So that

We can now substitute this into the input equation to eliminate Ii

which gives

the minus sign in the last equation means is that the output voltage is phase inverted with respect to the input voltage, which is a characteristic of the common emitter amplifier. This is illustrated graphically by the plot of Vo against Vi shown below.

Some practical considerations

1. Note that in the equation

the term 2 10^–4 Vo contributes little. In other words, we could ignore the

term due to hre with very little error. This will be true for all the circuits

we shall encounter and we will make this approximation in all subsequent

work. This permits a simplified equivalent circuit

(ii) Any load connected to the output of the common emitter circuit will

appear in parallel to hoe. This will include the collector resistor Rc as well

as any other external load connected. We can, if we wish, represent all

such loads by the single equivalent load RL as

(iii) In many applications the transistor's output resistance is very much

greater than the load, i.e. 1/hoe >> RL. Under such conditions, hoe can

also be omitted from the equivalent circuit and an even simpler model

could be drawn as

(iv) In practice, all the parameters will vary with supply voltage, frequency,

collector current and temperature. For our purposes, though, we will not

worry about such variations and in any given application assume the

parameters to be constant.

Let’s consider the next circuit:

FIGURE upward shows a simplified h parameter equivalent circuit of a common

emitter amplifier. Determine:

(i) the voltage gain Vo/Vi

(ii) the current gain IL/Ii

(iii) the power gain Po/Pi

Take hie = 1500 Ωand hfe = 200. Rc = 4.7 kΩ, RL = 2.2 kΩ

The two resistors on the output can be represented by the single resistor RL

Ω

The circuit reduces to that shown

for which we can write

This will yield the voltage gain as

Current gain

The current gain can be found from the equations

The power delivered to the input is given by Pi = Vi Ii and that produced at

the load is given by Po = Vo IL. Thus, the power gain is given by the ratio:

Note that the modulii of the voltage and current gains are used.

In this example, therefore,

CIRCUITS’ FREQUENCY RESPONSES

We have shown how an electronic device such as an amplifier can be represented by a two port 'black box' model. In this model, though, only the resistive components of the original amplifier circuit were considered, the effects of the capacitors in the circuit being ignored. In point of fact, it is usually valid to ignore the circuit capacitances of an amplifier over its range of operating frequencies. The value of the capacitors will have been chosen to permit such an assumption. But how do we know how far the operational range of the amplifier extends? Or, to reverse the question, how do we determine the value of the capacitors, given the range of frequencies over which the amplifier must operate? It will be shown that an amplifier's frequency response has its low, mid and high ranges. The low range can be modelled by the 'black box' loaded by a series RC circuit, the mid range by a simple resistive load and the high range by a parallel RC load. The effect on voltage, current and power gains for each of these conditions will be investigated.