Maker Pro
Maker Pro

A Lock-In Riddle

H

Haude Daniel

Jan 1, 1970
0
Hello folks,

this may be slightly off-topic in this group, but I know that there are
enough gurus around here that can help me with this problem.

We're doing a STM spectroscopy measurement (molecules adsorbed on a
substrate) and obtain the local density of states by measuring the
differential conductance as a function of applied voltage using a lock-in
amplifier.

For those that aren't interested in physics: we're measuring the I(V)
characteristics of a black box that has two wires sticking out of it.

Applying this technique to a theoretical system that exhibits a perfect
step function characteristic (i.e., something that doesn't conduct at all
below a certain voltage and has uniform conduction above it) would result
in a peak which naturally is broadened by the fact that the modulation
voltage is finite. It's easy to see that the width of the base of the peak
equals the p-p amplitude of the modulation signal.

Using a 10mV(eff) modulation voltage we cannot expect to see any features
in our differential conduction function that exhibit a peak width of more
than about 25-30 mV.

But we do. We see a 3mV FWHM peak above the molecules at zero bias.

From the way I understand lock-in, this is theoretically impossible. Maybe
my reasoning is flawed, or we're dealing with some kind of strange
artefact. The lock-in is a SR830 digital unit; we're currently trying to
reproduce the experiment with an analog lock-in which is a bit difficult
due to the poorer input filtering capabilities.

Another lab in our group has seen the same effect on an all-metal sample
as well. At modulation frequencies of a few hundred Hz we exclude any
dynamic effects, that is to say, we assume the I(V) function of the
tip-sample system under investigation to be time-invariant.

Thanks for any suggestions,
--Daniel
 
Haude said:
Hello folks,

this may be slightly off-topic in this group, but I know that there are
enough gurus around here that can help me with this problem.

We're doing a STM spectroscopy measurement (molecules adsorbed on a
substrate) and obtain the local density of states by measuring the
differential conductance as a function of applied voltage using a lock-in
amplifier.

For those that aren't interested in physics: we're measuring the I(V)
characteristics of a black box that has two wires sticking out of it.

Applying this technique to a theoretical system that exhibits a perfect
step function characteristic (i.e., something that doesn't conduct at all
below a certain voltage and has uniform conduction above it) would result
in a peak which naturally is broadened by the fact that the modulation
voltage is finite. It's easy to see that the width of the base of the peak
equals the p-p amplitude of the modulation signal.

Using a 10mV(eff) modulation voltage we cannot expect to see any features
in our differential conduction function that exhibit a peak width of more
than about 25-30 mV.

But we do. We see a 3mV FWHM peak above the molecules at zero bias.

From the way I understand lock-in, this is theoretically impossible. Maybe
my reasoning is flawed, or we're dealing with some kind of strange
artefact. The lock-in is a SR830 digital unit; we're currently trying to
reproduce the experiment with an analog lock-in which is a bit difficult
due to the poorer input filtering capabilities.

Another lab in our group has seen the same effect on an all-metal sample
as well. At modulation frequencies of a few hundred Hz we exclude any
dynamic effects, that is to say, we assume the I(V) function of the
tip-sample system under investigation to be time-invariant.

Thanks for any suggestions,

Have you actually modelled what happens when you apply a DC biassed
10mV rms sine wave across a 3mV step in diifferential conductance
versus voltage?

Is there a capacitance associated with the changing conductance? You
might be seeing a phase shift - your digital lock-in incorporates a
phase-sensitive detector.
 
H

Haude Daniel

Jan 1, 1970
0
Have you actually modelled what happens when you apply a DC biassed
10mV rms sine wave across a 3mV step in diifferential conductance
versus voltage?

Yes. You get a cos(arcsin(U/Up))-shaped peak centered around the step.
Up = 1.41*Umod(eff)
Is there a capacitance associated with the changing conductance?

