measuring attoamps
David Wipf
Hello,
I am considering an experiment that would involve measuring aA
levels of dc current (perhaps even down to < 1 aA) in an
electrochemical cell. I know that I can't directly measure that
current level but I think that I can chop the signal at 10 Hz or
less. I have a pretty good setup to do the measurment, I can use
either a Keithely 428 current amp or a 6512 electrometer to
transduce the current to a voltage. I will use an SRS 850 digital
lock in to demodulate the signal. I was thinking that a 0.1 Hz
modulation and a 100 or 1000 S time constant might get me down to
the low aA level. The source will have an impedence of between
1k to 100k (it depends on the electrode I use)and will be at room
temp. Obviously I will shield the system and do my best to prevent
interference from the chopping system. I expect that the system
will have a background current of 1 to 10 pA and a noise level of
around 0.1 pA maximum.
My questions are,
Can I do this without resort to heroic extremes (such as cooling
my amplifier with LN2?
Should I use the 6512 or 428 to do the measurement?
Any other suggestions on Lock-in use that I am missing.
Finally, can someone suggest a model system that I could use to
demonstrate to myself (and to the granting agency) that I can really
measure aA without actually setting up the experiment. I am trying an LED
shining on a photodiode, I am getting signal at the .1 fA level but I can't
get less signal (and I think that the signal I am getting is due to leakage
current not photons, unless 0.004 V rms produces light in a red LED)
I know that I am being vague about the system and the chopping method
but I am not able to disclose more at this time.
Thanks for any help.
--
David Wipf
Department of Chemistry, Mississippi State U
wi...@ra.msstate.edu
James Meyer
On Sat, 16 May 1998 15:39:27 -0500, "David Wipf" <wi...@ra.msstate.edu>
wrote:
>Hello,
> I am considering an experiment that would involve measuring aA
>levels of dc current (perhaps even down to < 1 aA) in an
>electrochemical cell.
Since 1 attoamp is the result of a charge displacement of 1
attocoulomb per second and since 1 single electron has a charge of .16
attocoulombs, you are expecting to measure the equivalent of 6.25
electrons per second going somewhere.
Does the term "snowball in a hot place" give you an idea of my
estimate of your chances of pulling this off?
Jim
Ian St. John
James Meyer wrote in message <6jleij$p...@bgtnsc02.worldnet.att.net>...
Good point Jim. But that just translates the problem to 'how can I measure
on the basis of counting electron's. Doesn't quantem physics do that? Course
it would involve custom silicon components..
James Meyer
On Sun, 17 May 1998 02:38:25 -0400, "Ian St. John" <ist...@sprint.ca>
wrote:
>
>James Meyer wrote in message <6jleij$p...@bgtnsc02.worldnet.att.net>...
>>
>> Since 1 attoamp is the result of a charge displacement of 1
>>attocoulomb per second and since 1 single electron has a charge of .16
>>attocoulombs, you are expecting to measure the equivalent of 6.25
>>electrons per second going somewhere.
>>
>
>Good point Jim. But that just translates the problem to 'how can I measure
>on the basis of counting electron's. Doesn't quantem physics do that? Course
>it would involve custom silicon components..
>
Sure. But setting up a whole quantum physics lab to measure
the output from one electro-chemical cell doesn't sound like an easy
job to me.
I'd be inclined to use the current to plate out some metal
over a *long* period of time and then simply weigh the plated out
sample. There is a direct relationship between each atom plated out
and the current through the plating bath. Even that would be plenty
hard enough to do with any accuracy.
Jim
Walter Gray
In article <6jn6tr$j...@bgtnsc02.worldnet.att.net>, notj...@worldnet.att.net (James Meyer) writes:
:On Sun, 17 May 1998 02:38:25 -0400, "Ian St. John" <ist...@sprint.ca>
If I haven't lost any powers of 10, this experiment would take
approx 50 million years to plate out 1 milligram of copper.
Walter
Disclaimer: My employer is not responsible for this stuff.
Please reply to: wag...@dra.hmg.gb as the address given above
sometimes doesn't work.
Bill Sloman
David Wipf wrote:
> Hello,
> I am considering an experiment that would involve measuring aA
> levels of dc current (perhaps even down to < 1 aA) in an
> electrochemical cell. I know that I can't directly measure that
> current level but I think that I can chop the signal at 10 Hz or
> less. I have a pretty good setup to do the measurment, I can use
> either a Keithely 428 current amp or a 6512 electrometer to
> transduce the current to a voltage. I will use an SRS 850 digital
> lock in to demodulate the signal. I was thinking that a 0.1 Hz
> modulation and a 100 or 1000 S time constant might get me down to
> the low aA level. The source will have an impedence of between
> 1k to 100k (it depends on the electrode I use)and will be at room
> temp. Obviously I will shield the system and do my best to prevent
> interference from the chopping system. I expect that the system
> will have a background current of 1 to 10 pA and a noise level of
> around 0.1 pA maximum.
The Johnson noise current on a 100k resistance at room temperature is
4^-13A per root Hz, or 400 fA. To get down to 1aA you have to reduce
this by 400,000, which implies reducing the badndwidth by 1.6^-11, which
implies a settling time of about 2000 years.
I think you have better think it out again.
Bill Sloman, Nijmegen
Daniel Haude
On 18 May 1998 12:30:51 GMT,
Walter Gray <wag...@taz.dra.hmg.gb> wrote:
> : I'd be inclined to use the current to plate out some metal
> :over a *long* period of time and then simply weigh the plated out
> :sample. There is a direct relationship between each atom plated out
> :and the current through the plating bath. Even that would be plenty
> :hard enough to do with any accuracy.
