Dynamic Range of ADC

Hi,

I am doing my design calculations for my project. I am considering using the following chip

http://www.intantech.com/files/Intan_RHA2000_series_datasheet.pdf

Its a 32 channel chip. The data sheet discusses how to use this chip with AD7980 on page 26. I am confused about the following issue

The amplifier has a fixed gain of 200 and the EEG signal range is from 1uV to 100uV . So, the "MUX_out" pin will have 1uV x 200 = 200uV and 100uV x 200 = 20mV.

The used reference voltage is 2.5 volts. The ADC input voltage range is set from 0 to 2.5 Volts but the input signal ranges from 200uV to 20mV. The resolution of the ADC is 16 bits. 2.5 / 65536 = 38uV/ LSB.

The input signal is not utilizing the full range of the ADC and the amplifier does not have more than 200 gain. It might reduce the accuracy. Should I use reference voltage like 1 Volt instead of 2.5V?

But why the data sheet is using 2.5V reference voltage?

Plus the rms noise of the amplifier is 2uV rms (12uV peak to peak). How can I take this into my calculations?

 

[Tim Wescott]

On Sun, 07 Feb 2016 09:31:20 -0800, davebrown227 wrote:

Hi,

I am doing my design calculations for my project. I am considering using
the following chip

http://www.intantech.com/files/Intan_RHA2000_series_datasheet.pdf

Its a 32 channel chip. The data sheet discusses how to use this chip
with AD7980 on page 26. I am confused about the following issue

The amplifier has a fixed gain of 200 and the EEG signal range is from
1uV to 100uV . So, the "MUX_out" pin will have 1uV x 200 = 200uV and
100uV x 200 = 20mV.

The used reference voltage is 2.5 volts. The ADC input voltage range is
set from 0 to 2.5 Volts but the input signal ranges from 200uV to 20mV.
The resolution of the ADC is 16 bits. 2.5 / 65536 = 38uV/ LSB.

The input signal is not utilizing the full range of the ADC and the
amplifier does not have more than 200 gain. It might reduce the
accuracy. Should I use reference voltage like 1 Volt instead of 2.5V?

But why the data sheet is using 2.5V reference voltage?

Plus the rms noise of the amplifier is 2uV rms (12uV peak to peak). How
can I take this into my calculations?

I would ask if anyone experienced with EEG signals can comment on how
they do it. Getting a gain of 1000 out of an amplifier chain is doable,
but fraught with problems.

The data sheet is using 2.5V for a reference either because that's the
standard of the chips that they're trying to compete with, or because
that's where the chip works best. They give a minimum reference voltage
of 2.4V, so presumably there's stuff inside the chip that's powered from
the reference, and it just won't work right with anything lower.

You take your 2uV RMS noise of the amplifier into account by saying
"gosh, this is extremely way higher than the signal I'm trying to
measure. Maybe I'm using the wrong amplifier here". Or you say "oh
well, I guess the performance of my system will be limited by the noise
of the amplifier!"

If the bandwidth of the noise is wider than the signal you're interested
in, you can sample each channel multiple times with that good fast ADC
and average the samples. As long as the sampling is about the same speed
as or slower than the noise bandwidth (and, of course, way faster than
you need) the averaged samples will be less noisy.

All in all, I think I'd be inclined to isolate the power supply of the
RHA2000 as much as I could, then amplify it's output by 10 or 100 with a
separate amplifier that was also isolated as best as I could, and see if
I could get a clean, higher-voltage signal to the ADC.

--
www.wescottdesign.com

 

Hi,

I got more info. on this amplifier. It generates 1V DC on all the channels all the time with or without any input. The +/- 100uV signal after amplification uses 1V as a baseline. That's why the company's recommended reference voltage is 2.5V.


So, +/- 100uV x 200 = +/- 20mV and noise +/- 16uV x 200 = 3.2mV. The output voltage would be +/ - 1.02V. I am using 16 bit ADC. (2.5) / 65536 = 38uV / LSB.

(1.2) / 38uV = 31579. So, there will be 65,536 - 31,579 = 33,957 unused states.

How can I calculate the accuracy?

 

Sorry , I made the mistake in my past post


(1.02) / 38uV = 26,842. So, there will be 65,536 - 26,842 = 38,694 unused states.


How can I calculate the accuracy?

