Passive RC Filter Loading?

tangodirt

I'm trying to use a second order passive RC filter to attenuate noise in a thermocouple signal. I've heard that using two passive RC filters in series can generate "loading" effects. I'm not sure exactly what this means, except for the output response is different from the theoretical response.

Here is my theoretical response for a 500Hz cutoff frequency:


What effects does loading have on the output response and how can I combat them?

berkeman

To get an ideal frequency response from two cascaded 1st order filters, you need a buffer stage between them with an infinite input impedance and a zero output impedance. Does it make more sense now?

yungman

Yes, like Berkeman said, if you don't have a buffer to isolate the two stages, they interact with each other and have a different response. I notice your frequency is quite low, My suggestion is looking into active low pass filter. Here are two articles:

http://en.wikipedia.org/wiki/Sallen%...93Key_topology

http://www.ti.com/lit/ml/sloa088/sloa088.pdf

They are very simple to implement, just calculate RC like your passive filter, then adjust the d.

NascentOxygen

To minimize loading effects, you should give the first stage a low impedance, and the second stage a high impedance.

e.g., if the first low pass stage comprises R kΩ + C μF, make the second stage 100·R kΩ + 0.01·C μF.

Other considerations, however, generally make this simple method impracticable, and you have to implement buffering.

tangodirt

Yes, this makes sense.

Can anyone clarify the means that cause the response change?

Also, how does the response change? Does filtering become more aggressive, less aggressive, or just different?

NascentOxygen

Have you got simulation software? You'll get a much better feel for what is going on if you simulate a couple of filter circuits with different components and see the responses. In particular, what I said earlier.

rbj

but you can still, without an op-amp buffer, put the two poles anywhere you want on the negative real axis, with a passive RCRC filter. if you want resonant (complex-conjugate) poles, you need either an inductor: RLC filter, or you need some gain (an op-amp circuit like the Sallen-Key that yungman pointed to).

rbj

what you're referring to here is filter analysis and filter design. it's about transfer functions and frequency response. this is sorta what you get in some electronics class or in a linear electric circuits course.

so where are you now in your EE curriculum?

tangodirt

I do not have simulation software, unfortunately. Perhaps I will just build the circuit and quantify the response experimentally.

I'm a mechanical engineer, although probably an electrical engineer at heart. I've worked with transfer functions, frequency response, etc., but it's all been theoretical. My transfer function approach is what generated the plot in my first post. Sadly, it doesn't account for the non-ideal response generated by stacking filters.

yungman

To help you further, you need to provide info on what are you going to use the output of the filter to drive. Your question ended at the output of the filter which is really an open ended question. The important question is the drive capability requirement. If you need to drive a low impedance load, then that will change the filter response if it is a simple RC, RLC filter.

That's the reason I suggested the active filter as you are working with lower frequency range where active filter behave very ideal. And also the output impedance is low so it can drive the next stage easily without affecting the response. Also, if you are a mechanical engineer and don't get into the details of the filter design, there are cook books that you can just follow how to design the Sallen-Key filter. Once you know how to set the d, the RC=1/(2πf) so you can implement it very easy.

If you use totally passive components, it might look simpler, but with relative low frequency, the value of the components tend to increase and the output loading start to become a big part of the filter and you end up having to put an op-amp buffer and you end up with a bigger circuit. With two stage passive circuit, calculation tends to be more complicated also.

rbj

okay, set up the circuit but with completely variable R's and C's. so there is R1 and C1 and R2 and C2. analyze the circuit. determine where your poles and zeros (well, i guess there aren't any zeros) are in terms of R1, R2, C1, C2.

then you have freedom to set two of those four parameters to something sorta arbitrary. pick a decent typical values for C1 and C2. they can even be equal.

then, given C1 and C2, and the two real pole values (they might be equal if you want them to be), and solve for R1 and R2. they won't be equal.

but if R2 >> R1 which means that C2 << C1, you can see that the effect of "loading" (which, for me, is just an issue of circuit generality) gets smaller and smaller. maybe to the point you can ignore it. but you don't have to ignore it or do this. just get on top of the math and what knobs you can twist.

rbj

to pick up on yungman, i would also suggest getting down to the basics and telling us what you wanna do.

so you have the output of a thermocouple and it's a little noisy and you want a LPF to reduce some of the noise, is that it? it appears that you're sorta fussy about how the LPF should appear in the frequency response. because even with this extra "loading", you still have a Low-Pass Filter. what are the specific needs you have of your LPF?

because this is electrical, and possibly without active parts (transistors or op-amps), then the output resistance of your thermocouple is important, and the input resistance of whatever you're connecting your output (of the LPF) to is also important. do you know these values?

one thing, if you're going to get real anal about this, if you have a passive LPF, besides the attenuation you expect to get at high frequencies, you will also have attenuation at DC because of these input and output resistances.

tangodirt

I'm simply trying to put together a small circuit to amplify and filter a thermocouple signal (including, at the worst case, the thermocouple response to a step function, i.e. air to boiling water). The setup previously used was very poor (thermocouple leads connected straight to high impedance ADC). The noise floor with this setup is something like 5-10C, not to mention poor resolution from such a small signal.

My goal is to put together a small PCB that takes in the thermocouple input on one end, and outputs an amplified and filtered response on the other. My current design (on paper) uses an Analog AD8497 for a type-K thermocouple with a 320Hz input stage, common mode filter (500Ohm, 0.1uF and 1uF differential RC filter) for attenuating RFI and other high frequency noise. I was planning to put an output stage filter on the output line of the AD8497 for filtering power-line (50/60Hz) noise and further attenuating any higher frequency noise (25Hz cutoff?).

yungman

Post the schematic on the circuit you have first. Then people can help you better.

jim hardy

you dont need fancy computer simulation. Simple ohm's law will do.


First, figure transfer function of simple RC low pass , 1/(rsc+1)

now add a second stage and observe your transfer function sprouts a lot more terms - because second RC is in parallel with first C.

Derive that function and at corner frequency, tabulate attenuation for various ratios of R1::R2...
observe that if second R >> first 1/sc at corner freq, you're not too far off a properly buffered response.

Rule of thumb in analog days was keep it a decade away. Somebody said that earlier. ...

I hope you go thru the exercise in algebra for it'll plant the concept firmly in your mind.

NascentOxygen

All well and good, Jim ..... until he connects the [undisclosed] load.

I think a proper active filter is the go.

tangodirt

The diagram is on a different machine. I will try to post it later.

Thought I addressed this, perhaps not. The load is a high impedance ADC.

Regardless, I did go back and rederive the transfer function using KCL. Kirchoff's current law adds the additional loading term into the equation. Based on my limited understanding, it seems as if the "loading" term is caused by two things: the phase shift of the circuit and the extra current being pulled through the first stage resistor by the second stage filter. Does that sound right?

Here's a plot of my model and my experimental data (I did build the circuits on a breadboard):


For whatever reason, the bode analyzer I used generated some funky numbers on the second order filter with no buffering and the decade separation. The dashed teal line shows this (discontinuities).

The buffered response was rather disappointing. I used a 741 Op-Amp Voltage Follower to separate the two and it did get marginally better response, but only below the cutoff frequency. The corner didn't appear to get any sharper (in contrast with the theoretical model).