ADS122C04: VREF and AVDD questions for ADS122C04

user4674678

Intellectual 490 points

Part Number: ADS122C04
Other Parts Discussed in Thread: ADS8332, ADS1118, ADS1115, ADS7042

Hi team,

I will use ADS122C04.

Q1: I need to get what power supply should be connected to the REFP pin?

I note the EVM user guide, the REFP pin is connected to AVDD. So, should I connected the REF pin to AVDD?

Q2: If the REF pin should be connected to AVDD, then I will select a LDO device or a DC-DC device that can output to AVDD,

so, which requirements should be met for the a LDO device or a DC-DC device? low noise or other requirements?

How much output current of the LDO device or the DC-DC device can meet the AVDD?

Q3: I note the gain can be selected from 1 to 128, This mean the input signal can be amplified by the internal op amp of ADC, the gain can be selected from 1 to 128, correct?

over 7 years ago

0 Bob Benjamin over 7 years ago

TI__Guru** 113765 points

Hi user4674678,

For question 1, you can use an external reference source but it is not required. The ADS122C04 has an internal reference of 2.048V. The internal reference value is chosen based on the minimum AVDD supply of 2.3V (where AVSS is ground). The external reference must be equal to or greater than 0.75V and less than or equal to AVDD-AVSS. One reason for connecting and external reference to AVDD is when attempting to measure a voltage up to the AVDD voltage. There is also an option to select the AVDD supply as the reference in the Configuration 1 register settings. See Table 20 on page 43 of the ADS122C04 datasheet. So there is the option for an external reference connection, but the pins can be left floating if not used.

For question 2, there is no need for this extra supply. If you tell me what you are attempting to measure along with the sensor output range I can better help you decide. In most cases you can just use the ADS122C04 internal reference. If you do use an external reference you should use a low noise reference. The output of the reference should be no more than AVDD +0.1V (as shown in Recommended Operating Conditiions table in section 6.3 on page 4) and the output current could be low as the reference input current is typically +/-10nA as shown in the Electrical Characteristics table on page 6 of the datasheet.

For question 3, there is an internal PGA that will amplify the input signal up to 128. If you need to use gain, then most likely you would not want to use the AVDD supply as the reference unless you are making a bridge type measurement common to pressure sensors and load cells where the measurement becomes ratiometric. Again, more information regarding the measurement you want to make will be helpful in advising you what to do.

Best regards,
Bob B

0 user4674678 over 7 years ago in reply to Bob Benjamin

Intellectual 490 points

Hi Bob,
Thanks for your reply.

Q1: You mean if I use the internal 2.048V as the reference, the AVDD voltage should be 2.3V?

If I use the internal 2.048V as the reference, can I use 3.3V for AVDD?

Q2: My application is as below:
The input signal is a nA grade current, then I use two stage op amp. The first stage is the transimpedance amplify circuit
and the second stage is the non-Inverting amplify circuit. Then the signa lwill be sent to ADS122C04. I also have a external
2.048V reference.

In my application, do I use internal 2.048v reference or the external 2.048v reference?

0 Bob Benjamin over 7 years ago in reply to user4674678

TI__Guru** 113765 points

Hi user4674678,

I may have been confusing in my first response. What I meant to say is that the internal 2.048V reference will work for all AVDD/AVSS supply voltage ranges for the ADS122C04 and there is no requirement for an externally supplied reference. So for Q1, yes you can use the 2.048V internal reference for AVDD of 3.3V.

For Q2 you did not specify the output voltage range of your sensor amplifier circuit. As far as I can tell you could use either reference, but it may be easiest to use the ADS122C04 internal reference. If you have an external 2.048V reference, then I will assume that the output amplifier swings about 2.048V. Is this correct? And if so, what is the voltage supplied to the op amps? It would be really helpful to know specific details regarding your circuit, such as a schematic and the actual input voltage range to be applied to the ADS122C04.

Best regards,
Bob B

0 user4674678 over 7 years ago in reply to Bob Benjamin

Intellectual 490 points

Hi Bob,
Yes, you are correct. In my circuit, VCC is 3.3V, and there is a 2.048V reference that is in the input circuit. So the max voltage is less than2.048V.
Maybe I will do not use ADS122c04 since its max sample rate is 2Ksps. In my circuit, there is another ADC LTC1867L that is ADI company device.
The datasheet is in the attach, It is 16bit and 175ksps sample date. Its package is too large. I want to find a replacement for LTC1867L.
The package of the replacement should be more small. 3mm*3mm will be the best. VCC is 3.3V and it should have also the internal reference
voltage. Would you help me to find the suitable device?

