
- •Network analysis
- •The transistor amplifier circuit
- •The black box model
- •A mathematical model of the box
- •The equivalent circuit of the z parameter model
- •The equivalent circuit of the y parameter model
- •The equivalent circuit of the h parameter model
- •I1 flowing through it
- •The h parameter model of the common emitter circuit
- •In other words, the assumed directions of Io and Vo contradict each other. Hence the need for the minus sign. This is a point to watch in all circuits where assumptions
- •So that
- •Some practical considerations
- •Note that in the equation
- •In this example, therefore,
- •The effects of frequency and the model
- •(II) Internal capacitances of the device within the box
- •Effects of a capacitive load on the amplifier
- •Very high impedance. However, they appear in parallel across the load and
- •The low frequency range
- •The high frequency range
- •Power gain (gp)
- •If we calculate the power gain as
Power gain (gp)
The power gain is defined as
There are, however, a number of ways in which we could evaluate power gain.
Consider the circuit below; note that hoe and Rc have again been shown separately.
We can define the power gain as
All these definitions are perfectly valid; which one is chosen depends upon
which aspect of the circuit's performance we wish to evaluate.
If we calculate the power gain as
then, having found GV and GI, finding the power gain is straightforward.
Note also
that
which
offers an alternative method
of calculating the power gain:
One final word of caution! In the above discussion on power gain we have
assumed that the load is resistive. If, however, this is not the case, then due
regard must be paid to the phase relationship between voltage and current.
Let’s consider the example.
For the equivalent circuit shown
CL = 10 nF hfe = 100
RL= 1khie= 1kGo= 10 –4 S CC = 10 F
calculate:
(i)the mid-band current gain (IL/Ii)
(ii) the mid-band voltage gain (Vo/Vi)
the mid-band power gain (power in load/power input)
(iv) the lower half-power frequency
(v) the upper half-power frequency
(vi) the bandwidth
(vii) The magnitudes of the current and voltage gains and the power gain
at = 220 103 rad s–1.
Mid-band current gain
Mid-band voltage gain
(iii)Power gain
(iv) Lower half-power frequency
f1=1/2z
(v)Upper half-power frequency
f2=2/2z
(vi)
(vii) As 220 103 rad s–1 > 2, we must use the high frequency equations for
GI and GV.
Current gain
Voltage gain
Power gain
Again, as
the power gain has been defined as
where
IL
is
the current through a resistive load, we can write