Subtractor circuit impervious to components tolerance?

[Zico]

Hi,

I'm wondering if there's some circuit configuration (presumably
an op-amp based design) that would subtract two signals (say,
two audio signals) in a way that is *100% and absolutely
unaffected* by components tolerance.

I do not mean using 0.1% tolerance resistors, or matched resistor
networks, or adding a potentiometer for fine adjustment. Those
solutions *minimize* the effect of components tolerance.

For example (more like an analogy): if I need a non-inverting
buffer/amplifier with absolutely precise gain, say 2, then I could
try the standard op-amp circuit, and use two identical resistors,
for a gain of 2. Problem is, gain *can not* be exactly two (well,
it can not be *expected* to be exactly two); I *can* obtain an
absolutely precise gain of *1* ... Connect output terminal
*directly* to the inverting input, and voilŕ --- this will give, from
any conceivable point of view (at least for every practical
purposes), an *absolutely exact* gain of 1; where I'm trying
to get is: the solution in this example goes beyond the highest
available precision components; it goes beyond the most
expensive and most precise matched pairs of resistors, etc.

So, my question: what about for a circuit that subtracts two
signals? Or, equivalently (and even better for audio signals),
a circuit that adds two signals + an inverting circuit with gain 1 ?
(the standard solutions I know for these two rely on components
precision/tolerance)

Thanks,
-Zico

 

[Bob Myers]

On 5/25/2010 5:14 PM, Zico wrote:

Hi,

I'm wondering if there's some circuit configuration (presumably
an op-amp based design) that would subtract two signals (say,
two audio signals) in a way that is *100% and absolutely
unaffected* by components tolerance.

First question: are you utterly opposed to using trimmers? Because
I can't think of any way to get to "100% and absolutely unaffected by
components" (if you really, really mean things like "100%" and
"absolutely") without some sort of adjustment capability.

I do not mean using 0.1% tolerance resistors, or matched resistor
networks, or adding a potentiometer for fine adjustment. Those
solutions *minimize* the effect of components tolerance.

Well, depending on how sophisticated you want to get, you CAN
pretty much null out any effect of tolerance to within the limits of
whatever measuring equipment you're trying to satisfy. But there's
nothing truly "perfect" in this here universe.

I've really tempted to point out that the simplest solution is to
just abandon all hope of doing this "perfectly" in the analog
domain, and simply do it digitally - which can be arbitrarily
"perfect," to the limit of your budget.

Bob M.

For example (more like an analogy): if I need a non-inverting
buffer/amplifier with absolutely precise gain, say 2, then I could
try the standard op-amp circuit, and use two identical resistors,
for a gain of 2. Problem is, gain *can not* be exactly two (well,
it can not be *expected* to be exactly two); I *can* obtain an
absolutely precise gain of *1* ... Connect output terminal
*directly* to the inverting input, and voilŕ --- this will give, from
any conceivable point of view (at least for every practical
purposes), an *absolutely exact* gain of 1;

Well, no, not really. Again, nothing's "perfect" or "exact." A
few moments thought should suggest to you several sources of
error (even though some are minute beyond belief) that prevent
it. But see, you've used "for every practical purpose" and
"absolutely exact" in the same sentence. Which do you really
mean?

Bob M.

 

[Phil Allison]

"Zico"

I'm wondering if there's some circuit configuration (presumably
an op-amp based design) that would subtract two signals (say,
two audio signals) in a way that is *100% and absolutely
unaffected* by components tolerance.


** A transformer can be used to subtract one signal from another IF the
ends of the primary are connected to each signal source - due to the fact
that windings respond only to the difference between the ends.



...... Phil

 

[Zico]

For example  (more like an analogy):  if I need a non-inverting
buffer/amplifier with absolutely precise gain, say 2, then I could
try the standard op-amp circuit, and use two identical resistors,
for a gain of 2.  Problem is, gain *can not* be exactly two (well,
it can not be *expected* to be exactly two);  I *can* obtain an
absolutely precise gain of *1* ...  Connect output terminal
*directly* to the inverting input, and voilŕ --- this will give, from
any conceivable point of view (at least for every practical
purposes), an *absolutely exact* gain of 1;

Well, no, not really.  Again, nothing's "perfect" or "exact."  A
few moments thought should suggest to you several sources of
error (even though some are minute beyond belief) that prevent
it.  But see, you've used "for every practical purpose" and
"absolutely exact" in the same sentence.  Which do you really
mean?