>
> If I haven't lost any powers of 10, this experiment would take
> approx 50 million years to plate out 1 milligram of copper.
Yes, but using gold instead of copper would reduce this time to about 16
million years. That's an improvement already (but then -- think of the
compound interest you'd get after 50 million years for the price
difference between copper and gold). Hard to decide.
But you might also plate gold onto an atomically flat crystal surface and
then count the adsorbed atoms per unit area with an STM. That'd give you a
turnaround time of maybe a few days instead of 50 million years. Again, an
improvement.
--Daniel
James Meyer
On 18 May 1998 12:30:51 GMT, wag...@taz.dra.hmg.gb (Walter Gray)
wrote:
>In article <6jn6tr$j...@bgtnsc02.worldnet.att.net>, notj...@worldnet.att.net (James Meyer) writes:
> : I'd be inclined to use the current to plate out some metal
> :over a *long* period of time and then simply weigh the plated out
> :sample. There is a direct relationship between each atom plated out
> :and the current through the plating bath. Even that would be plenty
> :hard enough to do with any accuracy.
>
>If I haven't lost any powers of 10, this experiment would take
>approx 50 million years to plate out 1 milligram of copper.
>
>
>Walter
Microgram measurements are routine in many not so
sophisticated labs. That would reduce the time to 50 thousand years.
Nanogram measurements need only a little more care. That's 50 years
already.
Now all one has to do is to put 500 little plating baths in
*series*. The same test current would do the plating in all 500 at
the same time and you could finish in a month or two.
Jim
James Meyer
On 18 May 1998 14:17:34 GMT, p17...@public.uni-hamburg.de (Daniel
Haude) wrote:
>But you might also plate gold onto an atomically flat crystal surface and
>then count the adsorbed atoms per unit area with an STM. That'd give you a
>turnaround time of maybe a few days instead of 50 million years. Again, an
>improvement.
>
>--Daniel
Excellent suggestion! Counting beats weighing for accuracy.
Jim "Weighs 30 stone." Meyer
Kevin G. Rhoads
You start by designing a GOOD Faraday cage for the entire experiment
to sit in. You have a high impedance measurement, most of your
interference will be E-field coupled not B-field coupled. But overlay
heavy-duty aluminum foil over the copper screening where you can,
it will help block air-currents (see a few points below)
You use a Keithley electrometer to measure your current, integrating over
several minutes. They'll try to sell you a digital. The analog unit costs
twice as much - it's harder to use -- buy the analog unit 'cause there's no
active oscillators inside it - the digital has a digital display & the
driver
for that is a noise source. (Digital =$2500, analog=$5000 that was a few
years ago, prices probably went up)
The Keithley runs on it's internal batteries during measurement, no line
cord!
If you are running any other equipment, try to eliminate it.
After you've eliminated all the equipment you can, eliminate some more.
Once there is nothing left but essential equipment, rethink essential --
eliminate some more.
Anything you can't eliminate, run from batteries. Buy a continuous duty
UPS to run it. Put it in ANOTHER separate Faraday cage. Preferably
eliminate it - it's a noise source.
Learn what single point grounding is. Use it.
NO VENTILATION. Air currents can carry more charge by
convection than you are trying to measure.
NO HUMIDITY. Humid air can CONDUCT more current
than you are trying to measure.
RUN AT NIGHT. LATE NIGHT. You will pick up crud
from any one else nearby. So run when few people or no people
are nearby.
TRY TO FIND a temporary lab space in the boonies. The farther
from civilization the better. Civilization == Noise source. (There
are experiments that Faraday did that cannot be reproduced anywhere
outside of a Faraday cage 'cause there's so much electrical noise
floating around.)
NO FLUORESCENT LIGHTS, especially not the new energy
saving kind. (see next) Incandescents w/o dimmer circuits, or
even kerosene lanterns. Kerosene lamps are probably better.
NO RADIO. Bring sci-fi or romance or western or gothic novels to read
while waiting out the integration times. The LO in the superhet receiver
in
your radio will play hobb with your measurement -- even when it is in a
Faraday cage.
Consider TWO Faraday cages, one within the other. Yeah, yeah a
Faraday cage totally shields (bullshit), usually double Faraday isn't
necessary. For what you want to do, it probably is.
Wear cotton (NO WOOL, NO SILK, NO NYLON) use PLENTY
of fabric softener (it really is a conductivity enhancer for clothing)
so YOU are not a source of stray charges.
Now, with all the above precautions, you will run into problems
but they should be solvable. NOT EASY, solveable. If you
are not daunted, ask again when you are farther along.
------------
We used dual Keithley's to measure the voltages in molten
NaCl/KCl mixtures with totally blocking (Mo) electrodes. The grad
student couldn't get decent measurements during the day --
too much noise pick up from the rest of the building. MIT's
physical plant ran in a dedicated power drop to run the 3-phase
furnace heater with 120 VAC pick offs for the instrumentation.
The regular power drops were just too dang noisy.
The dual Keithley's front-ended for a Tek (digital) scope. But
we weren't going after atto-amps, low pico- and high femto-
but not atto-. Or we would have had to lose the digitizing Tek
scope. As it was the data was noisy enough to require extensive
post filtering . . .
--
Kevin G. Rhoads, Ph.D.
T_Rhoads@NO_SPAM.Classic.MSN.COM
KRhoads@NO_SPAM.CmpNetMail.com
Walter Gray
In article <slrn6m0gnt....@insitu.physnet.uni-hamburg.de>, p17...@public.uni-hamburg.de (Daniel Haude) writes:
:On 18 May 1998 12:30:51 GMT,
: Walter Gray <wag...@taz.dra.hmg.gb> wrote:
:
:> : I'd be inclined to use the current to plate out some metal
:> :over a *long* period of time and then simply weigh the plated out
:> :sample. There is a direct relationship between each atom plated out
:> :and the current through the plating bath. Even that would be plenty
:> :hard enough to do with any accuracy.
