Analogue input voltage

Hi

I have a Mitsubishi PLC with a 2 Channel analogue input card.
I want to use the A/D card for connecting 2 remote potentiometers, with which the user can adjust timer values.

The card can be input with 0-10V or 4-20mA and I would like to know what the simplest (or best) method is for generating a reliable reference voltage (or current). Mitsubishi seemed reluctant to offer any solid advice about this, first suggesting that I should use a separate power supply and then (when I pressed them for a method of using the 24VDC linear PSU which is already in the panel) they said it should be possible to use resistors to divide the voltage but they would not provide any real world example of the way to do this.

After a lot of trial and error (bear in mind that I know practically nothing about electronics) I came up with the attached circuit…

It seems to work but I’m concerned that it may not be reliable (could it lead to damage of the card etc).

Also, can anyone explain how to generate the 4-20mA input reference and what is the difference between that and the voltage input?

Schematic you provided would work fine for voltage inputs
if you swap the 1k and 4k resistors or use supply voltage
lower that 24VDC. Check out the little analog circuit
tester in download section of www.mrplc.com

ok here is the link [url]http://www.mrplc.com/dl/index.php?action=view&view=entry&entryid=94[/url]

Why don't you try this hook-up?

panic mode said:

ok here is the link [url]http://www.mrplc.com/dl/index.php?action=view&view=entry&entryid=94[/url]

Click to expand...

Well, almost...

HERE'S a working link.

beerchug

-Eric

Let's analyze your circuit. There is not much electronics involved, just plain old Ohm's Law.

Across the 24 VDC power supply you have a voltage divider, consisting of two legs. The first leg is 1 kOhm. The second leg contains three resistors in parallel: 4 kOhm plus 2 potentiometers 1 kOhm each. If you remember the rule for the parallel resistors, the total resistance of this leg is 1/(1/4 + 1/1 + 1/1) = 0.44 kOhm (note that with parallel connection, the total is always smaller than the smallest of the components).

24 Volts will be distributed between the legs in proportion with each leg resistance. So, for the lower leg, the maximum voltage will be 24 * (0.44 / (1 + 0.44)) = 7.33 V . Which is well within the input voltage range.

Strictly speaking, the internal resistance of the analog inputs must be a part of the calculation too. However, since it is usually in hundreds of kiloOhms range, we can omit it - it will not affect the result significantly.

As an alternative, look at the diagram above. If you apply the same calculations, you will see that for each input the maximum voltage will be 24 * (1 / (1.4 + 1)) = 10 V. Just lets you use a bit more of the analog input range.

Don't forget to check the resistors for the wattage! The current through both legs is 24V / (1kOhm + 1.4kOhm) = 10 mA; the wattage is 24V * 10 mA = 0.24 W. You should use resistors rated no smaller than 1/4 Watt, otherwise they will be very hot!

Even if some components fail, you will never get more than 24 Volt from this circuit. Although the analog input range is 0 - 10 V, it most likely can withstand much more without any damage.

The milliamp signal can be produced by a special type of electronic circuit called "current source". Unlike the regular "voltage source", its internal resistance is very high. The only reason to use 4-20 mA signal I know of is its better ability to withstand noise/ voltage spikes and so on - especially if the analog signal wires are long. I don't think you care much about noise in this kind of an application...

Arghhhhh, you are right. I quickly posted without even
looking at the rest of the circuit (I didn't notice
that potentiometers were only 1kOhm). shutit

Thanks for all your replies

I'll study them and check out the links when I get home from work tonight.

Cheers