CHAPTER 4 - Impedance Matching
174
4.3 Matching two unequal resistive impedances
In the case of two unequal resistive terminations we cannot carry out a simple resistive
matching as the one shown in Figure 4.3-1.
Figure 4.3-1 Resistive matching between a source of 50ΩΩ and a load of 4Ω Ω
It is quite easy to see why. The current flowing through this circuit will be equal to V
S
/100Ω. Now
since voltage is proportional to resistance we will have almost half of the source voltage dropped
across the matching resistor leaving very little voltage, and hence little power, to be delivered to the
load.
Also let us take a look at the impedances seen from source and load end. When we look into
the matching network from the source end, we will see an overall impedance of 50 Ω However
when we look back from the load end into the matching network we will see an impedance of 96 Ω!
This means that, if some of the power is reflected by the load into the matching network, it will see a
high impedance, and bounce back towards the load again. This may create oscillations which, in the
case of amplifier circuits for instance may be fatal!
Ideally we would like our matching network to dissipate as little power as possible. Reactive
elements such as capacitors and inductors are therefore ideal candidates! However bear in mind
that they do suffer from the effects of parasitics and that their real-life equivalents are lossy to some
extent.
In this section we will first look at L-section matching which makes use of two reactive
elements to create a match between two unequal resistive terminations. We will then look at the
frequency response of matching networks and ways to improve the Q.
The topics illustrated in this section are demonstrated with the aid of Microwave Office
simulations in video 4.2 and video 4.3
V
S
R
L
= 4 Ω
R
S
= 50 Ω
46 Ω
Matching
Network
50 Ω 96 Ω
Conquer Radio Frequency
174 www.cadence.com/go/awr
