Passive RC Filter Loading?

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In summary, using two passive RC filters in series can generate "loading" effects that can affect the output response. It is important to provide a low impedance input to the first stage to minimize these effects.
  • #1
tangodirt
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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:

34oc3f9.png

What effects does loading have on the output response and how can I combat them?
 
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  • #2
tangodirt said:
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:

34oc3f9.png

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

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?
 
  • #3
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%E2%80%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.
 
  • #4
tangodirt said:
how can I combat them?
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.
 
  • #5
berkeman said:
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?

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?
 
  • #6
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.
 
  • #7
berkeman said:
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.

yungman said:
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.

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).
 
  • #8
tangodirt said:
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?

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

so where are you now in your EE curriculum?
 
  • #9
NascentOxygen said:
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.

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

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

so where are you now in your EE curriculum?

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.
 
  • #10
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.
 
  • #11
tangodirt said:
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.

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 sort of 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.
 
  • #12
to pick up on yungman, i would also suggest getting down to the basics and telling us what you want to 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 sort of 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.
 
  • #13
rbj said:
to pick up on yungman, i would also suggest getting down to the basics and telling us what you want to 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 sort of 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.

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?).
 
  • #14
Post the schematic on the circuit you have first. Then people can help you better.
 
  • #15
you don't 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.
 
  • #16
jim hardy said:
you don't 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.
All well and good, Jim ... until he connects the [undisclosed] load.

I think a proper active filter is the go.
 
  • #17
yungman said:
Post the schematic on the circuit you have first. Then people can help you better.

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

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

I think a proper active filter is the go.

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):

2z40a38.png

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).
 
  • #18
jim hardy said:
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. ...

Yes, I did run the experiment again with the separated resistance and capacitance values. The experimental results showed a second order corner, but was marginally worse than the first order filter. I'm assuming this is because of poor component quality, but perhaps there is something else I am missing.

The theoretical response matched the buffered response almost perfectly, which I find very interesting.
 
  • #19
tangodirt said:
Thought I addressed this, perhaps not. The load is a high impedance ADC.
"high" as in "won't measurably load the preceding filter stage"? Then that's good!

Should I surmise that your green plots are of two identical stages, cascaded? Looks like you should now be able to come up with something satisfactory for filtering those high frequencies.
 
  • #20
NascentOxygen said:
"high" as in "won't measurably load the preceding filter stage"? Then that's good!

The ADC they use has an input impedance of greater than 10 GΩ, so I do not expect filter loading, although I would assume an active filter would entirely eliminate any concerns.

NascentOxygen said:
Should I surmise that your green plots are of two identical stages, cascaded?

Yes, I suppose I did not explain this very well. The first order filter uses a nominally 0.1uF capacitor and a 3.3kΩ resistor. The second order filters used the same 0.1uF capacitors, but with a 2kΩ resistors. For the decade separation filter, the same 2kΩ resistor was used, but the second stages used a 20kΩ resistor with a 0.01uF capacitor. The models used the measured values for resistance and capacitance.

NascentOxygen said:
Looks like you should now be able to come up with something satisfactory for filtering those high frequencies.

Now that you mention it, there is probably a nice IC containing higher order filters, right? Looks like Maxim makes a few. Can anyone recommend any IC's to accomplish this? I'm fine with passive resistors, but if there's a nice filter out there that already exists, why reinvent the wheel.
 
  • #21
tangodirt said:
The ADC they use has an input impedance of greater than 10 GΩ, so I do not expect filter loading, although I would assume an active filter would entirely eliminate any concerns.
10GΩ is fine. :smile: You could easily scale up to be 1:100 as I suggested.
Yes, I suppose I did not explain this very well. The first order filter uses a nominally 0.1uF capacitor and a 3.3kΩ resistor. The second order filters used the same 0.1uF capacitors, but with a 2kΩ resistors. For the decade separation filter, the same 2kΩ resistor was used, but the second stages used a 20kΩ resistor with a 0.01uF capacitor.
With the first resistor just 3.3kΩ, then you need to consider the impedance of whatever is driving this, because its impedance (hopefully resistance) becomes part of the first R in your filter calculations. So the input to the filter comes directly from the thermocouple? Do you know what typical impedance the manufacturer quotes that to be?
 

1. What is a passive RC filter loading?

A passive RC filter loading is a type of circuit design that uses resistors and capacitors to filter out unwanted signals or frequencies in an electrical system. It is called "passive" because it does not require an external power source to function.

2. How does a passive RC filter loading work?

A passive RC filter loading works by using the properties of resistors and capacitors to block certain frequencies while allowing others to pass through. The resistor acts as a voltage divider, reducing the amplitude of the signal, while the capacitor acts as a frequency-dependent impedance, allowing certain frequencies to pass through while blocking others.

3. What are the advantages of using a passive RC filter loading?

Some advantages of using a passive RC filter loading include its simplicity, low cost, and reliability. It also does not require an external power source, making it ideal for low-power applications. Additionally, it can be used in a variety of circuits to filter out different types of unwanted signals.

4. What are the limitations of a passive RC filter loading?

One limitation of a passive RC filter loading is that it can only attenuate signals, not amplify them. It also has a limited frequency range and may not be effective for filtering out very low or very high frequencies. Additionally, the components used in the filter may introduce some noise into the system.

5. How do I choose the right components for a passive RC filter loading?

The choice of components for a passive RC filter loading depends on the desired frequency response and the impedance of the circuit. Capacitors with different values can be used to target specific frequencies, while resistors can be chosen to achieve the desired attenuation. It is important to consider the limitations and trade-offs of each component when designing a passive RC filter loading.

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