参数资料
型号: LT1969CMS#PBF
厂商: Linear Technology
文件页数: 6/20页
文件大小: 0K
描述: IC OP-AMP ADJ CURRNT DUAL 10MSOP
标准包装: 50
放大器类型: 通用
电路数: 2
转换速率: 200 V/µs
增益带宽积: 700MHz
电流 - 输入偏压: 1.5µA
电压 - 输入偏移: 1000µV
电流 - 电源: 7mA
电流 - 输出 / 通道: 700mA
电压 - 电源,单路/双路(±): 4 V ~ 13 V,±2 V ~ 6.5 V
工作温度: 0°C ~ 70°C
安装类型: 表面贴装
封装/外壳: 10-TFSOP,10-MSOP(0.118",3.00mm 宽)
供应商设备封装: 10-MSOP
包装: 管件
14
LT1969
APPLICATIO S I FOR ATIO
WU
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wasted in the termination resistor. Second, the signal is
halved so the gain of the amplifer must be doubled to have
the same overall gain to the load. The increase in gain
increases noise and decreases bandwidth (which can also
increase distortion). Third, the output swing of the ampli-
fier is doubled which can limit the power it can deliver to
the load for a given power supply voltage.
An alternate method of back-termination is shown in
Figure 7. Positive feedback increases the effective back-
termination resistance so RBT can be reduced by a factor
of n. To analyze this circuit, first ground the input. As RBT =
RL/n, and assuming RP2>>RL we require that:
Va = Vo (1 – 1/n) to increase the effective value of
RBT by n.
Vp = Vo (1 – 1/n)/(1 + RF/RG)
Vo = Vp (1 + RP2/RP1)
Figure 7. Back-Termination Using Positive Feedback
+
1969 F07
RF
RBT
RP2
RP1
RG
Vi
Va
VP
Vo
RL
RF
RG
1 +
RL
n
=
Vo
Vi
= 1 –
1
n
FOR RBT =
()
RF
RG
1 +
()
RP1
RP1 + RP2
RP1
RP2 + RP1
RP2/(RP2 + RP1)
()
1 + 1/n
Eliminating Vp, we get the following:
(1 + RP2/RP1) = (1 + RF/RG)/(1 – 1/n)
For example, reducing RBT by a factor of n = 4, and with an
amplifer gain of (1 + RF/RG) = 10 requires that RP2/RP1
= 12.3.
Note that the overall gain is increased:
V
RR
R
nR
R
o
i
PP
P
FG
P
=
+
()
+
() +
()
[]+
()
[]
22
1
12
1
11
1
/
//
/
A simpler method of using positive feedback to reduce the
back-termination is shown in Figure 8. In this case, the
drivers are driven differentially and provide complemen-
tary outputs. Grounding the inputs, we see there is invert-
ing gain of –RF/RP from –Vo to Va
Va = Vo (RF/RP)
Figure 8. Back-Termination Using Differential Positive Feedback
+
RBT
RF
RG
RP
RG
RL
–Vi
Va
–Va
Vi
–Vo
Vo
+
RBT
1969 F08
RF
RL
n
=
Vo
Vi
n =
1 –
2
FOR RBT =
RF
RP
RF
RP
+
RF
RG
1 +
1 –
RF
RP
1
()
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