LSM6DS3TR's INT1 cannot detect interrupt

piaoroumi

LSM6DS3TR's INT1 cannot detect interrupt [Copy link]

 

  Read WHO_AM_I and get 69, configure accelerometer: gYroscopeWrite(CTRL9_XL, 0x38);//XL configure gYroscopeWrite(CTRL1_XL, 0x60);//XL configure gYroscopeWrite(INT1_CTRL, 0x01);// Then move the board, INT1 does not interrupt. Tried 2 boards, both are the same. Can you please tell me what the problem could be? The registers are written and read correctly.

littleshrimp

What interrupt do you want INT1 to output?

piaoroumi

littleshrimp posted on 2018-11-2 17:11 What interrupt do you want INT1 to output?

When accelerometer data is ready, the state of INT1 changes

piaoroumi

How can I configure the registers to simply enable INT1 to interrupt when the accelerometer has data?

littleshrimp

piaoroumi posted on 2018-11-2 17:32 When the accelerometer data is ready, the state of INT1 changes

You can take a look at this post, there is a complete projecthttps://en.eeworld.com/bbs/forum ... 0&highlight=LSM6DSl

piaoroumi

Thanks, Brother Shrimp. I'll look into it!

littleshrimp

I did a test. If you only configure the LSM6DS3 register according to your configuration method, there is indeed a situation where the interrupt cannot be triggered. Then I added a reset of LSM6DS3 before powering on in the code, and then configured the corresponding registers. The interrupt was normal. Similarly, in the case of power-on without resetting, the output register was read once after the register was configured, and the interrupt was also normal. You can try these two methods. The problem may be that after the device is configured, the data is ready, the INT1 pin outputs a high level, and the microcontroller does not read the interrupt in time due to reset and other reasons. When the interrupt pin is running again, it is high and the rising edge interrupt cannot be triggered again. In addition to the above two methods, you can also try to use high-level interrupt or rising edge interrupt to judge the pin level test code:

1. static uint8_t who_am_i; //WHO_AM_I (0Fh) HAL_I2C_Mem_Read(&hi2c1,0xd4,0x0f,I2C_MEMADD_SIZE_8BIT,&who_am_i,1,0x100); static uint8_t reg; //CTRL3_C (12h) 0x01 SOFT REST reg = 0x01; #warning "1. Reset sensor at power on" // HAL_I2C_Mem_Write(&hi2c1,0xd4,0x12,I2C_MEMADD_SIZE_8BIT,,1,0x100); HAL_Delay(500); //CTRL3_C (12h) 0x44 BDU IF_INC reg = 0x44; // HAL_I2C_Mem_Write(&hi2c1,0xd4,0x12,I2C_MEMADD_SIZE_8BIT,,1,0x100); //CTRL9_XL (18h) x,y,z en reg = 0x38; HAL_I2C_Mem_Write(&hi2c1,0xd4,0x18,I2C_MEMADD_SIZE _8BIT,,1,0x100); //CTRL1_XL (10h) reg = 0x60; HAL_I2C_Mem_Write(&hi2c1,0xd4,0x10,I2C_MEMADD_SIZE_8BIT,,1,0x100); //INT1_CTRL (0Dh) reg = 0x01; HAL_I2C_Mem_Write(&hi2c1,0xd4,0x0d,I2C_MEMADD_SIZE_8BIT,,1,0x100); #warning "2. Read the output register once after configuration" // HAL_I2C_Mem_Read(&hi2c1,0xd4,0x28,I2C_MEMADD_SIZE_8BIT,buffer,6,0x100); /* USER CODE END 2 */ /* Infinite loop */ /* USER CODE BEGIN WHILE */ while (1) { /* USER CODE END WHILE */ /* USER CODE BEGIN 3 */ if(lsm63se_drdy_flag) { lsm63se_drdy_flag = 0; HAL_I2C_Mem_Read(&hi2c1,0xd4,0x1e,I2C_MEMADD_SIZE_8BIT,&status,1,0x100); if(status & 0x01) { HAL_I2C_Mem_Read(&hi2c1,0xd4,0x28,I2C_MEMADD_SIZE_8BIT,buffer,6,0x100); } } } /* USER CODE END 3 */