:>
:> If I haven't lost any powers of 10, this experiment would take
:> approx 50 million years to plate out 1 milligram of copper.
:Yes, but using gold instead of copper would reduce this time to about 16
:million years. That's an improvement already (but then -- think of the
:compound interest you'd get after 50 million years for the price
:difference between copper and gold). Hard to decide.
:But you might also plate gold onto an atomically flat crystal surface and
:then count the adsorbed atoms per unit area with an STM. That'd give you a
:turnaround time of maybe a few days instead of 50 million years. Again, an
:improvement.
Or the current could be used to transport DNA fragments by electrophoresis.
The deposited fragments could be amplified by a factor of 10^20 using the
PCR. The resulting goop could then be weighed on kitchen scales.
(If anyone would like to patent this method, contact my lawyers.)
Walter Gray
In article <01bd82da$17bd3840$LocalHost@stupidwin95>, "Kevin G. Rhoads" <T_Rh...@NoSpam.CLASSIC.MSN.COM> writes:
[]
:Wear cotton (NO WOOL, NO SILK, NO NYLON) use PLENTY
:of fabric softener (it really is a conductivity enhancer for clothing)
:so YOU are not a source of stray charges.
Alternatively, get a friend to shave off your body hair and rub you
all over with purified lanolin. It's not really necessary, but it
beats hell out of reading a novel.
David Wipf
Bill Sloman <slo...@sci.kun.nl> wrote in message
<35603293...@sci.kun.nl>...
>
>
> David Wipf wrote:
>
> > Hello,
> > I am considering an experiment that would involve measuring aA
> > levels of dc current (perhaps even down to < 1 aA) in an
> > ...
> > lock in to demodulate the signal. I was thinking that a 0.1 Hz
> > modulation and a 100 or 1000 S time constant might get me down to
> > the low aA level. The source will have an impedence of between
> > 1k to 100k (it depends on the electrode I use)and will be at room
> > temp. Obviously I will shield the system and do my best to prevent
> > interference from the chopping system. I expect that the system
> > will have a background current of 1 to 10 pA and a noise level of
> > around 0.1 pA maximum.
>
> The Johnson noise current on a 100k resistance at room temperature is
> 4^-13A per root Hz, or 400 fA. To get down to 1aA you have to reduce
> this by 400,000, which implies reducing the badndwidth by 1.6^-11, which
> implies a settling time of about 2000 years.
>
> I think you have better think it out again.
>
> Bill Sloman, Nijmegen
>
Could you explain how you calculated a Johnson Noise CURRENT? I think
I can see what you did. The Vnoise is calculated for a 100k resistor as
4x10^-8 V Hz^-1/2. That's fine. Then I think you divided Vnoise by 100k to
get a Inoise of 4x10^-13 A Hz^-1/2. Which you then used to find the desired
BW for the aA level. For comparison that would be 40 pA at a 10kHz BW
This can't be correct. Let's Assume a 1 Ohm resistance. Now Vnoise is
1.3x10^-10 V Hz^-1/2. Using your method, I would get Inoise of
1.3x10^-10 A Hz^-1/2. That would be 13 nA at a 10 kHz BW????
Infinite noise current in superconductors?
Remember that current will be measured in this case flowing into a virtual
ground.
I think Shot Noise is more appropriate.
for 1 aA DC (a lovely concept) I find noise levels of
6x10^-19 A Hz^-1/2. 0.1 Hz would be sufficient.
I don't think that the source impedance is a big problem.
Thanks
Dave
David Wipf
David Wipf
David Wipf
Bill Sloman <slo...@sci.kun.nl> wrote in message
<35603293...@sci.kun.nl>...
>
>
> David Wipf wrote:
>
> > Hello,
> > I am considering an experiment that would involve measuring aA
> > levels of dc current (perhaps even down to < 1 aA) in an
>
David Wipf
Walter Gray <wag...@taz.dra.hmg.gb> wrote in message
<6jp9lr$gcd$2...@trog.dra.hmg.gb>...
> In article <6jn6tr$j...@bgtnsc02.worldnet.att.net>,
notj...@worldnet.att.net (James Meyer) writes:
> : I'd be inclined to use the current to plate out some metal
> :over a *long* period of time and then simply weigh the plated out
> :sample. There is a direct relationship between each atom plated out
> :and the current through the plating bath. Even that would be plenty
> :hard enough to do with any accuracy.
>
> If I haven't lost any powers of 10, this experiment would take
> approx 50 million years to plate out 1 milligram of copper.
>
>
> Walter
>
> Disclaimer: My employer is not responsible for this stuff.
> Please reply to: wag...@dra.hmg.gb as the address given above
> sometimes doesn't work.
Plating out a metal and "measuring" them is an idea that I had. Actually
you can use a Quartz Crystal Microbalance (QCM) to measure subnanogram
amounts. This cuts our aA integration time to under 100,000 years. Still a
bit
long for the average Ph.D. project.
However, (assuming Cu(II) ions) plating copper ions for 10 Hrs gives
1.7 x 10-17 g ( or 0.19 aMol). If I then oxidize the copper metal to Cu(II)
over a
0.1 s time period, a measurable 0.4 pA will flow.
