Homework Help: Circuit drawing of two floodlights with relays

1. Jun 2, 2013

Hrvoje

1. The problem statement, all variables and given/known data

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.

3. 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.

2. Jun 2, 2013

Staff: Mentor

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.

3. Jun 3, 2013

CWatters

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.

4. Jun 6, 2013

Hrvoje

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?

5. Jun 6, 2013

CWatters

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.

6. Jun 7, 2013

Staff: Mentor

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 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.

7. Jun 8, 2013

CWatters

Could also make a small transformer by winding the wire feeding the main lamp around a ferrite.