Maintaining Signal When Relay contacts change

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Nsmith

Member
I am working on a project that evolves error checking using relays. Assume two relays both with a set of NO and NC contacts are wired as shown below. Under normal operation assume both coils of the two relays will be actuated simultaneously. Applying a 24v signal on the left side then monitoring the resulting signal on the right side there will be a loss of the signal as the relays change state due to bouncing and subtle time differences in the relays ability to change state.

What would be the easiest way to eliminate any loss of signal (provided the loss is not over 250ms) as these relays change state? Using discrete components excluding any IC chips.

ex.JPG
 

rcwilson

Senior Member
Location
Redmond, WA
Have the circuit output feed a 24V relay with a diode in series with a resistor across the relay coil. Connect the diode so no current flows in the resistor when coil is energized. If the signal goes away the coil current continues to flow through the resistor until the magnetic field in the coil collapses. (Usually that energy goes into the arc of the contact breaking the coil current.)

With proper selection of the resistor, you could get a .25 second delay. Don't know what that delay does to your error checks.
 

realolman

Senior Member
Have the circuit output feed a 24V relay with a diode in series with a resistor across the relay coil. Connect the diode so no current flows in the resistor when coil is energized. If the signal goes away the coil current continues to flow through the resistor until the magnetic field in the coil collapses. (Usually that energy goes into the arc of the contact breaking the coil current.)

With proper selection of the resistor, you could get a .25 second delay. Don't know what that delay does to your error checks.

That's very interesting. I did not realize there would be enough energy in the collapsing field to hold the relay from dropping out for that long using a diode and resistor.

I'll be durned.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
120225-1425 EST

Using a back-biased diode across a DC relay coil will lengthen the drop-out time. If you add resistance in series with the diode it shortens the drop-out time.

Using a diode only and a Potter & Brumfield KUP 24 V DC relay the drop-out time was still very short. At the moment I do not have an easy way to measure this time. My guess is less than 0.1 seconds.

With a 2800 ufd capacitor across the KUP coil I get about 3 seconds for drop-out. 300 ufd would be about 0.3 seconds, and 25 ufd seemed longer than the shunt diode, but that is a crude judgement.

Different relays used with DC excitation and a shunt diode may provide very different drop-out times.

.
 

ronaldrc

Senior Member
Location
Tennessee
Looks like I'm going to be the only one to ask.

But what are you testing if you don't want to break the circuit at any time when the relays switch?

The only application I can see is maybe testing the difference in the conductance
of the two conductors between terminals #32 and #31 and #44 and #43.

I am just an electrician so be easy with me.

Thanks :)
 

justdavemamm

Senior Member
Location
Rochester NY
The datasheet for the KU series of relays have a pickup time of 15ms and a drop out time of 10ms, (excluding bounce). The OP has said that less than 250ms is OK.

To get a relay that has no break in conduction in the way the OP has drawn, use a relay that has Late Break contacts or rewire with Form C contacts that are Make Before Break, vs the two Form A/B contacts shown.

More info / comment from the OP is needed for further help.
 

Nsmith

Member
The relays being used cannot be changed, they are force guided PCB mount. I have been trying to use a capcitor across the contacts to maintain the signal for a period of time. However I am having trouble keep the signal clean enough so a loss does not occur.
 
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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
120322-1132 EDT

Nsmith:

Can you provide a schematic of your circuit. What I can see in your first post tells me nothing of the circuit.

A capacitor across a contact may not do anything for you. It will depend upon on how it fits into the actual operation of the logic.

.
 

kwired

Electron manager
Location
NE Nebraska
120322-1132 EDT

Nsmith:

Can you provide a schematic of your circuit. What I can see in your first post tells me nothing of the circuit.

A capacitor across a contact may not do anything for you. It will depend upon on how it fits into the actual operation of the logic.

.

That is what I was thinking, hopefully there is more to the circuit that may make sense, what is shown so far makes the contacts seem pointless. As it is shown in the OP the only time the load of that line is off is during transition of the contacts.

My best guess without knowing more about the needs of the controls is some kind of time delayed bypass of these contacts to allow transition. I really don't see what a capacitor will do. When across a closed contact the capacitor will never charge. When across an open DC circuit it will charge and current will stop flowing. When across an open AC circuit it will be an impedance in the circuit and will have a voltage drop across it, but the circuit will never be totally open either at this point somewhat defeating whatever purpose the contact is there for in the first place.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
120324-2019 EDT

Nsmith:

I may have finally seen your circuit from the first post, but I am not sure. Blue lines on a black background do not show.

Is the desired circuit represented by
OUT = (CR12 and CR18) or (/CR12 and /CR18)?

What is the object of this circuit? It appears to be a NOT XOR circuit.

Truth table for NOT XOR

0 0 = 1
0 1 = 0
1 0 = 0
1 1 = 1

Truth table for XOR, binary adder

0 0 = 0
0 1 = 1
1 0 = 1
1 1 = 0

I do not know how delaying change of state on drop out of the relay would be of any value. If you want make before break contacts, that is an entirely different operation that you can not solve with delay on the relay coil. Also relay dropout delay does not solve the problem on energizing the a relay.

What is the real problem you want to solve?

.
 

SG-1

Senior Member
If your objective is to have continuity all the time, then you need a logic circuit that will only allow a relay to drop out when both are energized. The de-energized relay would seal in the energized relay. Or supervisory relays could perform the function. In other words energizing one relay would de-energize the other.

Since we do not know the purpose for the circuit this type of arrangement could negate it's function.
 
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