Dave
Mississippi State University
Department of Chemistry
Bret Cannon
Have you considered the problem with the Johnson noise in the feedback resistor
of your current to voltage circuit in your electrometer? A 10^12 ohm resistor
will give you 130 aA/root Hz current noise due to Johnson noise. This
resistance coupled with a capacitance of 10 pF gives RC=0.1 second. This is
about the most you can tolerate if you are going to modulate at 0.1 Hz. If you
want to reach 1 aA at a SNR of 1 you need to reduce the bandwidth to (1/130)^2
Hz = (1/17,000) Hz. This also ignores the whole issue of 1/f noise.
You might want to look at
http://www.national.com/rap/Story/Index/0,1009,0,00.html , which is a collection
of articles written by Bob Pease at National Semiconductor, a couple of which
deal with measuring femtoamp bias currents in amplifiers.
Bret Cannon
David Wipf wrote:
> Bill Sloman <slo...@sci.kun.nl> wrote in message
> <35603293...@sci.kun.nl>...
> >
> >
> > David Wipf wrote:
> >
> > > Hello,
> > > I am considering an experiment that would involve measuring aA
> > > levels of dc current (perhaps even down to < 1 aA) in an
> > > ...
> > > lock in to demodulate the signal. I was thinking that a 0.1 Hz
> > > modulation and a 100 or 1000 S time constant might get me down to
> > > the low aA level. The source will have an impedence of between
> > > 1k to 100k (it depends on the electrode I use)and will be at room
> > > temp. Obviously I will shield the system and do my best to prevent
> > > interference from the chopping system. I expect that the system
> > > will have a background current of 1 to 10 pA and a noise level of
> > > around 0.1 pA maximum.
> >
> > The Johnson noise current on a 100k resistance at room temperature is
> > 4^-13A per root Hz, or 400 fA. To get down to 1aA you have to reduce
> > this by 400,000, which implies reducing the badndwidth by 1.6^-11, which
> > implies a settling time of about 2000 years.
> >
> > I think you have better think it out again.
> >
> > Bill Sloman, Nijmegen
> >
> Could you explain how you calculated a Johnson Noise CURRENT? I think
> I can see what you did. The Vnoise is calculated for a 100k resistor as
> 4x10^-8 V Hz^-1/2. That's fine. Then I think you divided Vnoise by 100k to
> get a Inoise of 4x10^-13 A Hz^-1/2. Which you then used to find the desired
> BW for the aA level. For comparison that would be 40 pA at a 10kHz BW
> This can't be correct. Let's Assume a 1 Ohm resistance. Now Vnoise is
> 1.3x10^-10 V Hz^-1/2. Using your method, I would get Inoise of
> 1.3x10^-10 A Hz^-1/2. That would be 13 nA at a 10 kHz BW????
> Infinite noise current in superconductors?
>
> Remember that current will be measured in this case flowing into a virtual
> ground.
>
> I think Shot Noise is more appropriate.
> for 1 aA DC (a lovely concept) I find noise levels of
> 6x10^-19 A Hz^-1/2. 0.1 Hz would be sufficient.
>
> I don't think that the source impedance is a big problem.
> Thanks
> Dave
>
slo...@sci.kun.nl
In article <6jv3le$smi$1...@nntp.msstate.edu>,
Looks fine to me. Essentially the Johnson noise is a power generated by the
resistor and it is V^2/R or I^2.R.
> For comparison that would be 40 pA at a 10kHz BW
> This can't be correct.
Why not?
> Let's Assume a 1 Ohm resistance. Now Vnoise is
> 1.3x10^-10 V Hz^-1/2. Using your method, I would get Inoise of
> 1.3x10^-10 A Hz^-1/2. That would be 13 nA at a 10 kHz BW????
> Infinite noise current in superconductors?
Just like you get infinite voltage noise in insulators ....
> Remember that current will be measured in this case flowing into a virtual
> ground.
That is what I'd expect in an electrochemical experiment.
> I think Shot Noise is more appropriate.
> for 1 aA DC (a lovely concept) I find noise levels of
> 6x10^-19 A Hz^-1/2. 0.1 Hz would be sufficient.
I imagine that you do think shot noise is more appropriate,
since this happy delusion allows you to persist with your idea, but
conduction exhibits shot noise if the individual charge carriers
don't influence one another, scarcely likely in an electrochemical
experiment.
More to the point, if the current you are measuring does exhibit shot
noise, you have to use a big enough load resistor to develop about
50mV for the shot noise to match the Johnson noise - 5x10^16R in your case -
which is rather higher than the 100k you mentioned.
First get rid of the parallel resistances in your experiment, then you can
start thinking about counting individual ions.
> I don't think that the source impedance is a big problem.
Go back to your thermodynamics texts, and think about what a 100k parallel
resistance at 298K is actually doing.
Sorry this isn't going to help.
Bill Sloman, Nijmegen
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
http://www.dejanews.com/ Now offering spam-free web-based newsreading
Roy McCammon
> > The Johnson noise current on a 100k resistance at room temperature is
> > 4^-13A per root Hz, or 400 fA. To get down to 1aA you have to reduce
> > this by 400,000, which implies reducing the badndwidth by 1.6^-11, which
> > implies a settling time of about 2000 years.
> I don't think that the source impedance is a big problem.
Thats really the crux of issue. If the source impedence
is an ordinary conductor, then its going to behave
like Bill says. It is possilbe to build active
circuits with 100K impedence, but less noise
than a 100K resister. So, just what is the 100K.
Opinions expressed herein are my own and may not represent those of my employer.
Chuck Parsons
> >
> >
> > David Wipf wrote:
> >
> > > Hello,
> > > I am considering an experiment that would involve measuring aA
> > > levels of dc current (perhaps even down to < 1 aA) in an
> >
> > 4^-13A per root Hz, or 400 fA. To get down to 1aA you have to reduce
> > this by 400,000, which implies reducing the badndwidth by 1.6^-11, which
> > implies a settling time of about 2000 years.
