# RFI/AC Filter Rejection for Instrumenation Amplifiers

I'm attempting to measure the voltage developed across a very long piece of cable (upto 500m bundle). The issue is that the cable picks up a lot of noise from nearby radio stations and also has a 50 Hz component which I assume comes from the mains.

I would like to have a filter before my instrumentation amplifier, which I use to amplify the developed voltage about 500x. What sort of low pass filter should I have before the inputs of the in.amp? Do I gain anything by using a active filter or would a simple RC filter do the job?

Note: I would like to reject all AC signals so they don't interfere with my measurement. I have tried a RC low pass filter with R = 10K and C = 22u. This gives a cut off frequency of 0.72 Hz. Is there any disadvantage of using such a aggressive filter (other than the fact that the capacitor will charge very slowly?)

• What's the purpose of D1? Why can't one end of the cable be tied to ground and this then become a single ended measurement? Sep 26, 2012 at 12:57
• @OlinLathrop I haven't yet picked a In.Amp but it seems most of them don't have inputs that can reach rails completely (even those who are marketed as rail to rail). They need to be above gnd by a few mV at least.
Sep 26, 2012 at 13:01
• Yes, many amps don't really sense all the way down to ground, often despite what the datasheet says. However, I'd fix that by making a small negative supply, not by making the signal differential and all the complications that go with that. A small charge pump can make a few volts negative at a few mA, good enough for many opamps. Sep 26, 2012 at 13:57
• @OlinLathrop That's a VERY neat solution! Does the -ve voltage need to be -5? I don't think so, I think even -1V would do. Though I reckon I will need to reference the output to ground as thats what the ADC would expect. I think I will read up on that and ask questions here if I need.
Sep 26, 2012 at 13:59
• Just a few volts negative is enough, which can easily be done with a ordinary charge pump running from 5V. After the transistor and diode drops, you might end up with -2.2V. That's plenty for most opamps. For "rail to rail" input opamps just a few 100 mV would help. Some truly go down to ground on the input. The simplest answer would be to find one of those. Sep 26, 2012 at 16:07

First, you don't want to get rid of all AC. If you did, the reading would never change. You need frequencies below some level to get thru, so you have to decide what that level is. You also should look at this in the time domain after deciding how long you are willing to wait for a reading and how accurate it must be. That will implicitly define some frequency response, but I think settling time is a more relevant way to think about it in this case.

Your problem is more complicated due to diode D1 in the circuit. I don't see why it is there and what it is doing for you. If you can simply replace the diode with a connection, then you have a single ended measurement instead of a differential one. As it is now, you have to worry about common mode and differential mode issues separately.

You want to do most of the filtering passively because that will work to much higher frequencies than what active electronics such as in the inamp can handle. This includes the common mode part of the signal. In the ideal case, that doesn't matter, but the inamp isn't ideal. Above some frequency, common mode signal will change faster than the active electronics in the inamp can compensate for it, and some will appear as differential mode signal on the output. Radio frequencies are likely well above what the inamp can deal with correctly.

Unfortunately, filtering the common mode part of the signal is tricky. Any assymetry in the filters results in a differential mode signal. If you absolutely must have D1 there for some reason, then I would filter each line separately with a single R-C filter at a few kHz. That's still well above any real signal, but low enough that the inamp should be able to take it from there. 1 kΩ followed by a 100 nF ceramic cap to ground should do fine. That is a low pass rolloff of 1.6 kHz, which is well above any signal you care about, but low enough to filter out the nasty stuff that will confuse the inamp. For example, 1 MHz would be down by over 50 dB.

Now that the two signals contain only frequencies the inamp can deal with, you can run these straight into the inamp. You can put another cap directly accross the inamp inputs as Steven suggested. This will work with both resistors on each of the lines as if they were in series. If you put another 100 nF cap there, then the low pass rolloff frequency would be 800 Hz.

You now have a nice single ended output with about 800 Hz rolloff and radio pickup eliminated. Here is where I would put a lower dominant filter that is adjusted as low as possible given your settling time constraint. Let's say you need the signal to settle to 1 part in 1000 and are willing to wait 2 seconds for that. For a single pole filter, the 1:1000 specifies 7 time constants. The time constant of the filter is therefore 2s/7 = 290 ms = R*C. If we pick 1 µF for C, then R would need to be 290ms / 1µF = 290 kΩ. Those are tractable values, although you will probably need another buffer amp after the filter. Just to see what this came out to in frequency space, 1 µF and 290 kΩ have a low pass rolloff of 550 mHz. That is way below the other low pass filters we put before the inamp, so we can ignore them for the purpose of settling time and ultimate bandwidth. Their purpose was only to limit the frequencies going into the inamp so that it works as intended.

• would you suggest I also get rid of the in. amp and use a single-ended amplifier? Or simply connect the in. amp's In- to gnd?
Sep 26, 2012 at 13:45

I would simply place a capacitor across the InAmp's inputs, or rather a couple of them in parallel, like for instance 10 nF, 100 nF and 10 µF. These should short-circuit any RF signals and 50 Hz noise picked up by the cable. Ignoring the 10 µF for HF (it won't perform that well at high frequencies), 110 nF is 0.14 Ω at 10 MHz, which should take almost all energy out of it. The 10 µF should take care of lower frequencies.

In theory the parallel capacitors should be just one larger capacitor, but their impedance rises again at higher frequencies due to ESL (Equivalent Series Inductance), and the poles are different for different capacitances, like this graph shows:

By placing different capacitances in parallel you take advantage of the poles for each of them.

• stevenvh, wouldn't the capacitors in parallel act like one capacitor?