Voltage drop in a high impedance circuit

[Bootalito]

So I'm trying to explain to my co-workers (automation company of which I am the most versed in how the industrial electronics work) why a 4-20mA current loop is used instead of a 0-10V signal in industrial automation.  I'm using an example that a 0-10V signal and a 4-20mA signal originate from the same panel, travel along identical shielded twisted pair wire, travel the same 200 feet (or just a generally long distance), and experience identical interference from the environment.  The 0-10V signal will be more noisier than the 4-20mA signal, even though the only difference between them is the type of signal.  This is because the 4-20mA transmitter is actively working against any voltage induced in the wire by the interference.  Most controls engineers just stop at "4-20mA loops are less noisy" and never really ask why if voltage signals are transmitted in the same type of wire in the same environment, etc, etc
OK.
So in explaining the above, I had to introduce the concept that there will be no voltage drop in a wire if no current is flowing, as many of them just made the statement that the 0-10V is losing voltage because of "voltage drop".  (Industrial Control analog input cards have sufficiently enough high impedance to consider the current negligible for the purposes of the conversion).  So I made a claim that if you put 10.0000V on a pair of wires and run this pair 70 miles and measure the voltage on the other side, you would get 10.0000V on the other side.  It would be noisy, but it would still be around 10V.  I of course immediately backtracked because a multimeter will induce some current even with a 10MOhm impendance, but again for the purposes of the conversation the point still stood. No current, no voltage drop.  Period.

So someone who has worked a lot with PoE mentioned that a general rule of thumb used in PoE architecture is that you get 2V drop per 100feet of wire.  I asked him what was the specified load that that this frule of thumb is applicable for.  He said, no load, you just get 2V drop per 100feet.  I asked him if you have 48.000V on one side of the PoE wire, and run it 100feet and then test the voltage with a high Independence multimeter you will get ~46V?  He said yes.  I said since there is no current flowing (open circuit) there would be no voltage drop.  (UPDATE: Has has since come around after playing with an online voltage calculator)

However, I don't want to mislead my co-workers.  So my question is this:  Is everything I stated above correct? 
And more specifically if you have 48.000V measured on a 22awg (16.14mR/foot) PoE line and measure it let's say 1000 feet later, with a 10MOhm multimeter, in the REAL world you would read 47.99984506V right? 0.01614R * 2000ft = 32.38R total wire resistance.  48V input / 10000000R multimeter impendance = 0.0000048A.  48V - (0.0000048A*32.38R) = 47.99984506V (in the REAL world)? Yes it would be noisy, but if you could say average the value over 10 minutes or use a low pass filter it would be that voltage right?  He's making me question my sanity!

[w2aew]

Keep your sanity about ya - it all comes down to simple ohms law for DC circuits like this.

[Benta]

I my opinion, you're using the wrong argument.
The important point with 4...20 mA loops is, that they are powered from the controller and can be passive sensors/devices.
0...10 V devices need their own power supply.

I know, I know, some 4...20 mA devices need more power than 4 mA, but that's beside the main point.

[Vtile]

Please put them to measure an 10 meg resistor connected series with 9V battery with regular handheld DMM in DC volt range and explain to you why the meter doesn't show 9 Volts. If you do want to be even more a magician do the same with DMM turned to the DC current range. Brainfart.

Btw 4..20mA is nominal range, it does have error ranges in both ends.

PS. Your current calculation is wrong, you assume that the total resistance is equivalent to the DMM 10 meg, but you do not take account the wire resistance, which adds up to total resistance so correct would be only 48V/10000032.28ohm=4.79998uA aprox. (with precision your typical DMM can't reach) and the the DMM would show 47.9998V so wire resistance is somewhere in 6th or 7th decimal (2am here, I can't bother to verify.  ).

Also the source resistance do have it own effect if the magnitudes of the network are unfavorable.

My edit were sloooooooooooooow. Time to get some sleep.   

[Brumby]

However, I don't want to mislead my co-workers.  So my question is this:  Is everything I stated above correct?

Pretty much.