> >
> >
> > Bill Sloman, Nijmegen
> >
> Could you explain how you calculated a Johnson Noise CURRENT? I think
> I can see what you did. The Vnoise is calculated for a 100k resistor as
> 4x10^-8 V Hz^-1/2. That's fine. Then I think you divided Vnoise by 100k to
> get a Inoise of 4x10^-13 A Hz^-1/2. Which you then used to find the desired
> This can't be correct. Let's Assume a 1 Ohm resistance. Now Vnoise is
> 1.3x10^-10 V Hz^-1/2. Using your method, I would get Inoise of
> 1.3x10^-10 A Hz^-1/2. That would be 13 nA at a 10 kHz BW????
> Infinite noise current in superconductors?
>
Unfortunately, Bill has given you the real scoop. I have measured aA
currents in CCD detectors, but that was without pAs of baseline current.
In regards to infinite noise current in superconductors;
Infinite noise voltage in insulators?
Many are familiar with placing a capacitor in parallel with a resistor
to reduce high frequency Johnson noise. For sufficiently high
frequencies this always happens because there is always parasitic
capacitance. One finds the average noise energy in the parallel
capacitance is kT/2.
For low resistance sources a similar phenomena occurs, the
series inductance filters the noise current. If you work it out
you will find the average energy stored in the magnetic field
is kT/2.
In the limit of large R stray capacitance prevents a "ultraviolet
catastrophe". In the limit of low R stray inductance prevents
the same thing.
David Wipf
slo...@sci.kun.nl wrote in message <6jvm0a$sdq$1...@nnrp1.dejanews.com>...
>In article <6jv3le$smi$1...@nntp.msstate.edu>,
> "David Wipf" <wi...@ra.msstate.edu> wrote:
>>
>> Could you explain how you calculated a Johnson Noise CURRENT? I think
>> I can see what you did. The Vnoise is calculated for a 100k resistor as
>> 4x10^-8 V Hz^-1/2. That's fine. Then I think you divided Vnoise by 100k
to
>> get a Inoise of 4x10^-13 A Hz^-1/2. Which you then used to find the
desired
>> BW for the aA level.
>> Let's Assume a 1 Ohm resistance. Now Vnoise is
>> 1.3x10^-10 V Hz^-1/2. Using your method, I would get Inoise of
>> 1.3x10^-10 A Hz^-1/2. That would be 13 nA at a 10 kHz BW????
>> Infinite noise current in superconductors?
>
>Just like you get infinite voltage noise in insulators ....
>
>> Remember that current will be measured in this case flowing into a
virtual
>> ground.
>
>That is what I'd expect in an electrochemical experiment.
Well, I don't see nA levels of current noise in my electrochemical
experiments because of the wire resistance, so somewhere we are not speaking
the same language.
In my mind, One can observe the effect of Johnson noise only by using a high
input
impedance amplifier, as the impedance of the amplifier decreases, the
voltage noise decreases because the overall circuit impedance drops.
Likewise, a current generated by
Johnson Noise would be shorted out by a low-impedance amplifier, such as the
virtual ground of a current feedback transducer.
>
>> I think Shot Noise is more appropriate.
>> for 1 aA DC (a lovely concept) I find noise levels of
>> 6x10^-19 A Hz^-1/2. 0.1 Hz would be sufficient.
>
>I imagine that you do think shot noise is more appropriate,
>since this happy delusion allows you to persist with your idea, but
>conduction exhibits shot noise if the individual charge carriers
>don't influence one another, scarcely likely in an electrochemical
>experiment.
>
Electrons crossing a tunneling barrier from the metal electrode to a redox
ionic species seems to fit the bill for shot noise. The redox ions in
solution don't exhibit the type of correlation electrons in a conductor do.
Also, the redox ions are "screened" by a large excess of inert supporting
electrolyte to prevent migration.
The ionic species undergoing electron transfer are moving only by diffusion.
>More to the point, if the current you are measuring does exhibit shot
>noise, you have to use a big enough load resistor to develop about
>50mV for the shot noise to match the Johnson noise - 5x10^16R in your
case -
>which is rather higher than the 100k you mentioned.
>
>First get rid of the parallel resistances in your experiment, then you can
>start thinking about counting individual ions.
>
What's that mean? There aren't any explicit parallel resistances in my
circuit.
See below. As I said, there will be a background current of about 1 pA,
which I will
discriminate against by the chopping method I mentioned. So I am
essentially talking about measuring a 1 ppm change in signal over time.
Certainly, there are "leakage" currents from other impurity ions, etc. but I
know that
that current will not be more than 1 pA.
>> I don't think that the source impedance is a big problem.
>
>Go back to your thermodynamics texts, and think about what a 100k parallel
>resistance at 298K is actually doing.
>
>Sorry this isn't going to help.
> Bill Sloman, Nijmegen
>
>-----== Posted via Deja News, The Leader in Internet Discussion ==-----
>http://www.dejanews.com/ Now offering spam-free web-based newsreading
For comparison, this is the basic circuit
______ Electrode-Solution Interface
| < ____\_____
| < Rs ' '
--- < Cd
- < |--| |--|
--- |_____| |---> to V-ground
- Vs | |
| |___Z___|
|
___
_
.
The Vs value will be about 500 mV,
Rs is the solution resistance. It's value depends on the supporting
electrolyte composition and the electrode geometry. For about a 1 M buffer
solution and the ultramicroelectrode I will use, Rs will be about 100k down
to about 1k. I think that
it most likely will be near 1k in the final experiment but I wanted to allow
for a higher value.
