參數(shù)資料
型號(hào): LT1498
廠商: Linear Technology Corporation
英文描述: 10MHz, 6V/μs, Dual Rail-to-Rail Input and Output Precision C-Load Op Amps(10MHz, 6V/μs,雙路滿幅度輸入和輸出,精密型,電容負(fù)載運(yùn)算放大器)
中文描述: 10MHz的,6V/μs,雙軌至軌輸入和輸出精度葷負(fù)載運(yùn)算放大器(10MHz的,6V/μs,雙路滿幅度輸入和輸出,精密型,電容負(fù)載運(yùn)算放大器)
文件頁(yè)數(shù): 13/16頁(yè)
文件大?。?/td> 264K
代理商: LT1498
13
LT1498/LT1499
APPLICATIO
S I
N
FOR
ATIO
U
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.
W
U
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
°
phase margin at 6dB
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 APPLICATIO
S
N
U
1A Voltage Controlled Current Source
1A Voltage Controlled Current Sink
+
1/2 LT1498
1k
500pF
t
r
< 1
μ
s
1498/99 TA03
Si9430DY
V
IN
V
+
R
L
100
0.5
1k
I
OUT
=
I
OUT
V
+
– V
IN
0.5
1k
V
IN
+
1/2 LT1498
500pF
1498/99 TA04
Si9410DY
V
+
V
+
R
L
I
OUT
100
0.5
1k
I
OUT
=
t
r
< 1
μ
s
V
IN
0.5
Figure 2b. LT1498 Large-Signal Response
V
S
= 5V
A
V
= 1
Figure 2a. LT1498 Small-Signal Response
1498/99 F02a
C
L
= 10nF
C
L
= 500pF
C
L
= 0pF
1498/99 F02b
V
S
= 5V
A
V
= 1
C
L
= 0pF
C
L
= 500pF
C
L
= 10nF
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