AD829 Analog Devices, AD829 Datasheet - Page 14
AD829
Manufacturer Part Number
AD829
Description
High Speed, Low Noise Video Op Amp
Manufacturer
Analog Devices
Datasheet
1.AD829.pdf
(20 pages)
Specifications of AD829
-3db Bandwidth
120MHz
Slew Rate
230V/µs
Vos
200µV
Ib
3.3µA
# Opamps Per Pkg
1
Input Noise (nv/rthz)
1.7nV/rtHz
Vcc-vee
9V to 36V
Isy Per Amplifier
6.5mA
Packages
DIP,LCC,SOIC
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AD829
Because the closed-loop bandwidth is a function of R
C
loop gain, as shown in Figure 41. To preserve stability, the time
constant of R
<65 MHz. For example, with C
small signal bandwidth of the AD829 is 10 MHz. Figure 40
shows that the slew rate is in excess of 60 V/µs. As shown in
Figure 41, the closed-loop bandwidth is constant for gains of
−1 to −4; this is a property of the current feedback amplifiers.
Figure 40. Large Signal Pulse Response of Inverting Amplifier Using Current
V
COMP
IN
Feedback Compensation, C
Figure 41. Closed-Loop Gain vs. Frequency for the Circuit of Figure 38
Figure 39. Inverting Amplifier Connection Using Current Feedback
(see Figure 39), it is independent of the amplifier closed-
*RECOMMENDED VALUE
OF C
<7pF
≥7pF
100%
90
10
0%
–12
–15
CABLE
COAX
15
12
–3
–6
–9
50Ω
100k
9
6
3
0
50Ω
COMP
GAIN = –4
–3dB @ 8.2MHz
GAIN = –2
–3dB @ 9.6MHz
GAIN = –1
–3dB @ 10.2MHz
V
V
R
R
C
C1 = 15pF
IN
S
L
F
COMP
F
= ±15V
= 1kΩ
= 1kΩ
and C
= –30dBm
FOR C1
0pF
15pF
C1*
5V
R1
= 15pF
COMP
IN4148
1M
needs to provide a bandwidth of
COMP
Compensation
C
15pF FOR THIS CONNECTION
FREQUENCY (Hz)
COMP
= 15 pF, C1 = 15 pF R
R
COMP
F
SHOULD NEVER EXCEED
0.1µF
2
3
= 15 pF and R
AD829
–
+
C
+V
COMP
7
10M
S
–V
4
S
200ns
5
F
0.1µF
= 1 kΩ, R1 = 1 kΩ
6
F
= 1 kΩ, the
F
and
R
1kΩ
100M
L
V
OUT
Rev. I | Page 14 of 20
Figure 42 is an oscilloscope photo of the pulse response of a unity-
gain inverter that has been configured to provide a small signal
bandwidth of 53 MHz and a subsequent slew rate of 180 V/µs;
R
response as a unity-gain inverter, this using component values
of R
Figure 43. Small Signal Pulse Response of Inverting Amplified Using Current
F
Figure 42. Large Signal Pulse Response of the Inverting Amplifier Using
= 3 kΩ and C
F
Current Feedback Compensation, C
= 1 kΩ and C
100%
90
10
0%
100%
90
10
0%
Feedback Compensation, C
20mV
5V
COMP
COMP
= 1 pF. Figure 43 shows the excellent pulse
= 4 pF.
COMP
COMP
= 4 pF, R
= 1 pF, R
F
= 1 kΩ, R1 = 1 kΩ
200ns
10ns
F
= 3 kΩ, R1 = 3 kΩ
Data Sheet