And more specifically if you have 48.000V measured on a 22awg (16.14mR/foot) PoE line and measure it let's say 1000 feet later, with a 10MOhm multimeter, in the REAL world you would read 47.99984506V right? 0.01614R * 2000ft = 32.38R total wire resistance.  48V input / 10000000R multimeter impendance = 0.0000048A.  48V - (0.0000048A*32.38R) = 47.99984506V (in the REAL world)?

Pretty close - but to be nitpicking, the current calculation I highlighted in blue is not, strictly speaking, correct.  You have not included the ESR of the 48V source, nor the 32.38ohm total wire resistance with the 10M meter impedance.  Your result will change, but with the presumption that the ESR is going to be a lot closer to the wire resistance than the meter impedance, it won't be by any amount that a 5 digit meter could discern.


Edit:    Just noticed Vtile beat me to it.

[Brumby]

(UPDATE: Has has since come around after playing with an online voltage calculator)

Sometimes people need to convince themselves.

[Bootalito]

I my opinion, you're using the wrong argument.
The important point with 4...20 mA loops is, that they are powered from the controller and can be passive sensors/devices.
0...10 V devices need their own power supply.

I know, I know, some 4...20 mA devices need more power than 4 mA, but that's beside the main point.

Yup that's another advantage for single ended analog input cards.  You do get less noise if transmitters are actively powered and the controller is using differential inputs. But using differential inputs usually cost twice as much and it's hard to justify it

[Bootalito]

Please put them to measure an 10 meg resistor connected series with 9V battery with regular handheld DMM in DC volt range and explain to you why the meter doesn't show 9 Volts. If you do want to be even more a magician do the same with DMM turned to the DC current range. Brainfart.

Btw 4..20mA is nominal range, it does have error ranges in both ends.

Haha. It took me way too long(60 sec) to figure out why wouldn't read 9 volts. I'm guessing it would read ~4.5v because it's just putting two 10 Meg resistors in parallel right?.... Damn it, I'm not a 100% percent sure. I got to go do it now

[Vtile]

Edit:    Just noticed Vtile beat me to it.

No I didn't, it took me 30 mins to get my calculator, sort the feets and one finger type my edited post.  I will just now drop from my chair..  

[Bootalito]

Right so just to be perfectly accurate is it for my own education is it?:

0.01614R * 2000ft = 32.38R total wire resistance.
For argument sake 1.00R power supply ESR
48V input / 10000033.38R = 0.00000479998A total circuit current
48V - (0.00000479998A*33.38R) = 47.99983978V

(Assuming you had a multimeter capable of that precision and the input impedence was precisely 10.000000MOhm...)

EDIT: Doh..yeah I see his numbers above....Thanks

[Vtile]

Did you figure out the 9V battery thing already?  ..what I'm still doing here...

[Bootalito]

bah..yeah I'll go do it now...

[Bootalito]

Well that is an interesting result...

Measured with Brymen BM867s (5 4/5 digit DMM) on 500,000 count mode, 10MOhm impendence w/ 60pF capacitance. :
Vbat = 9.5932
Vbat in series with a 9MOhm resistor (didn't have a 10) = 8.0V +- 0.1 (not steady)

It has something to do with the buffer op amp not getting enough current (little as it may be) because the current is being severely limited by the external resistor in series.  If the base of the transistors inside the op amp can't get enough current then the buffer amp won't be able to function correctly (for some reason).  Then...something, something, something, the buffer output has a lower voltage.  But why specifically?....

[IanMacdonald]

Not sure if you already appreciate this, but the 4-20mA range is based on the 4-20psi pneumatic control signals that used to be used before electronics. Still are some places, for example where there is a gas explosion risk. 

The reason for using current is that volts drop does not affect accuracy, and the reason for a non-zero origin is so that a disconnected sensor shows up as a fault instead of a zero quantity. 

[IanB]

The reason for using current is that volts drop does not affect accuracy, and the reason for a non-zero origin is so that a disconnected sensor shows up as a fault instead of a zero quantity.

And greater immunity from noise.

