How to measure bi-directional small signal currents

Hi All,

I want to measure small current pulses (For applied voltage pulses) in the range of 10 to 300 uA. Will you have any idea on that ?. The frequency of the pulses will not exceed 5 kHz and the amplitude is between 1 to 3 V Maximum. I think differential amplifier is fine in this case but I want to make sure before desinging the circuit. I can measure the corresponding voltage with a NI card and convert that back to current.

Pubudu.

Reply to
Pubudu
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So, what is this about? Battery charger? Your post seems to have "lost something in translation". Its hard to tell just what you are asking for here.

Anybody here got a clue?

Luhan

Reply to
Luhan

I think Pubudu has just been a bit careless with his description. For example, is he applying a stiff voltage source and seeking the resulting current into the load?* Or is he applying small current pulses as he says, and wants to measure the voltage drop? Or is he perhaps applying some hybrid between current and voltage pulses?

  • These techniques can go under the heading "source-measure" and are a lead-in to a field of interesting instruments and circuits.
--
 Thanks,
    - Win
Reply to
Winfield Hill

I also don't understand your question exactly, but maybe you want something like this. This array of opamps translates a current into voltage and has zero voltage drop, so you can measure without any influence and also on the high side as long as the compliance is met. ___ +-|___|------+ | R | I | |\\ | o----->----+-|+\\ | ___ ___ | >--+----)-+-|___|-+-|___|-+ | +-|-/ | | | R1 | R1 | | | |/ .-. | | | === | | | | | | | GND | | | |R | | | | | '-' | | | |\\ Ue=0 | | | | +-|+\\ Uout | +-------+ | | | >--+----o | | | | | +-|-/ | | | .-. | | | |/ | Uout=2*R*I | | | | | | | | | | | |R | | | | | | |\\ '-' | | | | | +-|-\\ | | | ___ | ___ | V | >--+----+-)-|___|-+-|___|-+ o-----

Reply to
Ban

Hi GUys,

Sorry for my writting and I really appreciate your response. My problem is this. I am applying a oscilating voltage pulses to a chemical cell and I want to study the current flow through cell. Voltage pulse frequency is between 1 Hz to 5 kHz and expected current is in between 10 to 300 uA. (Cell impedance is between 10kOhms and 100 kOhns approximately) Voltage signal is applied as like this (using two analog channels in NI card): First pulse : Ch0 = 1 V and Ch1 = 3 V. Second pulse Ch0 = 3 V and Ch1 = 1 V Third pulse : Ch0 = 1 V and Ch1 = 3 V. Fourth pulse Ch0 = 3 V and Ch1 = 1 V Fifth pulse : Ch0 = 1 V and Ch1 = 3 V so on . . . . . .

Ch0 and Ch1 are connected to the two cell terminals.

Reply to
Pubudu

Than I think this is exactly what you will need. If you use a current shunt, it will have a voltage drop and to have a good sensitivity the shunt has to be quite large. So the voltage is not constant, but varies. With high frequencies there will be also a lowpass behaviour distorting the pulses. this thing has zero impedance, so the upper frequency is much higher and pulses are not distorted. OP27 for the front-end and some INA as a differential amplifier are recommended.

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ciao Ban
Apricale, Italy
Reply to
Ban

Hi Ciao Ban I really appreciate your reply. Thank you very much. I will go head and make it. Again thank you very much.

Pubudu

Reply to
Pubudu

Are you making a "Pulsed Amperometric Ion Sensor" ?

RR

Reply to
RRogers

That's a cute circuit (nice drawing, too). Where'd you find it?

--
 Thanks,
    - Win
Reply to
Winfield Hill

THX, I used it in a project when I was at university, maybe 1976. My main subject was measurement and instrumentation. I think it came out of some book/script. But my memory is fading. :-( At least I have a nice collection of useful circuits at hand, serving me well throughout my career and is still growing.

--
ciao Ban
Apricale, Italy
Reply to
Ban

If those pulses are square pulses then the common mode voltage could have significant components up to at least 5x5KHz. Getting a good common mode rejection at high frequency can be difficult and it is always worth while investigating the possibility of not having a CMV in the first place.

It looks like you have a two-terminal cell, which needs a +/- 2V drive and have perhaps chosen those +1V to +3V differential signals because your NI card does not do bipolar voltages.

The circuit below will provide a +/- 2V drive to the cell from your NI card, and produce a grounded I-V conversion.

3 _ R1 R2 _| |_---/\\/\\---+-/\\/\\--+ All R's = 10k, 0.1% (say). 1 | | V1 +--|-\\ | OP amps powered from +/- | >-+ supply rails. 0v--|+/ | +------------------------>ADC-3 R3 | R7 V2 +-/\\/\\--+- --/\\/\\-- -+ 3_ _ R4 | R5 | R6 |_| ---/\\/\\---+-/\\/\\--+----[CELL]---+-/\\/\\--+-->ADC-1 1 | | | | +--|-\\ | +--|-\\ | | >-+-+ 0v--|+/ | 0v--|+/ +------------------------>ADC-2

The left hand end of the cell has the +/- 2V drive on it. The right hand is at 0V, into the virtual earth of an I-V converter. R6=10k gives ADC-1 a scaling of 3V/300uA. That is a bipolar output. R7 is an optional resistor that biasses ADC-1 into being positive-only. The amount of bias varies according to where V1 is, but this can be easily dealt with in the software. An ADC-3 input is useful for measuring the bias.

ADC-2 is an optional measurement that gives the exact voltage on the cell during each pulse-polarity. Useful for (say) software-adjusting V1/V2 for zero dc-bias on electrochemical cells.

--
Tony Williams.
Reply to
Tony Williams

Does a configuration of back-to-back NIC inverting current mirrors with differential pick-off ring a bell? Although analytically the same, cross-coupling the feedback must yield synchronization advantages:

View in a fixed-width font such as Courier.

. . . . i ->

. .---[R]---. . | | . | |\\ | . i >--------+--|+\\ | . | >---+ . .--|-/ | . | |/ | . | | . | i -> | . +---[R]---' . | . |

Reply to
Fred Bloggs

This looks like a lot more sensitive than you need, but it's quite interesting:

formatting link

Reply to
Ben Bradley

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