Leakage current detection (>100 nA, high voltage)

Thread Starter

Blimb

Joined Oct 19, 2017
12
Hello everyone!

I need to build a leakage current detector for a variable source voltage between -100 and 100 VDC.

I am using a -100 to 100 voltage DC source as a gate voltage for a home-built FET, that should in principle not source any current. However, as some of my FETs are broken, some will have a leakage current to the ground. I would like to be able to sense this current. It is not necessary to have a value, just a 0 V or 5 V output for no leak (<100 nA) or leak (>100 nA) current, respectively. (choosing the cutoff point would be good too)

I think I would need a high-side current sense amplifier (as it is leaking, I do not have access to the ground side), but I am not sure which component I could use. Also, I am unsure about what to do with the negative voltages, because I need to be able to sense the current over the entire range of -100V to 100V.

It would be great if you could point me in the right direction.

Cheers,

Bart
 

ebeowulf17

Joined Aug 12, 2014
3,139
Hello everyone!

I need to build a leakage current detector for a variable source voltage between -100 and 100 VDC.

I am using a -100 to 100 voltage DC source as a gate voltage for a home-built FET, that should in principle not source any current. However, as some of my FETs are broken, some will have a leakage current to the ground. I would like to be able to sense this current. It is not necessary to have a value, just a 0 V or 5 V output for no leak (<100 nA) or leak (>100 nA) current, respectively. (choosing the cutoff point would be good too)

I think I would need a high-side current sense amplifier (as it is leaking, I do not have access to the ground side), but I am not sure which component I could use. Also, I am unsure about what to do with the negative voltages, because I need to be able to sense the current over the entire range of -100V to 100V.

It would be great if you could point me in the right direction.

Cheers,

Bart
You've made FETs that can handle +/-100V gate voltage?! I thought 20-30V was the usual limit.

Maybe I'm just thinking of MOSFETs, or just smaller ones I've had experience with.
 

wayneh

Joined Sep 9, 2010
16,228
The only challenging part (for me, because I don't know any better), is the 0.1µA requirement. It's pretty easy to detect a very small current using a low-ohms shunt resistor and a comparator to watch the voltage across that shunt. If the voltage rises above some level, the comparator triggers and there you go.

But 100nA across a 1Ω resistor (and that's a high ohms value for a shunt) is only 100nV. It's pretty easy to trigger on the swing of, say, 1mV. I suppose you could use another op-amp to amplify the tiny voltage across the shunt.

It would take some gyrations to power the op-amp and comparator circuits and reference them properly but I think it's doable.
 

Thread Starter

Blimb

Joined Oct 19, 2017
12
You've made FETs that can handle +/-100V gate voltage?! I thought 20-30V was the usual limit.

Maybe I'm just thinking of MOSFETs, or just smaller ones I've had experience with.
They're graphene FETs, its just a function of the size of the dielectric (SiO2) between the gate and the graphene.

The only challenging part (for me, because I don't know any better), is the 0.1µA requirement. It's pretty easy to detect a very small current using a low-ohms shunt resistor and a comparator to watch the voltage across that shunt. If the voltage rises above some level, the comparator triggers and there you go.

But 100nA across a 1Ω resistor (and that's a high ohms value for a shunt) is only 100nV. It's pretty easy to trigger on the swing of, say, 1mV. I suppose you could use another op-amp to amplify the tiny voltage across the shunt.

It would take some gyrations to power the op-amp and comparator circuits and reference them properly but I think it's doable.
In principle, the resistance to ground should be very large (infinite?) unless the device is defective. Therefore, I don't need a 1Ohm resistor as shunt, but I think I could just use 1 kOhm or even more (is there any reason not to if I don't have to source any current?). If there is a current leak I would not need the output voltage to be precise.

The problem I'm having is that I need the current detection to be high-side, because it is leaking to ground where I cannot detect it.

I was thinking on something like this: http://tinyurl.com/y9j9ognj (which is probably very naive, because I'm not very knowledgeable about electronics!)

I put a 1G load in there just for demonstration, and used a triangle on the voltage source just to show what happens at different voltage. In this design I don't know how to correctly power the opamps (can I just use a HV opamp on the first, and a 5V rail to rail driven on the second?), and it (obviously) doesn't work on the negative voltages. How can I solve these issues?
 

Thread Starter

Blimb

Joined Oct 19, 2017
12
Just a concept (the transistors are selected for low leakage -below 2 na):
View attachment 146080

Edit: That rectangle between the two capacitors is a very low quiescent current -5V regulator, not labeled and not drawn properly.
Thank you for your reply. I don't quite understand how the circuit is supposed to work, so I am having trouble knowing which resistor/zener values to use. Also, is the output value precisely 10V (I guess 2x zener) below the supplied voltage? Could you give me some more information (or a link to an explanation?)

