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Singapore
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 « on: February 11, 2011, 06:58:38 am » Bigger Smaller Reset

hmm I'm building a shield to condition the output of a standard 4-20mA sensor for the Arduino, and I would like to make it safe for my poor Arduino.  I have the Zener set up as shown below, but I'm still getting the full 9 volts at "Arduino Input Pin" (it isn't yet actually attached to my Arduino!). It is supposed to be a 5.1v zener, so why isn't it saturating and conducting when at 9v?

PS: this works just fine with and without the protection diode- but I just want to prevent 9+ volts from damaging my analog input.  Also, there is of course a P/D resistor on the input pin not shown.
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 « Reply #1 on: February 11, 2011, 07:34:30 am » Bigger Smaller Reset

You should be asking yourself where the 9 volts comes from and you could then solve the problem.

A 4-20 signal should feed into a 250ohm resistor (R in your drawing) to produce a 1-5 volt signal for feeding into your arduino.

If the unstated value of R is greater than 250 ohms then the sensor will drive the voltage higher in an endeavour to ensure that the current range remains within the 4 to 20 scale.

I suspect your value of R is much too large.

Having said that, the zener should have conducted to clamp the voltage so I suspect you've blown it and it no longer works.

jack
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Singapore
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 « Reply #2 on: February 11, 2011, 08:26:54 am » Bigger Smaller Reset

Oops! didn't label it- it is 180ohm- making the signal work out to .72v to ~3.6v for the input.  I want to protect the input with the diode though, in case the resister fails/becomes loosened from the circuit due to my non-artistic soldering work, or the sensor power voltage increases (since it is regulated external to the Arduino- taken from Vin pin). I'm not sure how I could have blown it, since it is new.  Perhaps it was mislabeled?

I am using it correctly, yes? black stripe side away from ground?

The 9v is the power wire for the sensor, which I have temporarily simulated with resistors to make a 4mA signal (at 9v) and a 20mA signal.
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 « Reply #3 on: February 11, 2011, 10:13:09 am » Bigger Smaller Reset

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I am using it correctly, yes?
No.

The diode should be from the input to ground, the resistor should be between the arduino input and the source.
I suspect you are getting the full 9v because you have blown up your diode. There is nothing (but the impedance of the source) to limit your current in the setup you had.
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 « Reply #4 on: February 11, 2011, 11:50:02 am » Bigger Smaller Reset

You really should draw out the complete circuit showing the sensor, loop source voltage wiring, load resistor and ground connections to the arduino. Also most 4-20ma sensors (usually called transmitters) required a minimum of 12vdc of loop voltage (called minimum compliance voltage) to operate through the full 4-20ma measurement range. Also a link to your sensor might be useful. 4-20ma current loops are not difficult to use, but one must account for the complete circuit. I worked with 4-20ma current loops for decades, so if you can provide all the details asked for above, I'm sure we can come to a solution for you.

Lefty
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 « Reply #5 on: February 12, 2011, 04:05:06 am » Bigger Smaller Reset

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The diode should be from the input to ground, the resistor should be between the arduino input and the source.

Grumpy, Since the output of the sensor will vary current, shouldn't I need to measure the voltage drop across the 180ohm resistor?

So, I have not received my sensor yet, and I am working on the hardware (and sketch) to read it.  I am simulating the low value sensor output (4mA) by running my 9v Vin through 2250ohm resistance, giving 4mA current.  Of course, the sensor won't work off 9v, but that won't matter for the design of the circuit since 4mA is 4mA right?

With this setup, how could I possibly have blown my Zener? Running the 9V through my simulated sensor at 2250ohm would only give .03W, below my .5W rated zener!?

I would like a protection diode on it in case for some reason the sensor malfunctions and outputs 28 or more mA of current, which will result in more than 5v at the input pin.
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 « Reply #6 on: February 12, 2011, 04:57:31 am » Bigger Smaller Reset

I fail to understand your circuit.  Where is your 9 volts coming from to drive the milliamps and where is the P/D resistor you mentioned in your opening query.

You really do need to provide the full information if you want informed and correct answers, rather than best guesses.

jack
 « Last Edit: February 12, 2011, 06:50:46 am by jackrae » Logged

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 « Reply #7 on: February 12, 2011, 05:49:41 am » Bigger Smaller Reset

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I am simulating the low value sensor output (4mA) by running my 9v Vin through 2250ohm resistance, giving 4mA current.
Now you said nothing of a 2.25K resistor, that saves the zener. If you are relying on the current limiting function of the sensors then you don't need anything in line with the zener, otherwise you do. I know it drops the voltage but you have to limit the zener current some how.