 

[George Herold]

On Sunday, February 7, 2016 at 12:31:22 PM UTC-5, davebr...@gmail.com wrote:

Hi,

I am doing my design calculations for my project. I am considering using the following chip

http://www.intantech.com/files/Intan_RHA2000_series_datasheet.pdf

The spec sheet is a bit sketchy in parts. 2 uV rms input noise, with
no BW or high frequency white noise contribution.
How much money is the IC?
How long has the company been around?

Its a 32 channel chip. The data sheet discusses how to use this chip with AD7980 on page 26. I am confused about the following issue

The amplifier has a fixed gain of 200 and the EEG signal range is from 1uV to 100uV . So, the "MUX_out" pin will have 1uV x 200 = 200uV and 100uV x 200 = 20mV.

Seems to me you want a lot more gain... but I don't know anything about EEG.
Have you tried asking the company that makes the IC?

George H.

The used reference voltage is 2.5 volts. The ADC input voltage range is set from 0 to 2.5 Volts but the input signal ranges from 200uV to 20mV. The resolution of the ADC is 16 bits. 2.5 / 65536 = 38uV/ LSB.

The input signal is not utilizing the full range of the ADC and the amplifier does not have more than 200 gain. It might reduce the accuracy. Should I use reference voltage like 1 Volt instead of 2.5V?

But why the data sheet is using 2.5V reference voltage?

Plus the rms noise of the amplifier is 2uV rms (12uV peak to peak). How can I take this into my calculations?

 

[Tim Wescott]

On Mon, 08 Feb 2016 19:12:20 -0800, davebrown227 wrote:

Hi,

I got more info. on this amplifier. It generates 1V DC on all the
channels all the time with or without any input. The +/- 100uV signal
after amplification uses 1V as a baseline. That's why the company's
recommended reference voltage is 2.5V.


So, +/- 100uV x 200 = +/- 20mV and noise +/- 16uV x 200 = 3.2mV. The
output voltage would be +/ - 1.02V. I am using 16 bit ADC. (2.5) /
65536 = 38uV / LSB.

(1.2) / 38uV = 31579. So, there will be 65,536 - 31,579 = 33,957 unused
states.

It's worse than that. You only have a 40mV span (1V +/- 20mV). So it's
more like 1000 used codes from the ADC, out of 65536.

> How can I calculate the accuracy?

First, define what you mean by "Accuracy".

Then, calculate how much error you expect.

If necessary for your definition, calculate how much your full-scale
signal swing is.

Then plug the latter two numbers into the definition, turn the crank, and
voila, you have an answer.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

> First, define what you mean by "Accuracy".

I read the following paper online on accuracy. It defines accuracy.

http://m.eet.com/media/1060300/FreescaleADC_018.pdf

It calculates ENOB to determine the accuracy. The lower the ENOB better the accuracy.

I think, in order to make the system more accurate , first we need to know how much it is accurate as is.

The reference is set for 2.5V and the output of the amplifier is +/- 1.02 V.

Then, calculate how much error you expect.

If necessary for your definition, calculate how much your full-scale
signal swing is.

Then plug the latter two numbers into the definition, turn the crank, and
voila, you have an answer.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

But then the range is restricted to +/- 20mV after amplification (200). And the recommended reference is 2.5V. So, the readings will be highly inaccurate. The data sheet does not say to remove the 1V at all from the input signal. Infact, the data sheet is using the 1V + 20mV signal to match the input range of the ADC without any amplification

 

[Bob Masta]

On Mon, 8 Feb 2016 19:12:20 -0800 (PST),
davebrown227@gmail.com wrote:

Hi,

I got more info. on this amplifier. It generates 1V DC on all the channels all the time with or without any input. The +/- 100uV signal after amplification uses 1V as a baseline. That's why the company's recommended reference voltage is 2.5V.


So, +/- 100uV x 200 = +/- 20mV and noise +/- 16uV x 200 = 3.2mV. The output voltage would be +/ - 1.02V. I am using 16 bit ADC. (2.5) / 65536 = 38uV / LSB.

(1.2) / 38uV = 31579. So, there will be 65,536 - 31,579 = 33,957 unused states.

How can I calculate the accuracy?

It looks to me like this chip is doing what all audio chips
do, in order to operate off a single supply. They AC-couple
the input using a positive DC reference. They expect you to
AC-couple the output to remove the reference. Then you would
not have any downstream ADC range issues.

Best regards,


Bob Masta

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