LTC1863L-LTC1867L.pdf

0 Bob Benjamin over 7 years ago in reply to user4674678

TI__Guru** 113765 points

Hi user4674678,

You really have not identified your requirements. We started out our discussion with the ADS122C04, which is a delta-sigma ADC with an I2C interface and 2 differential (or 4 single-ended) inputs. This is much different than the then LTC part which is a SAR ADC with an SPI interface with twice as many input channels.

I know your operating voltage and package size while preferring an internal reference. How many inputs do you require? What is required communication interface? What resolution do you require? And what conversion rate is required?

We only have a few options for 3x3mm and multi-input channel devices. If you really need to have a SAR device that can operate at faster speeds we do have the ADS8332 that is similar to the LTC part in a 4x4mm package. However this device does not have an internal reference.

Best regards,
Bob B

0 user4674678 over 7 years ago in reply to Bob Benjamin

Intellectual 490 points

Hi Bob,

I would like to explain my application.

The input signal of LTC1867L is the output of the op amp. VCC is 3.3V for my whole circuit system.

The input signal is maybe 1na~1uA. Then I use the transimpedance amp to convert the current signal to the voltage signal.
After this, I use the non-inverting circuit to amplify the voltage signal. That mean I use the 2 stage amplify circuit.
At last, the output signal is input to the input of LTC1867L.

My requirements are as below:

VCC is 3.3V. The package should be less than 3mm*3mm.

The reference should be 2.048V that can be the internal reference or the external voltage.

The communication interface can be I2C or SPI. The input channel number can be 1 or 2 or more. I do nothave no special request

for the input channel numbers.

Now I cannot get my input signal frequency, the questions are as below.

My questions are below:

Q1: I do not know how to get the output signal frequency of the op amp.

If I get the output signal frequency of the op amp, I can select the ADC that the sample rate is higher 2 times than

the output signal frequency of the op amp. Is this correct?

Q2: Since the output signal of the op amp is DC signal. Is the output signal frequency of the op amp equal 0HZ?

If the output signal frequency of the op amp equal 0HZ, then I can select a very low sample rate and 20bit or 24bit
ADC to replace LTC1867L, correct?

Can I use the oscilloscope to capture the output signal frequency of the op amp?

Q3: You may select the 20bit or 24bit device for the replacement. VCC is 3.3V, The package should be less than 3mm*3mm.

The reference should be 2.048V that can be the internal reference or the external voltage.

For all requirements, would you help me to select a suitable device?

Q4: I have a voltage conversion sample question.

For example, if I use a 12bit ADC and the reference is 2.048V, and the ADC read the data is 3000.

a. What is the input analog voltage value? Can you give me a calculation formula?

b. If the input signal is 2.5V, what is the value of the digital for 2.5V analog signal? Can you give me a calculation formula?

2.5*Vref/(2*12-1=4095)=2.5*2.048/4095?

0 Bob Benjamin over 7 years ago in reply to user4674678

TI__Guru** 113765 points

Hi user4674678,

See my responses below.

Best regards,

Bob B

user4674678 said:
Hi Bob,

I would like to explain my application.

The input signal of LTC1867L is the output of the op amp. VCC is 3.3V for my whole circuit system.

The input signal is maybe 1na~1uA. Then I use the transimpedance amp to convert the current signal to the voltage signal.
After this, I use the non-inverting circuit to amplify the voltage signal. That mean I use the 2 stage amplify circuit.
At last, the output signal is input to the input of LTC1867L.

My requirements are as below:

VCC is 3.3V. The package should be less than 3mm*3mm.

The reference should be 2.048V that can be the internal reference or the external voltage.

The communication interface can be I2C or SPI. The input channel number can be 1 or 2 or more. I do nothave no special request

for the input channel numbers.

Now I cannot get my input signal frequency, the questions are as below.

My questions are below:

Q1: I do not know how to get the output signal frequency of the op amp.

If I get the output signal frequency of the op amp, I can select the ADC that the sample rate is higher 2 times than

the output signal frequency of the op amp. Is this correct? [Bob] There are many slow moving signals that are close to DC such as temperature sensors or strain sensors where there are physical properties of the sensor or sensing conditions preventing fast rates of changes. Also, the conditions of the measurement may not require a fast response in the measurement. In these cases it is not necessarily important to quantify a specific change over time in a very small period of time (for example the measurement may be in milliseconds as opposed to nanoseconds.) There are other measurements that appear to be near DC, such as a voltage or current measurement, but the static condition may not be of much interest compared to measurement of a vibration or transient event. In this type of measurement there may be more interest in the change over time in nanoseconds as compared to milliseconds. So the rate of measurement should be the frequency of interest to capture times 2 based on the Nyquist theorem. If the measurement is of the DC type, then the measurement is based on the time interval desired for the measurement to take place such as once a second.

