Radio Frequency Integrated Circuit Design - Webs

374 Radio Frequency Integrated Circuit Design
Example 10.3 Class E Amplifier
Design a class E amplifier that delivers 200 mW from a 3-V power supply at
2.4 GHz. Assume an ideal transistor and aim for a Q of 3 for the output circuit.
Specify device ratings and components.
Solution
Using Po ≈ 0.577V 2
CC /R results in R = 26�. The maximum transistor voltage
is vC,max = 3.56 � 3 = 10.68V. If this large voltage is not permissible (and it
is quite likely that it is not), the power supply voltage may need to be reduced.
From (10.28),
I dc = VCC
1.734R =
3
= 66.6 mA
1.734 � 26
ic,peak = 2.861I dc = 190.6 mA
B = 0.1836 0.81Q
R �1 +
Q 2 0.1836 1 + 0.81 � 5
=
+ 4� 26 � 3 2 � = 0.00755
+ 4
Hence C = 0.50 pF. Since Q = 3, C o has a reactance of 78� and is
therefore 0.85 pF. Using (10.27),
X = 1.10Q
Q − 0.67
1.110 � 3
R = 26 = 37.2�
3 − 0.67
L o therefore has a reactance of 37.2 + 78 = 115.2� and is thus 7.64 nH.
Ideally, the RFC should have a reactance of at least 10R and thus should be
at least 17 nH, which would likely need to be an off-chip inductor.
This circuit was simulated using a process with f T that is 25 times higher
than the operating frequency. With numbers calculated as above, and choosing
a transistor size that has optimal f T at about 66 mA, the results in Figure 10.25
were obtained. For simplicity, the input was a ±1.5V pulse waveform through
a 50-� source resistance. In a real circuit, a more realistic input waveform
would have to be used.
It can be seen that the output transistor collector voltage has not gone
down to zero when the transistor switches on. The problem is the parasitic
output capacitance of the very large transistor. As a result, the output power is
only about 77 mW and dc power is of the order of 100 mW. As a first-order
compensation, the capacitor C can be reduced to compensate. With this done,
the results are as shown in Figure 10.26.