Some practical considerations

(i) Note that in the equation

the term 2 ´ 10^–4 V o contributes little. In other words, we could ignore the

term due to h re 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 h oe. This will include the collector resistor R c 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/ h oe >> R ¢L. Under such conditions, h oe 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 V o/ V i

(ii) the current gain I L/ I i

(iii) the power gain P o/ P i

Take h ie = 1500 Ω and h fe = 200. R c = 4.7 kΩ, R L = 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