参数资料
| 型号： | AD810ANZ |
| 厂商： | Analog Devices Inc |
| 文件页数： | 4/16页 |
| 文件大小： | 0K |
| 描述： | IC CURR-FDBK AMP VIDEO LP 8-DIP |
| 标准包装： | 50 |
| 应用： | 电流反馈 |
| 电路数： | 1 |
| -3db带宽： | 80MHz |
| 转换速率： | 1000 V/µs |
| 电流 - 电源： | 6.8mA |
| 电流 - 输出 / 通道： | 60mA |
| 电压 - 电源，单路/双路(±)： | 5 V ~ 36 V，±2.5 V ~ 18 V |
| 安装类型： | 通孔 |
| 封装/外壳： | 8-DIP（0.300"，7.62mm） |
| 供应商设备封装： | 8-PDIP |
| 包装： | 管件 |
| 产品目录页面： | 768 (CN2011-ZH PDF) |
AD810
REV. A
–12–
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
is all that is needed to
achieve an extremely flat gain response.
Figures 44 to 46 illustrate the outstanding performance that can
be achieved when driving a 1000 pF capacitor.
AD810
RF
RG
7
3
2
3
0.1F
+VS
6
–VS
4
0.1F
RL
VIN
VO
CL
RS (OPTIONAL)
RT
1.0F
Figure 43. Circuit Options for Driving a Large
Capacitive Load
0
1
10
100
–3
–6
–9
3
6
9
G = +2
VS = ±15V
RL= 10k
CL = 1000pF
RF = 750
RS = 11
RF = 4.5k
RS = 0
CLOSED-LOOP
GAIN
–
dB
FREQUENCY – MHz
Figure 44. Performance Comparison of Two Methods for
Driving a Large Capacitive Load
2k
1k
4k
3k
1
10
100
1000
LOAD
CAPACITANCE
–
pF
FEEDBACK RESISTOR –
0
VS = ±15V
GAIN = +2
RL = 1k
VS = ±5V
Figure 45. Max Load Capacitance for Less than 1 dB of
Peaking vs. Feedback Resistor
90
100
0%
5V
100nS
V
IN
VOUT
Figure 46. AD810 Driving a 1000 pF Load,
Gain = +2, RF = 750
, R
S = 11
, R
L = 10 k
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
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