simple output filter question

Here's a question that has been bothering me for a while.

I want to make a simple lowpass output filter. With time constants from maybe 1 ms to about max 3 sec.

I like to cascade two of them. (two pole filter) But this is only tangentially related to the question.

If you just make a simple RC roll off with an opamp buffer, do you like the inverting or non-inverting configuration?

Vin----RRR-+-----|\\ C |+\\ C | >+--- C +-|-/ | | | |/ | G +-----+ N D

Non-inverting

+--RRR--+ +--CCC--+ | | Vin----RRR---+---|\\ | |-\\ | | >+--- +-|+/ | |/ G N D

Inverting.

Advantages of inverting config. constant input impedance no ground current (except for opamp bias)

Advantages of non-inverting config. less noise (but it's an output filter.. I couldn't care less about the Johnson noise.) it looks simpler.

Personally I like the inverting, but I worry that I'm missing something obvious.

George H.

Reply to
George Herold
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"George Hairoil" <

( snip actual question)

** The fact the op-amp has to drive a possibly large value cap in one case but not the other ?

.... Phil

Reply to
Phil Allison

You may want to get a copy of the "Active Filter Cookbook" by Don Lancaster. It mostly makes use of the non-inverting form. The main benefit of the non-inverting form is the ease of adjusting the damping when you use equal-component Sallen-Key filters for 2 or more poles.

If your thought about cascading is to make a

2-pole filter out of two 1-pole filters, you will probably be very disappointed. You need an actual 2-pole topology (and design equations, including selection of Butterworth, Bessel, Chebychev, etc) to get decent results. The book makes it simple for beginners.

(If you just cascade two identical single-poles, consider that the corner frequency is now at -6 dB instead of -3 dB.)

Best regards,

Bob Masta DAQARTA v4.51 Data AcQuisition And Real-Time Analysis

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Reply to
Bob Masta

case

Hi Phil, I've never had any problems using opamps as integrators with the cap tied to the inverting input. Very different situation if the cap was to have one end grounded.

George H.

Reply to
George Herold

I.E, a Buffer. The configuration is simply buffering the RC circuit. It has no wasted power. The cap can only charge and discharge back into the source.

An inverting amplifier with frequency attenutation. In this case it is an inverting buffer of the RC filter.

The non-inverting op amp is more ideal. You can load the input stage with small R in the inverting op amp config. Almost in all case they are equivalent except for the inversion. The inverting op amp gives the ability to have gain.

With the inverting op amp your cap's grounded pin is at virtual ground. If the op amp was ideal we could write it as

In this case we see that it is simply an inverting buffer fed in the cap. This effectively is just making the input stage have very large drive capabilities and then filtering. For large currents though this could cause C to get hot and create non-linear problems. With the non-inverting configuration We can choose R very large and reduce the size of C considerably. This is not the case with the inverting configuration for the same output drive. Also the input will be loaded slightly with the inverting case since it is not isolated from the output.

We can rewrite the above circuit again as

| C | GND

where Buff is the inverting buffer. (Of course here I have increased the output impedance)

While the non-inverting configuration is

| C | GND

As you can see the non-inverting case is better in just about all regards. Also note that I made the assumption that the op amp was ideal for the inverting case. In the process you can see that one looses the fact that the ouput is NOT isolated from the input in the inverting case.

So, the point is that you should use non-inverting configuration unless you need to invert.

Reply to
Jon Slaughter

Thanks Bob, I've read some of Don's book. For two pole filters I really like the state variable filter. (I'm less a fan of the Sallen- Key and or VCVS. After all the opamps are often cheaper than the capacitors.) For an output filter I like the option of having both a single pole and double pole available. The single pole is nice if you want to use the ouput as part of a control loop. (As you probably know.) I've made a few of these and have done it both ways.

As an interesting side note, if you cascade tow of them with one inverting and the other non-inverting and both using the same dual opamp I find that the opamp (current and/or voltage) offsets tend to cancel.

George H.

Reply to
George Herold

I like the inverting connection because it keeps the amplifier(s) from operating with common mode voltage. This may be an issue depending on the actual voltages used in your output filter plus the CMR of the amplifier(s) used.

If you want a two pole filter, why not use a single amplifier configuration instead of cascading two simple filters? Secondly, cascading two R-C filters with "real" poles has a very low non-resonant "Q" and therefore has a very sloppy response. I would use a Butterworth or Chebychev alignment to get a sharper cut off and a flatter pass band response. Other filter alignments are also possible with a single amplifier configuration.

You mentioned two time constants, does this imply that you want some sort of variable filter? If you define the filter you want, I can provide the component values to realize it. Bob

Reply to
Bob Eld

One nice thing about S-K filters is that they can be very DC accurate. The resistors don't affect the gain, which is basically 1.000... at DC.