Not a changing one. Of course there's the parallel capacitance of the
tunneling gap and then some.

robert
 
H

Haude Daniel

Jan 1, 1970
0
Applying this technique to a theoretical system that exhibits a perfect
step function characteristic (i.e., something that doesn't conduct at all
below a certain voltage and has uniform conduction above it)

Clarification: "uniform conduction" should mean: constant current
independent of voltage.

robert
 
M

Mark

Jan 1, 1970
0
Haude said:
Clarification: "uniform conduction" should mean: constant current
independent of voltage.

robert

I don't totally understand your system but....

a lock in amp is a PLL (phase locked loop) and a PLL is a narrowband
device...

sounds like you are applying a sine wave excitation to a non-linear
device and trying to measure the I-V characteristic....the non
linearity will distort the sine wave and the distortion is the data you
want to measure, but distortion is harmonics and the lock in amp being
narrow band will reject harmonics...

so I don't know what you will get...

why use AC excitation to measure I/V, why not DC?

Mark
 
Mark said:
I don't totally understand your system but....

a lock in amp is a PLL (phase locked loop) and a PLL is a narrowband
device...

sounds like you are applying a sine wave excitation to a non-linear
device and trying to measure the I-V characteristic....the non
linearity will distort the sine wave and the distortion is the data you
want to measure, but distortion is harmonics and the lock in amp being
narrow band will reject harmonics...

Some lock-ins will detect and measure higher harmonics - there are a
couple of physical techniques that depend on finding the second
harmonic content of the an output. I'm damned if I can remember what
they are ...

Threre ws an improved GaAs crystal puller that was going to detect the
contact angle between the melt and the growing crystal by modulating
the heat input - and thus the height of the solid-liquid boundary - and
detecting the second harmonic content in the "weight"of the crystal
(which includes the surface tension force acting on the circumference
of the crystal. Don't know if it ever worked.
 
H

Haude Daniel

Jan 1, 1970
0
a lock in amp is a PLL (phase locked loop) and a PLL is a narrowband
device...

Well, it's a synchronous rectifier (demodulator), not a PLL.
sounds like you are applying a sine wave excitation to a non-linear
device and trying to measure the I-V characteristic...
Right.

the non
linearity will distort the sine wave and the distortion is the data you
want to measure, but distortion is harmonics and the lock in amp being
narrow band will reject harmonics...

Well, the idea is to make the modulation so small that the system looks
linear at that particular offset. Of course the step function I discussed
is a gross violation of this premise; it was just a device to estimate the
smallest possible feature width obtainable by this technique.
so I don't know what you will get...

The differential conductance.
why use AC excitation to measure I/V, why not DC?

Because the signal is buried in lots of noise.

Mind you, this is a well-established technique. We've been using it for
years. It's just that these tiny peaks have been showing up occasionally
and we don't want to ignore them any more.

The question is:

If, using lock-in technique, I find features that are much narrower than
the pp excitation voltage, one of the following statements must be true

1) The presumption that a lock-in amplifier cannot yield features that are
narrower than about 2.5*Umod(eff) is false. (Of course it can detect
narrower features, but they will result in broader peaks)

2) The impedance characteristics of the system being measured are not
time-invariant but change in sync with Umod.

3) The lock-in amplifier introduces a measurement artefact.

I'm somewhat puzzled. If 1) holds, I'd really appreciate an explanation.
Also some LIA manuals would have to be rewritten.

2) would be odd because this has been observed on different tunneling
systems, both metallic and semiconductor.

3) Unlikely, because the peak is reproducibly seen over certain locations
of the sample and not over others. Also it is always at exactly zero bias
voltage, about which the AC-coupled LIA knows nothing.

--Daniel
 
Haude said:
Well, it's a synchronous rectifier (demodulator), not a PLL.


Well, the idea is to make the modulation so small that the system looks
linear at that particular offset. Of course the step function I discussed
is a gross violation of this premise; it was just a device to estimate the
smallest possible feature width obtainable by this technique.


The differential conductance.


Because the signal is buried in lots of noise.

