参数资料
型号: ATF-54143-BLK
厂商: AGILENT TECHNOLOGIES INC
元件分类: 小信号晶体管
英文描述: C BAND, GaAs, N-CHANNEL, RF SMALL SIGNAL, HEMFET
封装: PLASTIC, SC-70, 4 PIN
文件页数: 3/17页
文件大小: 212K
代理商: ATF-54143-BLK
11
The values of resistors R1 and R2
are calculated with the following
formulas
R1 =
Vgs
(2)
p
I
BB
R2 =
(V
ds – V
gs) R1
(3)
p
Vgs
Example Circuit
V
DD = 5 V
Vds = 3V
Ids = 60 mA
Vgs = 0.59V
Choose I
BB to be at least 10X the
normal expected gate leakage
current. I
BB was chosen to be
2 mA for this example. Using
equations (1), (2), and (3) the
resistors are calculated as
follows
R1 = 295
R2 = 1205
R3 = 32.3
Active Biasing
Active biasing provides a means
of keeping the quiescent bias
point constant over temperature
and constant over lot to lot
variations in device dc perfor-
mance. The advantage of the
active biasing of an enhancement
mode PHEMT versus a depletion
mode PHEMT is that a negative
power source is not required. The
techniques of active biasing an
enhancement mode device are
very similar to those used to bias
a bipolar junction transistor.
INPUT
C1
C2
C3
C7
L1
R5
R6
R7
R3
R2
R1
Q2
Vdd
R4
L2
L3
L4
Q1
Zo
C4
C5
C6
OUTPUT
Figure 2. Typical ATF-54143 LNA with
Active Biasing.
An active bias scheme is shown
in Figure 2. R1 and R2 provide a
constant voltage source at the
base of a PNP transistor at Q2.
The constant voltage at the base
of Q2 is raised by 0.7 volts at the
emitter. The constant emitter
voltage plus the regulated V
DD
supply are present across resis-
tor R3. Constant voltage across
R3 provides a constant current
supply for the drain current.
Resistors R1 and R2 are used to
set the desired Vds. The com-
bined series value of these
resistors also sets the amount of
extra current consumed by the
bias network. The equations that
describe the circuit’s operation
are as follows.
V
E = V
ds + (Ids R4)
(1)
R3 =
V
DD – V
E
(2)
p
Ids
V
B = VE – VBE
(3)
V
B =
R1
V
DD
(4)
p
R1 + R2
V
DD = I
BB (R1 + R2)
(5)
Rearranging equation (4)
provides the following formula
R2 =
R
1 (V
DD – VB)
(4A)
p
VB
and rearranging equation (5)
provides the following formula
R1 =
V
DD
(5A)
9
I
BB
(1 +
V
DD – VB
)p
V
B
Example Circuit
V
DD = 5 V
Vds = 3V
I
ds = 60 mA
R4 = 10
V
BE = 0.7 V
Equation (1) calculates the
required voltage at the emitter of
the PNP transistor based on
desired Vds and Ids through
resistor R4 to be 3.6V. Equation
(2) calculates the value of resis-
tor R3 which determines the
drain current Ids. In the example
R3 = 23.3
. Equation (3) calcu-
lates the voltage required at the
junction of resistors R1 and R2.
This voltage plus the step-up of
the base emitter junction deter-
mines the regulated Vds. Equa-
tions (4) and (5) are solved
simultaneously to determine the
value of resistors R1 and R2. In
the example R1=1450
and
R2 = 1050
. R7 is chosen to be
1k
. This resistor keeps a small
amount of current flowing
through Q2 to help maintain bias
stability. R6 is chosen to be
10k
. This value of resistance is
necessary to limit Q1 gate
current in the presence of high
RF drive level (especially when
Q1 is driven to P1dB gain com-
pression point).
相关PDF资料
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ATF-54143-BLKG C BAND, GaAs, N-CHANNEL, RF SMALL SIGNAL, HEMFET
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