Question 19.5: Determine the small-signal behaviour of the circuit in Examp......

For a more detailed view of the AC characteristics of the circuit, we turn again to a small-signal equivalent circuit.

The presence of the emitter resistor makes the equivalent circuit more complicated than that given in Figure 19.16. The emitter is connected to ground through the emitter resistor R_{E} , but the base resistors R_{1} and R_{2} and the collector resistor R_{C} are shown connected to ground as these resistors are connected either to ground or to V_{CC} (which is at ground potential for AC signals). The small-signal voltage gain and the input and output resistance can be calculated directly from the equivalent circuit as follows.

Small-signal voltage gain

The analysis of Example 19.4 can be performed with reference to the equivalent circuit, producing an identical result. The equivalent circuit perhaps illustrates more clearly why the voltage gain of the amplifier is negative. If one ignores the effects of i_{b} and considers a current flowing from the current source, through R_{E} and then back through R_{C}, it is clear that the magnitude of the current flowing in each resistor is equal. However, as the current is flowing in opposite directions in the two resistors, the polarity of the voltage across each will be reversed. As the voltage across R_{E} is approximately equal to the input voltage, the output voltage will be inverted with respect to the input. Also, as the same current flows through both R_{C} and R_{E} , it is logical that the voltage gain will be the ratio of the resistor values. Therefore, as before

voltage gain = \frac{v_{o}}{V_{i}} \approx – \frac{R_{C}}{R_{E}}

From this expression for the voltage gain of the amplifier it would seem that using a very small value for R_{E} would produce a very high gain. Taking this to its logical conclusion, we might expect that taking a vanishingly small value for R_{E} would result in an extremely high-gain circuit. A moment’s thought should make it clear that the gain of the resulting circuit cannot exceed that of the common-emitter amplifier of Figure 19.7, which is simply -g_{m}R_{C} , so a more complete expression for the gain is required.

If we ignore the effects of the output resistance 1/h_{oe} (this is normally large compared with R_{E} and R_{C}) and the base current (which is normally small compared with I_{E}) then the output voltage is given by

as the current in R_{C} is equal to that produced by the current generator. The negative sign here reflects the fact that a positive current from the current generator produces a negative output voltage.

The input voltage to the amplifier is equal to the sum of the voltages across the base–emitter junction v_{be} and the emitter resistor v_{e} . Therefore

If we again ignore the effects of h_{oe} and i_{b}, then v_{e} is given by

and combining this with the previous expression gives

This may be rearranged to give

and combining this with the expression above for v_{o} gives

Thus, the voltage gain of the amplifier is given by

voltage gain = \frac{v_{o}}{V_{i}} = -\frac{g_{m}R_{C}}{1  +  g_{m}R_{E}} = – \frac{R_{C}}{R_{E}  +  1/g_{m}}

It can now be seen that, if R_{E} is zero, the gain becomes equal to -g_{m}R_{C}, as for the common-emitter amplifier. However, when R_{E} is much greater than 1/g_{m} the gain tends to -R_{C}/R_{E}, as before.

Note that 1/g_{m} is equal to r_{e}, so the gain may be written as

voltage gain = -\frac{R_{C}}{R_{E}  +  r_{e}}

and the gain is approximately equal to -R_{C}/R_{E}  when R_{E} ⪢ r_{e} and equal to -R_{C}/r_{e} in the special case where R_{E} = 0 (when we have the common-emitter circuit).

Small-signal input resistance

From the equivalent circuit, it is apparent that the input resistance is formed by the parallel combination of R_{1}, R_{2} and the resistance seen looking into the base of the transistor. This last term is not simply the sum of h_{ie} and R_{E} because of the effect of the current generator. When a current i_{b} enters the base of the transistor, the current that flows through R_{E} is the sum of the base current and the collector current. As the collector current is equal to h_{fe}i_{b} , the emitter current is given by

When a current flows through a resistor, a voltage drop is produced and the ratio of the voltage to the current determines the resistance of the component. When i_{b} flows into the base of the transistor, a much greater current flows in the emitter resistor, producing a proportionately larger voltage drop. Therefore, the emitter resistor appears much larger when viewed from the base. In fact, the emitter resistor appears to be increased by a factor of (h_{fe} + 1). The input resistance h_{ie} is not amplified in this way as the current passing through this is simply i_{b}. Therefore, the effective input resistance seen looking into the base of the transistor is

r_{b} = h_{ie} + (h_{fe} + 1)R_{E}     (19.21)

and the input resistance of the amplifier is

r_{i} = R_{1}//R_{2}//r_{b}    (19.22)

It is interesting to look at the relative magnitudes of the three components of the input resistance, as defined in Equation 19.22. R_{1} and R_{2} are base-bias resistors and we noted when considering the DC performance of the circuit that these should be chosen such that the current flowing through them is large compared with the base current. This limits their maximum values, which typically might be a few kilohms or a few tens of kilohms. The magnitude of r_{b} is given in Equation 19.21. We know from Section 19.5.2 that a typical value for h_{ie} might be a few kilohms. As R_{E} will also typically be a few kilohms and h_{fe} will be perhaps 80 to 350, it would be reasonable to use the approximation
r_{b} = h_{ie} + (h_{fe} + 1)R_{E} \approx h_{fe}R_{E}

From this it can be seen that r_{b} will generally be several hundred kilohms and will usually be large compared with R_{1} and R_{2}. Therefore, as the three resistors are in parallel, the effects of r_{b} will often be negligible and it will be the parallel combination of R_{1} and R_{2} that will determine the input resistance of the amplifier.

If this is so

and in this circuit

From this it is clear that, to achieve a high input resistance, it is desirable to make R_{1} and R_{2} as high as possible. This is in conflict with the biasing considerations discussed earlier, as it is necessary to ensure that the current flowing through R_{1} and R_{2} is large compared with the base current to ensure that the effects of the latter may be ignored. These opposing requirements are resolved by a compromise, where R_{2} is typically chosen to be about ten times the value of R_{E} .

An advantage of this circuit – in comparison with the simple common-emitter amplifier – is that the input resistance is determined by the passive components within the circuit rather than by the transistor. This makes the circuit much more predictable and less affected by the characteristics of the active device used.

Small-signal output resistance

In Example 19.2 we considered the output resistance of a simple common-emitter amplifier and deduced that this was equal to the parallel combination of R_{C} and 1/h_{oe} . The addition of the emitter resistor R_{E} in the series feedback amplifier places an extra resistance in series with 1/h_{oe} , as shown in the equivalent circuit. The addition of this resistance makes calculation of the output resistance somewhat more complicated. However, as 1/h_{oe} is normally much greater than R_{C} , it follows that 1/h_{oe} in series with an additional resistance will also be greater than R_{C} and the resistance of the parallel combination is dominated by R_{C} . Therefore, in this example

As with the input resistance, the output resistance is determined by the passive components within the circuit rather than the transistor. This improves the predictability of the circuit by reducing its dependence on device characteristics.