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 I o is the current through Z. But in the case of low and mid-band gains we have defined the gain as where I o is the current through R L.

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 G I(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 + G o R L), this will give us an expression for G I(HF) in terms of the mid-band gain.

To tidy things up, we will define w2 as

Then

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

bandwidth of the amplifier is given by f 2 – f 1. This is illustrated in the

frequency response below