A simple way to convert 0-20ma into 0-10volt

Ops... R1 = R2 = 500 ohm

"Marco T." schreef in bericht news:eauu4a$it4$ snipped-for-privacy@nnrp.ngi.it...

One resistor is enough:

input 0-20ma------|------------ output about 0-10v R2 = 500 | ground

In message , dated Fri, 4 Aug 2006, Frank Bemelman writes

One resistor is never enough. Use thousands.

And a PIC.

And a 555. (;-)

I'm trying to simulate the circuit using spice. I have used a current generator to simulate the input, but in this way if I don't use also R1 and voltage source I don't obtain the desired output.

Which circuit should I use to simulate the 0-20ma input?

Many Thanks Marco

"Marco T." schreef in bericht news:eav239$k54$ snipped-for-privacy@nnrp.ngi.it...

Are you using AC or DC current? And what is the desired voltage output, AC or DC?

"John Woodgate" schreef in bericht news: snipped-for-privacy@jmwa.demon.co.uk...

Perhaps Bill can scribble something that uses a 741 as well ;)

I'm using a current generator with current fixed to 20ma.

I would obtain a dc voltage of 10volt.

Many Thanks Marco

"Marco T." schreef in bericht news:eav2sp$kga$ snipped-for-privacy@nnrp.ngi.it...

DC current?

And so you should. 20mA @ 500 ohm is 10V.

Yes, DC current.

However _beware_ that there may be a 'caveat' in the drive capabilities of the 4-20mA source. Most have a 'maximum resistance' specification, into which they can maintain drive. For many of the systems, it is something like 420R. The reason is that the drive itself is often only powered from

Best Wishes

Lots of industrial controls use 4-20 ma for the two wire sensor. 4 is used instead of zero because that gives a way to detect an open connection to the sensor or malfunctioning/out of range sensor. A shorted wire/out of range sensor indicates a greater than 20 ma output. The idea is sometimes called a "supervised loop."

The control consists of a battery or power supply to put 10-30 VDC on the line. To convert to a voltage you just supply a single fixed resistor and put it on the return of the sensor to ground and measure the voltage at the junction of the return and ground. You want ten volts? use ohms law to find the resistor that will take.

In theory, a current source will output whatever voltage it needs to to maintain the current in the loop. In practice the power supply voltage is the limiting factor.

Many sensors today use four wires with a two wire current loop - the extra pair supplys power to the sensor so that it can amplify and massage the signal at the sensor head.

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I agree, an op-amp is need. How about this:

In message , dated Fri, 4 Aug 2006, linnix writes

If an op-amp is working as an amplifier, there is, to a good approximation, NO voltage between its inputs. This is because of the huge open-loop gain. For any output voltage V, the voltage between the inputs is V/huge, which is very nearly zero.

Well, it does force the amp to work at a noise gain of at least 10, which will help stability some. Hanging a wire on a summing junction is not guaranteed to produce good behaviour.

Cheers,

Phil Hobbs

In message , dated Fri, 4 Aug 2006, Phil Hobbs writes

It usually produces the local pop music radio stations, for your entertainment while you wonder why. (;-)

No, it will max out at Vcc.

Most of the current will flow to ground via the input R, not the feedback R.

OK, we need the split the input current to avoid saturating the op-amp too much. The values are just approxmation (in ratios), you have to figure-out the exact values.

But this is not open-loop.

for 0..20mA -> 0..10V

r2 = 500 ohms, r1 not used. Bye. Jasen

With an FPGA for 'logic glue'.

Back up and debate PIC versus AVR, and end up with a 68HC11. :)