INA103: How to read noise spec?

INA103 input refered noise, Vn_RTI, is a function of gain and expressed as:  

The input noise spectral density of 1nV/rt-Hz is extremely low and the only way to further lower the total integrated noise is to lower the effective bandwidth by adding a low-pass filter at the output. Keep in mind that placing the INA103 in gain of 100 will require RG=60ohm, which itself generates 1nV/rt-Hz noise.  Thus, even if the noise of INA103 itself is zero (Vn_input=0 and Vn_output=0), you cannot get lower than the resistor, RG, thermal noise.

Having said that, considering that the input voltage offset of INA103 is as much as (250+ 5000/G) [uV], why do you consider 1uV noise to be a problem?

Marek,
thank you for the clarification. Actually, the signals we are looking to measure often run down to about 0.1 microvolts. We can do lots of signal processing, filtering, etc. to improve SNR but we need to start with as clean a signal as possible. If you have alternate suggestions for this rather odd case, we'd be more than open to hear them.
Thanks!
Pierre

Pierre,
I'm skeptical that you need to measure 0.1uV signal at 1MHz – such precision is usually require for low frequency signal and thus the only way to achieve this is by filtering and averaging the signal. Any device with sub nV/rt-Hz noise would require a very high quiescent current and we do not have anything like this in our porfolio.

Marek,

We of course expect to need to do some averaging and filtering. However, measuring this signal will enable a new class of medical imaging technologies. I'm at this point open to any suggestions as to how to capture the signal that has the lowest possible noise levels before any averaging, etc.  If the quiescent current needs to be high to achieve this, that should not be a major issue.  If you'd be willing to have an offline conversation about this, I would be most grateful.

Thanks,

Pierre

Pierre,

The need for high quiescent current to achieve low noise comes from the fact that the input voltage noise magnitude is inversely proportional to the current biasing the input differential pair - this means, a lower noise level would require designing of a new part and not anything you could do externally within your application. With an existing product, all you can do is filtering and averaging. Therefore, lowering the noise will require filtering – what is the required signal bandwidth for the medical imaging product you are working on? What is the signal gain and supply voltage? It would be the best if you could share your current schematic of the application.

Marek,

thank you. We don't have a schematic yet - still trying to choose the appropriate parts for the front end, and design around them. Signal bandwidth is 200Khz-5MHz, and differential input is preferred. At the sampling rates we're looking at, 16 bit A/D is about the best we can find, and this makes likely amplification 3000x to 10000x.

Pierre
Tucson