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AFE7444EVM: AFE7444 max data rate

Prodigy 230 points
ganesh singh1

Replies: 5

Views: 135

Part Number: AFE7444EVM

Hi ,

I am working with AFE7444EVM.. 

I have one question is 

1. if I want to bypass the decimation filters in ADC chain or Interpolation filters in DAC chain. What is the maximum data rate i can transfer from JESD204B to FPGA.

Like if ADC is 3GPS and each sample is 14 bit.  Can we transfer all data to FPGA and if yes in which Mode of LMFS? 

Thanks again.

With regards,

Ganesh

Prodigy 230 points
ganesh singh1
  • Expert 4870 points
    Vijayendra Varma Siddamsetty

    Hi Ganesh,

    It is not possible to bypass the decimation filters in AFE7444. All possible modes of operation are shown in section "7.4 Device Functional Modes" in the datasheet.

    If you need to use ADC in bypass mode, you can request access to AFE7422 secure folder at https://www.ti.com/licreg/docs/swlicexportcontrol.tsp?form_id=288371&prod_no=AFE7422&ref_url=asc_dc_hsc

    In AFE7422, configuring ADC in bypass mode is possible. LMFS and other mode details are in AFE7422 datasheet. 

    Regards,

    Vijay

  • Prodigy 230 points
    ganesh singh1

    In reply to Vijayendra Varma Siddamsetty:

    Thanks for your reply 

    I have done some calculation for Mode 3

    Mode Max Signal Bandwidth MHz  DAC Clock( MHz) DAC Sampling rate(GSPS) N (Tx resulution) in Bits  Interpolation JESD204B data Rate(MSPS) LMFSHd L M F S Data Rate(Gbps)
    2 800 8847.36 9 16 9 983.04 84111 8 4 1 1 39.3216
    3 800 8847.36 9 12 9 983.04 44320 4 4 3 2 117.9648
    4 600 8847.36 9 16 12 737.28 44210 4 4 2 1 58.9824

    I have Used  formula 

    Total data=Datarate*Num_Converters*Num_Octets*10bits/8bit 

    =983.04*4*3*10 = 117.9648 Gbps 

    Here in mode 3 it's coming 117.9648 Gbps which i think cant be mapped to 4 lanes as the limit is 15 Gbps per lane.

    Can you tell me where I am wrong?

    With regards,

    Ganesh Singh 

  • Expert 4870 points
    Vijayendra Varma Siddamsetty

    In reply to ganesh singh1:

    Hi Ganesh,

    Total lane rate = (Data rate)*Num_converters*Num_bit_per_sample*(10b/8b)

    For mode 3, Data rate = (8847.36 MSPS)/9 = 983.04 MSPS

    Number of bits per sample (bit resolution) = 12

    So Total Data rate = (983.04 MSPS)* 4* 12* (10/8) = 58982.4 Mbps = 58.9824 Gbps

    When mapped to 4 lanes, lane rate = (58.9824 Gbps) / 4 = 14.7456 Gbps

    Regard,

    Vijay

  • Prodigy 230 points
    ganesh singh1

    In reply to Vijayendra Varma Siddamsetty:

    Dear Vijendra,

    Thanks for it. But I want to know the significance of F, especially when it's more than 4. How the calculations for it.

    I have used (https://e2e.ti.com/blogs_/b/analogwire/archive/2014/09/24/jesd204b-determining-your-link-configuration

    There is 

    Datarate*Num_Converters*Num_Octets*10bits/Octet=

    193.75Msps*2*2*10=7.75Gbps Total throughput

    I am not able to understand the F calculations particularly.

    With regards,

    Ganesh Singh.

  • Expert 4870 points
    Vijayendra Varma Siddamsetty

    In reply to ganesh singh1:

    Hi Ganesh Singh,

    In the formula, 

    Total lane rate = Datarate*Num_Converters*Num_Octets*10b/Octet

    Num_Octets is number of octets per sample. This is equal to F in cases with S=1. Otherwise this is F/S. (Note that this doesn't work n cases where Hd = 1. In this case, you can use 'octets per sample' definition instead).

    Regards,

    Vijay

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