Class B, PA - Huji cse Moodle 2014/15
Download
Report
Transcript Class B, PA - Huji cse Moodle 2014/15
Classification of PAs: linear vs. switching
a. class A,
b. class B,
c. class AB,
d. class C
1
RF PA types
• Linear PAs – Classes A, B, AB, C(?)
• Switching mode PAs– Classes D, E,
F2, F3, S
Class A Power Amplifiers
Highest Lin. But ≤ 1/2
vIN θ
VDD
RFC
vD θ
vIN θ
VTH
vOUT θ
iOUT θ
iD θ
vD θ
iD θ
vD θ
VDD
vOUT θ
θ
iOUT θ
vOUT θ
0
θ 0
iOUT θ
Vom
η Drain
2
iD θ
θ
VDD
RFC
θ
θ
2
2
P
2 R Vom 1
RFout
2
2
PDC
2
VDD
2VDD
R
Class B, PA
Operates ideally with zero IQ. DC power is small
≤ 78.53%
single ended and push-pull versions
3
4
5
CLASS-AB & B
• low IQ – no IQ for vin=0
•Crossover Distortion
• Non-linear effects
6
CLASS C
Vdd
vO
iD
Vin
RLoad
Vt
t
iD
t
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.
7
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.
8
General structure
9
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.
10
PA’s Summary
11
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.
ID
I.Find PL, max in class A operation.
Imax
II.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.
IV.Determine the attainable efficiency in case of class A operation
at 1/2 PL, max.
V.Repeat Section I for a fixed VDD=3.4 V.
2
VDS
PA
example
VDS VDD
I D IQ
RL
ROL
PL , RF
13
2VDS VDD
I max
I max VDD Vmin
4
PA
example
I. The maximum RF power that can be delivered to the load
PL ,max
1 1
1
V ptp I ptp
2 2
2
1
1
Vmax Vmin I max 10 2 1 1W
8
8
II. The optimal value of the load resistance:
V 10 2
RL
8
I
1
PA
example
III. The drain efficiency for PL,max of part I.
pL ,out
1W
1
A V 33.3%
I Q VQ 0.5 6
3
The maximum value of drain efficiency is obtained for Vmin=0
1 Vmax Vmin
1
2 Vmax Vmin Vmin 0
2
15
PA
example
IV. 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
1
2
W
1
V ptp I ptp
8
1
Vmax Ron I max I max
8
I max 0.44 A, RL 20.8, 42%
16
PA
example
V. In this case, Imax=1A should be maintained.
So, Vmin=2V, and Vmax= 2VDD-Vmin=23.4 – 2 =4.8 V
1 1
1
PL ,max V ptp I ptp
2 2
2
( 4 .8 2 )
0.35W
8
17
Matching Networks (MNs)
Objective: to optimize coupling of RF power to load.
RF power
supply
Zg
Zin
MN
Zout
Zant
For optimum power transfer, we require:
Zg = Zin*
18
and
Zout = Zant*
Antenna
DC Resistive Circuits
the concept of available power
RS
RL
ES
When RL = RS, the power delivered to the load is a maximum:
2
PL ,max
ES
4 RS
PL,max is called the available power of the generator.
19
AC Circuits
ZS
VOC
Z S RS jX S ,
For Z L Z S* ,
20
ZL
Z L RL jX L
PL PL ,max
2
VOC
8RS
AC Circuits
Notes:
21
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.
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
22
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.
100
vs
1000
Solution
Let's assume vs = A cos t
Available power is obtained for a load of RL = 100.
23
For RL = 100, the input (at the load terminals)
current and voltage are:
A cos t
iL
2 RL
1
v L A cos t ,
2
Thus:
A 2 cos 2 t
PL (t ) i L (t )v L (t )
4 RL
PL t
24
max
A2
A2
8RL 800
Actual power:
10 A
1000
cos t
vL A cos t
1000 100 11
A
A cos t
cos t
iL
1000 100 1100
A2
cos 2 t
PL (t ) iL (t )vL (t )
1210
A2
PL t Actual
2420
PActual 800
0.33 4.8dB
2420
Pmax
25
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
Z
26
jX C RL
jX C RL
For XC=333
Z 100 j300 ohm
This is a series combination of a 100-ohm resistor and a j300-ohm capacitor.
A
1000
-j333
Please note that Rp > Rs
100
-j300
B
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.
27
L - Network Design Procedure
QS Q P
28
QS
XS
RS
QP
RP
XP
RP
1
RS
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.
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.
29
Solution
The need for dc path dictates an inductor in the series leg.
1000
1 3
100
X S QS RS 3 100 300 ohm (inductive)
QS QP
RP 1000
XP
333 ohm (capacitive)
QP
3
Component values for f 100 MHz :
X
300
L S
477 nH
6
2 100 10
1
C
4.8 pF
X P
30
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.
31
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.
32