I was hoping that the example/analogy would make it "perfectly
clear" (again with the "perfect" attribute!! ) what I meant by
"absolutely exact".

The conditioned statement (the "for every practical purpose")
was to acknowledge that, well, if we get philosophical, then we
would have to go and count electrons, and then yes, the signal
would not be *absolutely exact* ... Even without going to the
extreme of "getting philosophical" --- there's noise; there's the
fact that the wire does not have *truly zero* resistance, and
the amplifier's input is not a *truly and absolutely open circuit*
(it does have a finite, measurable input impedance) ... So the
gain is not *truly* one...

But again, no sane practical point of view would really constitute
an objection to the fact that the gain in that example is "truly" or
"exactly" one... (without having to rely on extra-precise
components,
and without the extra trouble of having to use trimmers).

Thing is, in that example, the "cleverness" of the design (or
the luck of getting away with a gain of 1, if you will) bypasses
the complete and direct dependence on the component's
tolerance --- it *avoids* having to use a trimmer when there's
another solution that gives you an even better result without
the trouble of placing a trimmer and adjusting it (and hoping
that it will remain adjusted --- not to mention the difference
in temperature-related variations in the different components)

So, basically, I was trying to find out whether I might be lucky
and there might be some "clever" configuration in which a
subtractor circuit could be obtained that is not affected by
any component's tolerance.

Incidentally, I had indeed considered going digital !! But the
only part that could arguably justify it is the subtraction of the
two signals --- the rest of the system is so trivial and so easy
to achieve with analog circuitry that a digital solution seems
somewhat overkill.

Thanks,
-Zico

 

[Tim Wescott]

On 05/25/2010 04:14 PM, Zico wrote:

Hi,

I'm wondering if there's some circuit configuration (presumably
an op-amp based design) that would subtract two signals (say,
two audio signals) in a way that is *100% and absolutely
unaffected* by components tolerance.

No, and no.

No, there isn't a subtracter circuit that doesn't depend on _some_
component tolerance somewhere in the circuit, and no, there is nothing
in our fallen and imperfect world that is 100% and absolutely anything.

I do not mean using 0.1% tolerance resistors, or matched resistor
networks, or adding a potentiometer for fine adjustment. Those
solutions *minimize* the effect of components tolerance.

For example (more like an analogy): if I need a non-inverting
buffer/amplifier with absolutely precise gain, say 2, then I could
try the standard op-amp circuit, and use two identical resistors,
for a gain of 2. Problem is, gain *can not* be exactly two (well,
it can not be *expected* to be exactly two); I *can* obtain an
absolutely precise gain of *1* ... Connect output terminal
*directly* to the inverting input, and voilŕ --- this will give, from
any conceivable point of view (at least for every practical
purposes), an *absolutely exact* gain of 1;

Take an op-amp with a 10MHz gain-bandwidth product. Make it into a
voltage follower, as you suggest. Gain is pretty close to -A / (s + A),
where A = GBW product. Feed a 16-bit ADC with that, and expect the
circuit to live up to the op-amp's advertised capability over audio --
because after all, the op-amp's good to 10MHz, right?. You'll be seeing
one LSB of error a bit above 150Hz.

Take that same op-amp, and assume it has an open-loop gain of A = 10^6,
which is in the ballpark for a lot of devices. Now feed a 24-bit ADC
(you can buy them off the shelf if you don't mind that they're not up to
audio speeds). Gain is pretty close to -A / (A + 1), for a 1ppm error
in your output _at DC_, or a whopping 17 LSB of error.

And I'm not even touching on the op-amp's bias.

where I'm trying
to get is: the solution in this example goes beyond the highest
available precision components; it goes beyond the most
expensive and most precise matched pairs of resistors, etc.

Yet what you're demonstrating is a charming naiveté.

So, my question: what about for a circuit that subtracts two
signals? Or, equivalently (and even better for audio signals),
a circuit that adds two signals + an inverting circuit with gain 1 ?
(the standard solutions I know for these two rely on components
precision/tolerance)

Just accept that until the last trump sounds, you are stuck with living
in a world where all circuits depend on their components, and even
op-amps don't live up to their Platonic ideals.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

 

[Tim Wescott]

On 05/25/2010 06:21 PM, Phil Allison wrote:

"Zico"

I'm wondering if there's some circuit configuration (presumably
an op-amp based design) that would subtract two signals (say,
two audio signals) in a way that is *100% and absolutely
unaffected* by components tolerance.


