Multiple current mirrors
C Egernet
I'm working on the concept for an instrument that needs to take a
number of current sources (photodiodes, 44 off) and find the ratio
between each of the 43 currents with the 44th (the largest). The
currents are in the range of, say, 100 nA to 10 uA.
Electronics design is a bit out of my skill set but I'd like to give
it a try.
I could try to A/D-convert as soon as possible and divide digitally
but I suspect that it is more sound to do the following:
Mirror the 44th current with BJTs and use Hobbs' laser noise canceller
to form the differences of the logarithms of the currents and A/D-
convert these.
Now, my question is: is it at all feasible to replicate a current so
many times with any kind of bandwidth and precision?
I will have to do this with discrete components so matching is going
to be a problem. I anticipate using small signal RF transistors to get
a reasonable beta, and having a calibration procedure to get rid of
remaining errors.
That said, I don't want calibrations to be a crutch for doing things
wrong in the first place.
The measurement bandwidth is going to be quite low (not fixed yet but
probably less than 200 Hz) but it feels right to do the ratioing out
to a few MHz.
Comments and suggestions are most welcome
Chris Egernet
Phil Hobbs
43 noise cancellers in one box? Cool.
Noise cancelling nanoamps isn't impossible, but it's hard to do well.
Manyfold current mirroring at megahertz bandwidths with only 100 nA is
going to be very tough. I'd suggest using a TIA plus and 43 biggish
resistors--big enough to drop a few volts at your comparison
photocurrent value--each one driving the emitter of a small RF BJT such
as a BFT25A, whose collectors goe to the noise canceller diff pairs.
Putting a similar BJT in series with the feedback resistor of the TIA
will get you a nice first-order temperature compensation and reduce the
nonlinearity. Keep the transistor dissipation down to something
reasonable--well below 1 mW--so that the temperature tracking doesn't
get screwed up. Also, make sure you use a diode-connected transistor
(CB shorted) for the compensation, because the BE diode doesn't have the
same characteristics as the transistor in normal bias.
As far as the noise cancellers go, the bandwidth of the integrating loop
needs to be wide enough to cover all the important modulation of the
photocurrent. This is because of the pronounced nonlinearity of the
logarithmic Ic vs Vbe characteristic of the BJTs. If you don't need to
be too close to the shot noise, you can use an LM13700 instead of the
diff pair--by using the linearizing diodes (see the data sheet), you can
get rid of the loop nonlinearity and have a straight ratio output rather
than a log ratio, which reduces the effect by a lot. That'll cost you
probably 8-10 dB in SNR, but you may not mind that--you'll still be
within shouting distance of the shot noise.
It's possible to do this other ways and get better performance, but it
takes real work. (I've been designing a follow-on to the original noise
canceller over the last 6 months or so, so I've been reminded of it.)
The temperature tracking of the mirror devices is the biggie. If you
keep their Ic and Vce the same, and mount them identically, it will
help, and of course you aren't going to get much heating down at a few
microwatts of dissipation.
If you don't need to noise-cancel all the photocurrents them at once,
you might want to look at multiplexing them into a single noise canceller.
The MAT04 that I used in my published circuits was discontinued earlier
this year, unfortunately.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net
C Egernet
wrote:
> 43 noise cancellers in one box? Cool.
When I first read about the noise canceller idea, I knew that I had my
hammer.
I just needed to find a suitable nail.
> Noise cancelling nanoamps isn't impossible, but it's hard to do well.
> Manyfold current mirroring at megahertz bandwidths with only 100 nA is
> going to be very tough.
In fairness, I would mirror the several uA current but still. Noise
cancelling
in a much wider bandwidth than I plan to measure in is maybe also the
wrong approach? It is essentially a DC signal I am after.
> I'd suggest using a TIA plus and 43 biggish resistors
This is a simple, idea even I can understand. Seems to make a lot of
sense.
> Putting a similar BJT in series with the feedback resistor of the TIA
> will get you a nice first-order temperature compensation and reduce the
> nonlinearity.
The temperature compensation sounds very neat, however, which
nonlinearity
are you referring to?
> ... you can get rid of the loop nonlinearity and have a straight
> ratio output rather than a log ratio, which reduces the effect by a lot.
