electrometer front end for dmm

Bit of a flaw in the logic. When the top diode is conducting in the forward direction, the bottom diode is conducting leakage current in the reverse direction, and vice-versa. So there is no zero current point somewhere between.

You already have the setup to measure leakage current. It is in an aluminum box you posted to dropbox. Instead of theorizing, why not measure it?

Reply to
Steve Wilson
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Obviously, that would be the way to go. But you missed whit3rd's statement about thermal coupling from the output stage back to the input.

Reply to
Steve Wilson

OK, I found the Bob Pease page. It's at

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There is a series of chapters on common mode which is not included in the above list. It is "Bob Pease Lab Notes 2005", at

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The referring page is

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It has a wealth of information, including Jim Williams Lab Notes at

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Reply to
Steve Wilson
[...]

More Bob Pease stuff. Here's an article from 1980 showing how to make stable millivoltmeters when the inputs are open, constant voltage drop picoammeters, and a bunch of other stuff. It is written around the LM11, but the same techniques could be applied to the LMC660/LMC64482 series parts. This is the first article I have come across that features Bob Pease and Bob Widlar as coauthors.

Applying a New Precision Op Amp

National Semiconductor Application Note 242

Robert J. Widlar Bob Pease Mineo Yamatake April 1980

Abstract: A new bipolar op amp design has advanced the state of the art by reducing offset voltage and bias current errors. Its characteristics are described here, indicating an ultimate input resolution of 10 uV and 1 pA under laboratory conditions. Practical circuits for making voltmeters, ammeters, differential instrumentation amplifiers and a variety of other designs that can benefit from the improved performance are covered in detail.

Methods of coupling the new device to existing fast amplifiers to take advantage of the best characteristics of both, even in follower applications, are explored.

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Reply to
Steve Wilson

"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...

The statement is true, but the practical significance is lacking.

Specifically, a linear difference in leakage currents corresponds to an exponential difference in voltages. So, trying to find the null might require +1/-30V supplies, or something like that.

All CMOS chips, when left to their own devices, will not actually have their inputs floating willy-nilly. They'll eventually get near a rail, because the difference in currents is not quite zero. It may require elevated temperature for this to become obvious (if the leakages are well matched already).

(And of course, the practical significance of this is small, because in the presence of not very much AC electric field, CMOS inputs can be biased back into the linear region where logic levels are unknown and current draw spikes.)

The only possible way to get practical supply voltages with balanced leakage is to thermally control one or both diodes with some sort of servo. Ironically, that's a hard one to do monolithically, unless you can cantilever your diode out on a MEMS structure.

Tim

--
Seven Transistor Labs 
Electrical Engineering Consultation 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

The statement ignores diode leakage.

Reply to
Steve Wilson

Reiterating ones' ignorance does not a statement disprove. :-)

Diode current is a continuous function of diode voltage. There is no magical discontinuity in the physical world where current switches over from what we call "conduction" to what we call "leakage". Those are just conventional labels for I > 0 and I < 0. Even if there were such a change, it wouldn't make any difference to the present case (assuming the laws of thermodynamics are present).

Tim

--
Seven Transistor Labs 
Electrical Engineering Consultation 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

You exclude zero in your paragraph.

There is a magical discontinuity at zero. In the forward direction, the current increases exponentially with voltage according to the diode equation:

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In the reverse direction, the leakage current is virtually independent of voltage up to the point where the diode breaks down:

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These are two different mechanisms.

The statement only talks about forward conduction. If the leakage was zero, the zero current point would be anywhere between VCC and Gnd.

Reply to
Steve Wilson

And this explains why you cannot set the bias current to zero in the LMC660/LMC64482 series parts. The diode leakages for the top and bottom diodes add together. However, they act as current sources, so they are independent of voltage.

This means changing VCC and VEE or changing the voltage at the input to the op amp will have no effect on the input bias current.

The reason this is true is because the actual bias current into the mosfet junction is close to zero, as Paul Grohe stated. So the only currents involved are the leakage currents into the protection diodes.

I have answered the questions I asked earlier. The mechanism that creates the input bias current is diode leakage. The currents flow to VCC and gnd.

