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It depends completely on your displays. (The manufacturer)
Contact the display (meter) manufacturer, and ask them.
You may very well need an amplifier to drive that many displays, and especially if
there is distance between meters involved.
Contact the display (meter) manufacturer, and ask them.
You may very well need an amplifier to drive that many displays, and especially if
there is distance between meters involved.
We just don't have enough information.
What's your application, what's the model number of your displays?
Typically in industrial applications, inputs are 4-20mA, 1-5VDC, here in the States anyway.
There are plenty of people on this site that will help you if you can give a little.
What's your application, what's the model number of your displays?
Typically in industrial applications, inputs are 4-20mA, 1-5VDC, here in the States anyway.
There are plenty of people on this site that will help you if you can give a little.
I am trying to display a depth reading on multiple displays. I have a pressure transducer with an output of 4-20ma or 0-10V DC,. I have 22 digital programmable panel meters that operate from 12 or 24 V dc. They can have an input of either 4-20ma or 1-10V dc.
I will calibrate each meter to show equivalent depth of 0 to 25 mts. The panel meter has input impedance of 10 Mohms. The meter type is a MB3800
I will calibrate each meter to show equivalent depth of 0 to 25 mts. The panel meter has input impedance of 10 Mohms. The meter type is a MB3800
I am not an expert on these, however this will not stop me!
If you use more than one meter on the current source, the meters will need to be connected in series. they may not like not being able to find their own ground. The input impedance will be determind by a resistor to generate a voltage, this will be much lower than 10M. Where does the current come from? If from the meter, then you will have 22 times the voltage required.
The advantage of the current source is that wiring resistance is of little importance.
The obvious way out of this is to go for the voltage source when any number of displays can be connected in parallel.
If you use more than one meter on the current source, the meters will need to be connected in series. they may not like not being able to find their own ground. The input impedance will be determind by a resistor to generate a voltage, this will be much lower than 10M. Where does the current come from? If from the meter, then you will have 22 times the voltage required.
The advantage of the current source is that wiring resistance is of little importance.
The obvious way out of this is to go for the voltage source when any number of displays can be connected in parallel.
A current loop works by encoding the signal on the current in the wire rather than the voltage across a pair of wires.
The idea is that the current remains constant in the wire regardless of the resistance and voltage drop in the wire.
All devices are connected in series (so they see the same current.
Not having a common earth for the signal is a feature and one that current loop devices will be able to handle.
The limitation is the maximum voltage the signal source is able to use. This determines the total resistance (and/or voltage drop) in the loop.
The main problem you will have is the output voltage of the 4-20mA source. To use 4-20mA on 22 meters you need to place all in series. Each display will drop a number of volts over the internal sensor resistance, and will 'eat' up the available overhead voltage available from the output. The sensor resistance is normally in the range of 100-200 ohm, and 20mA over 200 ohm will drop 4V off the overhead. For 22 displays this could be as much as 88V drop plus the normal 10V needed in the bottom. I've never heard of a 4-20mA output with 100v overhead, so I think I can safely say that this is not the way to go.
To solve your problem, you could put in a current to voltage converter, or 4-20mA to 2-10V driver amplifier to source all meters in parallel, with 2-10V inputs. You may find that the displays is not isolated and may interfere with each other. Some meters has an isolated input and this is no problem then.
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Gorgon's first paragraph hits the nail on the head regarding using a current loop and connecting the meters in series.
If the meters have a 10 megohm input resistance in voltage mode, connecting 22 of them in parallel will give a combined input resistance of 450 kilohms, which is still pretty high. That's the way to go, in my opinion. Set everything to voltage mode and connect the meter inputs in parallel.
If the meters have a 10 megohm input resistance in voltage mode, connecting 22 of them in parallel will give a combined input resistance of 450 kilohms, which is still pretty high. That's the way to go, in my opinion. Set everything to voltage mode and connect the meter inputs in parallel.
You could if you made the displays yourself, and used very good quality sensor resistors, but the OP never say he wants to make the displays himself, and most standard shelf displays use a much higher sensor resistance, as mentioned before.
That said, most 4-20mA drivers use 24V as drive voltage. This will result in a headroom of 14V above the normal 10V offset. In cases where the output driver is supplied from a separate voltage supply, you may even have all, or most of the 24V range as headroom.
I've had a look at several displays and found that 50-100ohm is a normal input sensor values, and the lowest I found was 33ohm, but most outputs limit the total resistance to 500 - 700 ohm. and this parameter is the limiting factor.
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Wiring a lot of current inputs in series and using a high voltage source will cause a problem if the meter inputs are simple current shunts.
Assume each meter has an input current shunt of 250 ohms, a typical value. At 20 mA the total series voltage drop will be 110V, so a 120V power source leaves 10V across the transmitter, which sounds good.
But at 4 mA loop current, the total voltage across the meters will only be 22V and the transmitter will have 98V across it, which is likely to damage it.
Assume each meter has an input current shunt of 250 ohms, a typical value. At 20 mA the total series voltage drop will be 110V, so a 120V power source leaves 10V across the transmitter, which sounds good.
But at 4 mA loop current, the total voltage across the meters will only be 22V and the transmitter will have 98V across it, which is likely to damage it.
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It's probably a reasonable way to go.
The alternative would be to convert the current into a digital representation and transmit that to the indicators, but I've never heard of an off-the-shelf panel meter that accepts data in digital form. You would have to design one, as well as the method of transferring the data to them all.
Panel meters with simple voltage input are produced in bulk, so they are cheap to buy. They're also easy to connect. Factoring in these practical advantages, I can't think of any better way to do what you want.
The alternative would be to convert the current into a digital representation and transmit that to the indicators, but I've never heard of an off-the-shelf panel meter that accepts data in digital form. You would have to design one, as well as the method of transferring the data to them all.
Panel meters with simple voltage input are produced in bulk, so they are cheap to buy. They're also easy to connect. Factoring in these practical advantages, I can't think of any better way to do what you want.
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