1. Classification of PAs: linear vs. switching
a. class A,
b. class B,
c. class AB,
d. class C
RF PA types
Linear PAs – Classes A, B, AB, C(?)
Switching mode PAs– Classes D, E, F2, F3, S

2. Class A Power Amplifiers Highest Lin. But  ≤ 1/2
RFC
RFC

3. Class B, PA Operates ideally with zero IQ. DC power is small
 ≤ 78.53%
single ended and push-pull versions

6. CLASS-AB & B Crossover Distortion Non-linear effects
low IQ – no IQ for vin=0
Crossover Distortion
Non-linear effects

7. CLASS C
It is biased so that the transistor conduction angle is significantly less than 180°.
Tuned circuit is mandatory, its Q determines the BW.
It is useful for providing a high-power CW or FM signal.

8. CLASS C
class C amplifier is NL- it does not directly replicate the input signal.
It requires one transistor- topologically similar to the class A except for the dc bias levels.
A tuned output (filter) is mandatory. Its Q determines the bandwidth of the amplifier.

9. General structure

10. CLASS C
Due to the RFC (large inductance RF coil), only DC current is drawn from the power supply.
When the transistor is on, the bias supply current flows through the transistor and the output voltage is approximately 90% of VCC.
When Q is off, the current from the power supply flows into the capacitor.
A major problem is the breakdown voltage and the leakage current reducing the efficiency.

11. PA’s Summary

12. PA
example
In this example the power handling of a MOSFET with, Imax=1A and Vmax=10V is examined. Let’s assume Ron= 2 for all values of VGS and ro=∞ so that the transistor’s I-V characteristics can be simplified as shown.
Find PL, max in class A operation.
Find RL required to achieve PL, max.
III. Determine the drain efficiency for
PL, max and find its maximum value,
and find the corresponding RL.
Determine the attainable efficiency in case of class A operation at 1/2 PL, max.
Repeat Section I for a fixed VDD=3.4 V.
Imax ID
VDS

13. PA
example

14. PA example The maximum RF power that can be delivered to the load
II. The optimal value of the load resistance:

15. PA example III. The drain efficiency for PL,max of part I.
The maximum value of drain efficiency is obtained for Vmin=0

16. PA example The drain efficiency for PL=0.5 PL,max
A new value of RL should be taken to satisfy the following equation the same Vmax

17. PA example V. In this case, Imax=1A should be maintained.
So, Vmin=2V, and Vmax= 2VDD-Vmin=23.4 – 2 =4.8 V

18. Matching Networks (MNs)
Objective: to optimize coupling of RF power to load.
RF power supply MN
Antenna
Zg Zin
Zout Zant
For optimum power transfer, we require:
Zg = Zin* and Zout = Zant*

19. When RL = RS, the power delivered to the load is a maximum:
DC Resistive Circuits
the concept of available power RS RL ES
When RL = RS, the power delivered to the load is a maximum:
PL,max is called the available power of the generator.

20. AC Circuits ZS 
VOC ZL

21. AC Circuits
Notes:
 Since reactances are frequency dependent, perfect impedance match is obtained only at a single frequency.
 At all other frequencies, the matching becomes progressively worse, and eventually non-existent.
 For broad-band applications, the match bandwidth must be increased.

22. The L networks
These are two-element MN consisting of one capacitor and one inductor arranged in an L-shape orientation.
There are 4 possible arrangements

23. Example
A 100 generator is connected to a resistive load of RL = 1000.
Find the ratio of the actual load power and the available power. 
1000
100 vs
Solution
Let's assume vs = A cos t
Available power is obtained for a load of RL = 100.

24. For RL = 100, the input (at the load terminals) current and voltage are:
Thus:

25. Actual power:

26. The matching network cancels this loss.
In this example, without impedance matching, about 4.8 dB of the available source power would be lost.
The matching network cancels this loss.
The parallel combination of the load resistor and the MN capacitor is calculated by

27. Please note that Rp > Rs
For XC=333
This is a series combination of a 100-ohm resistor and a j300-ohm capacitor. A
100
-j300 
-j333
1000 B
Please note that Rp > Rs
To complete the Z-match, all we must do is to add an equal and opposite (i.e. +j300 ohm) reactance in series with the source resistor.

28. L - Network Design Procedure
QS = Q of the series leg.
QP = Q of the shunt leg.
RP = The shut resistance.
RS = The series resistance.
XP = The shunt reactance.
XS = The series reaction.

29. Example
Design an L-shape matching network for ZS=100 ohm and ZL=1000 ohm at 100 MHz.
Assume that a dc voltage must also be transferred from source to the load.

30. Solution
The need for dc path dictates an inductor in the series leg.

31. Example
Design a circuit to match a source of ZS=100 +j126 ohm to a load consisting a parallel combination of a 1000-ohm resistor and 2 pF capacitor, at f=100MHz.

32. Design for a parallel capacitor and a series inductor.
Solution
First, we'll totally ignore the reactance and simply match the real part of ZS to the real part of ZL.
Design for a parallel capacitor and a series inductor.