Do active filters even care about input/output impedances?

[billcalley]

Hi All,

I've been looking at a lot of active filter design software and
design formulas, and none seem to care too much about the input and
output impedances that the filter will "see" when placed in a circuit.
When designing *passive* filters, we would typically specify these
impedance values as a matter of course (normally 50 ohms). Why don't
active filter design programs even ask what the input and output
impedances are that the filter will have to work with, nor state what
it is after the circuit is synthesized? Does it even matter; or will
the frequency response and gain just not be affected by most normal
values? Or is it assumed that the active filter will be placed between
certain impedance values? If not, then how can I tell what the
"optimal" impedance values should be for an active filter? This has me
baffled!

Thanks,

-Bill

 

[Andrew Holme]

billcalley wrote:

Hi All,

I've been looking at a lot of active filter design software and
design formulas, and none seem to care too much about the input and
output impedances that the filter will "see" when placed in a circuit.
When designing *passive* filters, we would typically specify these
impedance values as a matter of course (normally 50 ohms). Why don't
active filter design programs even ask what the input and output
impedances are that the filter will have to work with, nor state what
it is after the circuit is synthesized? Does it even matter; or will
the frequency response and gain just not be affected by most normal
values? Or is it assumed that the active filter will be placed between
certain impedance values? If not, then how can I tell what the
"optimal" impedance values should be for an active filter? This has
me baffled!

Thanks,

-Bill

The output of an active filter usually comes from the output of an op-amp.
This can drive any load impedance within reason. It doesn't change the
output significantly. In this respect, it's no different from an op-amp
inverting or non-inverting amplifier.

The input needs to be driven by a source impedance that's significantly less
than the input impedance of the filter, if it's not to upset the response.
The design software may assume zero source impedance i.e. a perfect voltage
source. If you have a resistor at the filter input, you can reduce it by
the size of the source impedance, since they are effectively in series. In
this way the non-zero source impedance is "absorbed" by the input resistor.

 

[Pooh Bear]

billcalley wrote:

Hi All,

I've been looking at a lot of active filter design software and
design formulas, and none seem to care too much about the input and
output impedances that the filter will "see" when placed in a circuit.
When designing *passive* filters, we would typically specify these
impedance values as a matter of course (normally 50 ohms). Why don't
active filter design programs even ask what the input and output
impedances are that the filter will have to work with, nor state what
it is after the circuit is synthesized? Does it even matter; or will
the frequency response and gain just not be affected by most normal
values? Or is it assumed that the active filter will be placed between
certain impedance values? If not, then how can I tell what the
"optimal" impedance values should be for an active filter? This has me
baffled!

All active filters require to be driven from a low impedance. This is kinda
'assumed to be known'.

Any active filter using op-amps or even a discrete active gain block will
normally have a low output impedance and therefore doesn't care much about
the load - within the normal limits of what the op-amp etc can drive.

Graham

 

[Reg Edwards]

Active circuits design assumes that the driving impedance is zero and
that the terminating impedance is infinite.

It is up to follow-on circuit designers to include analysis of the
effects of the surrounding impedances in which a paricular circuit is
embedded.

If you are not capable of doing this then you are not qualified to do
the job anyway. Pass it on to somebody who knows what he is doing.
----
Reg

 

[Pooh Bear]

Reg Edwards wrote:

Active circuits design assumes that the driving impedance is zero and
that the terminating impedance is infinite.

It is up to follow-on circuit designers to include analysis of the
effects of the surrounding impedances in which a paricular circuit is
embedded.

If you are not capable of doing this then you are not qualified to do
the job anyway. Pass it on to somebody who knows what he is doing.

Hi Reg,

most ' kids out of college / university ' aren't taught to do much that's
useful or practical these days that you and I would probably have taken
for granted.

A sad reflection on the state of education.

Graham

 

[billcalley]

Thanks guys -- this has really cleared up my questions on active
filters.

Best Regards,

-Bill

 

[John Larkin]

On Sun, 07 Aug 2005 14:34:33 +0100, Pooh Bear
<rabbitsfriendsandrelations@hotmail.com> wrote:

most ' kids out of college / university ' aren't taught to do much that's
useful or practical these days that you and I would probably have taken
for granted.

You can't learn everything in four years. I think it's better to stick
to fundamentals (math, physics, circuit theory, signals+systems,
control theory) in school, and learn the practical stuff on the job.
Heck, most profs don't know the practical stuff anyhow.

