d.c. emitter resistance RE is bypassed by the bypass emitter capacitor CE. Therefore, the total a.c. emitter resistance is re′ . The voltage gain of such an amplifier at no-load is given by ;
The voltage gain of a standard CE amplifier is quite large. However, the drawback of the circuit is that its voltage gain changes with emitter current IE, temperature variations and transistor replacement. For example, if emitter current IE increases, the a.c. emitter resistance re′ decreases. This changes the voltage gain of the amplifier. Similarly, when the temperature varies or when a transistor is replaced, the a.c. current gain β changes. This will also result in the change in voltage gain. In order to stabilise the voltage gain, the emitter resistance RE is partially bypassed by CE. Such an amplifier is called a swamped amplifier.
Fig. 10.36 (i) shows the emitter leg of a standard CE amplifier while Fig. 10.36 (ii) shows the emitter leg of swamped amplifier. In swamped amplifier, the resistance RE is split into two parts viz. RE1 and RE2. Only RE2 is bypassed by CE while RE1 is not.
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