External Memory Interfaces Intel® Cyclone® 10 GX FPGA IP User Guide

ID 683663
Date 3/29/2021
Public

A newer version of this document is available. Customers should click here to go to the newest version.

Document Table of Contents

9. Optimizing Controller Performance

When designing an external memory interface, you should understand the ways available to increase the efficiency and bandwidth of the memory controller.

The following topics discuss factors that affect controller efficiency and ways to increase the efficiency of the controller.

Controller Efficiency

Controller efficiency varies depending on data transaction. The best way to determine the efficiency of the controller is to simulate the memory controller for your specific design.

Controller efficiency is expressed as:

Efficiency = number of active cycles of data transfer/total number of cycles

The total number of cycles includes the number of cycles required to issue commands or other requests.

Note: You calculate the number of active cycles of data transfer in terms of local clock cycles. For example, if the number of active cycles of data transfer is 2 memory clock cycles, you convert that to the local clock cycle which is 1.

The following cases are based on a high-performance controller design targeting an FPGA device with a CAS latency of 3, and burst length of 4 on the memory side (2 cycles of data transfer), with accessed bank and row in the memory device already open. The FPGA has a command latency of 9 cycles in half-rate mode. The local_ready signal is high.

  • Case 1: The controller performs individual reads.

    Efficiency = 1/(1 + CAS + command latency) = 1/(1+1.5+9) = 1/11.5 = 8.6%

  • Case 2: The controller performs 4 back to back reads.

    In this case, the number of data transfer active cycles is 8. The CAS latency is only counted once because the data coming back after the first read is continuous. Only the CAS latency for the first read has an impact on efficiency. The command latency is also counted once because the back to back read commands use the same bank and row.

    Efficiency = 4/(4 + CAS + command latency) = 4/(4+1.5+9) = 1/14.5 = 27.5%