ADS1261: Data converters forum

Bin Huang

Part Number: ADS1261

Hello,

We are using ADS1261 to sample magnetometers in our design by configuring the sensors as differential inputs. Our code can read/wright registers and fetch samples from the sensors. To validate if the chip works correctly, we have test code to setup as below and read the differential signal of either: (AVDD-AVSS)/4 - DVDD/4, or DVDD/4 - (AVDD-AVSS)/4. Unfortunately we cannot get expected values.
1. In our design: AVDD = 5.0VDC; AVSS = gnd; DVDD = 3.3VDC. Reference is set to internal 2.5VDC. All the three voltage pins work correctly with the right readings from a multimeter.
2. The ADC is set to:
[MODE0] = 0b01001000: 1200 SPS, sinc1
[MODE1] as default, 0b00000001: chop mode off, continuous conversions, 50us delay
[MODE3] as default, 0x00h
[REF] = 0b00010000: 2.5V Internal Reference
[OFCAL0] as default, 0x00h
[OFCAL1] as default, 0x00h
[OFCAL2] as default, 0x00h
[FSCAL0] as default, 0x00h
[FSCAL1] as default, 0x00h
[FSCAL2] as default, 0x40h
[IMUX] as default, 0xffh
[IMAG] as default, 0x00h
[PGA] = 0b00000000: gain 1
[INPMUX] = 0b11001101: Positive_Input_Multiplexer=(AVDD-AVSS)/4=5.0VDC/4; Negative_Input_Multiplexer=DVDD/4=3.3VDC/4.
[INPBIAS] as default, 0x00h
3. The test code is applied RREG commands to read back all the registers with 3 bytes to validate the setups. All readings are right as below:
RREG ID: FF=0xff; B1_20h=0x20; B2_8xh=0x81
RREG STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05
RREG MODE0: FF=0xff; B1_22h=0x22; B2_48h=0x48
RREG default MODE1: FF=0xff; B1_23h=0x23; B2_01h=0x01
RREG default MODE2: FF=0xff; B1_24h=0x24; B2_00h=0x00
RREG default MODE3: FF=0xff; B1_25h=0x25; B2_00h=0x00
RREG REF: FF=0xff; B1_26h=0x26; B2_10h=0x10
RREG default OFCAL0: FF=0xff; B1_27h=0x27; B2_00h=0x00
RREG default OFCAL1: FF=0xff; B1_28h=0x28; B2_00h=0x00
RREG default OFCAL2: FF=0xff; B1_29h=0x29; B2_00h=0x00
RREG default FSCAL0: FF=0xff; B1_2ah=0x2a; B2_00h=0x00
RREG default FSCAL1: FF=0xff; B1_2bh=0x2b; B2_00h=0x00
RREG default FSCAL2: FF=0xff; B1_2ch=0x2c; B2_40h=0x40
RREG default IMUX: FF=0xff; B1_2dh=0x2d; B2_ffh=0xff
RREG default IMAG: FF=0xff; B1_2eh=0x2e; B2_00h=0x00
RREG PGA: FF=0xff; B1_30h=0x30; B2_00h=0x00
INPMUX: FF=0xff; B1_31h=0x31; B2_cdh=0xcd
RREG default INPBIAS: FF=0xff; B1_32h=0x32; B2_00h=0x00
4. Then the START command is issued.
5. Then the RDATA command and RREG to get STATUS command are applied to read the differential signal + status at 1KHz. The debug printings are
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0x7f; B3_MID=0xff; B4_LSB=0xff; result= 8388607
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0x7f; B3_MID=0xff; B4_LSB=0xff; result= 8388607
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0x7f; B3_MID=0xff; B4_LSB=0xff; result= 8388607
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05

There are 2 questions for the result:
1) 0x7fffff = 8388607 is 2.5V, which is not the expected value: 5.0V/4 - 3.3V/4 = 0.425V.
2) It seems there is overflow since we always get 0x7fffff. But STATUS of 0x05 isn't indicated it.

