would to measure milliamps with optascope

dechief michel

For my job,(engineer on heartmoving materials) i have to take some measurement as hydraulic pressure, hydraulic flow ,voltage and current.
For the three first, i d'ont have any problem to us my optascoppe.But i need to have a screen with at first the flow(analogic signal true a specific sensor,no problem)and on to the same screen(second signal) the amount(curve) of milli amps generated from the main computer to drive the hydraulic pump(range from 0 to 30 mA)
Did you have any sugestion to interface my optascoppe for ability to take measurement of ampere(or mA)
Thank you by advance
Michel Dechief
Technical inspector
Liebherr Belgium
heartmoving dpt

Paul Baker

By placing a very small·value resistor in the line you can detect the current by measuring the voltage across the resistor. The type of resistors used are called current sensing resistors and are fractions of an ohm. If the produced voltage is too small to measure with the optascope, you can use an op amp to boost the voltage to a more easily measured voltage.

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·1+1=10

dechief michel

Hy PAUL, AT FIRST MANY THANKS FOR YOUR QUICK REPLY.
You said a small resistor, if you know than i take measurement on to 24 volt and i need to test from 0 to +/-20 mA witch kind of resistor you suppose.
Sorry for my bad aknowledgment in to your langage, normally i speek french but it's a great opportunity for me to discuss about something whos interrest me in to another langage.
Thanks by advance for your answer
With frienship,
Michel

Paul Baker

Your english is good enough for me to understand.

Here is a link to a catalog page of current sensing resistors: http://dkc3.digikey.com/PDF/T053/1165.pdf

Since you are measuring a current loop (0-20mA) you'll likely want a radial type (wires sticking from both ends) and the wattage isn't important.

As an example, say you are using a 0.1Ω resistor, 20 mA flowing through it would show 2 mV across it's terminal, this is too low to measure properly, so using an operational amplifier whose·positive (+)·terminal is tied on the side leading to the source of the current loop and the negative (-) terminal tied to the side leading to the sink of the current loop, the output will be I=V*10/G, where I is the current flowing through the current loop, V is the measured output voltage from the op-amp, and G is the Gain you have set the op-amp to. The 10 is 1/R, so if you decide to go with another value the general equation is I=V/(RG).

Ask for further clarification on things you do not understand.


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Doug Pientak

HI Michel,
I was asked by the Parallax tech support team to give some suggestions to help you out.· But Puals suggestions are perfect.·· As for the oscilloscope the 100,200 and 500mV ranges all use the lower 4Vpp range.· So this will give you the best resolution for looking at small signals like this.·
Doug Pientak
Parallax Engineering

Paul Baker

Thanks for the clarification Doug, I deleted a line about reducing the range because I couldn't verify through the documents if the resolution scaled accordingly, now I know its dual scale.

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dechief michel

Thanks very mutch Paul and doug for informations.
Certainly i try that next day.
It's very greet for me to receive this kind of information because i'm not a electronic specialist but i have to use it in to my job.
Paralel of that, if you need some help in to a mechanical design or hydraulic project, please don't have hesitation i'm your man.
Many thanks again from Belgium
Michel

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1+1=3 because the first 1 remember of his historical coeficient Interrest Paul?????