Determining Bandwidth of any circuit

I am getting familiar to SNR, noise calculations, thanks to this forum, however I often come across Bandwidth of the system to calculate resistor noise, to get rms noise from nV/rt-Hz, etc. I have a pressure sensor (wheatstone) connected to ADC through an Opamp and I have a single RC LPF between Op-amp and ADC. So should I take cutoff of LPF as bandwidth? And on what factor should the cutoff of my LPF depend since the sensor is just giving DC differential output.

Op amp I am using is MCP6v07, in its datasheet I see a large spike at 10Khz in noise density graph, should I choose my LPF cutoff to be much lower than 10KHz.

I set my ADC to sample at 19.2KHz since it's datasheet says it to be optimal sampling frequency, Should My LPF cutoff depend on ADC sampling rate.? Also is it this ADC sampling rate my BANDWIDTH?

I am so confused. Thanks for any pointers.

• If the acceptable noise power (V^2/R) is given,then you can solve to Bw the equation V^2=noise power density(W/Hz)*Bw*R – GR Tech Sep 23 '14 at 17:47

And on what factor should the cutoff of my LPF depend since the sensor is just giving DC differential output.

Your application is a very sensitive Wheatstone bridge and, if the signal you are looking for is basically DC, then you want your filter cut-off frequency to be as low as possible in order to reduce noise from the op-amp amplifier. But, in reality you can't have a LPF with a DC cut-off frequency because nothing will ever change and, the component sizes will be infinite so you have to re-examine your requirements and possibly 10 Hz might be a good filter cut-off.

You are sampling at 19.2kHz but that is now irrelevant to your design - you could sample at 100Hz and get the same performance if 10 Hz is your low-pass filter. Remember, the LPF does two things: -

1. Gets rid of unwanted self-generated noise from your op-amp amplifier (this is your main problem)
2. Prevents aliasing (this won't be a problem because nothing will get through a 10 Hz filter that would cause aliasing when you sample at 19.2kHz)

In your previous question I reckoned your op-amp had a noise of 60 nV / $\sqrt{Hz}$ but, if you restrict your bandwidth to 10Hz, the sum of all the noises will be over a bandwidth that is 16Hz (believe it or do the math! link) therefore, your equivalent noise at the input to your op-amp will be $\sqrt{16}$ x 60nV = 240nV. This is then multiplied by your op-amp gain (say 10) to give you a real figure of 1.2 micro volts into the ADC.

In your previous question it was 10 micro volts because I had assumed the BW to be 16kHz.

Remember also that the op-amp noise will rise (per Hz) as frequency falls and that in the DC to 10Hz range there will be another figure in the data sheet for the op-amp that covers this area. I'm not sure about the MCP6v07 and how well it's "auto-zero" feature works well at eradicating this LF noise so you'll need to check. However, if I looked at the ADA4528 (because I use it similarly to you) it has only 97nVp-p noise in the 0.1Hz to 10Hz bandwidth and this is a really good figure for an op-amp, made so by the auto-zero feature. It appears that the MCP6v07 is 1.7 micro volts p-p for comparison.

Is this good-enough? - I can't tell you because I don't know what gain the op-amp is needed to be set at and I don't know your requirements - I can only make comparisons.

Noise Equivalent Bandwidth - for a low pass filter the NEB depends on the order of the filter: -

Noise bandwidth = 3dB cut-off frequency $\times \dfrac{\frac{\pi}{2n}}{Sin(\frac{\pi}{2n})}$ where n is the order of the filter. For n = 1 this reduces to Fc x pi/2

• Why would the bandwidth be 16 Hz if you restrict it to 10 Hz? (In other words, what late-night math are you referring to?) – alex.forencich Sep 23 '14 at 8:46
• @alex.forencich it's noise we're talking about and the noise above 10Hz to infinity when all added together effectively is like turning the single order LPF into a brickwall filter of about 1.6x the bandwidth. See pg 9 and 10 of this: google.co.uk/… – Andy aka Sep 23 '14 at 8:48
• Ah I see, a correction for the rolloff. I kinda figured it was something along those lines. – alex.forencich Sep 23 '14 at 9:27
• Thank you @Andyaka. BTW was just curious to know how you arrived to that 60nV/rtHz. – Sajid Sep 23 '14 at 10:20
• @Sajid Table 1-2 in the data sheet gives figures for noise and in the previous question (and not knowing your BW) I kind of averaged the values given for 100kHz and 2.5KHz but, in retrospect I think the noise you will be fighting is the 1.7uVp-p in the line above in that table. To convert this back to RMS it is usual to divide p-p by 6.6 to get an estimate of RMS equivalent (that's a whole new story involving the distribution of gaussian noises!!) – Andy aka Sep 23 '14 at 10:30