John

Reply to
John Larkin

"George Herold" "Phil Allison"

I've never had any problems using opamps as integrators with the cap tied to the inverting input.

** Integrators are not filters.

Very different situation if the cap was to have one end grounded.

** The cap IS grounded at one end.

The minus input is a virtual earth.

..... Phil

Reply to
Phil Allison

No! This is not equivalent (even assuming ideal op amps) to the previous inverting single pole filter. The original circuit has a pole which is formed by the capacitor and the feedback resistor. For more info see:

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The second circuit is simply an inverting amplifier with a capacitor loading its output. Its response is simply -Rf/Rin. There is no pole (assuming ideal op amps).

To see the difference, look at the currents around the inverting input to the op amp. In the original circuit, you have current through the input resistor which is being balanced by the current through both the feedback resistor and the current through the capacitor. In the second circuit, the current through the capacitor is not present at this node. Without the current through the capacitor at the inverting input node, you do not have the pole.

Dan

Reply to
Dan Coby

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Great another =91vote=92 for inverting. Thanks Bob E., I forgot about the common mode rejection.

I can certainly calculate my own time constants, but thanks for the offer. For the longest times I like the 10uF film caps from Panasonic.

George H.

Reply to
George Herold

Are the state varible filters worse in trems of DC accuracy?

George H.

Reply to
George Herold

Hi Phil, I hope you don't start swearing at me. But 'virtual earth' is different than real earth. If you made a circuit like John S. drew... (but you have to move the feedback resistor too so that it's resetting the cap to ground.) Then you might have problems. But integrators work just fine.

The inverting lowpass looks like an integrator with a resistor reset. (At least that's one way to think about it.)

George H.

Reply to
George Herold

"George Hairoil s going Bald "

Great another ?vote? for inverting.

** The truth of a matter cannot be determined by taking a vote.

And you are trying to CONTROL the answers you get.

Very BAD usenet idea - marks YOU as a TROLL. !!!

Thanks Bob E., I forgot about the common mode rejection.

** It's a total red-herring.

I can certainly calculate my own time constants, but thanks for the offer. For the longest times I like the 10uF film caps from Panasonic.

** Better calculate what *slew rate* a typical op-amp can deliver into a 10uF cap.

Might not be much.....

..... Phil

Reply to
Phil Allison

Thanks Dan, I wasn't going to say anything. I think his 'approximate' circuit would be a bit better if there was a resistor to ground on the output also... but still the circuit as originally drawn is the one I'm interested in.

George H.

Reply to
George Herold

"George Hairoil is Bald "

** Fraid it is not - pal.

The signal voltage at a virtual earth is virtually zero.

The op-amp in your "inverting" schem as the entire capacitance as a load to be driven.

YOU ARE MISSING THE POINT BY A MILE !!

..... Phil

Reply to
Phil Allison

Hi Phil, I guess in the inverting config. it's the first resistor that is limiting the current that the opamp is 'asked' to deliver. (After all this is an output stage and there is some 'similar' opamp driving the input resistor.)

Both the above circuits work 'almost' identically. (Well one does invert the signal.) My question is which one you like better. I think I indicated right away that I like the inverting config.... though this seems a not as simple as the non-inverting.

Sorry for the 'vote' remark.

George H.

Reply to
George Herold

Hi Phil,

"> The signal voltage at a virtual earth is virtually zero."

Oh really? You must have measured this. 'virtuallly zero' may only be 1mV or so but it's still 1mV. How about trying both circuits? They are both quite simple.

George H.

Reply to
George Herold

"George Hairoil is Bald "

Hi Phil, I guess in the inverting config. it's the first resistor that is limiting the current that the opamp is 'asked' to deliver.

** Irrelevant to the output ability of an op-amp driving a 10uF cap as a load.

Both the above circuits work 'almost' identically.

** Not true at all.

One is an active filter, the other is a **passive filter** followed by a buffer.

(Well one does invert the signal.) My question is which one you like better.

** Liking does not come into it - you silly old bugger.

This was your Q:

" Personally I like the inverting, but I worry that I'm missing something obvious."

So, do you want the BLEEDING OBVIOUS pointed out to you AGAIN and AGAIN or are you just gonna move the goalposts around until you get the answer you wanted all along.

Eh - Mr. Troll ?????

..... Phil

Reply to
Phil Allison

"George Hairoil is Blatant TROLL "

** Which is virtually zero - you pathetic ASS !!!

How about trying both circuits?

** How about you STOP TROLLING and go drop dead.

** Still way over your tiny, pointed head.

..... Phil

Reply to
Phil Allison

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