Mind you, this is a well-established technique. We've been using it for
years. It's just that these tiny peaks have been showing up occasionally
and we don't want to ignore them any more.

The question is:

If, using lock-in technique, I find features that are much narrower than
the pp excitation voltage, one of the following statements must be true

1) The presumption that a lock-in amplifier cannot yield features that are
narrower than about 2.5*Umod(eff) is false. (Of course it can detect
narrower features, but they will result in broader peaks)

2) The impedance characteristics of the system being measured are not
time-invariant but change in sync with Umod.

3) The lock-in amplifier introduces a measurement artefact.

I'm somewhat puzzled. If 1) holds, I'd really appreciate an explanation.
Also some LIA manuals would have to be rewritten.

2) would be odd because this has been observed on different tunneling
systems, both metallic and semiconductor.

3) Unlikely, because the peak is reproducibly seen over certain locations
of the sample and not over others. Also it is always at exactly zero bias
voltage, about which the AC-coupled LIA knows nothing.

That sounds like good physics.
From an instrumental point of view, have you measured both the in-phase
and quadrature outputs from your experiment at the excitation
frequency, twice the excitation frequency and three times the
excitation frequency?

The quadrature outptus may tell you something about any reactive
component in the response of your specimen, and the higher harmonics
could say something about non-linear interactions - second harmonic
about even order terms in a power law response, third harmonic about
odd order terms (square law versus cube law componenets etc).
 
K

Ken Smith

Jan 1, 1970
0
Is there a capacitance associated with the changing conductance?

Another thought along the same sort of lines:

You may have lossy capacitances.

If the insulation of the wiring etc does not make good capacitors, you may
have a lossy capacitance which will cause an in-phase compoent in the
current that passes throught it. This effect will change with frequency so
you should be able to check for it.

Stray capacitances may be interacting with the experimental area resulting
in what looks like a lossy capacitor across the terminals. When the
lossiness of this capacitor changes, it changes the in-phase component of
the current.
 
M

Mark

Jan 1, 1970
0
2) The impedance characteristics of the system being measured are not
time-invariant but change in sync with Umod.

Let me try to understnad better what you are doing,,,

applying a small AC excitation current to a DUT (device under test) and
using a lock in amp (PLL based synchronous demodulator) to recover the
low level AC signal and measure its amplitude and from this info you
can deduce the ___DIFFERENTIAL ___ resistance of the DUT.... and at the
same time you are applyting a larger DC current to the DUT to move the
operating (bias) point around at which you which to measure the small
differential resistnace... is this correct?

What if the DUT has a step function non-linearity or maybe an I/V
transfer with hysteresis or other "memory effect" , at some DC bias
points , would that cause the effect you are seeing?

Mark
 
K

Ken Smith

Jan 1, 1970
0
While people are thinking about them: I need a lock-in for, frequencies
from 50KHz to 750KHz. Are there any suggestions of ones to look at or
avoid from the group.

Other characteristics I'd like:

(1)
I'd prefer the two channel type that allows phase shifts to be seen
directly.

(2)
My experimental setup will be sensitive to EMI, basically from DC to 1MHz
so low radiation in that band is desired. I already have to live with
lost of 60Hz so I have designed around that problem. As a result, 60Hz
and its harmonics aren't a huge problem to me.

(3)
Because of (2) I worry about anything that needs a PC to control. I want
to avoid switching power supplies as much as posible,

(4)
My input signal may be frequency modulated. I need the lock-in to be able
to follow the modulation. The FMing will be by about 3% at a rate up to
about 10KHz.
 
H

Haude Daniel

Jan 1, 1970
0
Let me try to understnad better what you are doing,,,

applying a small AC excitation current

a voltage, actually
to a DUT (device under test) and
using a lock in amp (PLL based synchronous demodulator) to recover the
low level AC signal

the current through the DUT (a tunnel junction)
and measure its amplitude and from this info you
can deduce the ___DIFFERENTIAL ___ resistance of the DUT

Exactly.