But getting a linear signal rather than a logarithmic signal will
move
problems to the A/D-conversion. I am just as interested in knowing
whether the signal from the 23rd photodiode, say, is 1% or 1.01%
as I am in knowing whether the 37th photodiode gives 100% or 101%
of the reference.
> The temperature tracking of the mirror devices is the biggie. If you
> keep their Ic and Vce the same, and mount them identically, it will
> help, and of course you aren't going to get much heating down at a few
> microwatts of dissipation.
Exactly. And the instrument is going to live in a carefully air
conditioned
environment with plenty of airflow. I plan to keep power dissipation
down.
Best regards,
Chris Egernet
Phil Hobbs
> On Jan 5, 8:37 pm, Phil Hobbs<pcdhSpamMeSensel...@electrooptical.net>
> wrote:
>> 43 noise cancellers in one box? Cool.
>
> When I first read about the noise canceller idea, I knew that I had my
> hammer.
> I just needed to find a suitable nail.
>
>> Noise cancelling nanoamps isn't impossible, but it's hard to do well.
>> Manyfold current mirroring at megahertz bandwidths with only 100 nA is
>> going to be very tough.
>
> In fairness, I would mirror the several uA current but still. Noise
> cancelling
> in a much wider bandwidth than I plan to measure in is maybe also the
> wrong approach? It is essentially a DC signal I am after.
>
>> I'd suggest using a TIA plus and 43 biggish resistors
>
> This is a simple, idea even I can understand. Seems to make a lot of
> sense.
>
>> Putting a similar BJT in series with the feedback resistor of the TIA
>> will get you a nice first-order temperature compensation and reduce the
>> nonlinearity.
>
> The temperature compensation sounds very neat, however, which
> nonlinearity
> are you referring to?
The nonlinear emitter impedance and temperature drift of the common-base
device that you need between the resistor and the emitters of the noise
canceller's diff pair. The feedback network of the TIA would be a
resistor plus a diode connected transistor, which would match those of
each of the split outputs. There would be some small offset voltage if
the currents weren't identical in the feedback and output branches, but
if you're really making 43 1:1 mirrors with 44 identical resistors and
44 transistors, they'd match very well, as long as you're dropping a few
volts in the resistors.
>
>> ... you can get rid of the loop nonlinearity and have a straight
>> ratio output rather than a log ratio, which reduces the effect by a lot.
>
> But getting a linear signal rather than a logarithmic signal will
> move
> problems to the A/D-conversion. I am just as interested in knowing
> whether the signal from the 23rd photodiode, say, is 1% or 1.01%
> as I am in knowing whether the 37th photodiode gives 100% or 101%
> of the reference.
Terrific. Most folks don't like the log output because the gain depends
on the signal level, but your application is very good. Watch out for
base current errors, though--you need transistors with really good beta
linearity for that job. Have a look at the HFA3046 array.
>
>> The temperature tracking of the mirror devices is the biggie. If you
>> keep their Ic and Vce the same, and mount them identically, it will
>> help, and of course you aren't going to get much heating down at a few
>> microwatts of dissipation.
>
> Exactly. And the instrument is going to live in a carefully air
> conditioned
> environment with plenty of airflow. I plan to keep power dissipation
> down.
>
Lots of airflow is going to produce a certain amount of low frequency
noise, from microphonics and temperature variations due to turbulence.
I'd want to put a bit of insulation around the transistors--a 1 mK
fluctuation will give you 2 uV of drift, which is ~100 ppm in collector
current--not particularly subtle.
Please keep me posted on how it's going--that's the largest noise
canceller instrument I've heard of.
C Egernet
> device that you need between the resistor and the emitters of the noise
> canceller's diff pair.
Okay, so it is the action of the common-base transistor I don't
understand:
The TIA will convert the comparison photodiode current into a voltage
which is shared among the 43 noise cancellers, right? The large-ish
resistors will then convert that voltage into currents for each noise
canceller, right? So what does the common-base BJT do? Act as a
current-follower, pinning the voltage of the emitters of the
differential pair?
Sorry for asking elementary questions, but I'd very much like to
understand this clearly.
> Lots of airflow is going to produce a certain amount of low frequency
> noise, from microphonics and temperature variations due to turbulence.
