I am using an AD8421 with a 100 ohm gain resistor. The datasheet for the AD8229 explains how to calculate the Johnson noise contributed by the gain resistor. I would also like to calculate the current noise contributed by the gain resistor, but I'm not sure how to do that. The underlying question is whether the extra cost of a metal foil would be justified on the basis of noise reduction, or whether a thin film resistor will provide comparable results. I need the maximum possible S:N ratio, but I don't want to use a $10 resistor unless it is clearly justified by lower noise.

Hi Robert,

I've talked to our product expert regarding your query and he provided me calculations showing that the AD8421 noise generates the same level as the excess noise by the thin-film resistor .Therefore there is no big advantage of buying metal foil resistor since the thin film noise is already at the same level as the in-amp and therefore it cannot be the dominant noise source. Please see below for the sample computations that our product expert provided and feel free to ask if you need some clarifications.

If you look at the AD8421, it has very low 1/f noise. And typical thin film resistors can have excess noise of about -20dB, so let’s compare. If you use a 100Ω gain resistor, it puts the AD8421 in a gain of 100. For easy math, let’s say the maximum output voltage is 10V for the system, meaning you have about 0.1V across the gain resistor at the maximum.

First, we can calculate the 1/f noise of the AD8421 in the same units as the resistor excess noise (dB referred to 1 µV/V in 1 decade). You see from the gain of 100 noise plot that the AD8421 noise at 1Hz is 7nV/√Hz and the wideband noise is about 3.5nV/√Hz. If you RSS subtract the wideband value (you don’t need to do this for products where 1Hz is well-below the corner frequency), the 1Hz noise just due to 1/f is about 6.1nV/√Hz. The RMS noise in any band in the 1/f region can be calculated by the 1/f noise at 1Hz, en1Hz * √(ln(f

_{H}/f_{L})). For 1 decade, f_{H}/f_{L}= 10, so it simplifies to en1Hz * √(ln(10)) = 6.1nV/√Hz * √2.3 = 9.26nVRMS per decade.If you relate this back to a resistor with 0.1V across it, you get about 0.093µV/V or about -20dB. This is on the same level as a thin-film resistor, so there’s no big advantage by buying the metal foil which is better than -40dB, because the thin film noise is already at the same level as the in-amp and therefore it cannot be the dominant noise source.

If you relate this back to a resistor with 0.1V across it, you get about 0.093µV/V or about -20dB.

Also kindly note that the 1/f noise doesn't contribute much if you’re measuring even slightly outside the 1/f region. You can do the math and find that if the -3dB bandwidth of your circuit is 1 decade beyond the 1/f corner, the wideband noise starts to dominate. And if the -3dB bandwidth of your circuit is 100 times the 1/f corner, you can typically ignore 1/f noise altogether. So even in cases where the resistor excess noise might be larger than the 1/f noise of the amplifier, it still might not matter for the total noise of the system if the bandwidth is relatively wide.

I hope this helps.

Best regards,

Emman