Jump to content
  • 0

I2C interfacing with Zybo Z7020


Niranjana

Question

Hi ,

I am trying to establish I2c protocol between the FPGA board and an external device(a TI board) in which the FPGA board acts as master and sends the data.

I created the Zynq processing IP and enabled I2C1 for the corresponding pins. I exported the .xsa file to XIlinx Vitis and coded by referring the example given in the examples directory for i2c_master_polled_example.c.

The code is as follows.

/******************************************************************************
*
* Copyright (C) 2009 - 2014 Xilinx, Inc.  All rights reserved.
*
* Permission is hereby granted, free of charge, to any person obtaining a copy
* of this software and associated documentation files (the "Software"), to deal
* in the Software without restriction, including without limitation the rights
* to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
* copies of the Software, and to permit persons to whom the Software is
* furnished to do so, subject to the following conditions:
*
* The above copyright notice and this permission notice shall be included in
* all copies or substantial portions of the Software.
*
* Use of the Software is limited solely to applications:
* (a) running on a Xilinx device, or
* (b) that interact with a Xilinx device through a bus or interconnect.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
* IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
* FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL
* XILINX  BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY,
* WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF
* OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
* SOFTWARE.
*
* Except as contained in this notice, the name of the Xilinx shall not be used
* in advertising or otherwise to promote the sale, use or other dealings in
* this Software without prior written authorization from Xilinx.
*
******************************************************************************/

/*
 * helloworld.c: simple test application
 *
 * This application configures UART 16550 to baud rate 9600.
 * PS7 UART (Zynq) is not initialized by this application, since
 * bootrom/bsp configures it to baud rate 115200
 *
 * ------------------------------------------------
 * | UART TYPE   BAUD RATE                        |
 * ------------------------------------------------
 *   uartns550   9600
 *   uartlite    Configurable only in HW design
 *   ps7_uart    115200 (configured by bootrom/bsp)
 */

#include "xparameters.h"
#include "xiicps.h"
#include "xil_printf.h"

#define IIC_DEVICE_ID		XPAR_XIICPS_0_DEVICE_ID

#define IIC_SLAVE_ADDR		0x55
#define IIC_SCLK_RATE		100000

XIicPs IicPsInstance;			/* Instance of the IIC Device */

static void i2c_init();
static void datasend();


static void i2c_init()
{
	XIicPs_Config *Config;
	/*
		 * Initialize the IIC driver so that it's ready to use
		 * Look up the configuration in the config table, then initialize it.
		 */
	Config = XIicPs_LookupConfig(IIC_DEVICE_ID);
	XIicPs_CfgInitialize(&IicPsInstance, Config, Config->BaseAddress);
	/*
	 * Perform a self-test.
	 */
	XIicPs_SelfTest(&IicPsInstance);
	XIicPs_SetSClk(&IicPsInstance, IIC_SCLK_RATE);

}

static void datasend()
{
	unsigned char sendbuffer[3] = {0xA,0xB,0xC};
	//uint8_t myReadBuffer[3] = {};

	//xil_printf("the value at receive buffer before datasend is %X\n",rcvbuffer[0]);


	XIicPs_MasterSendPolled(&IicPsInstance ,sendbuffer, 3 , IIC_SLAVE_ADDR );

	xil_printf("Transferred successfully\n");

}

int main()
{
	i2c_init();
	datasend();

    return 0;
}

I am unable to receive the data in the slave.. what am I missing?

image.thumb.png.04a383e40285e8c7e4ce079057caecac.png

Link to comment
Share on other sites

0 answers to this question

Recommended Posts

There have been no answers to this question yet

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...