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Use bleeder resistors in audio opamp design?

G

Grant

Jan 1, 1970
0
....
Leapfrog design is well known. You can verify the filter with spice.

Stuff that, spent most of today learning Eagle :(
Regarding something that will work the first time, you have the wrong
philosophy. The design has to work every time. Leapfrog filters have
the lowest sensitivity to components of any filter design.

So what? I've not heard of them, therefore they have no track record
for this particular application, where the topology I'm using does have
a track record.
The biggest source of error in an active filter is the caps, and you
want to add more caps than is needed?

No, I really don't. So I didn't, now.
Lowpass leapfrog design is pretty easy. You need to pick a prototype
LCR filter which ends in a cap shunting the load resistor. When you
draw the signal flow graph, this will become obvious.

I like nice symmetrical swap the Rs and Cs to get the complimentary
function with S-K.

Grant.
 
G

Grant

Jan 1, 1970
0
On Thu, 24 Mar 2011 17:48:18 -0700 (PDT), "[email protected]"


Out-of-band is often problematic... my way has no integrator with gain

Run the math, you'll like it :)

Drops right into leap frog, AND saves one OpAmp per "lump".

I'm so glad to have dropped away from your target audience :)

Grant.
 
P

Phil Allison

Jan 1, 1970
0
"Grant"
Anyway, as I was uncomfortable with series caps I ordered some 10n today
for where I was going to series up to get lower value.


** My god you are one, fucking king size moron.

The worst kind of PITA bloody troll.

Cos cunts like you almost blend right in.




..... Phil
 
G

Grant

Jan 1, 1970
0
....
Localized Q isn't necessarily a standardized term, so I will
elaborate. Every op amp node is essentially a filter output. You can
use the traditional definition of Q (i.e. relate delta F to center
frequency), even though for the ladder the filter order will be higher
than second order. [My recollection is every node of a leapfrog design
has the same denominator due to Mason's Rule, but don't quote me on
that.] If you compared the Qs, the cascade of biquads would have
higher Qs than the localized leapfrog Qs. A high Q section is prone to
shaping the noise at the center frequency. Much worse, the PSRR of the
filter is worse with high Q.
All that said, much of the datacom filtering was done as a cascade of
biquads and all passes. It was only after sophisticated software was
developed that leapfrog (i.e. ladder) filters were used in high volume
designs.- Hide quoted text -
- Show quoted text -

Thanks Miso, that's interesting.  What sort of frequencies are you
talking about?  For the two pole (Q=.707) S-V filters I've built I see
a few percent of Q-enhancement at 100kHz.  (8 MHz GBW)

George H.
SV as in state variable? At two poles, there really probably isn't
much difference in leapfrog versus SV. You would only see the
difference in high order filters. You would have to measure all the
components just to make sure it isn't some tolerance issue that makes
the Q appear higher.

Measure to what sort of tolerance, for 2 pole Butterworth response
(Q=.707) filters?

Grant.
 
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