> I'd want to put a bit of insulation around the transistors--a 1 mK
> fluctuation will give you 2 uV of drift, which is ~100 ppm in collector
> current--not particularly subtle.
It's good that you mention it. It seems that in physical science, it
is always easy to make a thermometer. The hard part is making
something that is either _not_ a thermometer or _only_ a thermometer.
> Please keep me posted on how it's going--that's the largest noise
> canceller instrument I've heard of.
So far I am only toying with the concept so I am increasingly
convinced that this is the way to go.
Best regards,
Chris Egernet
Phil Hobbs
>> The nonlinear emitter impedance and temperature drift of the common-base
>> device that you need between the resistor and the emitters of the noise
>> canceller's diff pair.
>
> Okay, so it is the action of the common-base transistor I don't
> understand:
You need to give the diff pair of the canceller a good quality current,
because otherwise the cancellation won't work as well--changing Vbe will
change the total current as well as the splitting ratio, which is Bad News.
The easiest way to do that is to stick the current into the emitter of a
transistor, bias the base someplace reasonable--somewhere _very_ quiet,
1 nV/sqrt(Hz) or less--and come out the collector...i.e. a common base
buffer.
>
> The TIA will convert the comparison photodiode current into a voltage
> which is shared among the 43 noise cancellers, right? The large-ish
> resistors will then convert that voltage into currents for each noise
> canceller, right?
Yup.
> So what does the common-base BJT do? Act as a
> current-follower, pinning the voltage of the emitters of the
> differential pair?
>
No, it's the other way up: low impedance in, high impedance out--the
collector goes to the diff pair emitters. You want the current not to
depend on the voltage at the emitters.
> Sorry for asking elementary questions, but I'd very much like to
> understand this clearly.
>
>> Lots of airflow is going to produce a certain amount of low frequency
>> noise, from microphonics and temperature variations due to turbulence.
>> I'd want to put a bit of insulation around the transistors--a 1 mK
>> fluctuation will give you 2 uV of drift, which is ~100 ppm in collector
>> current--not particularly subtle.
>
> It's good that you mention it. It seems that in physical science, it
> is always easy to make a thermometer. The hard part is making
> something that is either _not_ a thermometer or _only_ a thermometer.
>
Yup.
>> Please keep me posted on how it's going--that's the largest noise
>> canceller instrument I've heard of.
>
> So far I am only toying with the concept so I am increasingly
> convinced that this is the way to go.
>
Good luck!
Jon Kirwan
<pcdhSpamM...@electrooptical.net> wrote:
><snip>
>The easiest way to do that is to stick the current into the emitter of a
>transistor, bias the base someplace reasonable--somewhere _very_ quiet,
>1 nV/sqrt(Hz) or less--and come out the collector...i.e. a common base
>buffer.
><snip>
>> So what does the common-base BJT do? Act as a
>> current-follower, pinning the voltage of the emitters of the
>> differential pair?
>>
>No, it's the other way up: low impedance in, high impedance out--the
>collector goes to the diff pair emitters. You want the current not to
>depend on the voltage at the emitters.
><snip>
What about Early effect here? With the common base arranged
collector feeding the diff pair emitters that are moving
around? That would seem to move the collector and thus some
difference due to the Early effect, to me. What am I
missing?
Jon
Phil Hobbs
The magnitude of the change. In general purpose NPNs, the Early voltage
is ~100V, whereas the emitters only move 50 mV at most, and the resistor
current sources won't be dropping anywhere near 100V. The OP can
probably use BF862 JFETs just as well, but I wouldn't lose too much
sleep over it, at least at the 60 dB cancellation level. If he needs
more than that, it'll start to matter, but so will a lot of other things.
Jon Kirwan
<pcdhSpamM...@electrooptical.net> wrote:
Got it. The diff pair bases can't be driven around that much
so their shared emitters won't move much, etc.
By the way, 100V seems to be kind of a "blind default" case
(unmeasured, it seems to me, but estimated by some gross
sweep of the hand) for modeling small signal NPNs like the
2N3904 or 2N2222. But I don't have much experience actually
measuring the parameter -- something I should remedy. The
specific NPNs the OP might use for the common base design
might have significantly smaller magnitude Early voltage,
though. Regardless, your point stands. I can't recall ever
seeing a model (as I said, I've not measured so I'm lacking
experience there) with an Early voltage less than 20V. And
even there movement in the tens of millivolts shouldn't be a
serious problem. (Maybe before I say that, I should see what
50mV movement does on a VAF=20V device against the 60dB
level, though. At my level of understanding, this is kind of
interesting to think about.)
