Datasheet
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M0087-01
S(1,1)
Frequency(100kHzto500MHz)
F2
F1
F1
Freq=50MHz
S(1,1)=0.967/ –13.241
Impedance=62.211 – j421.739
F2
Freq=400MHz
S(1,1)=0.273/ –59.329
Impedance=58.132 – j29.510
0
100
200
300
400
500
600
700
800
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0 50 100 150 200 250 300 350 400 450 500
Magnitude of Zin -- W
f -- Input Frequency -- MHz
I
F1
F1
Frequency = 50 MHz
Mag(Zin1) = 426.302
F2
F2
Frequency = 400 MHz
Mag(Zin1) = 65.193
Using RF-Transformer Based Drive Circuits
ADS6424
ADS6423
ADS6422
SLAS532A – MAY 2007 – REVISED JUNE 2007
APPLICATION INFORMATION (continued)
Figure 66. ADC Input Impedance, Zin
For optimum performance, the analog inputs must be driven differentially. This improves the common-mode
noise immunity and even order harmonic rejection. An example of input drive using RF transformers is shown in
Figure 67 and Figure 68 .
The single-ended signal is fed to the primary winding of the RF transformer. The transformer is terminated on the
secondary side. Putting the termination on the secondary side helps to shield the kickbacks caused by the
sampling circuit from the RF transformer ’ s leakage inductances. The termination is accomplished by two resistors
connected in series, with the center point connected to the 1.5 V common mode (VCM pin). The value of the
termination resistors (connected to common mode) has to be low (< 100 Ω ) to provide a low-impedance path for
the ADC common-mode switching current.
Figure 67 shows a configuration using a single 1:1 turns ratio transformer (for example, WBC1-1) that can be
used for low input frequencies up to 100 MHz.
Copyright © 2007, Texas Instruments Incorporated 41
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