6. For the above test, we change [INPMUX] = 0b11001101 to 0b11011100 for:
Positive_Input_Multiplexer=DVDD/4=3.3VDC/4; Negative_Input_Multiplexer=(AVDD-AVSS)/4=5.0VDC/4.
The debug printings are:
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0xde; B3_MID=0xc6; B4_LSB=0xc8; result= -2177336
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0xde; B3_MID=0xdb; B4_LSB=0x0b; result= -2172149
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0xde; B3_MID=0xc9; B4_LSB=0x95; result= -2176619
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0xde; B3_MID=0xd3; B4_LSB=0x16; result= -2174186
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0xde; B3_MID=0xd5; B4_LSB=0xdf; result= -2173473
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05

The result is around -2177336 (guess noise introducing changes).
The expect value is: (3.3V/4 - 5.0V/4) X 0x7fffff / 2.5V = -1426063.
The sample value seems partly reasonable even it's not close to the expected.

7. For the above test with Steps 1~6, we change [PGA] = 0b00000001 to have gain of 2 for another test.
1) for step 5 we get debug printings as
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0x7f; B3_MID=0xff; B4_LSB=0xff; result= 8388607
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x25

Although result of 0x7fffff = 8388607 has no change, PGAL_ALM bit of STATUS is set to high to indicate overflow.

2) for step 6 we get debug printings as
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0xbd; B3_MID=0x97; B4_LSB=0xc1; result= -4352063
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0xbd; B3_MID=0xb5; B4_LSB=0xdc; result= -4344356
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0xbd; B3_MID=0x92; B4_LSB=0x84; result= -4353404
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05
RADTA: B0_FF=0xff; B1_12h=0x12; B2_MSB=0xbd; B3_MID=0xb1; B4_LSB=0x03; result= -4345597
STATUS: FF=0xff; B1_21h=0x21; B2_05h=0x05

Result of -4352063 is roughly doubled value as of results in step 6. This is expected since the gain is doubled.

Summarized questions to ask your help please:
1. Why cannot get the sample values when [INPMUX] = 0b11001101?
2. Why cannot get the accurate sample values when [INPMUX] = 0b11011100?

Regards,

Bin

over 3 years ago

0 Bryan Lizon86 over 3 years ago

TI__Guru 52316 points

Hi Bin,

Can you try setting the INPMUX to 1100 1100b or 1101 1101b? You should not be mixing the two signals together, this is not how the internal measurements work.

Setting INPMUX to 1100 1100b or 1101 1101b should provide you with ~1.25 V and ~0.825V, respectively.

-Bryan

0 Bin Huang over 3 years ago

Prodigy 10 points

Hi Bryan,

Thank you for your quick answers! I changed INPMUX to 1100 1100b and 1101 1101b, and got 1.258 V and ~0.826V as expected, I have 2 other questions:

1. How can I understand that configuring INPMUX to 1100 1100b should get ~1.25V? I mean that both Positive_Input_Multiplexer and Negative_Input_Multiplexer are connected to (AVDD-AVSS)/4. Should the differential input be zero at this situation in theory? In my application, INPMUX will be set to 0x012, 0x034, and 0x056 as differential inputs to read 3D magnetometer3.

2. How can I configure INPMUX to get -1.25 V and -0.825V?

Bin

+1 Bryan Lizon86 over 3 years ago in reply to Bin Huang

TI__Guru 52316 points

Hi Bin,

The measurement of AVDD and DVDD is differential, similar to the image I have shown below. The ADC is measuring an internal resistor divider that is established between AVDD and AVSS, and measuring the center resistor to track the voltage at AVDD. The actual circuit is different from the one below, but the same principles apply.

In practice then, if AVDD = 5V and AVSS = ground, then AINP is ~3.125V and AINN is ~1.875. The differential voltage between AINP and AINN is then 1.25V, which is what the ADC measures and reports back (DVDD operates similarly). This is why setting INPMUX = 1100 1101b did not provide the results you were expecting: neither the voltage at AINP or AINN = AVDD/4, it is the differential voltage between AINP and AINN that equals AVDD/4

There is no way to configure the ADC mux to provide the voltages you are requesting. You could provide these voltages externally and then measure them with the ADC, but the AVDD and DVDD measurements will always be positive.

-Bryan

0 Bin Huang over 3 years ago in reply to Bryan Lizon86

Prodigy 10 points

Hi Bryan,

Now I understand. Thank you very much for your help!

Bin