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Join Date: Jun 2012
Posts: 14

Current to Voltage Converter

06/14/2012 8:22 AM

hello, am m mirza.

I have a small question here,

may I know what can be done to improve the current to voltage converter measurements in a circuit.?

am thinking that adding resistance would increase the mesurements.

so is my idea correct ?

any suggestions from you people will be appreciated.

thanks,

regards,

m mirza.

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#1

Re: current to voltage converter

06/14/2012 9:07 AM

Can you provide some details of this circuit? A functional block diagram is a good start and a schematic is even better.

Some questions for you: What are the voltage measurements being used for? What is the range of current that is being measured? How do you know that you are not already measuring the voltage accurately?

What makes you think that adding resistance would increase the measurements???

As you provide more details, we can provide more answers. Otherwise, its a shot in the dark.

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#4
In reply to #1

Re: current to voltage converter

06/14/2012 2:11 PM

the current range being measured is very low form -10pA to 10 pA;

the voltage range is something around -2048 mV to 2048 mV.

hope this may help you a little.

so the schematic below is a potentiostat for an electrochemical array.

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#5
In reply to #4

Re: current to voltage converter

06/14/2012 2:48 PM

Your schematic is not very legible in the CR4 format. It does tell me that you are trying to measure these signals and have another current or power source available.

You're correct that ±10^-11 amperes is quite a low amount of current to measure. My initial recommendation is that you instead of making your own design you use some true instrumentation like the products made by Kiethley. I greatly recommend that you register there to get a PDF download of their book "Low Level Measurements Handbook". This handbook is not a sales brochure filled with just their product line. It explains the most common pitfalls and errors one finds in trying to design a system to read any signal at such small currents. (Well it is a little biased to their products because they already did a lot of work in this realm.)

Another advantage of using a test instrument is that you will have a reliability and accuracy pedigree that references back to a calibration standard.

If you still must make your own task specific circuitry for this, then I would recommend starting with a pair of electrometer grade operational amplifiers as the front end of a differential amplifier circuit. (The handbook explains this topology.) This will permit high common mode noise rejection along with good control of sensor impedances. I recommend this also by implication of the numbers you've presented. ±2.048V that correlates to ±10pA implies signed 10 bit data acquisition where the LSB will be about 5 femto-amperes of current. This is not a trivial circuit.

Oh, for the sake of full disclosure. I regularly use Kiethley instruments. I do not work for them.

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#2

Re: current to voltage converter

06/14/2012 9:09 AM

Broad question.

  • In industrial control loops using 4-20mA signals, a precision resistor of 250Ω wired in series in the loop will give a 1-5V signal of similar accuracy to the resistor accuracy.

Does that help?

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#3

Re: current to voltage converter

06/14/2012 9:58 AM

You do have a good generic idea here. An ideal current source driving its current into a higher resistance will produce a higher voltage. However, there are many hidden assumptions and complications in that statement that people forget.

First, there will always be a limit to how high of a resistance one can place in a circuit for multiple reasons. One of these reasons is that parasitic resistances of insulators and capacitor leakage can easily become alternate paths for your current that you are trying to measure. Next, the input impedance and biasing currents of your measuring instrument can induce significant errors when resistance values get too high. Let's not forget that your current source will not be ideal, too. Your current source must operate in a region to be linear while also itself having some finite output impedance. Then there's the aspect that baffles many novices, thermal noise. The noise produced by heat can produce a larger error component with large resistors trying to measure tiny currents.

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