Maker Pro
Maker Pro

Measuring sub picoamp DC by electrical or chemical means

B

Bill Penrose

Jan 1, 1970
0
Integration is a common way to amplify small unidirectional
currents. The process adds much less noise than a high
value feedback resistor in a current to voltage converter.
The trick is producing a low leakage current reset
mechanism. A glass encased reed relay is one of those ways.

Do you have a link (or a paper reference)? I'm interested.

DB
 
P

Paul

Jan 1, 1970
0
Don't you plan to use a current follower?http://en.wikipedia.org/wiki/Current-to-voltage_converter
The input impedance is close to zero, provided you can reference the
signal to ground. There are op amps with <15 fA input bias current.
The output voltage is equal to the product of the input current x the
feedback resistor.


You are talking about the preamp, correct? The preamp cannot have
appreciable voltage offset or bias current relative to the voltage or
current levels one is trying to measure. So what I was saying is that
my 0.5 uV fed into the INA116PA op-amp would cause 5E-22 amps of
current due to the op-amps extremely high resistance. So the highly
sensitive INA116PA op-amp could not detect 0.5 uV. I think Keithley
did an amazing job, as the Keithely electrometer you pointed out can
measure down to 200uV.


The hard part is getting the very high value resistors. The even
harder part is defeating all the leakage currents and stray noise
sources.


A MOSFET would do trick with their ~ 10T ohms and higher resistance
from base to source. I believe an IRF540 has close to 9T ohms from
base to source.
 
P

Paul

Jan 1, 1970
0
Dear Paul:





Just wondering, could you put the current into a small capacitor
at the input of one opamp (forming a differential), and take that
output and put it into the input of an opamp rigged to integrate?
That way, the input impedance would not be an issue. You'd just
need to short the input cap every so often (while isolating the
input to the integrator).

David A. Smith




Integration is a common way to amplify small unidirectional
currents. The process adds much less noise than a high
value feedback resistor in a current to voltage converter.
The trick is producing a low leakage current reset
mechanism. A glass encased reed relay is one of those ways.- Hide quoted text -

- Show quoted text -


Thanks David and John. That's actually a great idea, and applicable to
this design. I've never tried that method. Google patents has an
interesting patent describing such a method-- Figure 2. They use a FET
to reset the cap, but I'd imagine John's idea of using a glass encased
reed relay would help reduce noise.

http://www.google.com/patents?id=9fAAAAAAEBAJ&dq=6011252


Regards,
Paul Lowrance
 
B

Benj

Jan 1, 1970
0
Paul said:
I'm trying to think of methods that may detect ~ 1/2 picoamp DC
without applying any appreciable *bias* current.

I have a passive analog amp meter that with the aid of a microscope
can accurately detect down to 100 pA. Close, but no cigar.

Any ideas is greatly appreciated.

The first idea is to go talk to a real electronic engineer and explain
your system to him!

Look. There is much MORE to "1/2 picoamp DC" than just current! One
needs to know the voltage and the source impedance of your equipment.
For example there are many Keithly meters that can measure extremely
low currents. However they ONLY work with relatively high voltage
sources. They use very large (high value) precision resistors to get a
reasonably measurable voltage from a tiny current. But in the
measuring range they have to develop a decent fraction of a microvolt
as an output signal. Some sources don't have this voltage available
and the whole thing fails.

The best way is probably to use a MOSFET (the kind with no protection
diodes!) and then simply use your source to charge the input
capacitance of the device as someone already suggested. The slope of
the charging voltage gives the current. If it's too sensitive you can
add input capacitance in parallel which increases the known accuracy.
Just just a manual discharge.

Get help.
 
B

Bill Penrose

Jan 1, 1970
0
Note that with a current input instead of a voltage input
you don't need R1.

Thanks, David and John. I'm familiar with the concept, but never
thought of it in terms of measuring charge.

DB
 
P

Paul

Jan 1, 1970
0
The first idea is to go talk to a real electronic engineer and explain
your system to him!

