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AD6640ST/PCB 数据表(PDF) 13 Page - Analog Devices

部件名 AD6640ST/PCB
功能描述  12-Bit, 65 MSPS IF Sampling A/D Converter
PDF  25 Pages
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制造商  AD [Analog Devices]
网页  http://www.analog.com
标志 AD - Analog Devices

AD6640ST/PCB 数据表(HTML) 13 Page - Analog Devices

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AD6640
–12–
REV. A
ENCODE
ENCODE
AD6640
R
T1–1T
SINE
SOURCE
Figure 11. Sine Source–Differential ENCODE
If a low jitter ECL clock is available, another option is to ac-couple
a differential ECL signal to the ENCODE input pins as shown in
Figure 12. The capacitors shown here should be chip capacitors
but do not need to be of the low inductance variety.
ENCODE
ENCODE
AD6640
ECL
GATE
0.1 F
0.1 F
–VS
510
510
Figure 12. Differential ECL for ENCODE
As a final alternative, the ECL gate may be replaced by an ECL
comparator. The input to the comparator could then be a logic
signal or a sine signal.
ENCODE
ENCODE
AD6640
0.1 F
0.1 F
–VS
50
AD96687 (1/2)
510
510
Figure 13. ECL Comparator for ENCODE
Driving the Analog Input
Because the AD6640 operates from a single 5 V supply, the
analog input voltage range is offset from ground by 2.4 V. Each
analog input connects through a 450
Ω resistor to the 2.4 V bias
voltage and to the input of a differential buffer (Figure 14). This
resistor network on the input properly biases the followers for
maximum linearity and range. Therefore, the analog source driving
the AD6640 should be ac-coupled to the input pins. Since the
differential input impedance of the AD6640 is 0.9 k
Ω, the analog
input power requirement is only –3 dBm, simplifying the drive
amplifier in many cases.
AD6640
450
2.4V
REFERENCE
AIN
0.01 F
450
BUF
BUF
BUF
AIN
VREF
0.1 F
Figure 14. Differential Analog Inputs
To take full advantage of this high input impedance, a 20:1 trans-
former would be required. This is a large ratio and could result
in unsatisfactory performance. In this case, a lower step-up
ratio could be used. For example, if RT were set to 260
Ω,
along with a 4:1 transformer, the input would match to a 50
source with a full-scale drive of 4 dBm (Figure 15). Note that the
external load resistor, RT, is in parallel with the AD6640 analog
input resistance of 900
Ω. The external resistor value can be
calculated from the following equation:
RT
=
1
1
Z
1
900
where Z is the desired impedance (200
Ω for a 4:1 transformer
with 50
Ω input source).
AIN
0.01 F
AIN
VREF
0.1 F
RT
1:4
ANALOG
INPUT
SIGNAL
AD6640
Figure 15. Transformer-Coupled Analog Input Signal
If the lower drive power is attractive, a combination transformer
match and LC match could be employed that would use a 4:1
transformer with an LC as shown in Figure 16. This solution is
useful when good performance in the third Nyquist zone is
required. Such a requirement arises when digitizing high inter-
mediate frequencies in communications receivers.
AIN
0.01 F
AIN
VREF
0.1 F
1:4
AD6640
–j125
+j100
ANALOG
SIGNAL
AT
–3dBm
Figure 16. Low Power Drive Circuit
In applications where gain is needed but dc-coupling is not
necessary, an extension of Figure 16 is recommended. A 50
gain block may be placed in front of the LC matching network.
Such gain blocks are readily available for commercial applications.
These low cost modules can have excellent NF and intermodulation
performance. This circuit is especially good for the “IF” receiver
application previously mentioned.
In applications where dc-coupling is required, the circuit in
Figure 17 can be used. It should be noted that the addition of
circuitry for dc-coupling may compromise performance in terms of
noise, offset, and dynamic performance. This circuit requires an
inverting and noninverting signal path. Additionally, an offset must
be generated so that the analog input to each pin is centered
near 2.4 V. Since the input is differential, small differences in
the dc voltage at each input can translate into an offset for the
circuit. The same holds true for gain mismatch. Therefore, some
means of adjusting the gain and offset between the sides should



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