AD810AR Analog Devices Inc, AD810AR Datasheet - Page 12
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AD810AR
Manufacturer Part Number
AD810AR
Description
IC CURR-FDBK AMP VIDEO LP 8-SOIC
Manufacturer
Analog Devices Inc
Datasheet
1.AD810ARZ.pdf
(16 pages)
Specifications of AD810AR
Slew Rate
1000 V/µs
Mounting Type
Surface Mount
Rohs Status
RoHS non-compliant
Applications
Current Feedback
Number Of Circuits
1
-3db Bandwidth
80MHz
Current - Supply
6.8mA
Current - Output / Channel
60mA
Voltage - Supply, Single/dual (±)
5 V ~ 36 V, ±2.5 V ~ 18 V
Package / Case
8-SOIC (0.154", 3.90mm Width)
No. Of Amplifiers
1
Bandwidth
80MHz
No. Of Pins
8
Settling Time
125ns
Operating Temperature Max
85°C
Peak Reflow Compatible (260 C)
No
Lead Free Status / RoHS Status
Contains lead / RoHS non-compliant
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Part Number:
AD810AR
Manufacturer:
ADI/亚德诺
Quantity:
20 000
Company:
Part Number:
AD810ARM-REEL7
Manufacturer:
AD
Quantity:
5 510
Company:
Part Number:
AD810ARM-REEL7
Manufacturer:
XILINX
Quantity:
5 510
Part Number:
AD810ARZ
Manufacturer:
ADI/亚德诺
Quantity:
20 000
AD810
CAPACITIVE LOADS
When used with the appropriate feedback resistor, the AD810
can drive capacitive loads exceeding 1000 pF directly without
oscillation. By using the curves in Figure 45 to chose the resistor
value, less than 1 dB of peaking can easily be achieved without
sacrificing much bandwidth. Note that the curves were
generated for the case of a 10 k load resistor, for smaller load
resistances, the peaking will be less than indicated by Figure 45.
Another method of compensating for large load capacitances is
to insert a resistor in series with the loop output as shown in
Figure 43. In most cases, less than 50
achieve an extremely flat gain response.
Figures 44 to 46 illustrate the outstanding performance that can
be achieved when driving a 1000 pF capacitor.
Figure 44. Performance Comparison of Two Methods for
Driving a Large Capacitive Load
Figure 43. Circuit Options for Driving a Large
Capacitive Load
–3
–6
–9
V
9
6
3
0
IN
1
R
G
R
T
2
3
3
AD810
+V
–V
R
R
FREQUENCY – MHz
7
4
R
F
S
S
S
F
= 4.5k
= 0
1.0µF
0.1µF
0.1µF
1.0µF
10
6
R
is all that is needed to
S
(OPTIONAL)
G = +2
V
R
C
S
L
L
= 10k
= ±15V
= 1000pF
C
R
R
L
F
S
= 750
= 11
R
L
V
O
100
–12–
DISABLE MODE
By pulling the voltage on Pin 8 to common (0 V), the AD810
can be put into a disabled state. In this condition, the supply
current drops to less than 2.8 mA, the output becomes a high
impedance, and there is a high level of isolation from input to
output. In the case of a line driver for example, the output
impedance will be about the same as for a 1.5 k resistor (the
feedback plus gain resistors) in parallel with a 13 pF capacitor
(due to the output) and the input to output isolation will be
better than 65 dB at 1 MHz.
Leaving the disable pin disconnected (floating) will leave the
AD810 operational in the enabled state.
In cases where the amplifier is driving a high impedance load,
the input to output isolation will decrease significantly if the
input signal is greater than about 1.2 V peak to peak. The
isolation can be restored back to the 65 dB level by adding a
dummy load (say 150 ) at the amplifier output. This will
attenuate the feedthrough signal. (This is not an issue for
multiplexer applications where the outputs of multiple AD810s
are tied together as long as at least one channel is in the ON
state.) The input impedance of the disable pin is about 35 k in
parallel with a few pF. When grounded, about 50 A flows out
Figure 45. Max Load Capacitance for Less than 1 dB of
Peaking vs. Feedback Resistor
V
Figure 46. AD810 Driving a 1000 pF Load,
Gain = +2, R
OUT
1000
V
100
IN
10
1
90
0
0%
100
5V
5V
1k
F
= 750 , R
V
S
= ±5V
FEEDBACK RESISTOR –
2k
S
= 11
V
S
= ±15V
3k
100nS
, R
GAIN = +2
R
L
= 1k
L
= 10 k
4k
REV. A