There are two types of current loop output, active and passive. What sort do you have as it makes a difference to if you have to provide power of if it provides power.
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 « Reply #8 on: February 12, 2011, 12:04:38 pm » Bigger Smaller Reset

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There are two types of current loop output, active and passive. What sort do you have as it makes a difference to if you have to provide power of if it provides power.

The OP is showing an example of a passive  current transmitter (sensor) in that he is providing the 'loop voltage' via his 9vdc Vin pin. Passive 'transmitters' are often called 2-wire transmitters in that they only have two terminals and can only be wired into a series loop that already has loop voltage applied. 4-wire transmitters are active current transmitter where 2 of the wires are providing a power source and 2 of the wires are a active 'constant' current where the actual constant current value is manipulated by the transmitter to represent the process measurement value.

The drawing showing a 5.1vdc zener diode as a overvoltage protection device should work as you want. But be aware that zener 'knee' voltage value are not real sharp and a 5.1volt zener may start to conduct a little as the 180 ohm resistor voltage drop approaches close to 5.0vdc.

Last point is what I mentioned before, that current transmitters are designed with a minimum and maximum loop voltage source. Typically something like 12-36vdc and will not be able to work at less then their minimum voltage specification.

Lefty

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 « Reply #9 on: February 12, 2011, 12:18:22 pm » Bigger Smaller Reset

Your resistor is in the wrong place and doing nothing useful. If you moved the resistor to the input line and changed the value to at least 1K it should work. I would also recommend changing the zener to 4.7 volt.
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 « Reply #10 on: February 12, 2011, 12:36:03 pm » Bigger Smaller Reset

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Your resistor is in the wrong place and doing nothing useful. If you moved the resistor to the input line and changed the value to at least 1K it should work. I would also recommend changing the zener to 4.7 volt.

I don't think you understand how industrial process control 2 wire 4-20ma current loop measurements work.

If you are talking about the 180 ohm resistor, it is of course correctly placed and is essential for the operation of this current measurement loop. The loop will have a constant current value between 4 and 20ma set by the variable resistance control inside the transmitter (sensor). At 4 ma the 180 ohm will have .72 volts dropped across that and that is what the Arduino will measure for a process 'zero' measurement value. At 20ma the resistor will have 3.6 volts dropped across that and that is what the Arduino analog input pin will measure to represent 100% process measurement value.

That resistor's function is to convert the loop current measurement value to a voltage measurement value that stays within the arduino's 0-5vdc analog input voltage range.

The OP desire to have a zener wired across that resistor is as a protection device in case something goes wrong in the sensor or sensor wiring such that greater then 5vdc was dropped across the 180 ohm resistor, which would exceed the maximum safe voltage for a arduino I/O pin and could damage the chip.

Lefty
 « Last Edit: February 12, 2011, 12:42:10 pm by retrolefty » Logged

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 « Reply #11 on: February 13, 2011, 11:41:16 am » Bigger Smaller Reset

Thanks Lefty! I think I may have a mislabeled few Zeners... perhaps I grabbed 9+ v Zeners instead.  I haven't had a chance to pick up some new ones that will hopefully conduct at 5.1v.

I also want to add some other things to make this circuit even better- a 10k protection resistor for the input pin, a 10k protection resistor for the Zener, and also a 1uF cap to smooth the sensor output.  Am I on track with the below? I will be datalogging with a gas sensor, so the output doesn't need to be super fast.  If this looks like a workable circuit to your expert eyes, I will be posting it along with my finished sketch to help anyone with 4-20mA sensors- my first contribution to the community! (yay!!!)
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 « Reply #12 on: February 13, 2011, 01:21:05 pm » Bigger Smaller Reset

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If this looks like a workable circuit to your expert eyes, I will be posting it along with my finished sketch to help anyone with 4-20mA sensors- my first contribution to the community! (yay!!!)

LOL, expert eyes aside, most looks good on your latest design. The resistor/cap on the analog input pin creates a low pass filter that can help with noise and jumpy signal. I only question your adding a series resistor with the zener diode. This would only protect the diode rather then the analog input pin because it would drop voltage across the resistor rather then clamp the voltage at the analog input pin to the zeners voltage rating.

As originally used the zener is acting like a sacrificial part and should not have that series resistor. I would use the largest wattage zener you can fit in and leave it as a poor man's fuse. I think your loop would limit worst case current in most cases to some value that won't destroy a larger wattage zener and still offer the voltage clamping protection you want to have.

Lefty
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