+- 100mA process current signal into a 0 V to 10 V output

Hi,

have a circuit that I will need help with. I am trying to design a circuit that the input source is +- 100 mA and convert that to a 0V to 10 V reference voltage so I could use that reference for inputting it on a program. I have attached a drawing that might explain it a abit more.

the load could be any value ie 300 ohms.
Thanks

What about a 100Ω pre-resistor?

If I do that I won't get my linearity correct ie 100ma equals to 10v , 50ma equals to 5 v, -100 ma equals to 0 volts

- - - Updated - - -

I need some sort of instrumention amplifier ?? Any ideas?

Logitech_A said:

... -100 ma equals to 0 volts

Click to expand...

No: -100mA equals to -10V .

Hi,
What i meant was :

+100mA = 10V
+50mA= 7.5
0mA= 5
..
.
.
-100mA= 0V.

because the values I will be using is 0 V to 10 V. so therefore I dont think i could use pre resisters of 100ohms.

What about this:

**broken link removed**

I'm a beginner to circuit design so I don't know what kind of problems that circuit will have, but it works well in the simulator at least. It uses a 50 ohm resistor (R5) to create +/- 5V from the +/- 100mA, then a voltage level shifter derived from here to add 5V and shift the level to 0 - 10V.

The values of R1-R4 should be the same, for a level shift with no scaling. The values of R1-R4 should be large enough that the voltage drop is negligible compared to R5. You'll hit +10V at +100mA input, but the minimum output will be 10V*R5/(R1+R3+R5) = 24.94 mV @ -100 mA input for 10K resistors. You could put a buffer on the input, I think, to eliminate that entirely, or choose larger resistors to reduce the magnitude.

You'll need to supply +5V through R3 for the level shift, and at least +10V to power the op-amp (the LM741 datasheet, at least, says the output voltage cannot exceed the supply voltage).

There are offset voltages and other sources of error here that I don't really understand, the circuit diagram above is for ideal components.

Hope that actually works and/or helps,
J

Thanks, i will try that.I have sent you a message regarding on what simulating program you are using so I could practise . Thank you again. i will keep you update

I just now found a neat online one called CircuitLab and put it in there with an LM741 op-amp and it seemed to work:

https://www.circuitlab.com/circuit/smkudj/100ma-10v/

Load it up in CircuitLab and run a DC Sweep simulation. If it saved my sim parameters with the circuit (it looks like it did), a -100mA -> +100mA sweep with voltage output graph should be created.

Don't forget to consider resistor tolerances if you actually build it. V1+V2 is the power supply. The output will clip at V2 so give your op-amp at *least* 10V supply, I gave 12V to be on the safe side. V1 is the voltage offset. The closer to exactly 5V the better, since this will offset the output, so depending on the quality of your voltage source, you may have to trim the 5V line with some resistors.

Another online circuit simulator I like is https://www.falstad.com/circuit/. That's the one I got the initial diagram I posted. It's really basic, but really easy to use with tons of example circuits and also lets you run and watch the simulation and visualize voltage, current, and scope outputs as it happens. CircuitLab lets you enter op-amp parameters though, the one at falstad does not.

Hope that helps,
J

- - - Updated - - -

Here is an improved version with a buffer on the input (to bring -100mA in closer to 0V out) and a 5V regulator supplying the offset voltage. I don't know enough about regulators to make any statements about the consistency of that 5V, though:

https://www.circuitlab.com/circuit/79ga9g/100ma-10v-2/

You may also want to put a fuse on the input to protect the op-amps if it's a risk. The 741 datasheet says max input voltage is either +/- supply voltage or +/- 15V, whatever is smaller, so that's +/- 12V here, which works out to +/- 240mA input current, so maybe either a 200mA fast blow fuse on one of the current input lines, or mount your op-amps in an IC socket so you can easily replace them if you blow them out.

J

Question!

Thus the measurement will accurate? since the current is small, and the opamp inverting/non-inverting terminal will require some input current, in the end, the SOURCE 100mA flow through the 50ohm is no longer exact 100mA. If the opamp input terminal current required is 1mA, then only 99mA flow through the 50ohm, and increase the offset current of the opamp, will the result not accurate finally?

the tolerance might more than 5% will it?

SOLUTION: use other opamp? or any another configuration? such as current mirror etc?

Good question; let's find out. Assume a constant current source, where any current drawn by the op amp would cause that much less current to flow through the 50 ohm resistor. The op amps I've looked at all have very small input current draw, in the nanoamp range. The **broken link removed** gives the following input bias current specs:

Typical @ 25°C = 80 nA
Max @ 25°C = 500 nA
Max @ 125°C = 1500 nA

That would translate to the following circuit output voltage ranges (in the 0-10V output):

Typical @ 25°C = +/- 4 μV
Max @ 25°C = +/- 25 μV
Max @ 125°C = +/- 75 μV

That would also correspond to the following tolerance on the 50 ohm resistor:

Typical @ 25°C = 49.99998 ohm - 50.00002 ohm
Max @ 25°C = 59.999875 ohm - 50.000125 ohm
Max @ 125°C = 49.999625 ohm - 50.000375 ohm

Meaning that even in the absolute worst case at 125°C, the op-amp input current corresponds to a +/- 0.0000075% change on the resistor, well below the component tolerance of any available resistor. In other words, at least for the LM741, the current draw is very negligible.

BTW 0.1% tolerance resistors are not prohibitively expensive for small quantities. Mouser sells a 50 ohm +/- 0.1% 1W resistor for $0.69 each.

J

When you need accurate Return Loss measurements for RF, it costs a lot more than $0.69 to buy these dummy loads. $$$ for Spectrum Analyzer, RL bridge or VNA use. ~$50 for 50Ω DC-6GHz 30W. Of course you can make one cheaper if you know how. -30dB RL expensive = 0.1%
Of course physical shape and stray capacitance, and lead inductance is everything and geometry must be more precise than 0.1% for this to work in RF.. But for LF< 30MHz $0.69 is cheap.

SunnySkyguy said:

When you need accurate Return Loss measurements for RF, it costs a lot more than $0.69 to buy these dummy loads.

Click to expand...

Thanks for the tip!

Question: The datasheet for the resistor I mentioned is here: https://www.vishay.com/docs/31021/cpf.pdf (PN CPF1-50R000-B-H-B14). It is a metal film resistor.

I read **broken link removed** that:

Film type resistors having a spiral construction do tend to exhibit the properties of inductors (which are basically spirally wound coils of wire) but this is not usually a problem until used at frequencies in the MHz range.

Click to expand...

Is there enough information on that datasheet to determine the actual frequency response of that resistor? The sheet says "excellent high frequency characteristics", but that doesn't seem to mean much.

Also, let's say I intended to use the circuit I posted for high frequency measurements. The other important considerations would then be... frequency response of all the resistors, op-amp slew rate, what else is important? I just read a great write-up on PCB layout design so I guess there'd be noise reduction considerations there too. Also, what's all this business with "decoupling capacitors" in high frequency designs?

Sorry, that was a lot of questions. I presume the original question was for lower frequencies (might be wrong), but I'm wondering about high frequency considerations for future reference.

Thanks!!
J

Hi,

I tried to build the design, however I am coming across a problem or two. the major problem is that as soon as I put the 300ohms load the output value changes, from 10V to 9.63V, (bare in mind that the value of the resistance changes from 0 to 400ohms) and I require it to stay stable no matter what the load of the circuit is.

Thank you