Anybody can learn the practical stuff and be a tech, but to design
serious electronics you need some theory.


John

 

[Chris Carlen]

John Larkin wrote:

On Sun, 07 Aug 2005 14:34:33 +0100, Pooh Bear
rabbitsfriendsandrelations@hotmail.com> wrote:


most ' kids out of college / university ' aren't taught to do much that's
useful or practical these days that you and I would probably have taken
for granted.

You can't learn everything in four years. I think it's better to stick
to fundamentals (math, physics, circuit theory, signals+systems,
control theory) in school, and learn the practical stuff on the job.
Heck, most profs don't know the practical stuff anyhow.

Anybody can learn the practical stuff and be a tech, but to design
serious electronics you need some theory.


John

Right on, John. The thing that limits me is theory, since I have a
Chemistry degree instead of EE. (Why did I do that?!?!)

Theory takes a lot of time to learn, and few people can do it on their
own outside of formal instructional environments.

There are things which just can't be done without it.

I'm back to a designing a new PLL motor control project, so I know.




--
_____________________
Christopher R. Carlen
crobc@bogus-remove-me.sbcglobal.net
SuSE 9.1 Linux 2.6.5

 

[Pooh Bear]

John Larkin wrote:

On Sun, 07 Aug 2005 14:34:33 +0100, Pooh Bear
rabbitsfriendsandrelations@hotmail.com> wrote:

most ' kids out of college / university ' aren't taught to do much that's
useful or practical these days that you and I would probably have taken
for granted.

You can't learn everything in four years. I think it's better to stick
to fundamentals (math, physics, circuit theory, signals+systems,
control theory) in school, and learn the practical stuff on the job.
Heck, most profs don't know the practical stuff anyhow.

Anybody can learn the practical stuff and be a tech, but to design
serious electronics you need some theory.

True. I guess. I'm suppose I'm unusual. I started learning theory and practice
from about age13 on out of personal interest.

Much of what I see taught gives me little confidence though.

Graham

 

[Mac]

On Sun, 07 Aug 2005 11:37:42 -0700, John Larkin wrote:

On Sun, 07 Aug 2005 14:34:33 +0100, Pooh Bear
rabbitsfriendsandrelations@hotmail.com> wrote:



most ' kids out of college / university ' aren't taught to do much that's
useful or practical these days that you and I would probably have taken
for granted.

You can't learn everything in four years. I think it's better to stick
to fundamentals (math, physics, circuit theory, signals+systems,
control theory) in school, and learn the practical stuff on the job.
Heck, most profs don't know the practical stuff anyhow.

Anybody can learn the practical stuff and be a tech, but to design
serious electronics you need some theory.


John

I agree with you. Another point is that universities are supposed to teach
fundamentals, not solely prepare people for work. Preparing people for a
particular trade is what trade schools are for.

I got my BSEE fairly recently (2000), and many times I heard students
complain that the EE curiculum wasn't practical enough. But the professors
(and, I guess, the accrediting folks) place a lot of emphasis on EE
fundamentals.

--Mac

 

[Ben Bradley]

On 6 Aug 2005 23:50:14 -0700, "billcalley" <billcalley@yahoo.com>
wrote:

Hi All,

I've been looking at a lot of active filter design software and
design formulas, and none seem to care too much about the input and
output impedances that the filter will "see" when placed in a circuit.

You're doing well to have observed this enough to ask the question,
OTOH if you really knew your circuits analysis (and had tough EE profs
like I did) you'd know how to analyze the circuits and decide what
impedances would affect the filter.
But there's a way you can redeem yourself...

When designing *passive* filters, we would typically specify these
impedance values as a matter of course (normally 50 ohms). Why don't
active filter design programs even ask what the input and output
impedances are that the filter will have to work with, nor state what
it is after the circuit is synthesized? Does it even matter; or will
the frequency response and gain just not be affected by most normal
values? Or is it assumed that the active filter will be placed between
certain impedance values? If not, then how can I tell what the
"optimal" impedance values should be for an active filter? This has me
baffled!

Read Don Lancaster's "Active Filter Cookbook." (Do Not Pass Go, Do
Not Collect $200) All your questions will be answered, and no, he's
not paying me to say this.
This is NOT an AC circuits analysis or design book, though it does
have the formulas and can allow a working engineer to fake a lot of
good knowledge in a lot of situations.

Thanks,

-Bill

-----
http://www.mindspring.com/~benbradley