Reply to
Steve Wilson

Can you please show me where this occurs in the following equation? :-)

I = Is * (exp(Vf/Vth) + 1)

Tim

--
Seven Transistor Labs 
Electrical Engineering Consultation 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

Isn't it just the reverse leakage (saturation) current of a diode? And they'd want to make that flat over the whole range. Right till the doide starts to turn on, and for pA that's a small voltage... less than 100mV at a guess.

Some silly ideas. What about temperature? maybe there's a reasonable temp where the current is minimal. or perhaps a temperature gradient across the inputs?

You could try balancing the current. How much does a reversed biased LED leak? What's a low leakage diode?

George H.

Reply to
George Herold

The gate current of some pHEMTs (e.g. the SKY65050) changes sign at a small negative bias. A disconnected gate will spontaneously go a couple of hundred mV below the source, and then stop.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

How does that work? One is a positive current, and the other is negative.

If the diodes are closely matched, the leakage should be close as well. Second order effects such as bias voltage might very well be enough to make them match exactly.

To leading order.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

It's minus one. Otherwise you have a magic battery.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

t impedance of a typical dmm is too low (10 Megaohm or so). Any suggestion s on a simple electrometer adapter for a dmm? I can't afford a new electro meter so I was hoping I could make a simple circuit to add to a dmm for the se measurements.

If your currents are in the fA range, you are in another regime than a pyroelectric crystal that puts out more power. Assuming the former, Paul Gr ohe and Bob Pease ("whats all this femtoampere stuff anyhow?") have examples of what you must do to get a measurement. Here is a list of what they do:

-space wiring if possible (TI datasheet actually suggests this.)

-scrub any insulators with solvent. (Grohe has a video on this.)

-use only polystyrene for insulators - epoxy too conductive.

-teflon has charge storage problems - it's too good.

Pease went to a *lot* of trouble. For an easy solution, space wire an elect rometer IC from TI. See if the crystal you are using has enough capacitance to hold a signal on the input of the fA leakage IC.

techlib.com has some interesting circuits using the 4117 jfet (obsolete.) B ut I think the overall design idea here is that you use an integrating cap as feedback on the 6001 op amp to persist a transient signal, if you must. And if you experimental set-up allows continuous current source, it can just trick le charge a low leakage cap. Measurements in the attoampere range are done thi s way.

I wouldn't even think of PCB's or insulators though. Too leaky.

jb

Reply to
haiticare2011

put impedance of a typical dmm is too low (10 Megaohm or so). Any suggesti ons on a simple electrometer adapter for a dmm? I can't afford a new elect rometer so I was hoping I could make a simple circuit to add to a dmm for t hese measurements.

Grohe

of

ctrometer IC from TI. See if the crystal you are using has enough capacitan ce to hold a signal on the input of the fA leakage IC.

But I

nd if

ckle

his way.

And of course, shield the circuit in a metal box, like an altoids tin. Electrometers will pick up fields in the room.

Reply to
haiticare2011

Oh - and another solution is to look for a Keithley 610BR electrometer on ebay. This is a tube version, and is under $100 usually. jb

Reply to
haiticare2011

And another thought to make things complete: IF you are in the fA regime, try to avoid an integrating cap as feedback on an op amp. This is because switches to zero them are leaky. Pease, who was a better circuit than I'll ever be, used a mechanical push-rod to zero his feedback cap.

Charles Wenzel at techlib.com has a possible way around this. He puts a 2n4117 jfet as an input stage on a capacitance. (est. 5 pF of a ion chamber.) This is connected to a simple DC amplifier, and read by a picaxe ADC at intervals.

The 4117 has very low gate leakage.

And here's the cute trick: When he wants to zero the capacitor (could be your xtal), he just brings the source low using the picaxe, and the inbuilt diode in the jfet just shorts the cap to ground. If your xtal needs more capacitance, a 5 pF PS to ground might suffice. If, OTOH, your signal is truly transient, then maybe his scheme could be used to switch (short out) a feedback cap o an interator.

I built and ran the circuit, which worked fine, except I should have space-wired more of it, so was beset by leakage.

The whole enchilada is on techlib.com

jb

Reply to
haiticare2011

You can do it by briefly inverting the power supplies with a mux and a couple of current-limiting resistors. The protection diodes work like a sampling bridge. I forget whose idea this was--it's brilliant. It's similar to the one you give below, but better since it doesn't have the diode drop uncertainty. You don't get back to exactly zero, of course, so you have to measure the reset value.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Interesting, thanks.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

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