simple output filter question

[George Herold]

Phil Allison wrote:

"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

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.

 

[Jon Slaughter]

Dan Coby wrote:

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.

This is exactly why I said virtual ground.

"With the inverting op amp your cap's grounded pin is at virtual

ground."

It makes a big difference. Else we are simply driving a capacitor. With a
low impedence voltage source. Only the internal resistance of that source
is at play in filtering.

Because the capacitor is frequency dependent we obviously can't simply tie
it directly to ground instead of virtual ground. If we think of the
capacitor as frequency dependent resistance then the gain of the op omp
configuration is simply frequency dependent. Because it's a gain parameter
it must be in the negative feedback loop to have any effect.

Again, there is a difference between ground and virtual ground and the
reason I explicitly mentioned it.

 

[Michael A. Terrell]

Jon Slaughter wrote:

Again, there is a difference between ground and virtual ground and the
reason I explicitly mentioned it.

True. Even if they have exactly the same voltage, the virtual ground
isn't connected to ground.


--
The movie 'Deliverance' isn't a documentary!

 

[Bob Masta]

On Fri, 16 Oct 2009 13:24:37 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com>
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.

The Lancaster book uses "equal component
Sallen-Key", which is a slightly different beast
from the standard S-K. It uses (as the name
implies) equal Rs and equal Cs to determine
frequency tuning, but the damping ("Q") is set by
the (non-inverting) gain.

So the gain isn't unity (for most cases), but in
exchange for that little detail it is *much*
easier to tune. It's also much easier to buy
parts, because everything is equal. The standard
S-K requires the Cs to be computed according to
the damping, with one of the Cs going up and one
down as the damping changes. With the
equal-component S-K, you only need to adjust the
gain of the stage... a standard trimpot can even
be used.

The upshot is that (IMHO) the standard S-K is best
for a production design where you are going to
build a bazillion and can order custom parts. The
equal-component S-K is best for one-offs and
limited production, since you can tweak the stage
gains to get the desired response.

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!

 

[Dan Coby]

Jon Slaughter wrote:

Dan Coby wrote:
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.

This is exactly why I said virtual ground.

I agree that you said "virtual ground". However you changed the circuit
to show the capacitor connected to "ground" not "virtual ground". That
is why I put my comments immediately below your circuit diagram.

"With the inverting op amp your cap's grounded pin is at virtual
ground."

It makes a big difference. Else we are simply driving a capacitor. With
a low impedence voltage source. Only the internal resistance of that
source is at play in filtering.

I agree that simply driving a capacitor (as your diagram shows) is not
the same thing as the original circuit. That is why I replied to your
posting.

Because the capacitor is frequency dependent we obviously can't simply
tie it directly to ground instead of virtual ground. If we think of the
capacitor as frequency dependent resistance then the gain of the op omp
configuration is simply frequency dependent. Because it's a gain
parameter it must be in the negative feedback loop to have any effect.

With an ideal op amp, i.e. one with no output impedance and infinite gain,
the capacitor, in the circuit that you drew, would have no effect upon the
gain of the circuit. The gain is simply -Rf/Rin.

With a real op amp, i.e. one with a non zero output impedance and probably
its own internal pole, the capacitor (in your circuit) would add a second
pole and could easily make the circuit unstable.

Again, there is a difference between ground and virtual ground and the
reason I explicitly mentioned it.

I agree that there is a big difference between ground and virtual ground.
Why did you then draw a circuit in which the capacitor is connected to
ground? It does not help the discussion.


Dan

 

[Jon Slaughter]

Dan Coby wrote:

Jon Slaughter wrote:
Dan Coby wrote:
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.

This is exactly why I said virtual ground.

I agree that you said "virtual ground". However you changed the
circuit to show the capacitor connected to "ground" not "virtual
ground". That is why I put my comments immediately below your
circuit diagram.

Well, I didn't use the term Virtual ground in the graphic because it was too
much trouble to draw all that stuff.

Again, this is why I explicitly said virtual ground right above it. Anyone
that is going to response should at least read what is said

">>>> With the inverting op amp your cap's grounded pin is at virtual

ground."

to me that makes complete sense even if I put the GND symbol. The reason is
that if it were a mistake on my part to use virtual ground then it would be
highly unlikely to insert a random word that made the situation make
complete sense as opposed to leaving off the term in the graphic.

Anyone reading the post with any care should have no problem understanding
what I was talking about. Even if thats not the case it shouldn't matter too
much. This is the internet. Also it's not that big a deal. If anyone was
confused before then they are not now

Sure it would have helped(since it couldn't have hurt) to add VGND or
something but I didn't... it's over and done. I guess when I start getting
paid to reply to people I'll be more cautious.

 

[John Larkin]

On Fri, 16 Oct 2009 19:59:12 -0700 (PDT), George Herold
<ggherold@gmail.com> wrote:

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.

Well, the DC gain is a function of resistor ratios. Cheap delta-sigma
ADCs have drifts of a few PPM per degree C, and you'll have to pay
dollars per resistor, four resistors typically, to build a
state-variable or biquad filter that doesn't wreck that.

John

 

[George Herold]

Phil Allison wrote:

"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

Hi Phil, I guess I do need the 'bleeding obvious' pointed out again
and again.