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Test project file: Use NUCLEO-L496ZG development board, STM32CubeMX to create a project, I2C communicates with LSM6DSL, CS and SA0 are connected to low level, CS is connected to high level, PA5 is connected to LSM6DS3_INT1 pin, rising edge interrupt, no pull-up and pull-down resistors lsm6ds3 nucleo-l496zg.rar (9.03 MB, downloads: 58)

2018-11-4 13:55 上传

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piaoroumi

I was busy with other things recently, and I found that Brother Xia has answered my questions again. I am really grateful. I will try it right away.

piaoroumi

Brother Xia, I have obtained the interrupt according to your configuration. And I configured it to generate an interrupt on a single click. Let me talk about my application scenario. We are working on the smart lock solution. Because the customer mentioned that there are already suppliers making gyroscopes to detect door opening and closing. So we also want to use this solution. The idea is: do a good job of calibration when the door is closed. Then when the door moves, the gyroscope is used to calculate the angle of the door to determine the door's open and close status. I would like to ask: Is this idea feasible?

piaoroumi

Brother Xia, I did a test and configured it using your method. I got the interrupt. Thank you very much. Our application scenario is this: when the door is closed, do the initialization. Then use the gyroscope to calculate the opening and closing angle of the door. Implement the automatic door closing function. I would like to ask, is this usage method feasible?

littleshrimp

piaoroumi posted on 2018-11-24 18:27 Brother Xia, I did a test. Configured it using your method. Got an interrupt, thank you very much. Our application scenario is this: When the door is closed, ...

It should be possible. When the gyroscope is working, the current will be larger. Battery power supply needs to be noted. In addition, the gyroscope calculates the door opening angle as a relative angle, which can be combined with acceleration to determine the door closing detection. Another method is to use a magnetometer, which is equivalent to putting a compass on the door. There is a fixed magnetic field around it. The infection can be solved by calibration to calculate the absolute angle of the door opening. The disadvantage is that it is afraid of dynamic magnetic field interference. For example, if someone shakes a magnet outside the sensor, it may misjudge. If necessary, these three can be combined

piaoroumi

littleshrimp posted on 2018-11-25 07:56 It should be possible. The current will be larger when the gyroscope is working. Battery power supply needs to be noted. The gyroscope calculates the angle of the door opening as a relative angle.

I have enabled the gyroscope now, and read the interrupt, and then found that interrupts will be generated all the time. I hope that there will be no interrupt when the door does not move. Only when it moves. Where can I set the angular velocity threshold? /* íó?Yò?3?ê??ˉ******** ************************/ void gYroscopeG_Init(void) { gYroscopeWrite(CTRL2_G, 0x1C);// gYroscopeWrite(CTRL6_C, 0x70); gYroscopeWrite(CTRL7_G, 0x43); gYroscopeWrite(INT1_CTRL, 0x02);//ê1?üíó?Yò??D??. }

littleshrimp

piaoroumi posted on 2018-11-26 11:28 I have enabled the gyroscope and read the interrupt, and found that it will always generate interrupts. I hope that there will be no interrupts when the door does not move...

Normally, you need to keep receiving data and then analyze the angle of the door opening. The lsm6ds3 has a 4D or 6D threshold to trigger the interrupt function, which should not meet your requirements

piaoroumi

I guess I can't use it. I can't stay awake all the time. It's a power-sensitive system.

devil'eye

piaoroumi posted on 2018-11-24 18:27 Brother Xia, I did a test. I configured it using your method. I got the interrupt. Thank you very much. Our application scenario is like this: when the door is closed, ...

Brother, I saw your door switch solution: when the door is closed, do the initialization. Then use the gyroscope to calculate the opening and closing angle of the door. Have you realized the automatic door closing function? I am doing a graduation project now, and I chose this example. I would like to ask you about the idea and implementation of this solution.

littleshrimp

devil'eye posted on 2018-12-27 14:20 Brother, I saw your door switch solution: when the door is closed, do initialization. Then use the gyroscope to calculate the opening and closing angle of the door. Automatic door closing...

There is long-term drift in the angle calculated by the gyroscope. It is recommended to use a magnetometer, or use a magnetometer in conjunction with a gyroscope. If you just want to determine whether the door is open, you can do it by judging the size of the angular velocity. When the door is not open, the angular velocity is close to 0. When the door is open, the angular velocity will increase. Because it is not an absolute data, if the door is opened slowly enough, it will not be detected.