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| 型号： | AD8079BR-REEL7 |
| 厂商： | Analog Devices Inc |
| 文件页数： | 10/12页 |
| 文件大小： | 0K |
| 描述： | IC AMP BUF DUAL LP LDIST 8SOIC |
| 标准包装： | 1,000 |
| 放大器类型： | 缓冲器 |
| 电路数： | 2 |
| 转换速率： | 800 V/µs |
| -3db带宽： | 260MHz |
| 电流 - 输入偏压： | 3µA |
| 电压 - 输入偏移： | 10000µV |
| 电流 - 电源： | 10mA |
| 电流 - 输出 / 通道： | 70mA |
| 电压 - 电源，单路/双路(±)： | 6 V ~ 12 V，±3 V ~ 6 V |
| 工作温度： | -40°C ~ 85°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 8-SOIC（0.154"，3.90mm 宽） |
| 供应商设备封装： | 8-SOIC |
| 包装： | 带卷 (TR) |
9
REV. A
AD8079
–7–
RESISTANCE
–
FREQUENCY – Hz
10k
1G
100k
1M
10M
100M
100
10
1
0.1
0.01
VS = ±5.0V
POWER = 0dBm
(223.6mV rms)
RbT = 50
RbT = 0
Figure 21. Output Resistance vs. Frequency
–44.0
–66.5
125
–61.5
–64.0
–35
–55
–59.0
–56.5
–54.0
–51.5
–49.0
–46.5
105
85
65
45
25
5
–15
JUNCTION TEMPERATURE –
°C
PSRR
–
dB
–69.0
–PSRR
+PSRR
2V SPAN
CURVES ARE FOR WORST
CASE CONDITION WHERE
ONE SUPPLY IS VARIED
WHILE THE OTHER IS
HELD CONSTANT.
Figure 22. PSRR vs. Temperature
PSRR
–
dB
FREQUENCY – Hz
0
–4
–84
30k
500M
100k
1M
10M
100M
–14
–24
–64
–34
–44
–54
–74
VIN = 200mV
–PSRR
+PSRR
Figure 23. PSRR vs. Frequency
THEORY OF OPERATION
The AD8079, a dual current feedback amplifier, is internally
configured for a gain of either +2 (AD8079A) or +2.2
(AD8079B). The internal gain-setting resistors effectively elimi-
nate any parasitic capacitance associated with the inverting in-
put pin, accounting for the AD8079’s excellent gain flatness
response. The carefully chosen pinout greatly reduces the cross-
talk between each amplifier. Up to four back-terminated 75
video loads can be driven by each amplifier, with a typical dif-
ferential gain and phase performance of 0.01%/0.17
°, respec-
tively. The AD8079B, with a gain of +2.2, can be employed as a
single gain-trimming element in a video signal chain. Finally,
the AD8079A/B used in conjunction with our AD8116 cross-
point matrix, provides a complete turn-key solution to video
distribution.
Printed Circuit Board Layout Considerations
As to be expected for a wideband amplifier, PC board parasitics
can affect the overall closed-loop performance. If a ground
plane is to be used on the same side of the board as the signal
traces, a space (5 mm min) should be left around the signal lines
to minimize coupling. Line lengths on the order of less than
5 mm are recommended. If long runs of coaxial cable are being
driven, dispersion and loss must be considered.
Power Supply Bypassing
Adequate power supply bypassing can be critical when optimiz-
ing the performance of a high frequency circuit. Inductance in
the power supply leads can form resonant circuits that produce
peaking in the amplifier’s response. In addition, if large current
transients must be delivered to the load, then bypass capacitors
(typically greater than 1
F) will be required to provide the best
settling time and lowest distortion. A parallel combination of
4.7
F and 0.1 F is recommended. Some brands of electrolytic
capacitors will require a small series damping resistor
≈ 4.7
for optimum results.
DC Errors and Noise
There are three major noise and offset terms to consider in a
current feedback amplifier. For offset errors refer to the equa-
tion below. For noise error the terms are root-sum-squared to
give a net output error. In the circuit below (Figure 24) they are
input offset (VIO) which appears at the output multiplied by the
noise gain of the circuit (1 + RF/RI), noninverting input current
(IBN
× R
N) also multiplied by the noise gain, and the inverting
input current, which when divided between RF and RI and sub-
sequently multiplied by the noise gain always appears at the out-
put as IBN
× R
F. The input voltage noise of the AD8079 is a low
2 nV/
√Hz. At low gains though the inverting input current noise
times RF is the dominant noise source. Careful layout and de-
vice matching contribute to better offset and drift specifications
for the AD8079 compared to many other current feedback am-
plifiers. The typical performance curves in conjunction with the
equations below can be used to predict the performance of the
AD8079 in any application.
V OUT =V IO × 1+
R
F
R
I
± IBN × RN × 1+
R
F
R
I
± IBI × RF
where:
RF = RI = 750 for AD8079A
RF = 750
, R
I = 625
for AD8079B
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