AFE7444EVM: AFE7444 max data rate

ganesh singh1

Part Number: AFE7444EVM
Other Parts Discussed in Thread: AFE7444, AFE7422

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

over 3 years ago

0 Vijayendra Varma Siddamsetty over 3 years ago

TI__Genius 10505 points

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

0 ganesh singh1 over 3 years ago in reply to Vijayendra Varma Siddamsetty

Intellectual 400 points

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

0 Vijayendra Varma Siddamsetty over 3 years ago in reply to ganesh singh1

TI__Genius 10505 points

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

0 ganesh singh1 over 3 years ago in reply to Vijayendra Varma Siddamsetty

Intellectual 400 points

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.

0 Vijayendra Varma Siddamsetty over 3 years ago in reply to ganesh singh1

TI__Genius 10505 points

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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AFE7444EVM: AFE7444 max data rate