Hi,

I am going through the data sheet of ADS1298. I have the following queries:

1. What are the basic difference between the different members of the ADS129x series?

2. The input referred noise is 4 μVPP (150 Hz BW, G = 6).

i) What does G stand for?

ii) Considering the max voltage =5.25V. A 24 bit ADC would require an noise which is less than 5.25/(2^24) V , which is equal to 0.3μ. Please explain with the calculations involved??

iii) What is the reference volatge for the sigma delta modulator ??

• Hello Pankaj,

1. The differences are in the number of channels. For example, the ADS1294 has 4 channels, but the ADS1298 has 8 channels. The devices that end in "R" have circuitry on channel 1 which allows for the measurement of respiration rate.
2. Regarding noise:
1. G stands for the PGA gain. There are several PGA gain settings on this device. To see noise for the various settings, refer to tables 1-4.
2. The noise inherent in the device is larger than the LSB voltage. The noise is measured by shorting the inputs together, taking a lot of samples, and taking the standard deviation of those samples. By the way, the LSB is calculated like Vref/(Gain*2^24) rather than supply/(2^24).
3. This device has a built-in internal reference voltage, but also allows for an external reference. For the data in the electrical characteristics table, the voltage for the reference varies but is always stated somewhere on the page. Typically, a 2.4 V reference is used.

Regards,

Brian

• In reply to Brian Pisani:

Dear Brian,

Thanks for the clarification.

I request you to kindly check the following calculations

1 LSB = Vref/(Gain*2^24)

Vref=5.25 V

Gain=6

then, 1 LSB = 5.25/(6*2^24) =0.05214 μV

But the input refered noise  in the datasheet is 4 μVPP

which is >> 1LSB.

Please correct me if my understanding is wrong.

• In reply to Pankaj Jha:

Pankaj,

That is the correct calculation of the LSB voltage, except that Vref = 2.4 V and not 5.25 V unless you are providing Vref = 5.25 externally.

In any case, there are multiple sources of noise in a signal chain. In this device alone there is the PGA and the ADC. Both contribute noise. The LSB voltage you calculated is the noise of the ADC, but it does not include the noise of the PGA. The PGA noise is larger than the LSB voltage and therefore it will dominate the noise of the signal chain. Does this make sense?

Brian
• In reply to Brian Pisani:

Pankaj,

Correction: The LSB voltage is 1 LSB (V) = 2*Vref/2^24. The factor of 2 comes from the fact that the range goes from Vref to -Vref.

Brian
• In reply to Brian Pisani:

HI Brian
I have two follow-up questions:

Vref=Vref = 2.4 V
1 LSB (V) = 2*Vref/2^24 = 0.286μV
But the input refered noise in the datasheet is 4 μVPP
which is >> 1LSB.
This still does not matchup. ???

(2) I agree to you that PGA also contributes noise. But, what is the use of having a high resolution (as high as 24 bit) ADC when the noise at the output of the PGA is much much larger than the ADC's 1 LSB. ???
• In reply to Pankaj Jha:

Pankaj,

The reason you would want the noise of an ADC to be larger than the LSB voltage is because generally engineers like to have data in byte multiples (8-bits, 16-bits, 24-bits, 32-bits, etc.). 16 bits would be too few to capture the dynamic range of the ADC, so we send it out in 24 bits, even if the last few bits are always noise.

This method of measuring and specifying noise is an industry standard. Even TI's competitors who manufacture 24-bit ADCs will have described noise in their datasheets in this way.

Regards,
Brian
• In reply to Brian Pisani:

Hi Brian,
Thanks a ton for the clarifications.
(1) You made a very interesting point -- "generally engineers like to have data in byte multiples ". Can you please elaborate a little on the importance of the same in the engineering community ???
(2) can you please describe the point about the dynamic range of the ADC with some numbers and calculations ??
• In reply to Pankaj Jha:

Hello Pankaj,

1. The length of 8 bits as it relates to engineering comes from computer architecture. All microprocessors, microcontrollers, digital signal processors, and even larger processors store binary data in multiples of 8 bits. I do not know exactly why they settled on 8 versus 4 or 16, but data is stored in bytes. Even the letters you are reading here are are stored in your PC's memory as 8-bit ASCII encoded values.

2. Dynamic range is just the ratio of the smallest signal an instrument can measure to the largest one it could measure. This is closely related to bit resolution.

For example, the ADS1298's dynamic range is measured as a ratio of the noise (which is the "smallest" signal - you could not measure something smaller than the noise) to it's largest possible signal, which is +/- the reference voltage divided by the PGA gain.

For simplicity, look at Table 1 in the datasheet and look at the rms noise for 500 sps data rate and PGA gain of 1. The noise is 2.1 uV. The reference voltage is 2.4V, but we can multiply it by 2 since you can measure between +2.4V and - 2.4V. The dynamic range (in dB) is then 20xlog_10(2x2.4/2.1e-6) = 127 dB.

Now, you have to chose how many bits to use to represent that dynamic range. You can calculate the largest possible dynamic range for any number of bits by finding the ratio between 1 LSB and the total number of possible combinations for a given length of a binary word. For example, in an 8-bit word, there are 2^8 = 256 possible combinations (i.e. 0x00, 0x01, 0x02, ..., 0xFE, 0xFF). For a 16-bit word there are 2^16 = 65,536 different possible combinations. But is it possible to represent both 2.1 uV and +/-2.4V with 65,536 combinations?

It is not. The largest possible dynamic range of 16 bits is 20xlog_10(65,536 combinations/1LSB) = 96 dB. Instead, we chose 24 bits, with which we could have a maximum of 144 dB of dynamic range. This is larger that 127 dB so we can use it to represent both 2.1 uV and +/- 2.4 V.

Regards,
Brian
• In reply to Brian Pisani:

Dear Brian,