** A transformer can be used to subtract one signal from another IF the
ends of the primary are connected to each signal source - due to the fact
that windings respond only to the difference between the ends.

I have one word: "plastics".

Oops -- wrong world.

I have two words: "capacitive coupling".

You may get good enough, you may get pretty darn good with a specially
constructed transformer with careful shielding between primary and
secondary, but you won't get "100% perfect".

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

 

[Phil Allison]

"Tim Wescott"
Phil Allison wrote:

"Zico"

I'm wondering if there's some circuit configuration (presumably
an op-amp based design) that would subtract two signals (say,
two audio signals) in a way that is *100% and absolutely
unaffected* by components tolerance.


** A transformer can be used to subtract one signal from another IF the
ends of the primary are connected to each signal source - due to the
fact
that windings respond only to the difference between the ends.

I have one word: "plastics".

Oops -- wrong world.

** Wrong universe in your case - dick brain.

I have two words: "capacitive coupling".

You may get good enough, you may get pretty darn good with a specially
constructed transformer with careful shielding between primary and
secondary, but you won't get "100% perfect".

** The OP did not ask for any such thing.

He only asked that the subtraction arrangement was good for audio band
signals and not affected by tolerances in passive components.

I figured he never considered an all passive solution like a transformer.

A good one has a CMRR of over 100dB across the band.



...... Phil

 

[Jasen Betts]

On 2010-05-25, Zico <zico@mailinator.com> wrote:

Hi,

I'm wondering if there's some circuit configuration (presumably
an op-amp based design) that would subtract two signals (say,
two audio signals) in a way that is *100% and absolutely
unaffected* by components tolerance.

If you need precise ratios use a transformer


--- news://freenews.netfront.net/ - complaints: news@netfront.net ---

 

[Tim Wescott]

On 05/26/2010 07:12 AM, Zico wrote:

So, my question: what about for a circuit that subtracts two
signals? Or, equivalently (and even better for audio signals),
a circuit that adds two signals + an inverting circuit with gain 1 ?
(the standard solutions I know for these two rely on components
precision/tolerance)

Just accept that until the last trump sounds, you are stuck with living
in a world where all circuits depend on their components, and even
op-amps don't live up to their Platonic ideals.

Though at this point I'm convinced that there is no circuit
configuration
like what I was looking for, I will clarify something (you know, at
the
risk of continuing to squeeze to death a pointless discussion )

Nicely put, this idea that "even op-amps don't live up to their
platonic
ideas" (and no, I'm not being sarcastic!) ... I should have
expressed
better what I intended with the analogy: two parts: (1) the
tolerance
and imperfections of an op-amp are below my threshold of what
would be considered "perfect" and "absolutely precise" (maybe there
is a linguistic self-contradiction here, I don't know) ...

Not so much a linguistic one as a problem with life. Experienced
engineers don't like terms like "perfect" and "absolutely precise",
because "perfect" for one person may be "absolute crap" for someone
else, or it may be "way too expensive and excessively over-engineered".
After you've been around the block a few times you learn that putting
in lots of effort to meet vague descriptions just leads to unhappiness.
And since no one wants to take responsibility for communications
breakdowns, and the grunt engineer is, by definition, not the boss, he
automatically loses the subsequent finger-pointing contest.

That is:
the
errors introduced by the op-amp and the resistance of the wires and
traces do not count.

"Do not count" in this case is just as bad as "perfect", though. Does
not count to who? Does not count to you? What doesn't count to you?
What are your thresholds? In numbers, please! Or if not in numbers,
then in something still vague but a specific application like "consumer
audio quality", "pro audio quality", "good enough for my home
thermostat", etc.

(2) Perhaps more importantly: the op-amp will be part of the circuit
anyways, so, by placing resistors, we're *adding* to the
imprecisions;
not only that, but compensating for those requires trimmers, which at
the very least is an extra/house-keeping part of the design and an
extra source of trouble (both at manufacture time and at operations
time)

So, this was sort of the main intent of the example/analogy ...

