# Best way to measure tiny currents

Discussion in 'General Electronics Chat' started by DedeHai, Jan 22, 2009.

1. ### DedeHai Thread Starter Active Member

Jan 22, 2009
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Hi everyone

A general question:

How do you measure really tiny currents accurately? I want to build a circuit that gives me a voltage that is proportional to a current in the range of e few micro Amps. So lets say the minimal current is 0.5μA.
I thought about taking a shunt resistor and amplify the voltage, but that would probably result in a lot of noise.
Any suggestions?

2. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Depends. If the current is going to ground and is mostly DC a simple op amp will do it. Something like this...

The resistor in this circuit is a multiplier. So if you have a 100KΩ resistor the output will be 1V/10µA.

3. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
In theory the insertion resistance of a curent meter is zero. In practice we try to keep it as low as possible. Otherwise its insertion into the current path can appreciably alter the currents flowing in the circuit.

Equally if you try the other route, as you suggest, you must overcome the problem that the shunt resistance must be at least ten times the resistor it is shunting or it may disturb the voltage levels in the circuit. This means that the current you draw will be an order of magnitude smaller than the one you are sensing; already at 0.5 micoamps in your case.

So tell us more about the circuit conditions, in particular what are the impedance levels you are trying to measure at?

And is this low frequency?

4. ### DedeHai Thread Starter Active Member

Jan 22, 2009
39
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Thanks for the replies. I try to be a little more specific:

I do charge a small high voltage capacitor (1kV - 3kV). The ESR of the capacitor is quite high, resulting in tiny currents. What I want to do is measure the charging and discharging currents.
So i can't assume a DC-like current. Also I want to analyse the "current shape" on an oscilloscope.
Having a high ESR already, a series shunt resistor of a few hundred ohms in the current-path won't significantly disturb my circuit (or will it?).
I thought of connecting the resistor to GND because of the high voltage on the positive side.
What circuit is best to amplify the voltage of the shunt resistor? I'm looking for a suited opamp-ciruit being as accurate as possible.

5. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
You want a differential amp circuit. That voltage is a killer though.

The op amp circuit I showed swings either way, but again, that voltage is a killer. If the cap is completely discharged you would have the full charging voltage across the measurement circuit, right?

6. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
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Would a current transformer around the charge lead be of use?

7. ### studiot AAC Fanatic!

Nov 9, 2007
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What do you understand by a shunt resistor?

Why can't you use direct measurement of current?

8. ### DedeHai Thread Starter Active Member

Jan 22, 2009
39
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What is the input impedance of the differential amp circuit? The capacitor has a series resistance of more than 2kohms, so the series resistance should take most of the voltage if the equivalent circuit of the input of the ifferential amp circuit equals a small resistor.

A current transformer is also a possibility I considered, but that solution does not measure any DC currents, so I would not be able to measure any leakage currents after the capacitor is charged.

Last edited: Jan 23, 2009
9. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
An op amp is a submember of the class of differential amps, so the impedance is quite high. If the power sources were totally disconnected, that circuit I showed might work. Problem with the amount of voltage you are dealing with is it qualifies as a static charge, and acts a lot like one if allowed, which means it is death for anything resembling a semiconductor unless the utmost caution is used.

Being a visiual sort of guy lets go through what I think you're going to do...

Putting this in a picture points out several issues. The big one in my mind is the surge current of 1A or so. You might be able to get around this by adding a switch around the amp meter, but what else is being assumed?

Assuming you are using a switch or two, 2KV is a very special breed.

Nothing beats a schematic to describe a circuit. It clarified several nagging things in my mind.

What is the minimum resolution you are aiming for? Least significant digit 1µA? pA?

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10. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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MICROAMPS? Shucks, that's practically a locomotive! If you REALLY want to talk about small, try some picoamps. This can be done with mosfet input op-amps....right off the shellf.

Eric

11. ### studiot AAC Fanatic!

Nov 9, 2007
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You didn't answer my questions, but I will bother again to help.

The conventional way to achieve what you seek is to add a modest value test resistor in series with the capacitor and measure the voltage across it.

You avoid the high voltage issue by placing the resistor in the earthy side of the circuit. the voltage developed across it is only a few volts at maximum.

The levels of current and voltage you describe are easily achieved by this method.

If you would like help proceeding with this method let us know.

