Instrumentation Amplifier Selection

SM,

You typically should calculate the noise refered to input, RTI, and then multiple it by the gain to refer to the output, RTO.

Doing so, for G=500, you will get: Vnoise RTI = [(8^2)+(80/500)^2]^0.5 = 8.0016nV/rt-Hz, which is in line with your Vnoise RTO = 500*8.0016 = 4000.8nV/rt-Hz

However, in order to calculate the total RMS noise, you must simply multiply the voltage spectral density by square-root of noise bandwidth.

Thus, if your noise bandwidth is 10kHz, Enoise RTO = 4000.8nV/rt-Hz *(10,000Hz)^0.5 = ~0.4mVrms, therefore close to 0.3mVrms RTO required by your customer.

Since in the gain of 500, the total noise is dominated by the input noise spectral density, this means that to meet customer RTO noise spec of 0.3mVrms, you must choose a device with Eni of 6nV/rt-Hz or lower.

It is difficult to address your other questions because you do not provide basic application requirements like supply voltage, input voltage range, IQ, Vos, drift, IB, etc. In order to obtain an effective bandwidth of 10kHz in gain of 500, you must have a minimum gain-bandwidth product, GBP, of 5MHz or higher.  Thus, some of the op amps you may consider are low voltage OPA365, OPA388, OPA376 (max Vs of 5V) or  OPA197, OPA189, and OPA828 with max 36V supply voltage.

To achieve +/-300V input common-mode voltage range, you would have to use the op amp in a difference configuration with properly scaled resistors (see INA149) but to meet 100dB minimum CMRR requirement would also necessitate matching of the resistors to within 0.001% or better, which will get expensive.

Marek,
Thank you very much for your insightful reply.
From the datasheets, CMRR is a function of the gain of the instrumentation amplifier and it goes up with the gain (ref. AD8422, unfortunately could not find an equivalent from TI) Our requirement is 100dB at a gain of 100. In such a scenario target CMRR would be achieved internally by the device itself and we do not need to worry about adding high precision resistors.
The complete DAQ chain (from sensor output to ADC) would have the instrumentation amplifier as the first stage, active low pass anti-aliasing filter as the second stage, an active single ended to differential converter as the third stage and finally the ADC itself. Each stage will add some noise so how do we calculate the total noise of the receive chain.
Supply voltage is going to be +/- 13V, input voltage range +/- 10V.
Thanks again,

If you would want to cover input common-mode voltage up to +/-300V, it would be necessary to build a difference amplifier with precision matched resistors.  AC CMRR is a function of gain but since CMRR=ΔVcm/ΔVos, dc CMRR in a difference amplifier is a function of matched resistors:  0.1% matching allows min CMRR of 54 dB while 0.001% results in min CMRR of 94dB.

In case of instrumentation amplifer you do not need to worry about resistor maching because they are already laser trimmed at the wafer level; however, instrumentation amplifer will not be able to cover common-mode input voltage range in hundreds of volts.  To cross-reference AD8422, please see INA821.

In circuit configuration involving multiple gain stages with similar noise figure, the noise of the first stage will dominate the total RTI noise since the noise of second and third stage gets devided by the gains of the previous stages.