參數(shù)資料
型號: ATF-54143-BLK
元件分類: 小信號晶體管
英文描述: C BAND, GaAs, N-CHANNEL, RF SMALL SIGNAL, HEMFET
封裝: PLASTIC, SC-70, 4 PIN
文件頁數(shù): 3/17頁
文件大?。?/td> 305K
代理商: ATF-54143-BLK
11
The values of resistors R1 and R2 are calculated with the
following formulas
R1 = Vgs (2)
p
IBB
R2 = (Vds – Vgs) R1 (3)
Vgs
Example Circuit
VDD = 5V
Vds = 3V
Ids = 60 mA
Vgs = 0.59V
Choose IBB to be at least 10X the normal expected gate
leakage current. IBB was chosen to be 2 mA for this
example. Using equations (1), (2), and (3) the resistors
are calculated as follows
R1 = 295W
R2 = 1205W
R3 = 32.3W
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 per
formance. 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.
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 VDD supply are
present across resistor 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
combined 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.
VE = Vds + (Ids R4)
(1)
R3 = VDD – VE
(2)
p
Ids
VB = VE – VBE
(3)
VB =
R1
VDD
(4)
p
R1 + R2
VDD = IBB (R1 + R2) (5)
Rearranging equation (4) provides the following
formula:
R2 =
R1 (VDD – VB) (4A)
VB
and rearranging equation (5) provides the following
formula:
R1 =
VDD
(5A)
9
IBB
(1+ V
DD – VB
)
p
VB
Example Circuit
VDD = 5V
Vds = 3V
Ids = 60 mA
R4 = 10W
VBE = 0.7V
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) calcu
lates the value of resistor R3 which determines the
drain current Ids. In the example R3=23.3W. Equation
(3) calculates the voltage required at the junction of
resistors R1 and R2. This voltage plus the stepup of
the base emitter junction determines the regulated
Vds. Equations (4) and (5) are solved simultaneously
to determine the value of resistors R1 and R2. In the
example R1=1450W and R2=1050W. R7 is chosen to
be 1kW. This resistor keeps a small amount of current
flowing through Q2 to help maintain bias stability. R6 is
chosen to be 10kW. 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).
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.
相關(guān)PDF資料
PDF描述
ATF-54143-TR1 C BAND, GaAs, N-CHANNEL, RF SMALL SIGNAL, HEMFET
ATF-54143-TR2 C BAND, GaAs, N-CHANNEL, RF SMALL SIGNAL, HEMFET
ATF-54143-TR1G C BAND, GaAs, N-CHANNEL, RF SMALL SIGNAL, HEMFET
ATF-54143-BLKG C BAND, GaAs, N-CHANNEL, RF SMALL SIGNAL, HEMFET
ATF-54143-TR2G C BAND, GaAs, N-CHANNEL, RF SMALL SIGNAL, HEMFET
相關(guān)代理商/技術(shù)參數(shù)
參數(shù)描述
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