Now, perhaps an interesting question that comes to mind now: are
the errors introduced by the imperfections of a reasonably good
audio-signals op-amp (say, an LF353?) worse than a 0.1% tolerance
in the resistors? (doesn't sound like --- 0.1% tolerance in a
configuration with linear dependence on the resistor value gives
an error 60dB below the signal ... An op-amp does better than
that ... right?

I suspect you'll find that some folks have objections to that one. IIRC
it has visible crossover distortion at audio frequencies on an
oscilloscope, which means that it'll definitely be audible. And if it's
the one I remember the distortion comes from the slew rate limit so it's
getting worse just as the amp's ability to compensate for it gets worse,
too.

And my point with the GBW stuff is that at only moderately high
frequencies op-amps _don't_ do better than that. My large point is that
the whole idealized, infinite gain, negative input equals positive input
stuff is a really good crutch to help you start getting your head
wrapped around how op-amps behave, but for many many applications you
just can't ignore their real characteristics and succeed.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

 

[Bob Myers]

On 5/26/2010 8:12 AM, Zico wrote:

Nicely put, this idea that "even op-amps don't live up to their
platonic
ideas" (and no, I'm not being sarcastic!) ... I should have
expressed
better what I intended with the analogy: two parts: (1) the
tolerance
and imperfections of an op-amp are below my threshold of what
would be considered "perfect" and "absolutely precise" (maybe there
is a linguistic self-contradiction here, I don't know) ... That is:
the
errors introduced by the op-amp and the resistance of the wires and
traces do not count.

Well, there's a REALLY big part of your problem so far, in terms
of getting an answer here - all you've ever said is "perfect," or words
to that effect, and still have let anyone in on just what would be
considered your "threshold." In short, tell us what you're really trying
to achieve, and you'll be a whole lot more likely to get some helpful
responses.

Bob M.

 

[John Larkin]

On Tue, 25 May 2010 16:14:05 -0700 (PDT), Zico <zico@mailinator.com>
wrote:

Hi,

I'm wondering if there's some circuit configuration (presumably
an op-amp based design) that would subtract two signals (say,
two audio signals) in a way that is *100% and absolutely
unaffected* by components tolerance.

I do not mean using 0.1% tolerance resistors, or matched resistor
networks, or adding a potentiometer for fine adjustment. Those
solutions *minimize* the effect of components tolerance.

For example (more like an analogy): if I need a non-inverting
buffer/amplifier with absolutely precise gain, say 2, then I could
try the standard op-amp circuit, and use two identical resistors,
for a gain of 2. Problem is, gain *can not* be exactly two (well,
it can not be *expected* to be exactly two); I *can* obtain an
absolutely precise gain of *1* ... Connect output terminal
*directly* to the inverting input, and voilŕ --- this will give, from
any conceivable point of view (at least for every practical
purposes), an *absolutely exact* gain of 1; where I'm trying
to get is: the solution in this example goes beyond the highest
available precision components; it goes beyond the most
expensive and most precise matched pairs of resistors, etc.

So, my question: what about for a circuit that subtracts two
signals? Or, equivalently (and even better for audio signals),
a circuit that adds two signals + an inverting circuit with gain 1 ?
(the standard solutions I know for these two rely on components
precision/tolerance)

Thanks,
-Zico

A flying capacitor circuit comes very close.

John

 

[Zico]

So, my question:  what about for a circuit that subtracts two
signals?  Or, equivalently (and even better for audio signals),
a circuit that adds two signals + an inverting circuit with gain 1 ?
(the standard solutions I know for these two rely on components
precision/tolerance)

Just accept that until the last trump sounds, you are stuck with living
in a world where all circuits depend on their components, and even
op-amps don't live up to their Platonic ideals.

Though at this point I'm convinced that there is no circuit
configuration
like what I was looking for, I will clarify something (you know, at
the
risk of continuing to squeeze to death a pointless discussion )

Nicely put, this idea that "even op-amps don't live up to their
platonic
ideas" (and no, I'm not being sarcastic!) ... I should have
expressed
better what I intended with the analogy: two parts: (1) the
tolerance
and imperfections of an op-amp are below my threshold of what
would be considered "perfect" and "absolutely precise" (maybe there
is a linguistic self-contradiction here, I don't know) ... That is:
the
errors introduced by the op-amp and the resistance of the wires and
traces do not count.