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12. ### italo New Member

Nov 20, 2005
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TO measure 5ua is I asme full scale is not feasable. asume a series resistance of 1 Ω which is not a shunt but a I*R to measure the current is big enough to interfere and does not provide the voltage necessary to measure it. you will need a hi stable chopper diferential amplifier to get any meaning full measurement. May i sugest to put a hi resistance divider across the circuit and then measure it with a scope . or do use a diferential amplifierwith a low gain. since the divider low end can be sclaed to 5ua for x volts

13. ### Wendy Moderator

Mar 24, 2008
20,772
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That's why I asked the question. Their accuracy is questionable, but all my DVM's with a 2ma scale show 1µa (least significant digit).

If he wants to measure pa all he needs is a 1MΩ resistor as a shunt for a DVM on the 2V scale. 1µa = 1V. If the meter is 10MΩ impedance then up it to 1.11MΩ.

Last edited: Jan 23, 2009
14. ### DedeHai Thread Starter Active Member

Jan 22, 2009
39
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Ok guys, thanks for the inputs. I think I have to clarify. Sorry if I'm not really specific, but i don't really know what my circuit looks like, since I did not yet have the time and equipment to measure it thoroughly.
The conductor of the capacitor is a carbon powder, resulting in the high resistance of that thing (between 2kΩ and 200kΩ, but thats really just estimated). The capacitance is in the range of pF, 1nF at most.

Maybe I got the term wrong, but as a shunt I understand a resistor that is actually in the current path and you measure the voltage drop over that resistor (measurement input with high impedance) to determine the current. Thats how I (and everyone around me) use the term.

why can't i use direct measurement of current? hm. sorry i don't get that question. the whole point IS to measure the current. and no, i cannot use my digital voltmeter, since that one is made for dc (or pure ac) only. i want to see the form of the current on an oscilloscope.

about the frequency: it is not high frequency, the whole charging process takes a few milliseconds. I charge and discharge with a frequency of maximum 5Hz (square signal).

The maximum current is NOT in the range of 1A (even if with the voltage and resistance given it might seem so). I'm not exactly sure what the current is, but I don't think it exceeds 1mA.
The capacitor is not static, it's a so called electro active polymer, changing shape with applied voltage. It actually is pretty dynamic, so the resistance and capacitance is voltage dependent.

The range of the current to measure is something like 0.1μA to 1mA.

I don't have a tool at hand to draw you a schematic, but i can describe in words:
it starts with the adjustable voltage source. connected to that is a series resistance(2k-200k), then capacitor(few pF), then resistor again (2k-200k) that one is connected to ground. there is another resistor, parallel to the capacitor, in the range of MΩ. get the picture?

15. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
This reply is all good stuff and clears up many points.

Strictly a shunt is in parallel with something. Shunts for a voltmeter (in parallel with its input) turn it into a current meter. The shunts may be regarded as voltage sensing resistors.

Conversely shunts for a curent meter, again in parallel with its input, use it as a voltmeter to measure the voltage across the shunt which is in series with (in the path of) the current to be measured.

So what do you think of my connection diagram? I would suggest a sensing resistor of around 100 ohms which would provide an output of 0.1 millivolts per microamp through it. This would also be small compared with the ESR of the capacitor, so not divide the HV significantly.

If this ouput is still too small for your voltmeter and scope you can increase the resistor, or connect an amplifier across it.

Since we are talking low frequency, a simple op amp follower should suffice, with a closeed loop gain of perhaps 100 to 1000.

Remember with an earthy side connection the HV at the other end of the cap is not an issue.

16. ### DedeHai Thread Starter Active Member

Jan 22, 2009
39
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So my initial idea of putting a resistor into the current path is feasable you say. I don't have much practical experience with analog circuits. I just was not so sure that an output voltage of 0.1mV can be amplified with a simple opamp without getting a lot of noise.
What's the best way to amplify that voltage? i think of a non-inverting amplifier with a gain of 1000, as you mentioned. Is there a simple way to keep the noise to a minimum and still achiefe high accuracy (dynamic wise)? like adding a low pass filter with a cutoff frequency of about 1kHz ?

17. ### John Luciani Active Member

Apr 3, 2007
477
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Why wouldn't the low-side shunt and instrument amp work? There are a variety of instrument amps you can get that have the gain and offset errors laser trimmed out.

Go to the Analog Devices website (www.analog.com) and take a look at the
gain and offset errors. You can get devices with pin selectable gains of 100 and 1000
which should be able to handle this application. TI (Burr-Brown) also has some
nice instrument amps.

(* jcl *)

18. ### hgmjr Moderator

Jan 28, 2005
9,030
214
If the introduction of the voltage of a forward biased diode into the measurement line is tolerable then this may be an application for a current-mirror. Of course measures to keep any sources of error to an absolute minimum would need to be taken. One way would be to use an IC package with dual NPN or PNP transistors. This would keep errors due to mismatched transistors to a minimum.

hgmjr