Hi,

I was wondering if anyone have a good and cheap solution to simulate an analoge input (4-20mA)?
At the moment I'm building a testboard for my FX2N, and this item has still to be solved.

I was thinking of using a 10K potentiometer with a fixed resistor to maximise the current, onlything is if I do so the analoge inputsignal is no longer linair.

So I'm open for some good ideas,

Carol

http://forums.mrplc.com/index.php?autocom=downloads&showfile=149

Here you go.

HMM? Why wouldn't it be linear?

Thnx Bob,

it looks good, and as it says in the textfile: this is one topic hard to find.
Can't wait to monday to purchuse some components and go right to it.

Thnx Carol

to Mickey:

perhaps "not linair" isn't the right expression. But with a fixed resistor it seems, in the lower part of 4-20mA, you have to turn 8 times on the potentiometer to gain a mA and in the upper part you just have to look at it and it shift quit a few mA in a blink.

Carol

Not really cheap, but have you considered a bench top power supply, most of them can vary voltage and current. You can always use it for other projects as well.

Here's one to think about: http://www.mpja.com/prodinfo.asp?number=14601+PS

i don't know why you think the output would be non-linear.
text in the article says it is linear and the version
I have built also checked out as liner...

Besides, this is extremly simple circuit with just few components
it begs to be analyzed (highschool material). In fact, I may
use it as an interview question (requires entry level engineering skill
such as finding, reading and understanding data sheet, checks the
most basic math skill and circuit analisys...):


this is not simple voltage divider made with passive components,
it uses voltage regulator chip.

datasheet for LM317L can be found using google of your choice
(nobody uses AskJeeves etc. anyway):
http://www.onsemi.com/pub/Collateral/LM317L-D.PDF

When voltage output is selected, R3 is not used (open circuit).
Rest of the circuit resembles classic regulator as seen in datasheet.
The voltage output becomes:

Vout=1.25*(1+R5/(R2+R4))+Iadj*R5.

where 1.25V is regulator output between Vout and Adj terminals.

Since R5 is the only variable (Iadj is fixed and it's low value which can
also be found in the datasheet), this is nothing but simple linear
function (y=a*x+b) where:

y=Vout

a=(Iadj+1.25/(R2+R4))

x= R5

b=1.25

Clearly, to get linear output, R5 must be linear...

Values of the resistors are chosen to produce desired output ranges
and make use of standard resistor values. I've decided it's better to
keep it simple, and what is point in making something simple if the
components have hard to find values or one has to perform calibration
that can be messed up anytime somone borrows the unit.
(initial idea was to just use regulator, pot and pair of trim-pots but
one could set trim-pots to min value and overload circuit. this was
abandoned and you can see result).


when current output is selected, same linear voltage output Vout is
applied to R3 which is clearly fixed value. This means that output current
of the circuit is:

Iout = Iadj+1.25V/(R2+R4)+Vout/R3

where 1.25V is again regulator output between Vout and Adj terminals.
Since Vout is the only variable in the equation, this can be represented
as standard linear function y=a*x+b. For example:

y= Iout

a=1/R3

x=Vout

b=Iadj+1.25/(R2+R4)

because b can never be zero (unless output is not connected), this circuit can't produce very low output (in the neighbourhood of 0mA).

As per attached datasheet, it is like below:
Vout=1.25*Vin ( 1 + (R2/R1)) + Iadj *R2

I dont get what are R5, R3 and R4.

Also I dont get how to calculate Iadj.

Any way I use LM317 usually to get regullar adjustable power supply, but in this case we are looking to 4-20 mA current source, so how we can build this regular current source for differance resistance load? (May only one device, 2 or three devices in seriese are connected)