AD829AR Analog Devices Inc, AD829AR Datasheet - Page 14
![IC VIDEO OPAMP LN HS 8-SOIC](/photos/5/66/56684/505-8-soic_sml.jpg)
AD829AR
Manufacturer Part Number
AD829AR
Description
IC VIDEO OPAMP LN HS 8-SOIC
Manufacturer
Analog Devices Inc
Datasheet
1.AD829ARZ.pdf
(20 pages)
Specifications of AD829AR
Slew Rate
230 V/µs
Rohs Status
RoHS non-compliant
Applications
Voltage Feedback
Number Of Circuits
1
-3db Bandwidth
120MHz
Current - Supply
5.3mA
Current - Output / Channel
32mA
Voltage - Supply, Single/dual (±)
±4.5 V ~ 18 V
Mounting Type
Surface Mount
Package / Case
8-SOIC (0.154", 3.90mm Width)
No. Of Amplifiers
1
Bandwidth
120MHz
No. Of Pins
8
Settling Time
65ns
Operating Temperature Max
85°C
Peak Reflow Compatible (260 C)
No
Input Bias Current
3300nA
Lead Free Status / RoHS Status
Contains lead / RoHS non-compliant
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Part Number:
AD829AR
Manufacturer:
ADI/亚德诺
Quantity:
20 000
Part Number:
AD829ARZ
Manufacturer:
ADI/亚德诺
Quantity:
20 000
Company:
Part Number:
AD829ARZ-REEL
Manufacturer:
NS
Quantity:
9 720
Part Number:
AD829ARZ-REEL
Manufacturer:
ADI/亚德诺
Quantity:
20 000
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
Feedback Compensation, C
IN
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
5V
C1*
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. H | 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