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參數(shù)資料
| 型號： | LT1228IS8#PBF |
| 廠商： | Linear Technology |
| 文件頁數(shù)： | 2/22頁 |
| 文件大?。?/td> | 0K |
| 描述： | IC CRNT FEEDBK AMP 100MHZ 8SOIC |
| 標(biāo)準(zhǔn)包裝： | 100 |
| 放大器類型： | 電流反饋 |
| 電路數(shù)： | 1 |
| 轉(zhuǎn)換速率： | 3500 V/µs |
| -3db帶寬： | 100MHz |
| 電流 - 輸入偏壓： | 10µA |
| 電壓 - 輸入偏移： | 3000µV |
| 電流 - 電源： | 6mA |
| 電流 - 輸出 / 通道： | 125mA |
| 電壓 - 電源，單路/雙路(±)： | 4 V ~ 36 V，±2 V ~ 18 V |
| 工作溫度： | -40°C ~ 85°C |
| 安裝類型： | 表面貼裝 |
| 封裝/外殼： | 8-SOIC（0.154"，3.90mm 寬） |
| 供應(yīng)商設(shè)備封裝： | 8-SO |
| 包裝： | 管件 |
LT1228
10
1228fd
APPLICATIONS INFORMATION
Substitutingintothe equationfortransconductancegives:
gm =
a
1.94R
=
10
R
The temperature variation in the term “a” can be ignored
since it is much less than that of the term “T” in the equa-
tion for Vbe. Using a 2.5V source this way will maintain the
gain constant within 1% over the full temperature range of
–55°C to 125°C. If the 2.5V source is off by 10%, the gain
willvaryonlyabout±6%overthesametemperaturerange.
Wecanalsotemperaturecompensatethetransconductance
withoutusinga2.5Vreferenceifthenegativepowersupply
is regulated. A Thevenin equivalent of 2.5V is generated
from two resistors to replace the reference. The two resis-
tors also determine the maximum set current, approxi-
mately 1.1V/RTH. By rearranging the Thevenin equations
to solve for R4 and R6 we get the following equations in
terms of RTH and the negative supply, VEE.
R4
=
RTH
1–
2.5V
VEE
and R6
=
RTHVEE
2.5V
Temperature Compensation of gm with a Thevenin Voltage
diode drops above the negative supply, a single resistor
from the control voltage source to Pin 5 will suffice in
many applications. The control voltage is referenced to
the negative supply and has an offset of about 900mV.
The conversion will be monotonic, but the linearity is
determined by the change in the voltage at Pin 5 (120mV
per decade of current). The characteristic is very repeat-
able since the voltage at Pin 5 will vary less than ±5%
from part to part. The voltage at Pin 5 also has a negative
temperature coefficient as described in the previous sec-
tion. When the gain of several LT1228s are to be varied
together, the current can be split equally by using equal
value resistors to each Pin 5.
Formoreaccurate(andlinear)control,avoltage-to-current
converter circuit using one op amp can be used. The fol-
lowing circuit has several advantages. The input no longer
has to be referenced to the negative supply and the input
can be either polarity (or differential). This circuit works
on both single and split supplies since the input voltage
and the Pin 5 voltage are independent of each other. The
temperature coefficient of the output current is set by R5.
LT1228 TA05
R4
1.24k
–15V
gm
5
4
R'
ISET
1.03k
Vbe
VTH = 2.5V
R'
R6
6.19k
Voltage Controlled Gain
Touseavoltagetocontrolthegainofthetransconductance
amplifier requires converting the voltage into a current
that flows into Pin 5. Because the voltage at Pin 5 is two
LT1228 TA19
R5
1k
R1
1M
V1
V2
IOUT
TO PIN 5
OF LT1228
50pF
R1 = R2
R3 = R4
IOUT =
= 1mA/V
–
+
R2
1M
R3
1M
R4
1M
(V1 – V2)
R5
R3
R1
LT1006
Digital control of the transconductance amplifier gain is
done by converting the output of a DAC to a current flow-
ing into Pin 5. Unfortunately most current output DACs
sink rather than source current and do not have output
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