Thank you so much,

Bart
 
A properly designed I-V converter should work. You can use a T-network to reduce the value of the resistors.
See: https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/lecture-notes/23_op_amps2.pdf

Triax cables would likely be mandatory. These have two shields. The outer being ground and the inner being guard. Guard is driven at the potential that is being sourced, so there is effectively no leakage between what your measuring and ground.

Your 100 V above ground and that might create some additional challenges unless you make it battery operated.

I don;t know of any easy sources for high value resistors. You can;t touch them with your hands and they usually have a glass envelope.

I did at one time build an I-V converter that was 10 uA full scale at +-10v out. It was designed at the other end of the scale though,

Work required me to set up a measurement system that did work at +-100 V with 4 switched samples as a function of temperature form -80C to +200 C.

When I had to measure really low currents to effectively "prove" that an etched gap was etched completely, I measured pico, nano, micro-coulombs over time and divided my the time to get current.

There's a lot of gotcha's in measuring low currents. Your additional glitch is a high side measurement.

The components used were from Keithley Instruments, now part of Tektronix. There's a low level measurements handbook which should be free and available online that you need to read.
 

DickCappels

Joined Aug 21, 2008
6,175
Thank you for your reply. I don't quite understand how the circuit is supposed to work, so I am having trouble knowing which resistor/zener values to use. Also, is the output value precisely 10V (I guess 2x zener) below the supplied voltage? Could you give me some more information (or a link to an explanation?)

Thank you so much,

Bart
First of all, this is not a schematic of a working circuit. It is only meant to show a possible circuit topology that can enable the detection of currents below 100 nanoamps when a load is driven by +100V and -100V.

A practical circuit might include features to protect the transistors from excessive voltage among others.

The +100V will be explained noting that the same operation occurs in the -100V circuit except the polarities are reversed.

The circuit makes use of two main principles: 1) Q1's collector current is a good indication of its emitter current, and 2) the comparator circuit and the optoisolator can be "floated" on the 100V power supply so the power supply needs of the circuit do not interfere with the leakage current being measured.

Voltage at the emitter of Q1 will be a fraction of one volt below the base voltage, so the voltage at the base should be set to achieve the desired test voltage.

The load resistor on the collector of Q1 will show a voltage drop that is proportional to the amount of current in the load plus a little extra extra from a few nanoamps from the transistor's collector cutoff current and the input bias current of the comparator.

The comparator drives an optical isolator to signal when the voltage across the resistor is above or below a chosen threshold.

The transistors used in this circuit will probably have to be hand selected for acceptable leakage current. The specifications for the transistors mentioned is 2 na to 100 na at 25°. I have checked a few of each type and find they are wonderfully close to 2 na and that is without any special preparation.

If you do not need an electrical signal to indicated when a particular threshold is exceeded, you can float a common digital voltmeter at +100V and -100V (values assigned only to note a reasonable range of voltages). Using a voltmeter with an input resistance of 10 Meg Ohms gives you sensitivity of 10 millivolts per nanoamp, so read the current directly from the meter, after testing and any necessary alignment. A capacitor across the meter's input terminals may be needed to block AC hum from disturbing the reading.

KeepItSimpleStupid offers some very good sounding advice in his post #7 and should be taken into account when designing the actual fixture. I hope you can do this without cabling because that would make your life a lot simpler. Similarly, working in a controlled environment to minimize thermal effects is a great idea.

You can get a lot of insight and inspiration into the tricks, techniques, and hazards of measuring low currents by reading the Bob Pease articles What's All This Teflon Stuff, Anyhow? and What's All This Femtoampere Stuff, Anyhow? which can be found online or in this collection of Pease articles on TI's website
https://www.ti.com/ww/en/bobpease/assets/www-national-com_rap.pdf
 

Thread Starter

Blimb

Joined Oct 19, 2017
12
Thanks alot both for your answers. Seems like it is going a little bit over my head at the moment, although it is very interesting and I will definitely read up on the things you have shared.

Would my life become considerably easier if I would increase the threshold current that I would like to measure to 1 µA? 10 µA? Basically.. I would just like to have a way to identify faulty devices automatically by the computer by means of a digital output signal.
 
That's definately a lot easier. With a 1 M resistor in the feddback path of an OP amp, you would get 1V out at -1 uA. The I-V converter produces inverts the output.

The biggest specs you have to worry about is Vos and Ib for the I-V converter. They need to be low. The transfer function is Vout =-I*Rf

Although I haven't used the LT6700 series http://www.linear.com/product/LT6700 they seem to be nice comparitors. They are available in single comparators too.
 

DickCappels

Joined Aug 21, 2008
6,175
Thanks a lot both for your answers. Seems like it is going a little bit over my head at the moment, although it is very interesting and I will definitely read up on the things you have shared.

Would my life become considerably easier if I would increase the threshold current that I would like to measure to 1 µA? 10 µA? Basically.. I would just like to have a way to identify faulty devices automatically by the computer by means of a digital output signal.
Getting in over your head is a good thing, unless it puts your project to risk.

Yes, the larger the threshold current the easier things will be.
 

