Hallo, I should acquire an analog signal in current and convert it in voltage.
The signal is in 0-20ma current range.
I thoguht to use the following circuit:
Voltage source 5volt | R1 input 0-20ma------|------------ output about 0-10v R2 | ground
R1 = R2 = 1k ohm
Any suggestion?
Many Thanks Marco Toschi
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
-- Thanks, Frank. (remove \'q\' and \'.invalid\' when replying by email)
In message , dated Fri, 4 Aug 2006, Frank Bemelman writes
One resistor is never enough. Use thousands.
And a PIC.
And a 555. (;-)
-- OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk 2006 is YMMVI- Your mileage may vary immensely. John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
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...
0-20mA suggests an industrial sensor of some sort. These give a DC current.Are you using AC or DC current? And what is the desired voltage output, AC or DC?
-- Thanks, Frank. (remove \'q\' and \'.invalid\' when replying by email)
"John Woodgate" schreef in bericht news: snipped-for-privacy@jmwa.demon.co.uk...
Perhaps Bill can scribble something that uses a 741 as well ;)
-- Thanks, Frank. (remove \'q\' and \'.invalid\' when replying by email)
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.
-- Thanks, Frank. (remove \'q\' and \'.invalid\' when replying by email)
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
12v, and with transistor drops etc., 10v may be 'pushing things'.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:
+------ 10K -----+ | | | |\\ | | | \\ | IN----+------|- \\----+---- OUT 1K | / +------|+ / | | / | / GNDIn message , dated Fri, 4 Aug 2006, linnix writes
20 mA in gets you 200 V out. Some op-amp! All the 20 mA going in MUST flow through the 10 k; the inverting input is a 'virtual earth' point. The 1 k does nothing.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.
-- OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk 2006 is YMMVI- Your mileage may vary immensely. John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
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. (;-)
-- OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk 2006 is YMMVI- Your mileage may vary immensely. John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
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.
+------ 10K -----+ | | | |\\ | | | \\ | IN----+------|- \\----+---- OUT | 1K | / 100 +------|+ / | 1K | / | | / GNDBut this is not open-loop.
for 0..20mA -> 0..10V
r2 = 500 ohms, r1 not used. Bye. Jasen
With an FPGA for 'logic glue'.
-- Paul Hovnanian mailto:Paul@Hovnanian.com ------------------------------------------------------------------ c (velocity of light in a vacuum) = 1.8x10^12 furlongs per fortnight
Back up and debate PIC versus AVR, and end up with a 68HC11. :)
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