# 4-20mA repeater output!

Discussion in 'General Electronics Discussion' started by sid2286, Aug 24, 2011.

1. ### sid2286

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Aug 24, 2011
hi, I'm trying to build a circuit wherein, i have a 4/20mA input...I calibrate using a micro-controller, display it on LCD and the same is further given to PLC. or external load. basically a loop indicator sort of thing!
I have a circuit with me...but for some reasons its not working!

the circuit is in the attachment !
it works without load resistor... however when i change the load it stops working!

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2. ### duke37

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Jan 9, 2011
Which is the load resistor, is it the sense resistor R6?
The op-amps do not have a common ground with the pic.
The current loop has no defined voltage so that the inputs to the op-amps may go outside the permitted levels. If the loop is floating, you could tie it down with a high value resistor each side R6 to ground.

3. ### Resqueline

2,848
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Jul 31, 2009
You'd need 200V to drive 20mA through a 10k resistor (dissipating 4W). Clearly that is wrong for a current loop.
R6 needs to be 1k ohms or less. Maybe it was even meant to be 10 ohms (10R). 1k = 20V, 100 ohms = 2V, 10 ohms = 200mV.
The op-amps have no gain, they only remove any common-mode voltage present, so the PIC measures whatever voltage there is across R6.
Choose a resistance that matches the wanted/programmed measuring range.

4. ### sid2286

102
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Aug 24, 2011
It was my mistake to show R6 as 10K it is 10E. How do I design the circuit? The Opamp I'm using is MCP609 which has a single supply, so i dont have to create a virtual ground in this case.
how do i define a circuit wherein, even if i change the load resistor i get a constant Tp1 and Tp2 for my 4-20mA input.
Please suggest a circuit or a modification i can try!

5. ### Resqueline

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Jul 31, 2009
I'm not quite sure I see what your needs are. R6 needs to be constant and correct for the PIC A/D & programming. 10 ohms seems like a sensible value.
If by load resistor you refer to some other device in the loop then the answer is that the current remains constant with varying resistance (within certain limits of course).
Feel free to post a block diagram/ drawing of the complete setup if this isn't clear.

6. ### sid2286

102
1
Aug 24, 2011
Yes you got me all right! Current remains constant accepted, but with varyin load(load resistance may vary from 0-250 ohms) my input to A/D changes.. And is not constant as you suggested.. Is there anything wrong with the circuit?

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,418
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Jan 21, 2010
That is exactly right. A constant current will produce a fixed voltage drop across a resistor, but the voltage drop depends on the resistance.

edit2: OK, I've only ever seen digital current loops. And I note that they may be analog. so in a 4-20mA current loop carrying an analog signal, the current can vary between 4mA and 20mA (0mA means the loop is broken)

So if the current loop is carrying an analogue signal you'll see a variable voltage.

Last edited: Aug 25, 2011
8. ### sid2286

102
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Aug 24, 2011
So does that mean that if I have a Zero load...that is my loop(i.e my load) is connected directly to the ground then there will be different voltage across TP1 and TP2 wrt ground. and when I connect say 50E load then there will be different reading across TP1 and Tp2 wrt ground.
so what kind of circuit will fetch me a constant TP1 and TP2 wrt ground irrespective of what my load is? so that I can program my A/D with PIC properly to have a good calibrate value.
Basically i want to display 0 P.V for 4mA and 100P.V for 20mA. irrespective of load, programming can be handled by me but for that I need a constant input to A/D for 4mA and 20mA again irrespective of load.

Please suggest something...it should be possible.

9. ### duke37

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Jan 9, 2011
Please give details of what is conncted to TP1 and TP2. Where is the load resistor?
Your diagram shows 4-20mA output from TP2 but there is no output from here, TP1 and TP2 are both inputs.

10. ### sid2286

102
1
Aug 24, 2011
first of all let me thank you all.

See I will explain it again!

I have a 4-20mA input which I need to give to A/D using a differential amplifier( Differential amplifier coz I thought it would work). I need a constant value to my A/D for 4mA and 20mA for me to calibrate(write a program using PIC) irrespective of whatever the load is R7. I have attached a file, please go through it.

The problem I'm facing is...everytime i change the load the input to A/D is different. So i need to find a way wherein even if the load is changed the input to the A/D will remain same for 4mA and 20mA.

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11. ### Resqueline

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Jul 31, 2009
Ok. Does it change for every, say 10 ohms step up on the load resistor, or does it keep constant - up to a certain R7 value - and then fall off?
Do you have the spec's of the current source?