Thread Starter

Blimb

Joined Oct 19, 2017
12
Thank you both again.

Is there any way to 'float' the opamp without having a voltage drop on the output? Also, is there a possibility of having the circuit work on the negative side as well, without the need to switch to a different circuit? If that will get too difficult then I can probably work around this, but it would be nice to not have to worry about differences in output voltage.

Can I just use a classic current sense circuit and have a zener connected from V+ on the opamp to 'V-', and a high resistor from 'V-' to gnd? If I use a rail-to-rail opamp it would give me a voltage between V+ and (V+ - zener) which I could then connect to a comparator and optoisolator to get a 5+ V signal out, is that right? Should I worry about anything there?

Bart
 
Is there any way to 'float' the opamp without having a voltage drop on the output?

Don't know if an isolated DC-DC converter can likely be used. These https://www.picoelectronics.com/node/13277 offer 100 M ohms @ 500V. That's the problem with high side stuff. ~ 100V, it should be 500 M ohms. That's 200 nA which is too much.
The I-V converter OP-amp approach does not have an appreciable drop of voltage when measuring. < 1 mV is achieveable. It's not dependent on input voltage. It must, however be able to source/sink the current measured if you want bi-polar output.

Battery operated is a possibility.


Also, is there a possibility of having the circuit work on the negative side as well, without the need to switch to a different circuit? If that will get too difficult then I can probably work around this, but it would be nice to not have to worry about differences in output voltage.
The I-V converter just INVERTS the measured current. The voltage drop is effectively Vos of the OP amp. The current Vos/Rf is an error term. Ib is an error term.

Say you selected a +-5V supply for the op amp and the output switch you needed was >1 V or <-1 V, you would just set two comparitors for that and OR (probably a Wire OR) the outputs. You can opto isolate the outputs.
 

Thread Starter

Blimb

Joined Oct 19, 2017
12
Okay, perhaps I am misunderstanding. How can I make a high-side IV converter? I thought an IV converter replaces the low side (ground) of the load by the - input of an opamp (+ grounded, feedback resistor from output to -). I don't have a ground connection (it is leakage current) so I cannot use that. If you mean something else, I would appreciate it greatly if you could supply a sample circuit :)

Bart
 
The "ground" of the I-V converter is the "common" of the I-V converter. In other words, what the output is referenced too.

It does not have to be connected to +-100 V ground, although it's better there. The power supply you use to power it is going to have a leakage spec and an isolation spec. A battery won't have one.
A keithley 480 pico ammeter can be floated 30 V above ground. Your power supply has to be able to be floated 100 V above and below ground with some safety margin.

Your best bet is likely a DC-DC converter. here https://www.trcelectronics.com/ecomm/pdf/mgfs15.pdf are some with a 1000 M ohm isolation.

So, you need a conevntional mains supply and a DC-DC converter to power a high side I-V converter.
the output is relative to common of that converter, so it's above your ground.
So, an opto isolator will get you an earth referenced logic output. The 4N25 https://www.noisebridge.net/images/1/17/4N26.pdf has an isolation resistance of 1e11 ohms.

So, you have an effective 1000 M || 1e-11 ohms plus all of the construction techniques. You may have to wire directly to the IC pins or use teflon iposts, guard terminals etc.

Just think like, if you took a battery powered multimeter and measured the output of the IV converter. It doesn;t matter where in the circuit the ammeter is placed. Low currents has different error sources and that's any other leakage paths. The actual leakage current is dependent on the potential. there will be much less when the potential is at ground as to 100 V.

BTW, I-V converters also have a capacitor across the feedback resistor. It limits the bandwidth and makes better stability. Capacitive loads sometimes cause problems. I used an LT1010 application note to fix that in my application.

The diagrams with the regular and T-network show the basic topology. Bypass caps and feedback caps aren;t shown.
 

Thread Starter

Blimb

Joined Oct 19, 2017
12
I'm sorry, but I don't quite understand what you're trying to tell me. I dont understand what a high side IV converter looks like. Could you perhaps show it with a simple circuit diagram?

Thanks for your continued help

Bart
 
In the link in this thread (this set to open ton page 7), https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/lecture-notes/23_op_amps2.pdf#page=7

there is the basic I-V converter.

The schematic shows a ground connection. it's the common for the converter, not earth. You just insert into a wire where I(in) is shown.
The inverting input to the op amp to the +-100 Volt source and (the "ground/common") connected to your device.

Done deal.

With a bi-polar supply to the OP-AMP, the output will be bi-polar and inverting.

it doesn't show the details of a good design or what you have to look for, but the basics are there.

Picture a device that uses two 9V batteries and has two binding posts. One connected at "ground" and the other to the inverting input.

You can put that anywhere. The high or low side. Just like you can a portable multi-meter. Loose the notion that the ground symbol is Earth. It's the 0 volt reference voltage that we defined for the I-V converter.

I'm not necessarily advocating that method because of the currents involved.
 
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