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Passive Filter Problem

J

Jim Thompson

Jan 1, 1970
0
I need to approximate a third-order Butterworth filter but without
benefit of active devices...

Corner frequency of 150MHz

Only passive components available are R's, C's (value less than 20pF),
L's (value less than 5nH).

How do I do that?

...Jim Thompson
 
T

Tim Wescott

Jan 1, 1970
0
Jim said:
I need to approximate a third-order Butterworth filter but without
benefit of active devices...

Corner frequency of 150MHz

Only passive components available are R's, C's (value less than 20pF),
L's (value less than 5nH).

How do I do that?

...Jim Thompson

Check the ARRL handbook -- they have filter tables. You could also do a
Google search on filter design software, it's out there. Frankly, with
that few poles I'd be tempted to just put a pi filter into SPICE and
dink with the values until it looked good.

If your L's are lossy you have to account for it in your design, or the
filter will be more rounded than you want.
 
J

John Larkin

Jan 1, 1970
0
I need to approximate a third-order Butterworth filter but without
benefit of active devices...

Corner frequency of 150MHz

Only passive components available are R's, C's (value less than 20pF),
L's (value less than 5nH).

How do I do that?

...Jim Thompson


Approximate?

For equal terminations (source and load), the values for a lossless
Butterworth filter are...


in------+---2H----+--------out
| |
| |
1F 1F
| |
| |
G G

which is for w=1, Rs=1, Rl=1

(I also have tables for asymmetric termination cases. I could snap a
pic of the page. Also have cases using lossy parts.)

So, you normalize this to w' (2*pi*150e6) and whatever impedance z'
you like.

C' = C / (w' * z')

L' = L * z' / w'

Donno if this works with your constraints. Looks dicey on L.

John
 
T

Tam/WB2TT

Jan 1, 1970
0
Jim Thompson said:
I need to approximate a third-order Butterworth filter but without
benefit of active devices...

Corner frequency of 150MHz

Only passive components available are R's, C's (value less than 20pF),
L's (value less than 5nH).

Don't hold your breath about the 5 nH. Didn't design the filter, but 50 nH
sounds more plausable for a 50 Ohm filter.

Tam
 
J

Jim Thompson

Jan 1, 1970
0
Approximate?

For equal terminations (source and load), the values for a lossless
Butterworth filter are...


in------+---2H----+--------out
| |
| |
1F 1F
| |
| |
G G

which is for w=1, Rs=1, Rl=1

(I also have tables for asymmetric termination cases. I could snap a
pic of the page. Also have cases using lossy parts.)

So, you normalize this to w' (2*pi*150e6) and whatever impedance z'
you like.

C' = C / (w' * z')

L' = L * z' / w'

Donno if this works with your constraints. Looks dicey on L.

John

Since the L's are pretty crappy (Q~=1 at 150MHz) I'd rather stick to
RC's only.

I know next to nothing about NON-active filters. What do lattice
filters do for you? (I am differential-in/differential-out.)

Thanks!

...Jim Thompson
 
J

Jim Thompson

Jan 1, 1970
0
Don't hold your breath about the 5 nH. Didn't design the filter, but 50 nH
sounds more plausable for a 50 Ohm filter.

Tam
[snip]

Source impedance can be as high as 3K or as low as (not very stable)
26 ohms.

Load impedance with be at least 5K.

(This is on-chip.)

...Jim Thompson
 
K

Ken Smith

Jan 1, 1970
0
I need to approximate a third-order Butterworth filter but without
benefit of active devices...

Corner frequency of 150MHz

Only passive components available are R's, C's (value less than 20pF),
L's (value less than 5nH).

How do I do that?

What can you drive this with.

Xl = 2 * pi * F * L = 2 * pi * 150E6 * 5E-9 = 4.7

You are either going to take a big amplitude hit or you need a stout
driver.

Xc = 1/ (2 * pi * F * C) = 1/(2 * pi * 150E6 * 20E-12) = 53

You are going to have to get L bigger or C bigger or say "That ugly RC
filter is close to a Butterworth with a straight face".
 
J

John Larkin

Jan 1, 1970
0
Since the L's are pretty crappy (Q~=1 at 150MHz) I'd rather stick to
RC's only.

I know next to nothing about NON-active filters. What do lattice
filters do for you? (I am differential-in/differential-out.)

Thanks!

...Jim Thompson



Q=1 is almost a resistor! I'm glad I have the Digikey catalog and not
some billion-dollar wafer fab.

You can't make a Butterworth out of just Rs and Cs, but you can make a
3-pole slushy R-C-R-C-R-C thing, which would roll off pretty ugly but
might suit your needs. Why Butterworth?

Diff, it would be


----r--+---r--+--r--+---
| | |
c c c
| | |
-----r--+---r--+--r--+--- ,


pretty nasty when you get into the gory details. If this is for data,
forget about it... you're talking swollen-shut-eye-diagram here.


John

Hey, what did you think of the Chandon?
 
T

Tom Bruhns

Jan 1, 1970
0
Jim Thompson said:
I need to approximate a third-order Butterworth filter but without
benefit of active devices...

Corner frequency of 150MHz

Only passive components available are R's, C's (value less than 20pF),
L's (value less than 5nH).

How do I do that?


Lowpass??

The freeware RFSim99 has a filter calculator as one of its (many)
tools that will do it for you, so long as you want equal input and
output resistances (e.g., want to operate from/into equal source and
load resistances). If you want unequal, you can apply Bartlett's
Bisection Theorem (http://members.tripod.com/michaelgellis/bartlett.html)
to the equal-resistance design. You'll probably be obliged to use a
bunch of those L's in series, or a bunch of the C's in parallel, or
both, since 1/(2*pi*20pF*5nH) is 500MHz.

