Analog Input With Potentiometer HELP!!

zmanvortex

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Join Date
Apr 2002
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Nebraska
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Ok, I am trying to get a 0-20ma or 4-20ma analog input off of a potentiometer into an analog input card.

I want to use the AutomationDirect F0-04AD-1. On the product page it specifically says it is for use with 24VDC power supply.

When I went to the spec page it showed input impedence of 125ohms at 1/8 watt.

Now correct me if I am wrong but 24VDC at 20ma would be almost 1/2 watt. What gives? I called tech support and they said that the 24VDC should be used to power a 0-20ma/4-20ma current generating device that should be limited to 1/8 watt output (which would mean 6.25volts at 20ma). I have a hard time believing that they would specify what voltage should be used to power the current generating device as long as it output the correct values.

What I had planned was using 24vdc with a 2k potentiometer and a 1k current limiting resistor in series with the wiper and the input of the analog card.

This would give me 0-20ma of current over the range of the pot but with the pot turned all the way up would also give 24vdc to the analog input which would result in almost 1/2watt.


I am just finding it hard to believe that it so hard to use a potentiometer with an analog input card.

Thanks.
 
For panels and permanent installations we always use a signal conditioner to convert the pot into 4-20 mA. Many people supply them, such as Red Lion, M-Systems, or Action Instruments.

For shop testing we use a pot in series with a resistor to give us the 4-20 mA with a 24 VDC source. And yes, it is 1/2 Watt. There isn't anything to do about that.

When sizing your resistors, remember that the input card has a 250 Ohm resistor in series with the current loop.
 
Thanks Tom. We are planning on using the F0-04AD-1 which is for the expansion slots on the 05 and 06 plcs. This has an internal resistance of 125ohms and 1/8 watt instead of 250ohms 1/2 watt.

The 250ohms is for the larger F2-04AD-1 for the 205 series rack. I also noticed this has a 1/2 watt allowance on the input which is why you were able to use the pot with the resistors.

Unfortunately I am only allowed 1/8 watt on the F0-04AD-1, so it makes it tough to use the supplied 24vdc.

We are looking at replacing an external adjustable timer with a potentiometer and analog card, but it would not be a viable option if our electricians had to do a bunch of solid state tricks to get it to work.
 
Hi zman,

The 24Vdc for the analog input card is the supply voltage only; It's not directly intended for the inputs. The actual input voltage is the result of input current multiplied by input impedance. So, 20mA into 125 Ohms gives you 2.5 Volts.

I must admit 125 Ohms as an input impedance sounds a bit odd, a more common value is 250 Ohms. Try to use a 5Vdc supply (or use a 3-legged voltage regulator like an 7805 and feed this with the 24Vdc) along with a standard 1kOhms carbon potmeter (0.1Watts). Add an extra 120 Ohms resistor in series with the analog input to make sure the input current does not exceed 20mA.
 
The 24Vdc for the analog input card is the supply voltage only



I guess I don't understand what you are saying here. The analog card logic is powered off of the rack and the inputs are simply a current loop. Supply voltage for what?

I was trying to utilize the 24vdc that we have available in the panel to get a 0-20ma current loop off the pot. I am able to do this with one 1k resistor in series with the wiper of the pot and the input card, but I exceed the 1/8 watt limit of the input card.

Keep in mind that our electricians have to wire this so I have to avoid soldering resistors or adding voltage control devices all over the place to get this to work.

No matter how you look at it, I am not going to be able to use the 24vdc into the card, so I will have to look at possibly going to the 0-10vdc analog card (little more expensive) and get a 10vdc power supply and run that through the pot.
 
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Must admit I'm more familiar with Siemens, and as fas as I know all of their analog modules need to be 24Vdc powered.

Anyhow, 20mA into 125 Ohms equals 2.5V. Therefore, consumption at 20mA is only 50mW. If you're allowed to dissipate 1/8Watts, it means you can raise the input current to 31mA without damaging the analog input.

If you insist on using 24Vdc to start from, you need (an) extra resistor(s) to limit the current (read dissipation).
 
First of all thanks for your help Sparkz.

Anyhow, 20mA into 125 Ohms equals 2.5V. Therefore, consumption at 20mA is only 50mW. If you're allowed to dissipate 1/8Watts, it means you can raise the input current to 31mA without damaging the analog input.

This suggest that I supply the correct voltage to maintain the correct wattage. Rather I figured that the voltage is fixed at 24vdc.

If you insist on using 24Vdc to start from, you need (an) extra resistor(s) to limit the current (read dissipation).

This is exactly what I have done. To get 20ma I would need a 1075ohm resistor to limit the current (24v / .020a = 1200ohms - 125ohm card impedence = 1075ohms). But once again this is over the maximum wattage allowed (24v x .020a = .48 watts).
 
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Ok, I have found a 5vdc power supply that we used on another machine.

I think I will use this 5vdc with a 1k pot and a 125ohm current limiting resistor in series with the wiper and analog input.

The 1k pot seems to give me the best linearity.

This will give me 0-20ma with a maximum of .1 watts.

5v \ 125ohm resistor + 125ohm card impedence = 20ma.

5v x .020a = .1 watts.

Not the solution I wanted but it will work. Thanks for the help guys.

I am still confused why the product page stated that this unit is designed for use with a 24vdc power supply.

Question for Tom. How much do the signal conditioners cost? This might be another option.
 
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Think you're mixing one or two things together...

Limiting the current using this 1075 Ohms resistor, also means you limit the voltage over the analog input! As I said before, 20mA into 125 Ohms means there's only 2.5V left over the analog input, the rest is dissipated in the 1075 Ohms resistor.

Try the attachment for a change... Remember NOT to connect the 24Vdc supply BEFORE connecting the analog input (or you'll burn the potmeter).
 
Oops

Forgot the attachment...

Current_limiter_0-20mA.JPG


Sorry, but the potmeter needs to be 1/2W as well if you want to use the full spectrum (0-20mA). ...Got the sketch off a 20mA fine tuning circuit.
 
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Yep, you are correct Sparkz. And they call me an electrical engineer.

I don't know why I missed that.

I just kept treating the current limiting resistor in series with the card impedence as one resistor and that is where I messed up.

I guess if you don't use it enough it goes away.

Thanks so much for the help.
 
You guys are getting your gozintos mixed with your gozoutas!

It doesn't matter if you use 500 VDC to power the circuit, 20 mA through the 125 Ohm resistor on the I/O card is only 2.5 VDC and 0.05 Watt. I don't know why they use 125 Ohms, but I've seen all kinds of crazy values from various manufacturers.

The wattage for any device is the voltage drop through that device times the current through that device. The power for the card logic in this case is coming off the backplane. In many other cards you need to use a 24 VDC power supply to power the logic in the card.

I usually pay asround $200 for the signal conditioner cards.
 
zmanvortex

I hope you don't need the Analog current to change in a linear fashion as the pot is turned. The current to the Analog Input flowing through part of the pot is going to make the response very non-linear. As a rule of thumb, for linear response, the resistance connected to the pot wiper should be at least 10x the resistance of the pot.

I'm with Tom, use a signal conditioned.
 
Think I just saw some gozintos, Tom...

... but can't get near to any gozoutas...


Did I forget to mention that response is nowhere near linear?
 
He is going to use a 5 volt supply now. That will help. For others trying this with 24DC, add a 5.1v zener and 4.7uf Tantulum cap across the pot. That will make it more linear by creating a simple 5v supply.
For true linearity, the final addition is an Op Amp after the pot wiper.
Am I getting too complex now? What I've done is explain the basics of the Signal Conditioner that Tom mentioned.
 

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