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Circuit Question

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integrator4410

Electrical
Aug 15, 2008
4
Hello,

We have a circuit which contains 5 flow switches in series(NC) which turns on one 120VAC relay. I have attached a diagram for what we currently have.

If anyone of the switches open, it turns on that light and the relay turns off, shutting the equpiemnt.

What we would like to do is add in 120VAC relays inplace of the light to give use more flexiblity. The problem is that we don't know how this will work in the circuit using the coil of the relay instead of the inductive light. The coils are 1.5K.

Thanks for any suggestions with this
 
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I have my doubts it will work.

For what you presently have the lamps have to light on enough less current that the relay can't remain pulled in. Going to relays it will probably be pretty hard to run other relays in lue of the existing one.

You could use Solid State Relays(SSR) instead of mechanical relays and that would work.

Do you have a specific relay you want to use? We also need the specification of the existing relay.

Keith Cress
kcress -
 
Hello

We don't have any specification on relays except for 120VAC coil. These are just standard relays SPDT.

Why would an SSR relay work but not the coil type.



Thanks

 
Because the input current of an SSR is so much lower. That means that its impedance is corresondingly higher and will probably no activate the relay. A parallel load (to the relay coil) may be necessary.

Gunnar Englund
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 

Based on the circuit, is it the low current and impedance of the light that makes this work.

If so and if I can match the current and impedance of the light with a SSR I shouldn't have any problems?

The light is around 47K resistance.

Thanks
again
 
I accept that your circuit is presently working. That tells me that the impedance of your lamps is high compared to the impedance of the relay. If this were not so, the current through the lamp would hold in the relay.
If you put an identical relay across one of the switches (Let's look at only one for now, it gets worse with more than one) When the switch opens, two identical relays will be in series.That means that each will get 50% voltage. depending on the drop out voltage of the relays, either both will drop out or both will hang in. Neither is the result that you want.
If you use the SSRs that itsmoked suggests, the high impedance of the SSRs will not pass enough current to hold in the relay.
You don't have to match the lamps impedance exactly, but you do need SSRs with an impedance that is high in relation to the electro mechanical relay.
itsmoked is more familiar than I with SSRs. Any advice he gives you should be good advice.
I think that Gunnar means that if the SSRs do pass enough current to operate the relay coil, a load in parallel with the relay coil will cure the problem.
Gunnar's advice is also usually very good.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
The drawing shows 6 flow switches, not 5. ;-)


Why not wire up each flow switch in series only with it's own relay coil (or input) - and power.

Then you can wire up the relay contacts (or outputs) to do whatever you want; including making individual indicator lights work and tripping off the equipment. Especially easy if the relays contacts are DPDT.

The circuit would then become boringly bog-standard with zero trickery.

Downside is that you might need more wire.

 
I can see why you want a light on every flow switch. Flow switches are notoriously unreliable.
VE1BBL has the answer, you must have the wires already, for the lights.
If you have a PLC you could wire from each contact to a 120V input then use some logic to display just the first open.
Roy
 
Thanks for all the help

This is one of the kind circuit.

We might look into placing a relay in series with each flow switch and use it's contact to series all them together.
 
Yes, do that. Put each switch to it's own relay and use the relay contacts connected to do what you need done.

Trying to put relays where you show lights in your drawing is just asking for problems.

 
If you mean SSR give it a shot. They actually have LEDs built in that will indicate which one is firing. They have covers that snap over them for user protection too if you mount them in a visible location. If you aren't driving more than a few amps with them you don't have to worry about heat dissipation.

I asked about quantity because SSRs bought in small quantiles are kinda pricey. If you were making a bunch of systems I'd have suggested soldering smaller thruhole pin based units into a circuit board as you could save $$, but for a one off, I'd just use the standard hockey puck units.

Keith Cress
kcress -
 
Juat one thing, SSRs typicaly require 3 - 32 Volts DC input.
You may be able to get high voltage AC Input but they are not so common. The low voltage DC are a lot safer if you have that available.
Roy
 
I still would not recommend doing it. Wire it normally like everyone will expect in the field which is one switch for each relay coil and then the relay contacts used as needed for the rest of the circuit.

The last thing you need is someone replacing a failed relay in the future with one that doesn't work just because you had to get "smart" with your wiring technique. Basically not understanding the circuit and why it was done that way and then finding a way to screw up the protection.

If you were designing a circuit board for this function then you could get fancy since no-one else would likely be modifying it down to the board level. But, with point-to-point control wiring it's not worth trying, especially when wiring it the more normal expected way really does not add anything extra to the circuit except some more wire.

 
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