OK Let's assume some generic DC signal with a bunch of white noise up
to 1MHz, after which it drops off quickly. I want to filter this
signal and feed it to a DMM or some other voltmeter. I'd like to be
able to measure the DC level with high accuracy as some other
parameter is changed. I can short the input and measure the DC
offset. If I want only single pole roll-off, would you recommend
using a passive filter with buffer, an inverting active filter, or
some other configuration.

Thanks,

George H.

 

[George Herold]

John Larkin wrote:

On Fri, 16 Oct 2009 19:59:12 -0700 (PDT), George Herold
ggherold@gmail.com> wrote:



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.

Well, the DC gain is a function of resistor ratios. Cheap delta-sigma
ADCs have drifts of a few PPM per degree C, and you'll have to pay
dollars per resistor, four resistors typically, to build a
state-variable or biquad filter that doesn't wreck that.

Ahhh, Thanks John, I guess I've just started using $0.20 0.1%
resistors, which seem cheap compared to the rest of the components (I
think I'm a man living in the past switches are the most expensive
component I use). If I can 'zero' my measurement do I care so much
about temperature drift?

I must admit that SV and biquad filters are a bit 'freaky'. Who
figured you could put two integrators in a row, close the loop and
everything would be OK. But they seem to work just great!

George H.

John

 

[John Larkin]

On Sat, 17 Oct 2009 20:29:14 -0700 (PDT), George Herold
<ggherold@gmail.com> wrote:

John Larkin wrote:
On Fri, 16 Oct 2009 19:59:12 -0700 (PDT), George Herold
ggherold@gmail.com> wrote:



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.

Well, the DC gain is a function of resistor ratios. Cheap delta-sigma
ADCs have drifts of a few PPM per degree C, and you'll have to pay
dollars per resistor, four resistors typically, to build a
state-variable or biquad filter that doesn't wreck that.

Ahhh, Thanks John, I guess I've just started using $0.20 0.1%
resistors, which seem cheap compared to the rest of the components (I
think I'm a man living in the past switches are the most expensive
component I use). If I can 'zero' my measurement do I care so much
about temperature drift?

The resistors in an SV or biquad don't contribute zero error (the
opamps take care of that for you) but they do donate gain error.

I must admit that SV and biquad filters are a bit 'freaky'. Who
figured you could put two integrators in a row, close the loop and
everything would be OK. But they seem to work just great!

Only one of the integrators is a real integrator; the other has
sort-of-local negative feedback. Two true cascaded integrators would
indeed be indeterminate and would saturate one way or another.

Here's a little 6-pole S-K filter. It uses chopper opamps so dc offset
is just a few microvolts and DC gain is 1.0 +- a PPM or so.

ftp://jjlarkin.lmi.net/Filter1.jpg

16 of them plug into this:

ftp://jjlarkin.lmi.net/Filter2.jpg


John

 

[George Herold]

George Herold wrote:

<snip>

Hey I wanted to thank you all for the stimulating discussion.

I think the obvious point I was missing. (Besides the active versus
passive difference.. Phil A.) Is that the inverting configuration's
accuracy depends on a resistor ratio. (As implied by John L.) Yeah,
this is probablly so obvious that no one thought to mention it. Duh!
<- (I'm mocking myself not any of you.)

In the passive (non-inverting circuit) I think there will also be
leakage / non-idealities in the capacitor. But I'd guess that with a
good film capacitor these errors will be less than the 0.1% of the
resistor ratio. (Any idea's what the biggest error would be.. say a
one second time constant formed with 100k ohms and a 10uF film cap.)


Concerning the temperature dependence of the resistors: If the
circuit was made with metal film resistors from the same roll then I'd
guess the resistor ratio would have much less temperature dependence
than the resistors themselves.

Say, there's an idea. To reduce the temperature dependence of the
gain of an inverting opamp, does anyone use a single R on the input
and then (say) ten of the same R's from the same roll as the feedback
resistor?

George H.

 

[George Herold]

John Larkin wrote:

On Sat, 17 Oct 2009 20:29:14 -0700 (PDT), George Herold
ggherold@gmail.com> wrote:



John Larkin wrote:
On Fri, 16 Oct 2009 19:59:12 -0700 (PDT), George Herold
ggherold@gmail.com> wrote:



John Larkin wrote:
The resistors in an SV or biquad don't contribute zero error (the
opamps take care of that for you) but they do donate gain error.


I must admit that SV and biquad filters are a bit 'freaky'. Who
figured you could put two integrators in a row, close the loop and
everything would be OK. But they seem to work just great!

Only one of the integrators is a real integrator; the other has
sort-of-local negative feedback. Two true cascaded integrators would
indeed be indeterminate and would saturate one way or another.

I worked through the 'math' of the SV filter once upon a time. And it
'behaves' just like a harmonic oscillator. And you are free to choose
any resonant frequency and damping. But still when you first look
at the circuit there are the two integrators... I just never would
have thought to put two of them in a row.

Here's a little 6-pole S-K filter. It uses chopper opamps so dc offset
is just a few microvolts and DC gain is 1.0 +- a PPM or so.

Wow. well I'll have to wait till monday till I can download the pics,
but how do you measure a DC gain to a few PPMs? I have a hard time
with just 0.1%

ftp://jjlarkin.lmi.net/Filter1.jpg

16 of them plug into this:

ftp://jjlarkin.lmi.net/Filter2.jpg


John

thanks,
George H.