Q2: Since the output signal of the op amp is DC signal. Is the output signal frequency of the op amp equal 0HZ? [Bob] I will try to keep this simple, but basically one can have the question whether or not there really is such a thing as a DC signal. The reason is noise. Even a battery has noise. The op amp itself will also have noise inherent to the device. But for what I believe is the context of your question, if a DC signal source is connected to the amplifier, then the output of the amplifier should also be DC.

If the output signal frequency of the op amp equal 0HZ, then I can select a very low sample rate and 20bit or 24bit
ADC to replace LTC1867L, correct? [Bob] This would be true assuming that you do not have a specific frequency of interest that you are looking for within the output of the sensor.

Can I use the oscilloscope to capture the output signal frequency of the op amp? [Bob] This may be possible.

Q3: You may select the 20bit or 24bit device for the replacement. VCC is 3.3V, The package should be less than 3mm*3mm.

The reference should be 2.048V that can be the internal reference or the external voltage.

For all requirements, would you help me to select a suitable device? [Bob] The smallest package we have for a 20 bit or greater resolution in a delta-sigma device is the ADS122C04 which is 3x3mm. We do have 16-bit devices in smaller packages such as the ADS1115 (I2C) or ADS1118 (SPI) that come in a 1.5mmx2mm package.

Q4: I have a voltage conversion sample question.

For example, if I use a 12bit ADC and the reference is 2.048V, and the ADC read the data is 3000.

a. What is the input analog voltage value? Can you give me a calculation formula? [Bob] Unfortunately I would need more information before I can make the calculation. The value of 1 LSB (or code or count) is related to the full-scale range of the ADC. Some ADCs will output in a straight binary while others output as binary 2's complement.

b. If the input signal is 2.5V, what is the value of the digital for 2.5V analog signal? Can you give me a calculation formula? [Bob] This is similar to the response above. I just don't have enough information.

2.5*Vref/(2*12-1=4095)=2.5*2.048/4095?

0 user4674678 over 7 years ago in reply to Bob Benjamin

Intellectual 490 points

Q1: If it is ADS7042 which is a 12bit ADC, the reference is 2.048V, and the ADC read the data is 3000, Vin=CODE*VREF/2^N, then VIN =
3000*2.048/4096=1.5V, correct?

Q2: I want to confirm the input signal voltage range.

For ADS7042, the input signal voltage range is 1LSB~(VREF-1LSB),correct? I refer to table 1 of the datasheet.

For ADS122C04, if VCC is 3.3V and the reference is 2.048V, the input signal voltage range is 0~(FS (223 – 1) / 223)? or -FS~(FS (223 – 1) ?

0 Bob Benjamin over 7 years ago in reply to user4674678

TI__Guru** 113765 points

Hi user4674678,

See my responses below.

Best regards,

Bob B

user4674678 said:

Q1: If it is ADS7042 which is a 12bit ADC, the reference is 2.048V, and the ADC read the data is 3000, Vin=CODE*VREF/2^N, then VIN =
3000*2.048/4096=1.5V, correct? [Bob] The ADS7042 uses AVDD as the reference voltage. So if AVDD is 2.048V, then your calculation is correct as the output of the ADS7042 is straight binary.

Q2: I want to confirm the input signal voltage range.

For ADS7042, the input signal voltage range is 1LSB~(VREF-1LSB),correct? I refer to table 1 of the datasheet. [Bob] Table 1 is a bit confusing. The input voltage range for the ADS7042 can be thought of as 0V (GND) to VREF (AVDD), and the total number of codes range from 0 to 4095.

For ADS122C04, if VCC is 3.3V and the reference is 2.048V, the input signal voltage range is 0~(FS (223 – 1) / 223)? or -FS~(FS (223 – 1) ? [Bob] The ADS122C04 outputs codes in binary two's complement (both positive and negative values). For a single-ended measurement only 1/2 of the full-scale range is used (the positive range only) where the output codes range from 0 to 8388607 (0x000000 to 0x7FFFFF) . The internal reference voltage is fixed at 2.048V, and a single-ended measurement can range from 0 to 2.048V using the internal reference. It is possible to use the analog supply as a reference by changing the setting in Configuration Register 1. This can change the measurement range from 0 to AVDD (up to 3.3V if AVDD = VCC = 3.3V). The value of one code (same as count or LSB) is Vref/(2^23-1) or 2.048V/(8388607) when using the internal reference or 3.3V/(8388607) when using the AVDD supply reference.

0 Bob Benjamin over 7 years ago in reply to Bob Benjamin

TI__Guru** 113765 points

Questions in this thread also relate to questions and responses in the following thread:

https://e2e.ti.com/support/data_converters/precision_data_converters/f/73/p/721699/2669393#2669393

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ADS122C04: VREF and AVDD questions for ADS122C04