Look. There is much MORE to "1/2 picoamp DC" than just current! One
needs to know the voltage and the source impedance of your equipment.
For example there are many Keithly meters that can measure extremely
low currents. However they ONLY work with relatively high voltage
sources. They use very large (high value) precision resistors to get a
reasonably measurable voltage from a tiny current. But in the
measuring range they have to develop a decent fraction of a microvolt
as an output signal. Some sources don't have this voltage available
and the whole thing fails.

The best way is probably to use a MOSFET (the kind with no protection
diodes!) and then simply use your source to charge the input
capacitance of the device as someone already suggested. The slope of
the charging voltage gives the current. If it's too sensitive you can
add input capacitance in parallel which increases the known accuracy.
Just just a manual discharge.

Get help.



Thanks for the tips. I'm having success simulating the LMC6001A op-
amp.

Paul Lowrance
 
D

dlzc

Jan 1, 1970
0
I'm trying to think of methods that may detect ~ 1/2 picoamp DC
without applying any appreciable *bias* current.

I wonder if, since you are talking about only 10^7 electrons, that you
could not apply this to a heated vacuum tube plate and *count* the
durned things?

David A. Smith
 
F

Fred Kasner

Jan 1, 1970
0
dlzc said:
I wonder if, since you are talking about only 10^7 electrons, that you
could not apply this to a heated vacuum tube plate and *count* the
durned things?

David A. Smith

First refer to the "Johnson Noise Limit".
I worked with DC circuits (for measuring very small amounts of thermal
changes) about a factor of 10 above that level. It took a nastily long
time for the system to get to a relatively quiet equilibrium before the
small emf produced in the DC circuit could be detected above the noise
level.
FK
 
M

Martin Brown

Jan 1, 1970
0
I would like to try a Keithley electrometer. Although one issue is the
input impedance. For example the Keithley 616 electrometer has "INPUT
IMPEDANCE: Greater than 2 x 10^14 ohms." Therefore, my 0.5 uV would
generate 0.5uV / 2E+14 = 2.5E-21 amps. So unfortunately there's no
chance of the Keithley electrometer detecting my 0.5 uV, which would
generate 2.5E-21 amps in the Keithley 616.

I think from what you have described you want the sort of current to
voltage converter that is typically used in mass spectrometry in
conjunction with Faraday cups to measure ion beam currents. ISTR the
ones I worked with were usually setup to give 1v out for 10pA current
in (but it was a long while ago) and then measured on Solartron 7060
voltmeters at 6x9s precision 1s integration under computer control.
The amplifiers had to be thermally managed and heavily insulated (and
they lived in a vacuum) - thermal gain calibration and baseline drift
was always an issue.

Breif description of MS hardware at: http://www.onafarawayday.com/Radiogenic/Ch2/Ch2-3.htm

They were good to sub-picoamp at around 10^-13A but needed careful
handling. The 10^11ohm resistors were exotic temperamental from batch
to batch and there had to be fancy feed forward compensation to allow
the amplifier to cope with transient signals reliably. I expect a
search of Faraday amplifier will get you some patents. Some of the
tricks were trade secrets - though you could try asking on
sci.electronic.design.

Regards,
Martin Brown
 
M

mike

Jan 1, 1970
0
John Popelish said:
Be sure to get one that has a magnetic shield over the coil, so you don't
induce voltage spikes in the input circuit when the coil switches.

Relays with the ends of the glass capsule sticking out of the coil forms
used to be common, but I don't see many, any more. You may have to buy
the switches and wind your own coils around them. Got any 40 AWG wire?

http://www.hamlin.com/specsheets/MDCG-4.pdf
Do you know of a distributor for these reed switches?
Mike
 
M

mike

Jan 1, 1970
0
John Popelish said:
mike said:
Do you know of a distributor for these reed switches?

http://www.digikey.com/
They list 409 hits for the search parameter [reed switch], including some
of the ones shown in that data sheet
I'll take a look, I liked the .2pf capacitance, I'm assuming that's across
the contacts open circuit?
Mike
 
Top