Cd is the electrode-solution "double-layer" capacitance. It's value will be
about 2-10 pF. I am ignoring its effect since we are at DC
Z is the "electron-transfer impedance". Under the conditions I envisage,
the driving force (i.e. the applied potential) will be large enough so that
Z is much smaller than Rs.
Roy McCammon
David Wipf wrote:
>
> ______ Electrode-Solution Interface
> | < ____\_____
> | < Rs ' '
> --- < Cd
> - < |--| |--|
> --- |_____| |---> to V-ground
> - Vs | |
> | |___Z___|
> |
> ___
> _
> .
> The Vs value will be about 500 mV,
> Rs is the solution resistance. It's value depends on the supporting
> electrolyte composition and the electrode geometry. For about a 1 M buffer
> solution and the ultramicroelectrode I will use, Rs will be about 100k down
> to about 1k. I think that
> it most likely will be near 1k in the final experiment but I wanted to allow
> for a higher value.
> Cd is the electrode-solution "double-layer" capacitance. It's value will be
> about 2-10 pF. I am ignoring its effect since we are at DC
> Z is the "electron-transfer impedance". Under the conditions I envisage,
> the driving force (i.e. the applied potential) will be large enough so that
> Z is much smaller than Rs.
I cannot quite figure out what is limiting the current.
Is it the tiny electrodes? The ions can't crowd into them?
I would guess there must be shot noise, but surely the ions have
thermal motions and also therefore give rise to thermal noise.
The ions in solution are heavier than electrons and are perhaps
damped by the electrolyte, so the thermal velocities would be lower
than in a metalic conductor, but there are also far fewer participating
carriers than in a metalic conductor, leading to higher noise. Surely
there must be some thermal noise along with the shot noise.
Roy McCammon
David Wipf wrote:
> For comparison, this is the basic circuit
>
> ______ Electrode-Solution Interface
> | < ____\_____
> | < Rs ' '
> --- < Cd
> - < |--| |--|
> --- |_____| |---> to V-ground
> - Vs | |
> | |___Z___|
> |
> ___
> _
> .
> The Vs value will be about 500 mV,
> Rs is the solution resistance. It's value depends on the supporting
> electrolyte composition and the electrode geometry. For about a 1 M buffer
> solution and the ultramicroelectrode I will use, Rs will be about 100k down
> to about 1k. I think that
> it most likely will be near 1k in the final experiment but I wanted to allow
> for a higher value.
> Cd is the electrode-solution "double-layer" capacitance. It's value will be
> about 2-10 pF. I am ignoring its effect since we are at DC
> Z is the "electron-transfer impedance". Under the conditions I envisage,
> the driving force (i.e. the applied potential) will be large enough so that
> Z is much smaller than Rs.
Please forgive my thick headedness, but its still not clear to me.
Vs is the cell that you want to measure?
Rs is the resistence due to the electrolyte that wets the electrode:
large electrode gives 1K and small electrode gives 100K?
If Z is small, Vs is 500mV, and Rs is 100K, what limits the current
to aA or pA levels?
That thing labeled Electrode-Solution Interface, is that the
electrodes?
slo...@sci.kun.nl
In article <6k1nda$76a$1...@nntp.msstate.edu>,
"David Wipf" <wi...@ra.msstate.edu> wrote:
>
>
> slo...@sci.kun.nl wrote in message <6jvm0a$sdq$1...@nnrp1.dejanews.com>...
> >In article <6jv3le$smi$1...@nntp.msstate.edu>,
> > "David Wipf" <wi...@ra.msstate.edu> wrote:
> >>
>
> >> Could you explain how you calculated a Johnson Noise CURRENT? I think
> >> I can see what you did. The Vnoise is calculated for a 100k resistor as
> >> 4x10^-8 V Hz^-1/2. That's fine. Then I think you divided Vnoise by 100k
> >> to get a Inoise of 4x10^-13 A Hz^-1/2. Which you then used to find the
> >> desired BW for the aA level.
>
> >> Let's Assume a 1 Ohm resistance. Now Vnoise is
> >> 1.3x10^-10 V Hz^-1/2. Using your method, I would get Inoise of
> >> 1.3x10^-10 A Hz^-1/2. That would be 13 nA at a 10 kHz BW????
> >> Infinite noise current in superconductors?
> >
> >Just like you get infinite voltage noise in insulators ....
> >
> >> Remember that current will be measured in this case flowing into a
> >> virtual ground.
> >
> >That is what I'd expect in an electrochemical experiment.
>
> Well, I don't see nA levels of current noise in my electrochemical
> experiments because of the wire resistance, so somewhere we are not speaking
> the same language.
The wire resistance is in series with your electrochemical cell, so you treat
the whole thing as one resistor having a rather higher resistance. In the
experiment you describe below, you say that the electrode interface has
a lower impedance (Z) than the resistance of the electrolyte (1k to 100k),
so it looks as if the electrolyte is the series resistance.
This is a rather surprising situation, and puts a very tight noise constraint
on the source for your 500mV Vs.
> In my mind, one can observe the effect of Johnson noise only by using a high
> input impedance amplifier, as the impedance of the amplifier decreases, the
> voltage noise decreases because the overall circuit impedance drops.
> Likewise, a current generated by Johnson noise would be shorted out by a
> low-impedance amplifier, such as the virtual ground of a current feedback
> transducer.
I don't think that your thought experiments are giving you realistic results.
If you connect one end of a high value resistor to the input of a high
input impedance amplifier whose input impedance is higher than that of the
resistor, the input-referred voltage noise you will see at the output of the
amplifier will be the root mean square sum of the following components, plus
a few others which one hopes can be made negligible by careful design and
construction.
