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ADC32RF80: some problems about Output data format?

Prodigy 190 points

Replies: 8

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Part Number: ADC32RF80

hi team:      

We have a block ADC32RF80EVM development board for use with the TSW14J56EVM. There are a few questions:


1. ADC32RF80 is 14 bits, the peak-to-peak value is 1.3V, and the full-scale count is 16384. But to use DDC, I set the 16-bit NCO frequency to 0, then one signal I is multiplied by 1, and the other Q is multiplied by 0. The output data I is  16-bit signals, the full-scale count bit is 65536, and the other Q  is 0. I want to know how the 16-bit data corresponds to the 14-bit ADC data. 14-bit data is multiplied by 16-bit 1 and the highest two bits should be It is 0, the full scale should be 16384, how is it 65536?  and When I input a signal with vpp=0.95V, the ADC sample data is almost 65536?

2. If i select the complex output, I is the signal, Q is 0. If you choose the real output, the output signal is not what I want, how is the complex to real module converted?

3. If it is fs=3Ghz, Divide-by-4 complex, Figure 97 shows the bandwidth  0.1*FS=300Mhz in adc32rf80.pdf, and OUTPUT BANDWIDTH (MHz) PER BAND is 600Mhz in Table 4. Why is it different? What is the meaning of the second bandwidth?


Thank you

  • 1. The 14-bit output from the ADC's quantizer is down-converted digitally where the other (NCO) input into the down-converter  is 16-bits. Therefore, rest of the digital signal path operates at 16-bit resolution.

    2. This is illustrated in figure 93 of the device datasheet.

    3. The 600MHz BW is complex. figure 97 shows the BW for either I or Q.

    regards,

    satish.

  • In reply to Satish Uppathil:

    thanks for the reply

    1. The 14-bit output from the ADC's quantizer is down-converted digitally where the other (NCO) input into the down-converter  is 16-bits. Therefore, rest of the digital signal path operates at 16-bit resolution.

    What I don't understand is how the 16-bit data corresponds to the input amplitude. The ADC32RF80 input signal Vpp = 1.3V, but when the input signal Vpp = 0.95V, the output data is almost 65536. For example, when the NCO frequency is 0, the ADC data is 16384, does the corresponding 16-bit data be 65536?

  • In reply to chi zhang2:

    The FS at the input should result in 16-bits max code at the output. This device doesn't support bypass mode and so, you'll have to set an NCO in addition to the decimation ratio.

    regards,

    satish.

  • In reply to Satish Uppathil:

    I have asked for more details from the development team and I'll get back to you.

    Regards,

    satish.

  • In reply to Satish Uppathil:

    When the NCO is set to Zero, the signal path will reduce to a simple real decimation chain. In this case, the multiplication by 2 in the last decimation stage needs to be disabled. 

    Two Scenarios of operating mode:

     1. Normal (non-zero) NCO: Here real input is converted to a complex base-band output. In this case, the last stage multiplication by 2 has to be enabled to ensure full scale real tone at ADC output maps to a full scale complex base-band tone at DDC output.

    2. Zero NCO: DDC reduces to a simple real decimation chain (Only I output will be non-zero). In this case the last stage multiplication by 2 has to be disabled to ensure full scale ADC output maps to full scale DDC output.

    You're seeing a full-scale output at lower input as this stage (multiplication by 2) is not getting disabled.

    Regards,

    satish.

  • In reply to Satish Uppathil:

    Thanks for the reply, there is another question, where to set the last stage multiplication by 2

  • In reply to chi zhang2:

    you can do that with a register write. highlighted below is the relevant register.

  • In reply to Satish Uppathil:

    Thank you very much for solving my problem