# measuring tiny currents (nA) with an oscilloscope

Discussion in 'The Projects Forum' started by Geza, Oct 14, 2009.

1. ### Geza Thread Starter New Member

May 18, 2009
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i found this text on a website and wondered if it is accurate:

"It's easy to connect an opamp as a current amplifier, just use the inverting amplifier configuration with the input resistor of zero. Then the output voltage is V = Iin x Rf. With a 1000 Mohm Rf, that is 1 volt per nA. The 1 volt output is then easily read by a AD converter and fed into a computer.

A lot of care is required to attain this accuracy, as PC board leakage can easily cause errors that are in the several nA range. You need guard traces, and perhaps special PC board material."

what I don't understand is I thought op-amps amplified voltage?

any help would be much appreciated.

Cheers

2. ### spacewrench Member

Oct 5, 2009
58
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The model for an op-amp is that the inputs sink/source no current, and the output goes up if the + input is at a higher potential than the - input, and down if the reverse is true. (If, for example, there's no feedback connection between the output and one of the inputs, then if the inputs are at a different potential, the output will swing all the way high or all the way low.)

In the case you describe, the (+) input is at 0V, the (-) input has a small current flowing towards it, and the output is fed back to the (-) input through a 1M resistor. Now, without the feedback, current flowing towards the (-) input will inevitably raise its potential, and would cause the feedback-less output to go all the way negative. However, with the feedback, the output only has to go negative enough to sink the current that is flowing towards the (-) input. If, for example, you have a 1uA current flowing to the (-) input, then the output will go to -1V, and the input current will go through the feedback resistor and into the output, leaving the (-) terminal at 0V, the same as the (+) terminal.

Theoretically, you could use an even larger feedback resistor to measure even smaller currents, but (as your article states) at those low levels, you'll run into other problems trying to measure things.

3. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
This will only work if the input-bias current of the op-amp is about a factor of 100 less than the current you are trying to measure. Google "current to voltage converter".

4. ### ELECTRONERD Senior Member

May 26, 2009
1,146
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Glad to have another technical engineer around these forums, welcome aboard Mike.

5. ### Geza Thread Starter New Member

May 18, 2009
7
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what i can't get my head around now is how to connect this op-amp up in series so that it does not sink the current I am trying to measure. How can I make it so that it can be used anywhere in a circuit? I understand that the current will enter the inverting input but then it is lost, no?!

sorry for being stupid!

cheers

6. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
Yes, a current-to-voltage converter can only operate if it is at the "grounded-end" of a series circuit. If you want to make a "floating" current measurement, you need to use either a "high-side current monitor" or a "low-side current monitor" which basically breaks the series circuit, inserts a low-value "shunt" resistor, and then amplifies the differential voltage between the two ends of the shunt resistor. Google "high-side current monitor", "low-side current monitor", "differential amplifier", and "instrumentation amplifier".

7. ### Geza Thread Starter New Member

May 18, 2009
7
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hey, sorry to bother you guys again. I have come up with this schematic but not sure if it's even right or how to tweak the components. I am just trying to make a basic current measurement circuit where the output can be read by a voltmeter or oscilloscope. I am trying to simulate this in Orcad PSpice as well.

any help would be much appreciated, i think i am really confusing myself.

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8. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
Are you doing it this way so you don't need a negative supply on the opamp?

The gain of the amplifier is (1+1k/1k) = 2. Why?

The 1K resistors are much too low in value; they unnecessarily load the opamp.

The 1 Ohm shunt is quite high. Why not use 10mOhm, and then use a gain of 100 in the amplifier. This would have a smaller impact on the loop you are measuring the current it.

9. ### Geza Thread Starter New Member

May 18, 2009
7
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ok, I have no idea what I am doing. All I want to do it measure the current in a simple circuit by having a voltage at the output which is proportional to the current in the device under test. I have attached another schematic. Sorry for not answering your previous questions I posted the wrong diagram. The op-amp has its own +12 -12v power supply.

many thanks

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10. ### Darren Holdstock Active Member

Feb 10, 2009
262
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There are some very good papers on the subject from Bob Pease (Electronic Design), Bonnie Baker (EDN) and Paul Rako (National Semiconductor and EDN). Go google "transimpedance" plus these authors.

Gain peaking due to the stray capacitance on the op-amp inverting input will limit the upper value of the feedback resistor, so start small and work your way up. But not as small as 5 milliohms - a classic PSPICE gotcha as it will take 5M as being 5m. You have to use meg or MEG, not M.

11. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
Ok. The feedback resistor should be 5MEG, not 5M which Spice takes as 5mΩ, or 0.005Ω. If you make that change, then the voltage at the output of the opamp will be -5V, instead of -5nV, or -0.00000005V

The 470 Ω load resistor is low for most opamps to drive. -5V across 470Ω means that the opamp has to sink 10ma. Unless you are trying to model an AD with such a low value of input resistance, why is it there?

The circuit will work, but here is what I would do. First make the 5meg resistor much lower, say 500K. To compensate, add a second opamp with a gain of -10, so you will still have a trans-impedance of 5*10^6 V/A, just like your intended circuit. The output will be +5V, rather than -5V. It will be more stable. It is sometimes hard to get precision 5meg resistors...

Last edited: Oct 21, 2009
12. ### Geza Thread Starter New Member

May 18, 2009
7
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thank you so much guys, it's beginning to get clearer in my head now! I thought something was funny and that 5M instead of meg explains it. thank you so much for your help, im sorry that I am an idiot!

13. ### Darren Holdstock Active Member

Feb 10, 2009
262
11
Not at all Geza, I still accidentally put M instead of MEG into PSPICE, it's an early clumsiness of the tool that haunts us all still. And I find that as I get older I just become more foolish at a higher level, so no worries there.

TI amps can be tricky, despite their apparent simplicity, so don't be surprised if strange things happen.