1) The Johnson noise from the test resistor, expressed as a voltage.
2) The input-referred voltage noise of the amplifier
3) The amplifier input current noise multiplied by the test resistance.
None of these values is directly dependent on amplifier input impedance
until the input impedance falls to a value comparable with the test
resistance. Once the input impedance of the amplifier is appreciably lower
than the test resistance, it makes more sense to root mean square sum
noise currents, which will include
1) The Johnson noise from the test resistor, expressed as a current.
2) The input-referred voltage noise of the amplifier divided by the
test resistance.
3) The amplifier input current noise.
Again, these don't depend directly on amplifier input impedance. If the
amplifier's low input impedance has been achieved by setting up an op amp
as a virtual-earth-input current-to-voltage convertor, you have to include
the Johnson noise generated in the feedback resistor.
> >> I think Shot Noise is more appropriate.
> >> for 1 aA DC (a lovely concept) I find noise levels of
> >> 6x10^-19 A Hz^-1/2. 0.1 Hz would be sufficient.
> >
> >I imagine that you do think shot noise is more appropriate,
> >since this happy delusion allows you to persist with your idea, but
> >conduction exhibits shot noise if the individual charge carriers
> >don't influence one another, scarcely likely in an electrochemical
> >experiment.
> >
>
> Electrons crossing a tunneling barrier from the metal electrode to a redox
> ionic species seems to fit the bill for shot noise. The redox ions in
> solution don't exhibit the type of correlation electrons in a conductor do.
> Also, the redox ions are "screened" by a large excess of inert supporting
> electrolyte to prevent migration. The ionic species undergoing electron
> transfer are moving only by diffusion.
Pity. If the charge carriers do influence one another, you get less noise
than predicted for classic shot noise.
> >More to the point, if the current you are measuring does exhibit shot
> >noise, you have to use a big enough load resistor to develop about
> >50mV for the shot noise to match the Johnson noise - 5x10^16R in your
> >case - which is rather higher than the 100k you mentioned.
> >
> >First get rid of the parallel resistances in your experiment, then you can
> >start thinking about counting individual ions.
> >
>
> What's that mean? There aren't any explicit parallel resistances in my
> circuit.
My mistake. But it looks as if you need to increase your series resistance
or impedance by some orders of magnitude.
> See below. As I said, there will be a background current of about 1 pA,
> which I will discriminate against by the chopping method I mentioned. So I
> am essentially talking about measuring a 1 ppm change in signal over time.
> Certainly, there are "leakage" currents from other impurity ions, etc. but I
> know that that current will not be more than 1 pA.
>
> >> I don't think that the source impedance is a big problem.
<snipped snort>
> For comparison, this is the basic circuit
>
> ______ Electrode-Solution Interface
> | < ____\_____
> | < Rs ' '
> --- < Cd
> - < |--| |--|
> --- |_____| |---> to V-ground
> - Vs | |
> | |___Z___|
> |
> ___
> _
> .
> The Vs value will be about 500 mV,
> Rs is the solution resistance. It's value depends on the supporting
> electrolyte composition and the electrode geometry. For about a 1 M buffer
> solution and the ultramicroelectrode I will use, Rs will be about 100k down
> to about 1k. I think that it most likely will be near 1k in the final
> experiment but I wanted to allow for a higher value.
> Cd is the electrode-solution "double-layer" capacitance. It's value will be
> about 2-10 pF. I am ignoring its effect since we are at DC
> Z is the "electron-transfer impedance". Under the conditions I envisage,
> the driving force (i.e. the applied potential) will be large enough so that
> Z is much smaller than Rs.
Such a low value for Z seems very odd. If it is equal to your lowest Rs,
of 1k, a 1uV change in Vs will produce a 1nA change in the current to be
measured, which is improbably large when compared with the 1aA current you
expect to measure or your 1pA leakage current.
1uV changes in voltage can be generated by the adventitious thermocouples
around you circuit under the influence of the sort of local temperature
fluctuations you see in a normal, nominally air-conditioned laboratory, which
is why most low level voltage sensing circuit tend to sport some sort of
draft shield.
Hope your ideas about the circuit are becoming clearer - I think mine are,
but since I don't know exactly what you have in mind I could be deluding
myself.
David Wipf
Roy McCammon wrote in message <35647DFE...@mmm.com>...
>Please forgive my thick headedness, but its still not clear to me.
>Vs is the cell that you want to measure?
>Rs is the resistence due to the electrolyte that wets the electrode:
> large electrode gives 1K and small electrode gives 100K?
>If Z is small, Vs is 500mV, and Rs is 100K, what limits the current
> to aA or pA levels?
>That thing labeled Electrode-Solution Interface, is that the
>electrodes?
>
Sorry for not being clearer originally and in followup what the circuit is.
Vs is a bias voltage that I put on the electrochemical cell. Rs is the
solution resistance and (for identical solution composition) will be larger
for small electrodes.
Rs ~= rho/4r for disk-shaped electrodes with radius r (rho is the solution
resistivity in
ohm-cm).
Z is a NONLINEAR impedance (my fault for not making that perfectly clear).
For a given redox species there is a redox potential where (theoretically at
least) there exists equal activity of the oxidized and reduced form of the
species at the electrode surface. For example, Ferrous and Ferric ion have
equal activities at an electrode biased to 0.77 V vs NHE.
The current that flows is limited by at least two processes. Initially,
there is a kinetic limitation on the rate of electron-transfer from the
electrode to the solution species. The kinetic expression is an exponential
function of the applied potential. Thus at potentials positive to the redox
potential an Oxidized species will not accept electrons at a significant
rate. At potentials negative to the redox potential the rate is very high.
At this point, the rate of mass transport becomes important and limiting.
