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DAC5687MPZPEP データシート(PDF) 53 Page - Texas Instruments

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部品番号 DAC5687MPZPEP
部品情報  16-BIT 500-MSPS 2쨈??쨈 INTERPOLATING DUAL-CHANNEL DIGITAL-TO-ANALOG CONVERTER (DAC)
PDF  75 Pages
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メーカー  TI1 [Texas Instruments]
ホームページ  http://www.ti.com
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DAC5687MPZPEP データシート(HTML) 53 Page - Texas Instruments

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IOUT1
4:1
IOUT2
100
100
RLOAD
50
AVDD (3.3 V)
AVDD (3.3 V)
S0033-02
Combined Output Termination
S0069-01
IOUTA1
1:1
RLOAD
50
IOUTA2
AVDD (3.3 V)
IOUTB1
IOUTB2
DAC5687-EP
SGLS333 – JUNE 2006
Figure 59. Driving a Doubly Terminated 50-
Ω Cable Using a 4:1 Impedance Ratio Transformer
The DAC5687 DAC A and DAC B outputs can be summed together as shown in Figure 60 to provide a 40-mA
full-scale output for increased output power.
Figure 60. Combined Output Termination Using a 1:1 Impedance Ratio Transformer into 50-
Ω Load
For the case where the digital codes for the two DACs are identical, the termination results in a full scale swing
of 2 VPP into the 50-Ω load, or 10 dBm. This is 6 dB higher than the 4:1 output termination recommended for a
single DAC output.
There are two methods to produce identical DAC codes. In modes where there is mixing between digital
channels A and B, i.e., when channels A and B are isolated, the identical data can be sent to both input ports to
produce identical DAC codes. Channels A and B are isolated when FMIX is disabled, the QMC is disabled or
enabled with QMC phase register set to 0, and CMIX is disabled or set to fDAC/2. Note that frequency
upconversion is still possible using the high-pass filter setting and CMIX fDAC/2.
Alternatively, by applying the input data on one input port only and setting the other input port to mid-scale
(zero), the NCO can be used to duplicate the output of the active input channel in the other channel by setting
the frequency to zero, phase to 8192 and NCO_GAIN = 1 and QMC gain = 1446. Assuming I(t) is the wanted
signal and Q(t) = 0, this is demonstrated by the simplification of the NCO equations in the Fine Mixer (FMIX)
section:
IOUT(t) = (IIN(t)cos(2π× 0 × t + π/4) – 0 × sin((2π× 0 × t + π/4)) × 2
(1 – 1) = I
IN(t)cos(π/4) = IIN(t)/2
½
53
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