Thanks,
Jon
Phil Hobbs
I looked at the 2N3904, which has a typical Early voltage of 140V as
measured by Google. ;)
Quicker things like the HFA3046 are down in the 20V range, which can be
very inconvenient...I had a beautiful noise canceller front end blown
out of the water by that problem.
It used a non-monolithic diff pair, and the temperature tracking was
done by changing V_CE of one half of the diff pair so that the power
dissipation in the two sides was identical, regardless of the current
split ratio. It would have worked great, except that V_A was about
12V.... (John L. pointed that out to me, the rotter. ;) )
Tim Williams
news:654ak5d9slgfgo3or...@4ax.com...
> By the way, 100V seems to be kind of a "blind default" case
> (unmeasured, it seems to me, but estimated by some gross
> sweep of the hand) for modeling small signal NPNs like the
> 2N3904 or 2N2222. But I don't have much experience actually
> measuring the parameter -- something I should remedy.
I just measured it the other day, as a matter of fact. (I'm finally to the
point in the curriculum where we're supposed to investigate and use
transistors... how droll!) I got 220V for this particular 2N3904 at 3.8mA
Ic. In contrast, the Multisim v10 Spice model has the unusually low value
of 74V.
Tim
--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
Jon Kirwan
<pcdhSpamM...@electrooptical.net> wrote:
Well, the LTSpice 2n3904 (Philips, it says) shows VAF=100.
YMMV, of course.
>Quicker things like the HFA3046 are down in the 20V range, which can be
>very inconvenient...I had a beautiful noise canceller front end blown
>out of the water by that problem.
Okay. So it _could_ be a problem. Glad to see I'm not
totally confused.
>It used a non-monolithic diff pair, and the temperature tracking was
>done by changing V_CE of one half of the diff pair so that the power
>dissipation in the two sides was identical, regardless of the current
>split ratio. It would have worked great, except that V_A was about
>12V.... (John L. pointed that out to me, the rotter. ;) )
>
>Cheers
>
>Phil Hobbs
Egads. Maybe I should be glad I picked up some monolithic
pairs of both polarities. (I haven't used them for diff
pairs, as just current mirrors.)
Thanks,
Jon
Jon Kirwan
<tmor...@charter.net> wrote:
>"Jon Kirwan" <jo...@infinitefactors.org> wrote in message
>news:654ak5d9slgfgo3or...@4ax.com...
>> By the way, 100V seems to be kind of a "blind default" case
>> (unmeasured, it seems to me, but estimated by some gross
>> sweep of the hand) for modeling small signal NPNs like the
>> 2N3904 or 2N2222. But I don't have much experience actually
>> measuring the parameter -- something I should remedy.
>
>I just measured it the other day, as a matter of fact. (I'm finally to the
>point in the curriculum where we're supposed to investigate and use
>transistors... how droll!) I got 220V for this particular 2N3904 at 3.8mA
>Ic. In contrast, the Multisim v10 Spice model has the unusually low value
>of 74V.
Can you describe more about how you did your measurement of
Va? I'm guessing you set Vbe to some voltage and Vce to
about the same and measured Ic at 3.8mA and then changed Vce
to some other higher voltage and measured Ic again, and then
computed the slope and divided that into 3.8mA to get your
figure. (Or multiplied, depending upon whether you computed
the slope as a resistance or conductance.) But I'm curious
if any of that is right. Could you describe the details?
In any case, it would seem also that calculations generating
large Va figures are often fraught with inaccuracies (not
that it matters for large Va values, since that means the
effect is nearing a negligible value, anyway.)
I'm curious if you tried this at a number of different Vbe's
to see if there was a basewidth modulation effect, too.
Thanks,
Jon
Tim Williams
> Can you describe more about how you did your measurement of
> Va?
Constant Ib. I can't imagine trying to take such a measurement at constant
Vbe, think of the runaway. However, I did not track change in Vbe vs. Vce,
so if it's defined in terms of constant Vbe, my data may be incorrect.
