Level Sensor and Remote display

[ADWSystems]

I'm working on a level sensor with a remote display. The sensor is a ratiometric hall effect with an OpAmp Buffer buffered output. The display is a LM3914 bargraph driver and a series of individual LEDs. The sensor and display will be atleast 25 feet apart. Based on that my concerns with the system are centered on powering it and transmitting the correct voltage signal from the sensor to the display (25-30 feet away).

As an additional feature, I would like to have only one power source and to be able to plug the wall wart power supply into either the sensor or display. I was thinking of using a 6-7V wall wart with 7805 regulators in both the sensor and display side. The cable between them would have wall wart power, system ground, and the level signal. I'm worried about the voltage loss between the sensor and display skewing the signal.

Any ideas on how to minimize the signal loss (voltage drop) would be greatly appreciated.

 

[ericgibbs]

hi,
The input impedance of the LM3914 is high, about 50nA typical, I would not expect any voltage drop thru a 25ft cable run.
Make sure the sensor output can drive the cable, is the cable screened/twisted etc.??

You may need a 27R on the output of the sensor in series with the cable, also grounding will be important.

 

[dougy83]

I'm worried about the voltage loss between the sensor and display skewing the signal.
Any ideas on how to minimize the signal loss (voltage drop) would be greatly appreciated.

Good old ohms law V = IR says that there will be no voltage drop across the wire unless there is current flowing through it. One might then naively say: keep the the current down & the voltage drop will be minimal; the problem here is not just the current through the signal wire (which is next to nothing), but the current through the reference/ground wire.

One solution would be to include another ground reference wire in the cable and use a differential amplifier at the display end to recover the actual signal.

Another solution would be to modulate the signal at the sensor (e.g. Voltage-frequency) and demodulate it at the display (e.g. freq-voltage).

A simpler solution is to send the signal as a current (rather than voltage) which won't suffer from the above problem and can be quite accurate. This would be my preferred method due to its simplicity and performance.


But then again, if you're using a bargraph display (which only has 10 units), I hardly think you're after any great deal of accuracy or precision...

 

[ADWSystems]

hi,
The input impedance of the LM3914 is high, about 50nA typical, I would not expect any voltage drop thru a 25ft cable run.
Make sure the sensor output can drive the cable, is the cable screened/twisted etc.??

You may need a 27R on the output of the sensor in series with the cable, also grounding will be important.

I'm looking at using a typical 3-wire shielded cable, with wall wart power, ground, and signal.

 

[ericgibbs]

I'm looking at using a typical 3-wire shielded cable, with wall wart power, ground, and signal.

hi,
As pointed out, the operating current will be flowing in the 0V return wire as well as the 'signal'.
With the LED's being local and the Wart say at the remote end, the sensor end, the local LM3914 and LED current will be flowing in the 0V line.

 

[ADWSystems]

Good old ohms law V = IR says that there will be no voltage drop across the wire unless there is current flowing through it. One might then naively say: keep the the current down & the voltage drop will be minimal; the problem here is not just the current through the signal wire (which is next to nothing), but the current through the reference/ground wire.

One solution would be to include another ground reference wire in the cable and use a differential amplifier at the display end to recover the actual signal.

This is interesting. I can read what you wrote, but can't picture the setup.

Another solution would be to modulate the signal at the sensor (e.g. Voltage-frequency) and demodulate it at the display (e.g. freq-voltage).

K.I.S.S. I'm thinking this is a little beyond my equipment. Not beyond me, just beyond the fact that I don't have an oscilloscope to troubleshoot that circuit.

A simpler solution is to send the signal as a current (rather than voltage) which won't suffer from the above problem and can be quite accurate. This would be my preferred method due to its simplicity and performance.

I'm thinking about this if I can find the right chip. Again something simple without a lot of features, just Vin and Iout (preferrably 4-20mA as I'm used to working with that industry standard signal)

 

[ADWSystems]

hi,
As pointed out, the operating current will be flowing in the 0V return wire as well as the 'signal'.
With the LED's being local and the Wart say at the remote end, the sensor end, the local LM3914 and LED current will be flowing in the 0V line.

