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I have a quick question and I was hoping someone here can help me out. I know alot of you are familiar with 24VDC systemns. I am trying to use a 24VDC to 12VDC relay to close a circuit.
The 24VDC (actually it closer to 29VDC) side will power up and close the relay on the 12VDC side. I need to figure out what the parasitic load for the relay will be. I understand that the relay will close the circuit using the power from the 24 VDC side, or at least that is what I am trying to do. I am
wondering since I am in a limited power availability situation how I can figure out how much amperage and voltage are required to close the relay and keep it closed.
To make things clearer, here is the circuit layout and a little background. In
my firehouse (volunteer) we have a audible alert systemn (very similar to a
klaxon) that is powered by a "24VOLT"DC power distro. I metered it out to 29VDC at the alarm.
What we are trying to do is have a old lightbar, that used to be on one of our old pieces of apparratus come on along when the Klaxon goes off. Powering the lightbar will be 110VAC/12VDC @ 20amps power inverter. The inverter will be constant on. I want to use the 12VDC side as the switched side through the relay.
Therefore when the house is alerted for an emergency the circuit for the Klaxon is closed created a 24-29VDC power to the relay, the relay now in the closed position will close the negative leg of the inverter and allow the lightbar to power up for duration of the alert.
What I am trying to make sure of is that the relay is not so parastic that it
causes a fuse to blow on the power supply for the house alert. I dont
understand if relays in a DC power systemn are amperage driven or voltage driven. I havent been up to find any formulas or anyone that knows about wether or not I am creating a problem. The idea has come up that if there is too much draw on the house alert systemn that I setup cascading relays and just use the constant energy from the inverter as the power source on a larger relay actually hooked up to the lightbar....
But I dont want to create such an exotic systemn, if its going to be
unecessary. So my question is, how do I calculate the parasitic load of the relay? The reason this is a problem is because I dont want to order the wrong parts just to find out that I need more parts, because what I would order for one design is different than what I would order for the other one.
The other suggestion that was given to me, was to use a contactor off the 24VDC alerting systemn and use the contactor as a switch for the supply to the inverter. 2 problems, electrical code and also putting a shock load on a
inverter. I can get away with a smaller inverter and keep it constant on as
oppossed to something that is going to have to go from zero to full service
when the contactor closes.
Any help or feedback is appreciated. I dont normally work in both the world of AC and DC at the same time, so I want to make sure I'm not overlooking something obvious.
The 24VDC (actually it closer to 29VDC) side will power up and close the relay on the 12VDC side. I need to figure out what the parasitic load for the relay will be. I understand that the relay will close the circuit using the power from the 24 VDC side, or at least that is what I am trying to do. I am
wondering since I am in a limited power availability situation how I can figure out how much amperage and voltage are required to close the relay and keep it closed.
To make things clearer, here is the circuit layout and a little background. In
my firehouse (volunteer) we have a audible alert systemn (very similar to a
klaxon) that is powered by a "24VOLT"DC power distro. I metered it out to 29VDC at the alarm.
What we are trying to do is have a old lightbar, that used to be on one of our old pieces of apparratus come on along when the Klaxon goes off. Powering the lightbar will be 110VAC/12VDC @ 20amps power inverter. The inverter will be constant on. I want to use the 12VDC side as the switched side through the relay.
Therefore when the house is alerted for an emergency the circuit for the Klaxon is closed created a 24-29VDC power to the relay, the relay now in the closed position will close the negative leg of the inverter and allow the lightbar to power up for duration of the alert.
What I am trying to make sure of is that the relay is not so parastic that it
causes a fuse to blow on the power supply for the house alert. I dont
understand if relays in a DC power systemn are amperage driven or voltage driven. I havent been up to find any formulas or anyone that knows about wether or not I am creating a problem. The idea has come up that if there is too much draw on the house alert systemn that I setup cascading relays and just use the constant energy from the inverter as the power source on a larger relay actually hooked up to the lightbar....
But I dont want to create such an exotic systemn, if its going to be
unecessary. So my question is, how do I calculate the parasitic load of the relay? The reason this is a problem is because I dont want to order the wrong parts just to find out that I need more parts, because what I would order for one design is different than what I would order for the other one.
The other suggestion that was given to me, was to use a contactor off the 24VDC alerting systemn and use the contactor as a switch for the supply to the inverter. 2 problems, electrical code and also putting a shock load on a
inverter. I can get away with a smaller inverter and keep it constant on as
oppossed to something that is going to have to go from zero to full service
when the contactor closes.
Any help or feedback is appreciated. I dont normally work in both the world of AC and DC at the same time, so I want to make sure I'm not overlooking something obvious.
posted by:
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Washington, D.C.
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