[Bootalito]

Not sure if you already appreciate this, but the 4-20mA range is based on the 4-20psi pneumatic control signals that used to be used before electronics. Still are some places, for example where there is a gas explosion risk. 

The reason for using current is that volts drop does not affect accuracy, and the reason for a non-zero origin is so that a disconnected sensor shows up as a fault instead of a zero quantity.

Yup they are still in use in older plants, however no one is installing pneumatic controls for new installations. There are three methods used now: 1. The current technology is using IS I/O (intrinsically safe) in class 1 (explosive) areas. IS I/O adds zener diodes to clamp the voltage to prevent sparks in the field. 2 Alternatively you can use XP (explosion proof) encloses, and  filled and sealed conduit. 3. Air purged boxes that create a positive pressure inside an enclosure to prevent explosive gasses from getting in.

Also there is no voltage drop in a 0-10V control signal as there is no current flowing.

[Bootalito]

The reason for using current is that volts drop does not affect accuracy, and the reason for a non-zero origin is so that a disconnected sensor shows up as a fault instead of a zero quantity.

And greater immunity from noise.

Yes, but....to be perfectly precise, they are not immune from noise, they just actively correct for it using feedback.
As per my original post I was teaching some of the other control systems engineers and presented them with a scenario where you have a voltage and a current signal experiencing the same amount of noise but the current signal is less noisy.  This isn't because the wires or shielding or current signal is inherently or magically better, its because current loops actively correct for any current induces in the wire by noise.

[tecman]

Today 4-20 ma is still the defacto standard in most industries.  The reason is simple:  Since the signal is a current source, any resistance in the wiring/connections does not impact the signal (to a point).  The signal source will have some voltage supply (typically 24 VDC).  The "receiving" end has a resistance.  The receiving resistance requires a "burden voltage" which is typically low.  So voltage drops between the transmitter and receiver is V supply, minus what is needed by the transmitter driver, munus the receiver burden voltage.  So typical is a 100 ohm in the receiver, at 20 ma it will have a drop of 2 volts (burden voltage).  If the transmitter needs, lets say. 2 volts for the driver, you would have 24v-2v-2v=20 volts of voltage drop and still work.  This corresponds to 1000 ohms drop.  While excessive, it shows that the signal integrity is quite robust.  Connections that have a varying resistance will also work in many cases.

Other reasons are fault detection and noise.  For fault detection, signals that are under 4 ma or over 20 ma are considered invalid and a fault in most systems.  Noise immunity is also quite good.  The circuit impedance is the receiver's resistance plus wiring resistance.  Usually a few hundred ohms, it is again quite robust against noise sources.  Another feature is the ability to wire more than one receiver in series in the circuit so that several devices can monitor a signal.  As long as the total burden voltage of the receivers is below the transmitters output capability (allowing for wiring loss) the signal loop will work.  Last 4-20 systems are available that can be used in intrinsically safe (IS) areas where explosion risks exist.

Again the advantages make it still today the default for analog signals in most industries.

paul

[Vtile]

Well that is an interesting result...

Measured with Brymen BM867s (5 4/5 digit DMM) on 500,000 count mode, 10MOhm impendence w/ 60pF capacitance. :
Vbat = 9.5932
Vbat in series with a 9MOhm resistor (didn't have a 10) = 8.0V +- 0.1 (not steady)

It has something to do with the buffer op amp not getting enough current (little as it may be) because the current is being severely limited by the external resistor in series.  If the base of the transistors inside the op amp can't get enough current then the buffer amp won't be able to function correctly (for some reason).  Then...something, something, something, the buffer output has a lower voltage.  But why specifically?....

That is actually interesting result, as one should expect the meter to show 4.5 volts give or take some decimals in case of 10 meg. resistor and 10 megaohm meter input impedance (~10 megs resistance). My calcular says that most obvious explanation is that the 10megaohm resistor were infact only 2 megaohms (skin on the finger tip is acting as parallel resistor?). Your result were that much a surprising result that I did read what that Brymen manual says about the input impedance and with a quick look it is a fixed. The lack of current sounds too far fetched especially since the meter did show 8 volts not 8 millivolts or other small number near zero.

