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Long (300m) run impedance transformation

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brice14

Civil/Environmental
Jun 2, 2010
21
thread240-218990

This question was asked and answered on the referenced thread. However, this engineering newbie didn't quite understand. I have a relay whose signal switch is about 300m from the relay. A tester (DC) "sees" the circuit as open, but the relay (240 AC) "sees" the circuit as closed. I am assuming this is the same issues as discussed in the reference thread (i.e. impedance transformation).

What I don't understand is how to solve the problem. From what I have read going DC for the signal circuit is the best solution. I would assume that I would put an AC/DC converter in my box and then change my relay to one that accepts a DC signal to switch an AC circuit?

Someone also recommended putting a resistor or capacitor to ground. If I understand this correctly, I would hook one leg of the signal circuit to a resistor and then just shunt that to neutral or ground? This is obviously a simpler solution, but does it make sense?

TIA
 
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You seem to have grasped the situation well.

DC solves the issue completely.

In the AC situation you cannot just "put a resistor to ground". You have to actually put it to ground or elsewhere in a specific spot of the system.

The problem is typically a relay in the control panel and then one side of the signal to the relay coil being run way over to a switch. Then right back to the control panel.

The long run with an AC signal appears to be a low impedance - even when the switch is open. It is the distributed capacitive coupling between the wires of the long run. The current that can flow is limited to the overall situation. If it is enough, what happens is once the switch is closed the relay coil is energized and the relay pulls in. The problem comes when the switch is opened. The low impedance is often enough to keep the coil pulled in.

The solution is a resistor across the coil in this case. So this low current is shunted so the coil can fall back out. So the value would vary, and you may have to pay attention to the power dissipation of the resistor.



Keith Cress
kcress -
 
If it is a really long run then solid state relays are really tolerant of cable volt-drop. If you use a 24V DC drive voltage you can lose about 20V in the cable loop before things get flaky. A burden resistor of maybe 1k[Ω] at the relay is worthwhile.


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Thanks to both! I now understand why a previous installation didn't have the same problem. I had what I think ScottyUK is referring to as a Solid State Relay (INTELIGENTE ZELIO SR38101FU 100/24OV 6E/45). The run was a lot longer but I didn't have the problem. So, I decided to "save" some money and just put in a regular low power relay on my second installation. Ha, we know how that goes.

That being said, is the solid state relay a "good enough" solution and if so, which is cheaper solid state or AC/DC conversion? The relay I used in the first installation cost me here in Costa Rica about $120US. How much is a suitable AC/DC converter?
 
On re-reading I think I misunderstood ScottyUK's advice. I now see he was talking about the different issue of voltage drop on long runs after having made the jump to a DC system.

Can anyone suggest why my AC system over a much larger distance worked when using a ZELIO SR38101FU 100/24OV 6E/45? It seems like it should have the same problem, but perhaps they've put some resistor or capacitor in the box to keep this sort of problem at bay.
 
Well, the SSR solution helps with volt-drop on either AC or DC, but DC is inherently better for long distances.

I used the DC SSR solution on a cable run roughly a mile in length, i.e. a loop approaching two miles in length, sucessfully. My predecessors had tried using 24V and 36V to drive 12V relays, paralelled cores, and all manner of tricks. The SSR made the rpoblems go away. I certanly don't advocate a 2 mile loop, but when some idiot has already installed it then it's a way out.


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SSRs tend to be more failure prone than mechanicals so I'd probably use a DC mechanical myself.

Of course a power supply has its own failure points but they will all be fail OFF. The mechanical relay tends to be fail-off too. Presumably the relay's point of use is energized = system running, de-energized = system off-safe-shutdown. So the aforementioned failures all result in a safe condition. With the SSR: It can fail shorted, so I wouldn't use it in a safety position.

Keith Cress
kcress -
 
I just wanted to finish here by thanking you guys for your input. It is invaluable to me to have your advice. I'm going to take the shortest route first and put a lamp across the A1 and A2 inputs of my AC circuit. If that fails, I'll transform to DC and stick with mechanical relay as ItSmoked suggested. If that fails, I'll tuck my tail and pay for the DC SSR that I obviously should have bought in the first place. Thanks again.
 
OK!

One point. You have never actually said what this is for. If it is indeed for a safety shutdown don't use a bulb as filaments fail and so your system will revert. If it isn't safety related - have at it.

Keith Cress
kcress -
 
With a handle "ItSmoked" I imagine you've seen a few safety hazards in your time. I'm not sure exactly what you mean by a safety shutdown. Basically, I have a water storage tank that I opted to manage the on/off switching with a float switch. But, the tank is about 300 meters from the pump station and the power. So, I ran a thin cable (14 gauge) to the float switch and then began all my troubles.
 
Oh yeah I've seen a few hazards, even plant leveling accidents. Your water tank issue probably doesn't have any of those issues, so a light bulb might only cause some water to be lost. Let us know how it works out.

Keith Cress
kcress -
 
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