12. ### sid2286

102
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Aug 24, 2011
It is not constant.. For 0 ohms there is different values and for 10 ohms is completely different values..it changes with different R7! The current source is a transmitter which gives 4mA and 20mA. No other specs:-(

13. ### duke37

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Jan 9, 2011
This is how I envisage the thing should work.
Diagram A shows the 4 to 20mA signal coming from an external source. No amp is needed, the voltage across the resistor will be 40 to 200mV and would go straight to the pic.

If the transducer needs to be in the lower part of the loop then the amp is necessary to reference the voltage across the 10R relative to ground. See diagram B. A second op-amp is not necessary. The inputs to the op-amp must not rise above its supply.

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14. ### Resqueline

2,848
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Jul 31, 2009
Ok, well, presuming you've connected the devices properly (in series) then evidently the so-called current source is not a current source.
Seriously, no spec's at all? The open circuit voltage is an important aspect, and having no common ground references (has to be floating) is another.

15. ### sid2286

102
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Aug 24, 2011
the current source is a 3 wire transmitter..it works on 24v and has a sensor attached. when there is a change in pressure/temperature the current changes, so basically for 4 mA is for normal condition and for maximum change it will be 20mA. but it can be anything (instrumentation device) which has a output of 4-20mA.

So i simply wanted to make a device wherein, when it is normal (when i receive 4mA) i should display zero on display. and for maximum change it should be (say 100). for that to happen I should get a same A/D input for whatever load(R7)and also my R7 should receive 4-20mA as input.

so basically my circuit should receive 4-20mA input display some reading and transmit the same 4-20mA as an output!
with the circuit i made, i'm able to transmit the current as an output(current that I receive), but due to changing load I'm not able to give my A/D a stable input.

thanks once again!
SId

Last edited: Aug 26, 2011
16. ### JoeyAVR

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Aug 25, 2011
> everytime i change the load the input to A/D is different

You have said your current source can vary between 4 and 20mA, but nothing else. is it a variable constant current source (i.e. you select the current and it stays the same, regardless of load within certain voltage limits).

If it isn't, changing the load resistance will change the current.

If it is, and the ADC is inconsistent with what you expect the current to be, then you are using a load outside the limits of the current source (possibly too great a resistance and it can't generate a high enough voltage to get the required current).

Joey

17. ### sid2286

102
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Aug 24, 2011
Thanks for 'Ohms law' suggestion! But i'm stil wondering what it has to do with differential amplifiers output.. Hope you 've a good look at the circuit and suggest something accordingly..

18. ### JoeyAVR

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Aug 25, 2011
Sorry, I posted without seeing the messages on page 2! My mistake, not yours.

The differential amp measures a voltage across a resistor.

The voltage is generated by the current from your transducer according to Ohm's Law.

You can use the equation V=IR to work out what the voltages at different parts of the circuit will be for different currents.

If you put the output into a 1K resistor to ground, the voltage at 4 mA will be:

V = I x R = 0.004*1000 = 4volts. At 20mA, it will be 20 volts.

Let's try a 220 ohm resistor.

0.004*220 = 0.88 volts

0.020*220 = 4.40 volts

Now those are both within the measuring range of your ADC which can measure from 0V = a count of 0, 5V = a count of 256. So no differential amp required.

At 4mA you will get a reading of 0.88/5 * 256 = 45 (no figures after the decimal point)

At 20 mA you will get a reading of 4.4/5 * 256 = 225

Subtract 45 from each number and you will get a result in the range 0 to 180.

You can work out the maths to get a result in the range 0-100 from here!

If the reading changes slightly, don't be surprised. these sensors are affected by temperature and most ADCs aren't stable in the bottom digit either. You can add capacitors to power supply, reference and input to improve this.

19. ### duke37

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Jan 9, 2011
As I pointed out earlier and as Joey says, you do not need an op-amp and you certainly do not need R7. The current source is intended to run into something approaching a short circuit and adding R7 will not help.

An accurate voltage source should feed a very high resistance.
An accurate current source should feed a very low resistance

The op-amp has a maximum permitted supply voltage of 5.5V and the inputs must not go above this, adding R7 possibly takes the input too high, use Ohms law!

20. ### sid2286

102
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Aug 24, 2011
Guys you are still not getting the appilication of the circuit!
As the name suggest 'loop indicator' i need to show the measure of current following..
So R7 is something which is not in my hand it can be anything whatever the enduser wants to connect.. So i want a circuit in which whatever the load R7 is the A/d input should be stable..