As for the slope, I measured Ic for Vce = 0...10V and got the
slope-intercept for points from 1-10V.
> In any case, it would seem also that calculations generating
> large Va figures are often fraught with inaccuracies (not
> that it matters for large Va values, since that means the
> effect is nearing a negligible value, anyway.)
That could very well be. Hey, I didn't get a negative number, so that's one
thing!
> I'm curious if you tried this at a number of different Vbe's
> to see if there was a basewidth modulation effect, too.
The Early effect is explained in terms of base width modulation from the B-C
depletion region, and has nothing to do with the B-E junction itself. I
don't know why Vbe might change (except to secondary effects like Ie and
improved beta).
C Egernet
> > current-follower, pinning the voltage of the emitters of the
> > differential pair?
>
> No, it's the other way up: low impedance in, high impedance out--the
> collector goes to the diff pair emitters. You want the current not to
> depend on the voltage at the emitters.
I think I get it now but just to reinforce my understanding:
The common-base transistor pins the voltage of the end of the resistor
at the BJT emitter, right?
I can understand that because the voltage drop over the resistor is
now well-defined and so will the current be.
The common-base BJT collector just faithfully sends the current
through to the differential pair emitters (assuming large beta, alpha
unity).
That leaves me with what you said originally, Phil:
> Also, make sure you use a diode-connected transistor
> (CB shorted) for the compensation, because the BE diode doesn't have the
> same characteristics as the transistor in normal bias.
Does this imply that a common-base BJT behaves a bit like a diode-
connected transistor?
about the non-linearity:
> The nonlinear emitter impedance and temperature drift of the common-base device
Can I think of it this way:
We drive the feedback path of the TIA, resistor and diode-connected
BJT, by a current source. There will be a Vbe voltage drop and a drop
over the resistor. This is the voltage signal out of the TIA, shared
to the noise cancellers.
At the noise cancellers, the current is regenerated by taking the
voltage and sending it through a similar resistor and the common-base
BJT to ground. Since the majority of the voltage drop is over the
resistor, the current into (out of, really) the emitter is roughly the
same as photocurrent so the Vbe is correct too. Because Vbe is
correct, the current ends up being not just roughly right but very
accurate.
Best regards,
Chris Egernet
Phil Hobbs
>>> So what does the common-base BJT do? Act as a
>>> current-follower, pinning the voltage of the emitters of the
>>> differential pair?
>>
>> No, it's the other way up: low impedance in, high impedance out--the
>> collector goes to the diff pair emitters. You want the current not to
>> depend on the voltage at the emitters.
>
> I think I get it now but just to reinforce my understanding:
>
> The common-base transistor pins the voltage of the end of the resistor
> at the BJT emitter, right?
>
> I can understand that because the voltage drop over the resistor is
> now well-defined and so will the current be.
>
> The common-base BJT collector just faithfully sends the current
> through to the differential pair emitters (assuming large beta, alpha
> unity).
>
Yes.
> That leaves me with what you said originally, Phil:
>
>> Also, make sure you use a diode-connected transistor
>> (CB shorted) for the compensation, because the BE diode doesn't have the
>> same characteristics as the transistor in normal bias.
>
> Does this imply that a common-base BJT behaves a bit like a diode-
> connected transistor?
A diode connected transistor is actually in its normal bias region--it
doesn't saturate until the collector drops ~200-300 mV below the base.
That's why it makes a good compensator for the BJT. Early effect will
make a bit of difference, as Jon pointed out, but it'll mainly be a
slight gain error--the tempco and Vbe nonlinearity will track pretty
well. You don't have to worry too much about dissipation-induced
thermals with only a few microamps, but if that were ever to be a
problem, just bias the CB stage's base at -0.8V or so compared with the
diff pair's bases.
>
> about the non-linearity:
>> The nonlinear emitter impedance and temperature drift of the common-base device
>
> Can I think of it this way:
>
> We drive the feedback path of the TIA, resistor and diode-connected
> BJT, by a current source. There will be a Vbe voltage drop and a drop
> over the resistor. This is the voltage signal out of the TIA, shared
> to the noise cancellers.