..... and at the
same time you are applyting a larger DC current

A voltage
to the DUT to move the
operating (bias) point around at which you which to measure the small
differential resistnace... is this correct?
yes.


What if the DUT has a step function non-linearity

Then I should see a peak in the dI/dV(V) curve that is about as wide as
the peak-peak amplitude of the modulation signal. What I'm seeing is a
peak that has only one-tenth the width.
or maybe an I/V
transfer with hysteresis or other "memory effect", at some DC bias
points , would that cause the effect you are seeing?

That's what I exclude because the effect has been seen on metal systems as
well which don't exhibit such slow charging effects.

--Daniel
 
K

Ken Smith

Jan 1, 1970
0
Well, it's a synchronous rectifier (demodulator), not a PLL.

Lock-ins typically have both in them. They lock the PLL onto the
reference or the signal its self and then use that to time the
demodulator.

[....]
Well, the idea is to make the modulation so small that the system looks
linear at that particular offset. Of course the step function I discussed
is a gross violation of this premise; it was just a device to estimate the
smallest possible feature width obtainable by this technique.

You don't have to assume that the system is linear. The non-linear effect
adds even harmonics to your output. With only a little added trickiness,
you can measure the "distortion" products this could give you better
results with the same basic set up. Here's basically the idea. We can
flesh it out if you want to do it.

(1)
Find (or you are) someone who knows which end of a soldering iron is
which.

(2)
For your few hundred Hz, use a synth that can either take or give you a
reference at, lets say, 10MHz.

(3)
Lets say your "few 100Hz" is really 100Hz

(4)
Your output is the 10MHz divided by (10MHz/100) = 100,000

100,000 = 2^5 * 5^5

Do the 5s first and then the 2s

(5)
Add some clever circuit to the reset pin of the counters so that we can
let them start at the right time in the cycle.

(6)
All the divide by two stages give you signals that can be used on the
lock-in to isolate the harmonic.
 
M

Mark

Jan 1, 1970
0
Then I should see a peak in the dI/dV(V) curve that is about as wide as
the peak-peak amplitude of the modulation signal. What I'm seeing is a
peak that has only one-tenth the width.

I do not understand your reply... If the step function change in
differential resistance is from say 0.1 Ohms to 1 Meg Ohm, would that
not create a very large retrun signal....
I don't understand what you mean by "width". Also I do not understan,
you are applying an excitation VOLTAGE and measuring an return
VOLTAGE??? If you want to infer the differentiual tesistance, you
need to have current in there someplace...no? i.e apply and
excitation current and measure the return voltage...or apply an
excitation voltage and measure a return current??? Perhaps you are
measuremting the return current indirectly by measuring the voltage
across a fixed reesistor that is in series with the DUT and excitation
voltage???

You are saying this is a well established procedure and I am not
questioning that, but if you would like us to help you, we need to
understand the well established procedure better.
That's what I exclude because the effect has been seen on metal systems as
well which don't exhibit such slow charging effects.

I don't know what you mean by "slow charging effect".
I'm envisioining your DUT as some kind of a spark gap or something like
a neon bilb that can break down and have all kinds of non-linear
behavior???

So does the effect change if you change the excitation frequency up or
down by say an order of magnitude?
 
A

Aristotle Eisenglas

Jan 1, 1970
0
Mark said:
Let me try to understnad better what you are doing,,,
applying a small AC excitation current to a DUT (device under test) and
using a lock in amp (PLL based synchronous demodulator) to recover the
low level AC signal and measure its amplitude and from this info you
can deduce the ___DIFFERENTIAL ___ resistance of the DUT.... and at the
same time you are applyting a larger DC current to the DUT to move the
operating (bias) point around at which you which to measure the small
differential resistnace... is this correct?
What if the DUT has a step function non-linearity or maybe an I/V
transfer with hysteresis or other "memory effect" , at some DC bias

Some people do make comparators with r-r outputs. What works
pretty well is a dual mode GSM + AMPS phone. One additional
requirement is that such a counter would be used to control Jews
and their geery behaviors. Anyone who wants to implement
algorithms should have a value 1/10th of the collector current
as base drive.

