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LT1499CS 数据表(PDF) 13 Page - Linear Technology

部件名 LT1499CS
功能描述  10MHz, 6V/us, Dual/Quad Rail-to-Rail Input and Output Precision C-Load Op Amps
PDF  16 Pages
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制造商  LINER [Linear Technology]
网页  http://www.linear.com
标志 LINER - Linear Technology

LT1499CS 数据表(HTML) 13 Page - Linear Technology

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LT1498/LT1499
APPLICATIONS INFORMATION
without oscillation at unity gain. When driving a heavy
capacitive load, the bandwidth is reduced to maintain
stability. Figures 2a and 2b illustrate the stability of the
device for small-signal and large-signal conditions with
capacitive loads. Both the small-signal and large-signal
transient response with a 10nF capacitive load are well
behaved.
Feedback Components
To minimize the loading effect of feedback, it is possible to
use the high value feedback resistors to set the gain.
However, care must be taken to insure that the pole formed
by the feedback resistors and the total input capacitance at
the inverting input does not degrade the stability of the
amplifier. For instance, the LT1498/LT1499 in a noninvert-
ing gain of 2, set with two 30k resistors, will probably
oscillate with 10pF total input capacitance (5pF input
capacitance + 5pF board capacitance). The amplifier has a
2.5MHz crossing frequency and a 60
°phasemarginat6dB
of gain. The feedback resistors and the total input capaci-
tance create a pole at 1.06MHz that induces 67
° of phase
shift at 2.5MHz! The solution is simple, either lower the
value of the resistors or add a feedback capacitor of 10pF
of more.
TYPICAL APPLICATIONS N
1A Voltage Controlled Current Source
1A Voltage Controlled Current Sink
+
1/2 LT1498
1k
500pF
tr < 1µs
1498/99 TA03
Si9430DY
VIN
V+
RL
100
0.5
1k
IOUT =
IOUT
V+ – VIN
0.5
1k
VIN
+
1/2 LT1498
500pF
1498/99 TA04
Si9410DY
V+
V+
RL
IOUT
100
0.5
1k
IOUT =
VIN
0.5
tr < 1µs
Figure 2b. LT1498 Large-Signal Response
VS = 5V
AV = 1
1498/99 F02a
CL = 10nF
CL = 500pF
CL = 0pF
Figure 2a. LT1498 Small-Signal Response
1498/99 F02b
VS = 5V
AV = 1
CL = 0pF
CL = 500pF
CL = 10nF



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