(2) Perhaps more importantly: the op-amp will be part of the circuit
anyways, so, by placing resistors, we're *adding* to the
imprecisions;
not only that, but compensating for those requires trimmers, which at
the very least is an extra/house-keeping part of the design and an
extra source of trouble (both at manufacture time and at operations
time)

So, this was sort of the main intent of the example/analogy ...

Now, perhaps an interesting question that comes to mind now: are
the errors introduced by the imperfections of a reasonably good
audio-signals op-amp (say, an LF353?) worse than a 0.1% tolerance
in the resistors? (doesn't sound like --- 0.1% tolerance in a
configuration with linear dependence on the resistor value gives
an error 60dB below the signal ... An op-amp does better than
that ... right?

Thanks,
-Zico

 

[John Fields]

On Wed, 26 May 2010 08:21:39 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 25 May 2010 16:14:05 -0700 (PDT), Zico <zico@mailinator.com
wrote:

So, my question: what about for a circuit that subtracts two
signals? Or, equivalently (and even better for audio signals),
a circuit that adds two signals + an inverting circuit with gain 1 ?
(the standard solutions I know for these two rely on components
precision/tolerance)

Thanks,
-Zico

A flying capacitor circuit comes very close.

---
To what?

 

[Tim Wescott]

On 05/26/2010 08:21 AM, John Larkin wrote:

On Tue, 25 May 2010 16:14:05 -0700 (PDT), Zico<zico@mailinator.com
wrote:

Hi,

I'm wondering if there's some circuit configuration (presumably
an op-amp based design) that would subtract two signals (say,
two audio signals) in a way that is *100% and absolutely
unaffected* by components tolerance.

I do not mean using 0.1% tolerance resistors, or matched resistor
networks, or adding a potentiometer for fine adjustment. Those
solutions *minimize* the effect of components tolerance.

For example (more like an analogy): if I need a non-inverting
buffer/amplifier with absolutely precise gain, say 2, then I could
try the standard op-amp circuit, and use two identical resistors,
for a gain of 2. Problem is, gain *can not* be exactly two (well,
it can not be *expected* to be exactly two); I *can* obtain an
absolutely precise gain of *1* ... Connect output terminal
*directly* to the inverting input, and voilŕ --- this will give, from
any conceivable point of view (at least for every practical
purposes), an *absolutely exact* gain of 1; where I'm trying
to get is: the solution in this example goes beyond the highest
available precision components; it goes beyond the most
expensive and most precise matched pairs of resistors, etc.

So, my question: what about for a circuit that subtracts two
signals? Or, equivalently (and even better for audio signals),
a circuit that adds two signals + an inverting circuit with gain 1 ?
(the standard solutions I know for these two rely on components
precision/tolerance)

Thanks,
-Zico

A flying capacitor circuit comes very close.

I've put on a flying capacitor circus or two (they usually involve
electrolytics, and errors in magnitude or sign). Does that count?

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

 

[whit3rd]

On May 25, 5:17 pm, Zico <z...@mailinator.com> wrote:

[given signals A and B, form a third signal A-B]

So, basically, I was trying to find out whether I might be lucky
and there might be some "clever" configuration in which a
subtractor circuit could be obtained that is not affected by
any component's tolerance.

Still, that's not clear. The transformer idea, for instance,
is good but limited to the 'ideal' high input impedance of
the unloaded transformer (and at DC, that always fails).

It seems to me, your question is really about the difference
equation having no reference to other units than the inputs,
i.e. no 'ohms' value is in the ideal difference equation.
The differencing amplifier built from op amps, however,
DOES have resistances (ratios of 'em, anyhow) all over
the operating equation.

The answer is, therefore, you can connect a floating voltmeter
from signal A to signal B, and it measures A-B without any
component-value elements in the equation (other than the
voltmeter innards). A transformer winding is a very good
approximation to that floating measurement circuit. So is
an op amp difference circuit, but it DOES depend on resistor
matching.

The use of floating (isolated, or battery-powered) meters and
sources is a very powerful technique for high precision work.

 

[stan]

Zico wrote:

I'm wondering if there's some circuit configuration (presumably
an op-amp based design) that would subtract two signals (say,
two audio signals) in a way that is *100% and absolutely
unaffected* by components tolerance.

