Circuit drawing of two floodlights with relays

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Discussion Overview

The discussion revolves around designing an electric circuit for two 500W floodlights using relays, where the secondary light activates if the primary light fails. Participants explore various methods and components to achieve this functionality, including relays, sensors, and alternative circuit designs.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory
  • Debate/contested

Main Points Raised

  • One participant suggests connecting an auxiliary relay in series with the primary light, but notes that the current from the 500W floodlight may exceed the relay's capacity.
  • Another participant proposes using a small resistor in parallel to bypass part of the current or considering transistors as an alternative.
  • A suggestion is made to use a light sensor that activates the auxiliary lamp when the primary lamp is off, although concerns about maintenance are raised.
  • Details about the floodlights indicate they operate on 220V AC and draw approximately 2.2A, prompting discussions about minimizing current draw for auxiliary relays.
  • One participant mentions the possibility of using an "under current" control relay that activates when current falls below a certain threshold, noting potential cost implications.
  • Another idea involves modifying a relay by replacing its wire with a thicker one to handle the current, though practicality is questioned.
  • A more complex approach is suggested involving a high-power LED and an opto-isolator, with a reference to an external circuit diagram.
  • Lastly, the idea of creating a small transformer by winding the wire feeding the main lamp around a ferrite is introduced as a potential solution.

Areas of Agreement / Disagreement

Participants express a variety of ideas and approaches, with no consensus on a single solution. Multiple competing views and methods remain under consideration.

Contextual Notes

Participants note limitations regarding the current capacity of relays and the need for additional components, but do not resolve these issues. The discussion includes various assumptions about the circuit design and component capabilities.

Who May Find This Useful

Individuals interested in electrical engineering, circuit design, or those working on similar homework assignments related to relay circuits and floodlight applications may find this discussion beneficial.

Hrvoje
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Homework Statement



I need to draw a simple electric drawing that will in reality present a circuit with two 500W floodlights connected in such a way that if the primary light dies out, the relay will connect the secondary light to the source. Only the primary light is supplied, while the secondary light is on the stand-by, waiting to be connected by the relay contactor in case of primary light dying out.

The Attempt at a Solution



I was thinking about connecting an auxiliary relay in series with the primary light, and the relay's normal closed contactor in series with the secondary light. In case primary light dies out the relay looses it's supply, closing it's normal closed contactor on the other light. But since the current drawn from 500W floodlight is too strong for the auxiliary relay it seems I can't connect it in this way.

I'm sorry but I'm writing this post from a satellite Internet connection from high seas so it's impossible for me to upload a drawing. Can you PLEASE post a drawing of the solution? If there is no way in creating such a circuit with relays only, can you please draw and explain which other elements the circuit would need.

Thank you very much.
 
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If your relay cannot handle the current, you could try to bypass a part of the current (with a very small resistor in parallel? With more identical relays without mechanical use?).

Some more details about the floodlights would be interesting. Which voltage and current do they use?
Transistors could be interesting.
 
Could you use an ordinary light sensor? The sort used to turn floodlights on at dusk? Just point it at the primary lamp. If that's working it will turn off the aux lamp. The need to keep the sensor clean is a dissadvantage.
 
The floodlights use 220V AC power supply. With the power of 500W they draw around 2,2A.
The idea is to have auxiliary relays connected to draw as little Amps as possible.

There must be a way to do it. If needed, additional elements can be used in a drawing. Can you PLEASE help me out?
 
Perhaps have a look at an "under current" control/monitoring relay which switches when the current is falls below a set value. Likely to be relatively expensive compared to a custom design but that depends on what your time is worth.
 
Hrvoje said:
I need to draw a simple electric drawing that will in reality present a circuit with two 500W floodlights connected in such a way that if the primary light dies out, the relay will connect the secondary light to the source.
Simple? In theory you could take the thin wire off the core of a relay and replace it with thicker (low resistance, low voltage loss) wire that carries 2A to operate the relay. Not sure how practical that would be.

Or, maybe you could find a high power 2.5A LED and insert it in series with the cable feeding the floodlight, and proceed to fabricate your own opto-isolator. Parallel the LED with a reversed diode to handle the other half-cycles. Here's the idea: http://aristarco.com.es/sites/default/files/dual_relay1.png[/color] The circuitry will need its own low-power DC supply. Probably easier to add a small (e.g., 2Ω) series resistor, and operate an off the shelf low power opto-isolator.
 
Could also make a small transformer by winding the wire feeding the main lamp around a ferrite.
 

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