Or, do it with mechanical resonators...

Cheers,
Tom
 
L

legg

Jan 1, 1970
0
Since the L's are pretty crappy (Q~=1 at 150MHz) I'd rather stick to
RC's only.

Is it practical to grab lead-frame pins in the L portion? This gives
some 'free' nH.

RL
 
K

Kevin Aylward

Jan 1, 1970
0
Jim said:
Since the L's are pretty crappy (Q~=1 at 150MHz) I'd rather stick to
RC's only.

I know next to nothing about NON-active filters.

You ceratinly dont, as your suggestion of trying to achive a passive
butterworth without inducters implies. Its a complete non starter. Only
goes to show one cant know it all, well, some of us cant anyway:)

Is a source follower buffer out of the question? This is a lot less
bother than a full amplifier, and will allow for the enhanced Q required
for a butterworth.


Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that John Larkin <jjlarkin@highSNIPland
THIStechPLEASEnology.com> wrote (in <fucj8011eobokeif6dvjlthpo9gsp98rqe@
4ax.com>) about 'Passive Filter Problem', on Fri, 23 Apr 2004:
You can't make a Butterworth out of just Rs and Cs, but you can make a
3-pole slushy R-C-R-C-R-C thing, which would roll off pretty ugly but
might suit your needs. Why Butterworth?

Diff, it would be


----r--+---r--+--r--+---
| | |
c c c
| | |
-----r--+---r--+--r--+--- ,


pretty nasty when you get into the gory details. If this is for data,
forget about it... you're talking swollen-shut-eye-diagram here.

Yes, RC ladder filters are very poor. You may be able to get acceptable
results by tapering from a low source impedance to a high load
impedance:
----r--+---3r--+--9r--+---
| | |
c c/3 c/9
| | |
-----r--+---3r--+--9r--+---

The factor 3 is just an example. You may be able to use a larger factor.
 
M

maxfoo

Jan 1, 1970
0
I need to approximate a third-order Butterworth filter but without
benefit of active devices...

Corner frequency of 150MHz

Only passive components available are R's, C's (value less than 20pF),
L's (value less than 5nH).

How do I do that?

...Jim Thompson

5nH too small for that freq to match to 50ohms, if that's the Z you want...

CAP 1 0 C=22pF
IND 1 2 L=61nH
CAP 2 0 C=22pF

___
-UUU-
| |
--- ---
--- ---
| |
- -







Remove "HeadFromButt", before replying by email.
 
M

maxfoo

Jan 1, 1970
0
Source impedance can be as high as 3K or as low as (not very stable)
26 ohms.

Load impedance with be at least 5K.

(This is on-chip.)

...Jim Thompson
--
Oh!!!

Just use a simple series R shunt C...

Frequency = 1/(2pi*R*C)






Remove "HeadFromButt", before replying by email.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that maxfoo <maxfooHeadFromButt@punkass
..com> wrote (in said:
5nH too small for that freq to match to 50ohms, if that's the Z you want...

CAP 1 0 C=22pF
IND 1 2 L=61nH
CAP 2 0 C=22pF

___
-UUU-
| |
--- ---
--- ---
| |
- -
Isn't it obvious that the impedance has to be chosen to match the
available parts? To meet the 5 nH limit (if the inductor had a
reasonable Q), the impedance would have to be 600 ohms or more. Then the
capacitors are just over 2 pF, or less.
 
C

Chris Carlen

Jan 1, 1970
0
Jim said:
Since the L's are pretty crappy (Q~=1 at 150MHz) I'd rather stick to
RC's only.

I know next to nothing about NON-active filters. What do lattice
filters do for you? (I am differential-in/differential-out.)

Thanks!

...Jim Thompson


Jim, you can't do a Butterworth with RCs since a Butterworth involves
pairs of complex conjugate poles.

I am still trying to figure out why you are asking this question, since
I can't seriously fathom that you don't already know this subject better
than most anybody around here.

Perplexed.
 
J

Jim Thompson

Jan 1, 1970
0
Jim, you can't do a Butterworth with RCs since a Butterworth involves
pairs of complex conjugate poles.

I am still trying to figure out why you are asking this question, since
I can't seriously fathom that you don't already know this subject better
than most anybody around here.

Perplexed.

Wishful thinking. Hoping for a R-C ladder or lattice that comes close
:)

...Jim Thompson
 
M

maxfoo

Jan 1, 1970
0
Isn't it obvious that the impedance has to be chosen to match the
available parts? To meet the 5 nH limit (if the inductor had a
reasonable Q), the impedance would have to be 600 ohms or more. Then the
capacitors are just over 2 pF, or less.
--

sure, but you don't have the luxury of excess real estate on a die.
the o.p. didn't tell us that until his 2nd post...









Remove "HeadFromButt", before replying by email.
 
K

Kevin Aylward

Jan 1, 1970
0
Chris said:
Jim, you can't do a Butterworth with RCs since a Butterworth involves
pairs of complex conjugate poles.

I am still trying to figure out why you are asking this question,
since I can't seriously fathom that you don't already know this
subject better than most anybody around here.

Jim is a *circuit* design expert. There is no requirement for someone
who is expert in designing 1000+ transistor circuits to know more
specialised mathematics, However, I think this one is a strange gap in
his knowledge. Maybe its because in the early 60's when he graduated,
filter synthesis was not covered much. Certainly, most specialised
filter experts would be hard pressed to bias up even a one transistor
amplifier. An "Electronic Engineer" is not very indicative as to just
what someone might know.


Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
 
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