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schematic

simulate this circuit – Schematic created using CircuitLab

I have a circuit of 4 light bulbs, with single switch on the positive terminal that will turn all the 4 bulbs at once. EDIT: Light bulbs circuit is parallel design. source is 12V DC and all bulbs are running 12V DC.

I want to be able to still have the original switch to turn on all 4 bulbs, with a new switch that overrides bulb #3 to keep it turning on. Essentially, when bulb #3 is switched on, the original switch ignores bulb #3.

To solve this I go with installing a diode at the positive terminal of the bulb #3. When original switch is on, electricity will go thru that diode and then turns the bulb on; when new switch is turned on, electricity will go to bulb #3 thru new line but will not go to the original circuit because of the new diode.

A friend said this might be able to be done with a transistor but he couldn't figure out the solution. Is that true we can use transistor for this? it's for automotive brake lights.

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  • \$\begingroup\$ What sort of bulb? What voltage? AC or DC? You'll need to give at least basic specifications. SPDT relay will probably work. \$\endgroup\$ – Tom Carpenter Feb 25 at 12:00
  • \$\begingroup\$ light bulbs circuit is parallel design. source is 12V DC and all bulbs are running 12V DC. Sorry for incomplete information. \$\endgroup\$ – Alvin Suryadi Feb 25 at 12:00
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    \$\begingroup\$ edit that into your question. \$\endgroup\$ – Tom Carpenter Feb 25 at 12:01
  • \$\begingroup\$ To avoid all confusion: draw a schematic, there's a schematic entry tool available for that when you Edit your question. \$\endgroup\$ – Bimpelrekkie Feb 25 at 12:02
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Given your control signal is just another switch, your diode approach in the question should work fine.

If the control signals came from two different supplies you would need a second diode in series with the override switch. However as both are fed from the same supply, this is not required.


An alternative solution would be an SPDT relay:

schematic

simulate this circuit – Schematic created using CircuitLab

When you turn the relay on, it will turn on lamp 4 regardless of the state of the switch. When the relay is off the lamp goes back to being controlled by the switch.

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  • \$\begingroup\$ Hey Tom, nice answer. I see OP edited the question after you put your answer, do you think that the diode has any drawbacks in an automotive scenario? \$\endgroup\$ – Vladimir Cravero Feb 25 at 12:45
  • \$\begingroup\$ Thankyou for helping.. this looks like just what I needed but not sure yet. Do you mean by Override is a switch connected to +12V? By the way, why do I need D1 here? \$\endgroup\$ – Alvin Suryadi Feb 25 at 12:46
  • \$\begingroup\$ @AlvinSuryadi D1 is what's called a flyback diode. Since the relay coil is inductive, the current can't change instantaneously, and without the path provided by D1, it can generate pretty high voltages, enough to damage whatever you're using to switch the override signal. \$\endgroup\$ – Hearth Feb 25 at 14:32
  • \$\begingroup\$ @Hearth - Flyback diodes are a good idea, but will not work as specified in this circuit. The diode must connect from the Override to +12. As shown here, all it will do is burn out the diode due to the voltage spike. Either that, or the Override circuitry gets fried before the diode. \$\endgroup\$ – WhatRoughBeast Feb 25 at 18:25
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    \$\begingroup\$ @WhatRoughBeast I'm not sure I understand your point - the flyback diode on the relay coil shown in my answer is in exactly the correct place. Connecting it between the control signal and 12V will not acheive anything. \$\endgroup\$ – Tom Carpenter Feb 25 at 18:59

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