If this "configuration" has components they will certainly have
tolerances and the resulting circuit will likewise have tolerances.

I do not mean using 0.1% tolerance resistors, or matched resistor
networks, or adding a potentiometer for fine adjustment. Those
solutions *minimize* the effect of components tolerance.

Sounds like you are looking for a pair of matching magic components
with a fixed gain over some unstated audio (simple telephony vs CD
quality) frequency range. But you aren't interested in tight tolerance
components. Sounds like a free lunch hunt. Actually that's not quite
right; sounds like a free gourmet lunch hunt. Good luck and if you
find it, please drop some crumbs.

 

[Zico]

On May 26, 6:15 pm, stan <smo...@exis.net> wrote:

Zico wrote:
I'm wondering if there's some circuit configuration (presumably
an op-amp based design) that would subtract two signals (say,
two audio signals) in a way that is *100% and absolutely
unaffected* by components tolerance.

If this "configuration" has components they will certainly have
tolerances and the resulting circuit will likewise have tolerances.

Well, again, maybe the components could have been "wires" or
traces, which can be reasonably taken as 0 ohms with zero
tolerance (from the point of view of the given application). Like
in the example --- I mean, there *is* the configuration of the
voltage follower that achieves a gain of one with higher accuracy
and lower housekeeping and trouble (compared to using trimmers
for fine-tuning).

Another (completely hypothetical) example: as you point out,
if there are components, there is tolerance; but imagine that
someone had come up with a clever configuration (again,
*hypothetically*) in which the output voltage is given by, say,

Vout = V1 (A + R1/R2) / (A + R2/R1) - V2

Where A is the open-loop gain of the op-amp.

(please don't ask me how I would achieve that --- *hypothetical
example* ... The thing is, just because I can not figure out a
way of doing it does not mean someone else can not)

You see, R1 and R2 could have 20% error, and still, the output
would be V1*1.0000004 - V2 (with A = 10^6)... So, yes, you're
going to tell me that "there, see? it's not *exact*" ... But using
components with 20% tolerance that have an net effect of a
0.00004% tolerance on the output.... well, sue me for calling
that "unaffected by components tolerance" ...

Ok, yes, yes, I did not specify "good enough for what" --- because
yes, if I feed that to a A/D converter with 24 *actual* bits of
accuracy, then yes, that A/D converter will be able to tell the
difference between that and the *exact* V1-V2 ... But then
again, that's what I get for using 20% components!! If in this
same hypothetical example I had used 1% resistors, I would
actually be at 0.000002% of the *exact* V1-V2; three times
more precise than the 24-bit A/D converter. (as opposed to
a circuit configuration that gives me a gain that directly depends
on, say, R2/R1, in which case, using resistors with 1% tolerance,
a 10-bit A/D converter could tell the difference!

components. Sounds like a free lunch hunt. Actually that's not quite
right; sounds like a free gourmet lunch hunt.

Well, if I was not aware of the existence of that voltage follower
configuration, then an amplifier with gain "exactly" 1 would have
sounded like a free lunch to me ... So, the thing is, I was posting
here because there may be some clever circuit configuration
that someone invented a while ago that I was not aware of.
You know, not knowing about something does not constitute
proof that that something does not exist!! Right?

Ok, I guess now the subject has been officially squeezed to
waaaaay beyond its death!!! So I will respectfully withdraw from
the discussion --- with thanks to all that replied!

Regards,
-Zico

 

[Bob Myers]

On 5/26/2010 7:23 PM, Zico wrote:

Well, if I was not aware of the existence of that voltage follower
configuration, then an amplifier with gain "exactly" 1 would have
sounded like a free lunch to me ...

No, that's generally referred to as a "wire"...;-)

Bob M.

 

[Zico]

Well, if I was not aware of the existence of that voltage follower
configuration, then an amplifier with gain "exactly" 1 would have
sounded like a free lunch to me ...

No, that's generally referred to as a "wire"...;-)

In my native language (by now you surely noticed that English
is not my native language), this would have made an excellent
joke/pun --- we have an expression/proverb that involves the
phrase "eating wire" ... So, you know, free *lunch* ... Eating
wire ... I would have almost thought that you intended the pun!

-Zico