simple output filter question

[George Herold]

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.

 

[Phil Allison]

"George Hairoil" <

( snip actual question)
Personally I like the inverting, but I worry that I'm missing
something obvious.

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



..... Phil

 

[Bob Masta]

On Thu, 15 Oct 2009 20:13:06 -0700 (PDT), George
Herold <ggherold@gmail.com> wrote:

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?

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
www.daqarta.com
Scope, Spectrum, Spectrogram, Sound Level Meter
FREE Signal Generator
Science with your sound card!

 

[Jon Slaughter]

George Herold wrote:

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

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.

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

Inverting.

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

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.

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

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

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

Vin----RRR---Buff---+--Vout
|

C
|
GND

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

While the non-inverting configuration is

Vin----RRR---+--Buff---Vout
|

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.

 

[Bob Eld]

"George Herold" <ggherold@gmail.com> wrote in message
news:8bc692bb-6a05-4b2c-8f7f-ccbfc6c885dd@2g2000prl.googlegroups.com...

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.

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

 

[George Herold]

On Oct 16, 12:15 am, "Phil Allison" <phi...@tpg.com.au> wrote:

"George Hairoil"

( snip actual question)

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

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

.... Phil

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.

 

[George Herold]

On Oct 16, 8:40 am, N0S...@daqarta.com (Bob Masta) wrote:

On Thu, 15 Oct 2009 20:13:06 -0700 (PDT), George

Herold <ggher...@gmail.com> wrote:
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?

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
             www.daqarta.com
Scope, Spectrum, Spectrogram, Sound Level Meter
           FREE Signal Generator
        Science with your sound card!

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.

 

[John Larkin]

On Fri, 16 Oct 2009 07:26:16 -0700 (PDT), George Herold
<ggherold@gmail.com> wrote:

On Oct 16, 8:40 am, N0S...@daqarta.com (Bob Masta) wrote:
On Thu, 15 Oct 2009 20:13:06 -0700 (PDT), George

Herold <ggher...@gmail.com> wrote:
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?

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
             www.daqarta.com
Scope, Spectrum, Spectrogram, Sound Level Meter
           FREE Signal Generator
        Science with your sound card!

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.)

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

 

[Phil Allison]

"George Herold"
"Phil Allison"

"George Hairoil"

( snip actual question)

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

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

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

 

[Dan Coby]

Jon Slaughter wrote:

George Herold wrote:
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



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.




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

Inverting.



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


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.


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


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

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:
http://en.wikipedia.org/wiki/Low-pass_filter#Active_electronic_realization

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

 

[Phil Allison]

"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

 

[Phil Allison]

"George Hairoil is Bald "

( snip actual question)

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

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


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.


Hi Phil, I hope you don't start swearing at me. But 'virtual earth'
is different than real earth.

** 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

 

[Phil Allison]

"George Hairoil is 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.....

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

 

[Phil Allison]

"George Hairoil is Blatant TROLL "

( snip actual question)

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

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


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.


Hi Phil, I hope you don't start swearing at me. But 'virtual earth'
is different than real earth.

** 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 !!


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

Oh really? You must have measured this. 'virtually zero' may only
be 1mV or so but it's still 1mV.

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

How about trying both circuits?


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

They are both quite simple.

** Still way over your tiny, pointed head.



...... Phil

 

[George Herold]

Bob Eld wrote:

"George Herold" <ggherold@gmail.com> wrote in message


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

Great another ‘vote’ 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.

 

[George Herold]

John Larkin wrote:

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

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

George H.

 

[Dan Coby]

George Herold wrote:

Dan Coby wrote:
Jon Slaughter wrote:
George Herold wrote:

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


+--RRR--+
| |
Vin---RRR---+---|\ |
|-\ |
| >+---+
+-|+/ |
| |/ C
G |
N GND
D
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:
http://en.wikipedia.org/wiki/Low-pass_filter#Active_electronic_realization

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

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...

Not really. Moving the capacitor totally changes the circuit. As A result
all the rest of his analysis does not add anything useful to your question.

but still the circuit as originally drawn
is the one I'm interested in.

Good luck with that. I have never had to worry aout the fine points of analog
filter design so I will not claim to be an expert. However since this is usenet
I will add a few comments.

The different circuits (inverting versus non-inverting) have different input
impedances. The constant input impedance of the inverting circuit often has
advantages.

Most op amps circuits work better (i.e. have fewer problems with drift, etc.)
if both inputs have the same input impedances. Both of your circuits have one
input with a zero impedance.

 

[George Herold]

Phil Allison wrote:

"George Herold"
"Phil Allison"
"George Hairoil"

( snip actual question)

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

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


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

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.

 

[George Herold]

Dan Coby wrote:

Jon Slaughter wrote:
George Herold wrote:


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


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


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:
http://en.wikipedia.org/wiki/Low-pass_filter#Active_electronic_realization

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

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.

 

[George Herold]

Phil Allison wrote:

"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

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.