Is it common to install a diode in series with a 4-20mA circuit? 1

[bdn2004]

To prevent backfeeds on a 4-20mA circuit is it common to put a diode in the loop? I know there is a voltage drop across the diode, and the voltage being relatively small, I wonder if this is a good idea?

 

[xnuke]

Usually the diode isn't there to prevent backfeeds, it's there to allow an ammeter to be be connected across the diode. While ammeters typically aren't connected across a component, in this case the now-parallel diode's resistance will then cause the current to shunt through the ammeter, allowing a current reading without requiring breaking the circuit to insert the meter.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
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[IRstuff]

There are two possible reasons for not providing such protection:
> The destination is supposed to be a purely passive device, and there is no expectation that such a situation could occur
> The addition of protective circuitry has the potential of introducing additional noise. For the diode example, the rectification feature of the diode makes it likely that coupled noise winds up being rectified, resulting in a non-zero mean, resulting in additional noise that can no longer be filtered.

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[bdn2004]

xnuke....

So all the current through the diode will divert through the ammeter? What is the resistance of the diode?
I guess this for troubleshooting?

 

[itsmoked]

It's not the diodes 'resistance' that allows the full shunt. It's the diode's forward voltage. That voltage is about 0.6V. Since the shunt inside the meter would never reach a voltage of 0.6V nothing of merit would rather go thru the diode, it all goes thru the meter.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[bdn2004]

Itsmoked,

Can you explain what that means...forward voltage? And perhaps the basics of an ammeter in a nutshell?
I'm not following how this works and why.


24vdc---+--[Diode]->+-->23.4V
| |
-----AM------

?

 

[itsmoked]

Forward voltage of a diode is the initial voltage required before a diode begins to conduct. It's like pushing a car. You have to push a certain amount before it will start rolling. The electrons and holes inside a PN junction (the heart of a diode) are not interested in conducting until they are impressed by the correct amount of forward voltage.

An ammeter is nothing more than a resistor, (generally of very low value), that is observed by a voltmeter. Ohms law describes what the voltmeter will see given any specific current.



Keith Cress
kcress -

http://www.flaminsystems.com

 

[IRstuff]

http://en.wikipedia.org/wiki/P–n_diode#Diode_law

TTFN
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[bdn2004]

So we are saying that when the meter leads go around the diode....there will 0V of potential difference in this loop. Sort of like a bird sitting on the high voltage line and lives to tweet another day. But since the diode requires at least 0.6V of potential difference all the current will divert through the meter?

 

[IRstuff]

The logarithm of the diode current can be seen to be exponentially dependent on the input voltage down to less than 0.2V Vf. But, since your circuit is running a minimum of 4mA for a valid signal, that's what requires the >0.6V:

http://www.physics.csbsju.edu/trace/CC.html

Diode resistance can be viewed as roughly the inverse of the diode current, so the parallel resistance of the diode will be at best 8ohms at 100mA, in the cited webpage, and if the current meter has a resistance of say 0.5ohm:

http://www.fluke.com/fluke/usen/wir...ke-CNX-3000-Wireless-Multimeter.htm?PID=74899

then at 100mA, the voltage burden of the ammeter will only by 50mV, where the diode resistance is many kohms.

TTFN
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[itsmoked]

So we are saying that when the meter leads go around the diode....there will 0V of potential difference in this loop. Sort of like a bird sitting on the high voltage line and lives to tweet another day. But since the diode requires at least 0.6V of potential difference all the current will divert through the meter?

Pretty much, with the exception that an ammeter has a burden and so has a finite voltage drop - not 0V.

Keith Cress
kcress -

http://www.flaminsystems.com