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Differential mode analog active filter?

  • Thread starter Vladimir Vassilevsky
  • Start date
J

John Popelish

Jan 1, 1970
0
john jardine wrote:
(snip)
The circuit is perfect using a couple of VCVS' (1Vpk inputs) but LTspice
"ideal" single pole opamps, 25ma limit, Vos=0, seem to develop an offset
output signal and 10% second harmonic. I've set something wrong but don't
know what.

I don't think there is any way to use this concept with non
inverting stages. I think each half must have a virtual
ground node. Multiple feedback filters have this, so they
work fine.
 
J

Jim Thompson

Jan 1, 1970
0
But with a loop gain of 1/4 I think.

It still be positive and will introduce squirrels... I'm an expert at
getting inadvertent squirrels ;-)

...Jim Thompson
 
J

John Popelish

Jan 1, 1970
0
Jim said:
It still be positive and will introduce squirrels... I'm an expert at
getting inadvertent squirrels ;-)

Haven't you ever been introduced to a squirrel you liked?
 
And you can synthesize BIG inductors ;-)

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

America: Land of the Free, Because of the Brave

I really don't see why you wouldn't do a leapfrog design. It is very
straightforward, plus the built in phase reversal availability would
come in handy. A number of books cover low pass leapfrog techniques.
Adding transmission zeros is a little tricky, but if you understand
signal flow graphs, it is also straightforward. Not totally obvious,
you will need to dynamic range adjust all the op amp nodes.
 
V

Vladimir Vassilevsky

Jan 1, 1970
0
John said:
I've been thinking about your differential input A/D converter and
wondered if you might be able to make use of a differential amplifier
with a differential output as a front end, something like this:


___ ___
in+ -|___|-+----|___|-+
| |
| |\| |
+----|-\ |
| | >--+ out-
| +-|+/
| | |/|
| |
| |
___ | | ___
in- -|___|----+-|___|-+
| | |
| | |\| |
| +-|-\ |
| | >--+ out+
+----|+/
|/|

This circuit removes the common mode part of the signal to the extent
that the two resistor pairs match in ratio. You can turn this into a
one pole low pass differential filter by paralleling each of the
feedback resistors with capacitors, or extend the concept into a pair of
MFB 2 pole Bessel filters.

You see, John, the whole idea of the differential filter is using just
one set of reactances between /+/ and /-/ paths. So the limited
tolerance of the capacitors will not impair the CMRR.

I came up with this variation on the single opamp subtractor, today, but
I don't remember ever seeing it before. Surely, something so simple and
useful has a name. Does anyone recognize it?


Vladimir Vassilevsky

DSP and Mixed Signal Design Consultant

http://www.abvolt.com
 
I've been thinking about your differential input A/D
converter and wondered if you might be able to make use of a
differential amplifier with a differential output as a front
end, something like this:

___ ___
in+ -|___|-+----|___|-+
| |
| |\| |
+----|-\ |
| | >--+ out-
| +-|+/
| | |/|
| |
| |
___ | | ___
in- -|___|----+-|___|-+
| | |
| | |\| |
| +-|-\ |
| | >--+ out+
+----|+/
|/|

This circuit removes the common mode part of the signal to
the extent that the two resistor pairs match in ratio. You
can turn this into a one pole low pass differential filter
by paralleling each of the feedback resistors with
capacitors, or extend the concept into a pair of MFB 2 pole
Bessel filters.

I came up with this variation on the single opamp
subtractor, today, but I don't remember ever seeing it
before. Surely, something so simple and useful has a name.
Does anyone recognize it?

Well, it took me a little pondering to figure out what was wrong with
this circuit. I had stumbled onto it while plinking around with
LTspice, an it seemed to work a miracle. It produced a perfect
conversion from single ended to differential (subtracting the inputs
from each other and producing symmetrical outputs centered on zero
volts) but it had no zero volt reference connection anywhere in it. I
couldn't figure out how it knew where ground was. So I tried to come
up with the simplified equations dscribing it, but there was only
one. Vi1-Vi2=Vo1-V02, with i and o indicating input and output nodes
of the two opamps. This matched the result, but again, offered no
clue how the circuit knwe where zero was, since there are an infinite
number of combinations that this equation satisfies.

So I went back to the simulator and added 3 mV of offset ot one
opamp. the outputs saturatedat the supply rails. Bingo. I had an
indeterminate circuit that only appeared to work, because themodel
used perfect opamps.

I can regain the operation I wanted, by adding a third voltage divider
between the inputs, and tieing the two + inputs to that, to get rid of
the implicit solution of their voltages, but that also adds a third
ratio tolerance to the solution. But that still might be good enough
to use as a speaker driver.

When things are too good to be true, they usually aren't.
 
I can regain the operation I wanted, by adding a third voltage divider
between the inputs, and tieing the two + inputs to that, to get rid of
the implicit solution of their voltages, but that also adds a third
ratio tolerance to the solution.

Oops. Make that a 3 resistor divider between the 2 inputs and ground,
with the two resistors to the two inputs twice the resistance of the
one to ground, so that the rresult is half of the common mode voltage
of the two inputs. I'm now up to 7 precision resistors. Sigh.
 
J

john jardine

Jan 1, 1970
0
Oops. Make that a 3 resistor divider between the 2 inputs and ground,
with the two resistors to the two inputs twice the resistance of the
one to ground, so that the rresult is half of the common mode voltage
of the two inputs. I'm now up to 7 precision resistors. Sigh.

The lack of reference point flummoxxed me. But all the same, I had a
worthwhile half hour.
 
J

joseph2k

Jan 1, 1970
0
Fred said:
Lessee 0.1% match preserves 60dB so ummm...0.001% match for 100dB CMRR?
Actually that penalty is paid only on transitions between single-ended and
fully differential. Just the same, for the desired performance "lab set"
capacitors and resistors would be the norm.
 
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