Communication Concepts AN758 - Manual (Page 3)

AR
C
HIVE INF
O
RMA
TI
O
N
PRODUCT TRANSFERRED T
O
M/A
–
COM
AN758
3 RF Application Reports
C1
= 4.65 Ω
X
I
= j 1.25 Ω
R2
R1
R
I
R
S
L2 = 55 nH
L1 = 4 nH
R
L
VCS1
Figure 2. Equivalent Base Input Circuit
Since a junction transistor is a current amplifier, it should
ideally be driven from a current source. In RF applications
this would result in excessive loss of power gain. However,
input networks can be designed with frequency slopes
having some of the current source characteristics at low
frequencies, where excess gain is available.
The complex base input characteristics of a transistor
would place requirements for a very sophisticated input
compensation network for optimum overall performance. The
design goal here was to maintain an input VSWR of 2:1 or
less and a maximum gain variation of + 1.5 dB from 2 to
30 MHz. Initial calculations indicated that these requirements
can be met with a simple RC network in conjunction with
negative collector-to-base feedback. Figure 2 shows this
network for one device. L1 and L2 represent lead lengths,
and their values are fixed. The feedback is provided through
R2 and L2. Because the calculations were done without the
feedback, this branch is grounded to simulate the operating
conditions.
The average power gain variation of the MRF428 from
2 to 30 MHz is 13 dB. Due to phase errors, a large amount
of negative feedback in an RF amplifier decreases the
linearity, or may result in instabilities. Experience has shown
that approximately 5 – 6 dB of feedback can be tolerated
without noticeable effects in linearity or stability, depending
upon circuit layout. If the amount of feedback is 5 dB, 8 dB
will have to be absorbed by the input network at 2 MHz.
Omitting the reactive components, L1, L2, C1, and the
phase angle of X
I
which have a negligible effect at 2 MHz,
a simple L-pad was calculated with R
S
= 2.77
Ω, and
R
L
= 4.65 2 + 1.25 2 = 4.81 Ω .
From the device data sheet we find G
PE
at 2 MHz is about
28 dB, indicating 0.24 W at R
L
will produce an output power
of 150 W, and the required power at R
S
= 0.24 W + 8 dB
= 1.51 W.
Figuring out currents and voltages in various branches
results in: R1 = 1.67 Ω and R2 = 1.44 Ω.
The calculated values of R1 and R2 along with other
known values and the device input data at four frequencies
were used to simulate the network in a computer program.
An estimated arbitrary value of 4000 pF for C1 was chosen,
and VCS2 represents the negative feedback voltage
(Figure 2.) The optimization was done in two separate
programs for R1, R2, C1 and VCS2 and in several steps.
The goals were: a) VCS and R2 for a transducer loss of
13 dB at 2 MHz and minimum loss at 30 MHz. b) R1 and
C1 for input VSWR of < 1.1:1 and < 2:1 respectively. The
optimized values were obtained as:
C1 = 5850 pF R2 = 1.3 Ω
R1 =2.1 Ω VCS2= 1.5 V
The minimum obtainable transducer loss at 30 MHz was
2.3 dB, which is partly caused by the highest reflected power
at this frequency, and can be reduced by “overcompensa-
tion” of the input transformer. This indicates that at the higher
frequencies, the source impedance (R
S
) is effectively de-
creased, which leaves the input VSWR highest at 15 MHz.
In the practical circuit the value of C1 (and C2) was
rounded to the nearest standard, or 5600 pF. For each half
cycle of operation R2 and R4 are in series and the value
of each should be
2
1.3 Ω
for VCS2 = 1.5 V. Since the voltage across ac and bd = V
CE
,
a turns ratio of 32:1 would be required. It appears that if the
feedback voltage on the bases remains unchanged, the ratio
of the voltage across L5 (VCS2) and R2R4 can be varied
with only a small effect to the overall input VSWR. To mini-
mize the resistive losses in the bifilar winding of T2 (Fig-
ure 3), the highest practical turns ratio should not be much
higher than required for the minimum inductance, which is
2 πf
4R
=
12.5
50
= 4.0 µH .
R = Collector-to-Collector Impedance = 12.5 Ω
f = 2 MHz
ac or bd will then be 1.0 µH, which amounts to 5 turns. (See
details on T2.) 25% over this represents a 7:1 ratio setting
VCS2 to 6.9 V.