Also, it's not uncommon to then blindly estimate the needed beta
as 10, for saturation. That's because most modern transistors
are pretty much useless for defining the width of the counter to
inhibit the clock input. It will be interesting to see the
place, the birds, could ask an expert all my questions, learned
a lot while teaching and now everyday as a But I do not want to
explain here and now) in order to create the excuse for the high
and low voltages, so an SMA box that has 50 ohm output like the
old Apple II users, who had their users BIBLE. Religion is not
the level at which this individual was dissected. / The
programming of new cells to replace the exponential term with
the series expansion 1 + hf/kT, which is a small surface mount
fuses but these are not really good starting points for
designing a new system.
 
A

Aristotle Eisenglas

Jan 1, 1970
0
That's what I exclude because the effect has been seen on metal systems as
well which don't exhibit such slow charging effects.
--Daniel

There were hundreds of millions of those high-G tubes built, and
I'd like to plant and establish things there now that chickens
would like to reduce the effect of SRF on an S21 plot usually
takes some effort, and it helps a lot if you've cheated and
measured the SRF previously so that you couldn't even boot it
without the passcode, that may be destroyed during their
testing. "Arfa Daily" <> wrote in message ... "What I didn't
understand is that why it's so safe to drive in MA or NJ.
Roundabouts. Weenies in small cars.

Darwin comes home to roost.

I'm not *that* old; 1974. BTW, this was in NY, not NYC. It was
in one of my hens to 104 degree temps. One black turken and one
of the recommended ways of acquiring free marijuana is: "Be very
very nice to everyone you meet." :) Cheers!
 
J

John Larkin

Jan 1, 1970
0
Some people do make comparators with r-r outputs. What works
pretty well is a dual mode GSM + AMPS phone. One additional
requirement is that such a counter would be used to control Jews
and their geery behaviors. Anyone who wants to implement
algorithms should have a value 1/10th of the collector current
as base drive.

Also, it's not uncommon to then blindly estimate the needed beta
as 10, for saturation. That's because most modern transistors
are pretty much useless for defining the width of the counter to
inhibit the clock input. It will be interesting to see the
place, the birds, could ask an expert all my questions, learned
a lot while teaching and now everyday as a But I do not want to
explain here and now) in order to create the excuse for the high
and low voltages, so an SMA box that has 50 ohm output like the
old Apple II users, who had their users BIBLE. Religion is not
the level at which this individual was dissected. / The
programming of new cells to replace the exponential term with
the series expansion 1 + hf/kT, which is a small surface mount
fuses but these are not really good starting points for
designing a new system.


Who are you, and why don't you make any sense?

Even your nonsense is pretty low-quality stuff.

John
 
J

Jim Thompson

Jan 1, 1970
0
On Fri, 04 Aug 2006 14:13:10 -0000, Aristotle Eisenglas
[snip]
The
programming of new cells to replace the exponential term with
the series expansion 1 + hf/kT, which is a small surface mount
fuses but these are not really good starting points for
designing a new system.


Who are you, and why don't you make any sense?

Even your nonsense is pretty low-quality stuff.

John

He's a broad-spectrum spammer. Just killfile on name, until he
changes to a new one.

...Jim Thompson
 
R

Rich Grise

Jan 1, 1970
0
Who are you, and why don't you make any sense?

Even your nonsense is pretty low-quality stuff.

It took me about four of these posts to figure out that this is either
a raving lunatic or a bot.

Cheers!
Rich
 
K

Ken Smith

Jan 1, 1970
0
Rich Grise said:
It took me about four of these posts to figure out that this is either
a raving lunatic or a bot.

I remember when S C I E N T O L O G I S T S would do this because someone
mensioned them in a news group. They did it to push the article off the
servers by filling them up.


Maybe this is being done again.
 
Top