Species can only be transported to the electrode at a certain rate. In my
experiment mass transport is solely by diffusion and the potentials will be
sufficiently large to maintain a "diffusion-limited" rate of transport.
Thus the exact value of Vs is not critical since small deviations will not
be sufficient to move out of the diffusion limited region.
So what limits the current to attoamp levels? Current that flows on this
diffusion-limited plateau is proportional to the redox species
concentration. So I am interested in very small concentrations of a
particular redox species. (from 1 to 1000 femto molar)
There is a background current in my experiment that arises from impurities
in solution and amplifier bias currents. But the background can be
precisely known for a given experiment.
The label electrode-solution interface refers to the nm-thin layer where all
the action happens in electrochemistry. The ions must diffuse to the
electrode solution interface and, if the potential is right, an electron can
be transfered. Other non-electroactive ions (the vast majority) are
attracted or repelled to the interface, but they cannot give up or accept
charge to the electrode and so they produce a charged electrical layer (the
"double-layer") this region of separated charge (solution and metal) is a
capacitor (thus Cd).
So, I think that my question is this (with preamble)?
I accept that fact that I will have noise, Johnson, Shot, 1/f, interference,
etc.
But since I know already that the noise is about 0.1 pA with a 1 Hz BW under
very similar conditions and the background is about 1 pA (which can be
partially bucked out), can I measure in the region of 1 to 100 aA of
current? Remember that I am chopping the solution concentration of the
desired species at a low rate (10 to 0.1 Hz) and using lock-in
amplification.
I think I can see now the problem I will have with 0.1 pA of noise. Getting
to 1 aA means that I need to reduce S/N by a factor of 100,000 and thus BW
is reduced by 10^10.
or 10^-10 s. To have a reasonable chance, I need to reduce my noise by a
factor of 30 (I think a factor of 10 is doable since I haven't really
exerted myself yet on that) and make my goal 30 aA. Now we are talking 10^-5
still a big BW reduction but just possibly.
Am I missing anything here? I appreciate all the tips for noise reduction,
note that I haven't really used any of them yet, and I am getting pretty low
noise already.
Thanks for all who replied, It really helps to stretch my mind and work out
the problem.
Daniel Lang
David Wipf <wi...@ra.msstate.edu> wrote in article
<6k7f1h$264$1...@nntp.msstate.edu>...
Shouldn't this be Z is much larger than Rs?
Otherwise your currents will be in the microamp range.
> >
> >Please forgive my thick headedness, but its still not clear to me.
> >Vs is the cell that you want to measure?
> >Rs is the resistence due to the electrolyte that wets the electrode:
> > large electrode gives 1K and small electrode gives 100K?
> >If Z is small, Vs is 500mV, and Rs is 100K, what limits the current
> > to aA or pA levels?
> >That thing labeled Electrode-Solution Interface, is that the
> >electrodes?
> >
>
> Sorry for not being clearer originally and in followup what the circuit
is.
> Vs is a bias voltage that I put on the electrochemical cell. Rs is the
> solution resistance and (for identical solution composition) will be
larger
> for small electrodes.
> Rs ~= rho/4r for disk-shaped electrodes with radius r (rho is the
solution
> resistivity in
> ohm-cm).
> Z is a NONLINEAR impedance (my fault for not making that perfectly
clear).
What is a typical value for Z?
Bill Sloman's point about the thermal noise is still valid but you need to
use Rs + Z, not just RS. If Z is sufficiently large (> 10**12 ohms),
attoamp accuracy may still be possible.
It also looks like the background current will contribute shot noise.
You will need to take this into account also.
> The label electrode-solution interface refers to the nm-thin layer where
all
> the action happens in electrochemistry. The ions must diffuse to the
> electrode solution interface and, if the potential is right, an electron
can
> be transfered. Other non-electroactive ions (the vast majority) are
> attracted or repelled to the interface, but they cannot give up or accept
> charge to the electrode and so they produce a charged electrical layer
(the
> "double-layer") this region of separated charge (solution and metal) is a
> capacitor (thus Cd).
Currents in the attoamp range can be measured in CCD detectors.
Of particular interest is the skipper amplifier developed by Jim Janesick.
A normal CCD sensor has an output well (gate of a mosfet) that is reset to
a fixed voltage, then the output voltage of the mosfet is measured. Next,
the charge to be measured is transferred to the output well and the output
voltage is measured again. Finally the signal is obtained by taking the
difference of the two measurements. The output well is then reset again
for
the next measurement. The procedure is called correlated double sampling
(CDS).
In the skipper CCD, after the first two measurements, instead of resetting
the output well, the charge is transferred into a holding well and the
output is measured again. The charge can be shuttled repeatedly between
the output well and the holding well repeatedly (typically 256 times).
This method has been used to measure charge with an error of less than
1 electron. Needless to say, the CCD must be cooled to -75 or -100 C.
Daniel Lang d...@hydra0.caltech.edu
James Meyer
On 24 May 1998 09:42:00 GMT, "Daniel Lang" <d...@hydra0.caltechx.edu>
wrote:
>In the skipper CCD, after the first two measurements, instead of resetting
>the output well, the charge is transferred into a holding well and the
>output is measured again. The charge can be shuttled repeatedly between
>the output well and the holding well repeatedly (typically 256 times).
>This method has been used to measure charge with an error of less than
>1 electron. Needless to say, the CCD must be cooled to -75 or -100 C.
>
>Daniel Lang d...@hydra0.caltech.edu
That's all well and good for charges that are developed
internally to the CCD by photo-electric means, but how would one
inject those charges from some external current source?
I'm still waiting for someone to suggest SQUIDS.
Jim