Page 1 of 1
Schematic question
Posted: Sun Nov 10, 2024 12:20 pm
by xbartx
I'm installing a contactor and using the Centroid schematic.
Looking at the schematic the contactor needs to be normally closed, right? Is it showing it closed?
Are contactors typically latching ?
-Brad
Re: Schematic question
Posted: Sun Nov 10, 2024 1:03 pm
by xbartx
I think I realized why this wasn't making sense, the schematic shows output 1 as normally open, it looks like it is changed in CNC12 to Normally closed.
Otherwise the contactor wouldn't get the 24V it needs. Does that sound right?
Re: Schematic question
Posted: Sun Nov 10, 2024 2:04 pm
by Ken Rychlik
Are you talking about an input power contactor? To make a contactor latching, get one with an extra pole. If you need 2 pole contactor get a 3 pole. Then you use a NO momentary button to activate the contactor. The extra leg of the contactor applies and holds the power on to it and is broken by a NC momentary button. There may be other ways, but that's the way I've always done them.
Re: Schematic question
Posted: Sun Nov 10, 2024 2:48 pm
by tblough
Contactors are normally open. The contacts close when the coil is energized.
They can have auxillary contact blocks installed that can be either NO or NC, but the main L1/T1, L2/T2, and L3/T3 contacts are NO.
Re: Schematic question
Posted: Sun Nov 10, 2024 3:13 pm
by xbartx
tblough wrote: ↑Sun Nov 10, 2024 2:48 pm
Contactors are normally open. The contacts close when the coil is energized.
They can have auxillary contact blocks installed that can be either NO or NC, but the main L1/T1, L2/T2, and L3/T3 contacts are NO.
So looking at the schematic, with Acorn Output relay 1 wired as normally open, the power supply for the contactor stays open and closes only with a fault. How does the motor get power?
What am I missing?
I have one like this.
Re: Schematic question
Posted: Sun Nov 10, 2024 3:35 pm
by RogDC
Relay 1 will be closed when the E-Stop is not activated providing power to the contactor coil closing the contactor. when the E-stop opens it will break the circuit for power to the contactor coil and it will de-energize opening the contactor.
Re: Schematic question
Posted: Sun Nov 10, 2024 4:32 pm
by xbartx
RogDC wrote: ↑Sun Nov 10, 2024 3:35 pm
Relay 1 will be closed when the E-Stop is not activated providing power to the contactor coil closing the contactor. when the E-stop opens it will break the circuit for power to the contactor coil and it will de-energize opening the contactor.
I guess I was just overthinking things, I wired it up as shown and it's working as it should.
Thanks for all the help!
Re: Schematic question
Posted: Mon Nov 11, 2024 1:06 pm
by Richards
[Edited schematic to show the correct latch circuit path]
Here's a schematic for a simple latching circuit. The latch (X terminals) can be an unused circuit on a relay or contatctor, or the auxiliary switch circuit on a contactor. ( The relay/contactor coil uses "high-side" switching similar to the high-side switch circuit on a wall switch. High-side switching is preferred over low-side switching because the coil carries no power unless it is on, similar to a light bulb in a lamp socket, making it safe to change the lamp if the lamp's switch is off. )
When the Start push-button is pressed, 24VDC passes through the relay/contactor's coil causing the coil to energize. The energized coil causes the X contact to close. When the X contact close, a new voltage path is created from +24V to the coil. The new (latching) path bypasses the Start switch. The coil stays energized (latched) until the Stop push-button is pressed. When the Stop push-button is press, no voltage passes to the coil and the coil de-energizes. When the coil de-energizes, the X contact opens. The schematic shows a DC circuit that uses a suppression diode to prevent voltage spikes when the coil is turned off. Note the orientation of the suppression diode.
The Line (L1) terminal is connected to the power going to the device. The Load (T1) terminal is connected to the device's (+) terminal if the device is DC or to the device's Line terminal if the device is AC.
Re: Schematic question
Posted: Mon Nov 11, 2024 2:06 pm
by centroid467
Richards wrote: ↑Mon Nov 11, 2024 1:06 pm
Here's a schematic for a simple latching circuit. The latch (X terminals) can be an unused circuit on a relay or contatctor, or the auxiliary switch circuit on a contactor. ( The relay/contactor coil uses "high-side" switching similar to the high-side switch circuit on a wall switch. High-side switching is preferred over low-side switching because the coil carries no power unless it is on, similar to a light bulb in a lamp socket, making it safe to change the lamp if the lamp's switch is off. )
Latch_circuit.png
When the Start push-button is pressed, 24VDC passes through the relay/contactor's coil causing the coil to energize. The energized coil causes the X contact to close. When the X contact close, a new voltage path is created from +24V to the coil. The new (latching) path bypasses the Start switch. The coil stays energized (latched) until the Stop push-button is pressed. When the Stop push-button is press, no voltage passes to the coil and the coil de-energizes. When the coil de-energizes, the X contact opens. The schematic shows a DC circuit that uses a suppression diode to prevent voltage spikes when the coil is turned off. Note the orientation of the suppression diode.
The Line (L1) terminal is connected to the power going to the device. The Load (T1) terminal is connected to the device's (+) terminal if the device is DC or to the device's Line terminal if the device is AC.
Mike,
Should the connection from X2 go between the two push buttons? Otherwise, I don't see how it would interrupt the current through the latch circuit and therefore the coil.
Re: Schematic question
Posted: Mon Nov 11, 2024 2:48 pm
by Richards
Yes. You're correct. That error got away from me. I've corrected the schematic and re-uploaded the file so that it's correct in the original post.