Yes.

So what you are saying is that I should have wall wart power and wall wart supply ground and signal and signal ground? The signal and power ground will be tied together at each end. Is this going to resolve the situation?


{Sensor Power}--{Sensor 7805}-----{Wall Wart Power}-----{Display 7805}--{Display Power}

{Sensor Ground}------- ^ -----------{Wall Wart Gnd}------------- ^ --------{Display Ground}

{Sensor Signal}-------------------------------------------------------------{Display Signal}

{Sensor Ground}------------------------------------------------------------{Display Ground}

 

[dougy83]

This is interesting. I can read what you wrote, but can't picture the setup.

It's like what you drew in the next post, except you don't connect the bottom ground signal to the ground at the display end; it gets connected to the -ve input of the differential amplifier.

I'm thinking about this if I can find the right chip. Again something simple without a lot of features, just Vin and Iout (preferrably 4-20mA as I'm used to working with that industry standard signal)

You don't need that chip - you can just set up an opamp (as a transconductance amp) to provide the volt-current conversion. The display end only needs a single resistor to convert back to voltage. If you do add the offset (e.g. 4mA for 4-20mA), you can remove it at the other end by setting the voltage at RLO of the LM3914.

Hope some of that is sensical.

 

[ADWSystems]

You don't need that chip - you can just set up an opamp (as a transconductance amp) to provide the volt-current conversion. The display end only needs a single resistor to convert back to voltage. If you do add the offset (e.g. 4mA for 4-20mA), you can remove it at the other end by setting the voltage at RLO of the LM3914.

Which chip don't I need?

I just found a block schematic for a transconductance opamp layout. It would appear that I need 2 250ohm matched resistors, an input that runs 1-5V, and an opamp that can handle a V+ of atleast 10V (probably 15V would be better) and 300mW, so as to provide 20mA over a minumum of 500ohms. 15V would give me 250ohms in the supply and return wires combine. I would need a 7815 as well as the 7805 in one side and the wall wart increased to 16-18V.

 

[dougy83]

Which chip don't I need?

"The right one", which I'm assuming you meant was a 4-20mA driver IC. You can of course use it, I really don't mind. I was just saying that you don't _need_ it; there are always other options - which may or may not be more attractive to your setup.

I just found a block schematic for a transconductance opamp layout. It would appear that I need 2 250ohm matched resistors, an input that runs 1-5V, and an opamp that can handle a V+ of atleast 10V (probably 15V would be better) and 300mW, so as to provide 20mA over a minumum of 500ohms. 15V would give me 250ohms in the supply and return wires combine. I would need a 7815 as well as the 7805 in one side and the wall wart increased to 16-18V.

I'm not sure what circuit you're talking about. What is the current range of your sensor? I take it you want to up the rail for a specific opamp?

 

[ADWSystems]

The sensor output has a 1mA max but runs 0.5 to 4.5V. I don't know what range of the sensor I will need (ie., how hi or lo I will need to go yet). The layout I found converted a voltage input to a current loop signal. Check here

To supply 20mA over two 250ohm resistors (one load and one reference) plus a 250ohm loop, the opamp needs to be able to put out +15V. But the 250ohm loop is high, so I won't need 15V out. Since 15Vin is a common PS voltage, that will work nicely then I can skip finding a rail-to-rail opamp for the job.

 

[dougy83]

That current source you linked to would require 2 wires in the cable for the signal; if you want to use the shielded 2-core, that circuit is not an option.

Here's a current source that only need the 1 wire. You don't need rail-rail opamps - half of an LM358 would be fine. You also don't require 15V rails. D1 is not required.

X axis of the graph is the input voltage, y axis is the voltage across the 110 ohm resistor at the display end. You'll see the output is linear, which is what you want. You'll alkso see that the output ranges from 0 - 1.245V for the 0-5 volt input; you just set up the LM3914 to accept the appropriate range using the REF, RLO & RHI pins.