Very high impedance. However, they appear in parallel across the load and

thus can also be neglected from any calculations.

Over the mid-frequency range, then, the behavior of the circuit is assumed to

be due entirely to resistive components. A suitable equivalent circuit is shown

below.

The current gain is given by

The only thing novel is the subscript 'MF' used in the symbol for current gain. Shortly, we will be expressing GI as a function of frequency and we need some means of distinguishing between the three frequency ranges over which the amplifier can operate. We will use the subscripts 'MF' for mid-frequency, 'LF' for low frequency and 'HF' for high frequency ranges.

The voltage gain is given by

this being simply a factor times the current gain value.

The low frequency range

In this range, the effect of the coupling capacitor cannot be ignored. The

reactance of the capacitance in parallel with the load will, however, be very

high at low frequencies and its effect can be ignored. The equivalent circuit

can be drawn as

If we represent C1 and RL by an impedance Z, then we can immediately write

the current gain as

The low frequency gain can be expressed in terms of the mid-band gain by

dividing top and bottom lines of the above expression by (1 + GoRL):

Remember

It is usual to express the term

as 1, the reason for which you will discover shortly.

Thus, the low frequency current gain can be written as:

The significance of 1

We have shown that at a frequency of 1, the current gain falls from its midband

value by a factor of √2. As power P = I²R, then at frequency 1 the

power gain will be half of its mid-band value.

The bandwidth of an amplifier is usually defined as the frequency range

between the half-power points. Thus, f₁=ω₁/2π represents the lower halfpower

frequency of the amplifier.

The high frequency range

Above the mid-band range the effect of the coupling capacitors can be ignored

but the capacitance across the load resistance will come into play. At high

frequencies, the equivalent circuit can be drawn as

Again, using the approach of representing the total load by Z, we can write

Equation (1) is based on above figure in which Io is the current through Z. But in the case of low and mid-band gains we have defined the gain as where Io is the current through RL.

To be consistent on our definitions of current gain, we need to modify

Equation (1) to define the gain as

This is easily done by using the 'ratio method' for determining the current in one branch of a parallel circuit.

Now we note that

In order to modify the current gain GI(HF) to the ratio we need to multiply

Equation (1) by the factor

Our modified current gain thus becomes

If we now divide the top and bottom lines of this last expression by (1 + GoRL), this will give us an expression for GI(HF) in terms of the mid-band gain.

To tidy things up, we will define 2 as

Then

The frequency f₂=ω₂/2π represents the upper half-power frequency. The

bandwidth of the amplifier is given by f2 – f1. This is illustrated in the

frequency response below