>
> At the noise cancellers, the current is regenerated by taking the
> voltage and sending it through a similar resistor and the common-base
> BJT to ground. Since the majority of the voltage drop is over the
> resistor, the current into (out of, really) the emitter is roughly the
> same as photocurrent so the Vbe is correct too. Because Vbe is
> correct, the current ends up being not just roughly right but very
> accurate.
Right.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
email: hobbs at electrooptical dot net
http://electrooptical.net
Jon Kirwan
<tmor...@charter.net> wrote:
>"Jon Kirwan" <jo...@infinitefactors.org> wrote in message
>news:iasak5pia917e27tn...@4ax.com...
>> Can you describe more about how you did your measurement of
>> Va?
>
>Constant Ib. I can't imagine trying to take such a measurement at constant
>Vbe, think of the runaway. However, I did not track change in Vbe vs. Vce,
>so if it's defined in terms of constant Vbe, my data may be incorrect.
>
>As for the slope, I measured Ic for Vce = 0...10V and got the
>slope-intercept for points from 1-10V.
Okay. As I understand things constant Vbe works without
runaway (say, for example, 600mV for Si types.) Since Ic
moves rapidly with small Vbe changes, though, it seems using
a constant current at the base is indeed easier and probably
better. So I think I'm with you on this.
>> In any case, it would seem also that calculations generating
>> large Va figures are often fraught with inaccuracies (not
>> that it matters for large Va values, since that means the
>> effect is nearing a negligible value, anyway.)
>
>That could very well be. Hey, I didn't get a negative number, so that's one
>thing!
Well, Va is on the negative side of the Vce axis. The main
thing, of course, is that you set up your equations right and
perform measurements appropriately.
>> I'm curious if you tried this at a number of different Vbe's
>> to see if there was a basewidth modulation effect, too.
>
>The Early effect is explained in terms of base width modulation from the B-C
>depletion region, and has nothing to do with the B-E junction itself. I
>don't know why Vbe might change (except to secondary effects like Ie and
>improved beta).
(Actually, I was thinking of modulating Vbe by changing your
Ib.) But there is a width in the emitter-base space charge
layer that is normally assumed to be negligible, but which
may be measurable for you. It's called the "Late Effect."
Va captures the Vbc basewidth modulation, and since that is
the topic here it should be obvious that there _might_ be
such a Late Effect, right?
Anyway, the best way to verify, of course, is to sit down and
_do_ it and see. If you get two different Va for different
Ib/Vbe settings, then it's certainly an indication, right?
I had wanted to someday try and play with both.
Jon
Tim Williams
> (Actually, I was thinking of modulating Vbe by changing your
> Ib.) But there is a width in the emitter-base space charge
> layer that is normally assumed to be negligible, but which
> may be measurable for you. It's called the "Late Effect."
:-p
That's true, even a forward-biased junction still has some V across it and
therefore some depletion region thickness. (It's dastardly difficult to
shove enough current density through silicon to "unbias" the junction --
IIRC, kiloamperes for a 1N914.)
Tim
P.S. Yeah, but was it named for Mr. Late? Is it actually Lat� and we're
saying it wrong? ;-)
Jon Kirwan
<tmor...@charter.net> wrote:
>"Jon Kirwan" <jo...@infinitefactors.org> wrote in message
>news:stock513nq6g9lr6g...@4ax.com...
>> (Actually, I was thinking of modulating Vbe by changing your
>> Ib.) But there is a width in the emitter-base space charge
>> layer that is normally assumed to be negligible, but which
>> may be measurable for you. It's called the "Late Effect."
>
>:-p
>
>That's true, even a forward-biased junction still has some V across it and
>therefore some depletion region thickness. (It's dastardly difficult to
>shove enough current density through silicon to "unbias" the junction --
>IIRC, kiloamperes for a 1N914.)
The procedure I've seen for seeing if Vb (or Spice's Var) is
important for the "normal" region is to plot ln(Ic) vs the
venerable (q/kT)*Vbe and see if the slope varies much from 1.
If it does depart, Var is important in the normal region.
>Tim
>
>P.S. Yeah, but was it named for Mr. Late? Is it actually Lat� and we're
>saying it wrong? ;-)
hehe. Early's 1952 paper was probably just later 'turned
around' on him!
Jon