It is normal voltage divider, just like your example of the voltage drop in wires, but more significant results and shows clearly the impact of the measuring device. Another variation is to put two identical resistors series lets still say 10 megaohms and measure the voltage over one of them. While the real potential difference is aprox. half of the source voltage (assuming the source ESR is insignificantly small) the meter however would show about 3.2 volts, in case of 10 megaohm input impedance and 9.5 volt source. This is because the second part of the divider drops to 5 megs. from the meters impact (parallel impedance* or current division which ever way is more familiar).

Edit. The reason I chose 10 megaohm resistors in my examples is because of the fact that most handheld DMMs do have input impedance ranging from 1 megs to 11 megaohms.

*inductance corrected to impedance. Sorry about silly mistake.

[Bootalito]

Well that is an interesting result...

Measured with Brymen BM867s (5 4/5 digit DMM) on 500,000 count mode, 10MOhm impendence w/ 60pF capacitance. :
Vbat = 9.5932
Vbat in series with a 9MOhm resistor (didn't have a 10) = 8.0V +- 0.1 (not steady)

It has something to do with the buffer op amp not getting enough current (little as it may be) because the current is being severely limited by the external resistor in series.  If the base of the transistors inside the op amp can't get enough current then the buffer amp won't be able to function correctly (for some reason).  Then...something, something, something, the buffer output has a lower voltage.  But why specifically?....

That is actually interesting result, as one should expect the meter to show 4.5 volts give or take some decimals in case of 10 meg. resistor and 10 megaohm meter input impedance (~10 megs resistance). My calcular says that most obvious explanation is that the 10megaohm resistor were infact only 2 megaohms. As this were that much a surprising result that I did read what that Brymen manual says about the input impedance and with a quick look it is a fixed. The lack of current sounds too far fetched especially since the meter did show 8 volts not 8 millivolts or other small number near zero.

It is normal voltage divider, just like your example of the voltage drop in wires, but more significant results and shows clearly the impact of the measuring device. Another variation is to put two identical resistors series lets still say 10 megaohms and measure the voltage over one of them. While the real potential difference is aprox. half of the source voltage (assuming the source ESR is insignificantly small) the meter would show about 3.2 volts, in case of 10 megaohm input impedance and 9.5 volt source. This is because the second part of the divider drops to 5 megs. from the meters impact (parallel inductance or current division which ever way is more familiar).

I checked the resistance of the 9MOhm resistor I used. It was ~9MOhm. I'd be curious to see the results of someone else's test. I don't have the background to say for sure what's happening. If anyone knows for sure, please chime in.

[Vtile]

Yes, but did you squeeze it (the resistor) between your fingers while you did the voltage measurement (something I didn't take account that someone might do it). Ie. When I measure resistance of my finger tip, ½" distance with my Keithley 197A it shows figures from 2.3 megaohms to four megaohms depending the applied pressure and spot. In high impedance measurement the leakage can show up in odd ways.

Epilogy: Note that in this measuring setup (meter in series) your meter acts more like a really sensitive current meter than voltage meter.  If your meter do have fixed 10 megaohm input impedance in DCV range and the smallest value it can show is 10 microvolts then ten microvolts shown in screen represents one picoampere of current through the meter. That is in magnitude where skin is an good conductor.

[IanB]

I checked the resistance of the 9MOhm resistor I used. It was ~9MOhm. I'd be curious to see the results of someone else's test. I don't have the background to say for sure what's happening. If anyone knows for sure, please chime in.

OK, here's a test I did. A 9 V battery, two meters each with ~11 M input resistance, one a BM896s, the other an RS 22-812. Individually the battery measured 9.90 V on the 869s, 9.89 V on the 22-812. I put both meters in series and measured, giving 4.94 V on the Brymen, 4.95 V on the 22-812. As expected the voltage was pretty much split in half given that both meters have about the same resistance.

In case it might be thought the second meter is not a passive resistance I dug out a 10 M resistor and put that in series with the Brymen. This time the meter read 4.97 V, which is again approximately half of the total voltage.