# Automotive wiring OR logic gate

I am wiring up taillights for my car and need to implement a OR logic gate, I understand how to do this using two diodes and a resistor, I don't know what spec components I will need.

The circuit I'm planning is this:

The intention is the top switch to only power the right bulb, and the left to power them both.

The circuit is running on 12v, and the bulbs are 21 watts. Any help would be appreciated.

Thanks

• Power=Voltage*Current so ~1.75Amps. Aim for at least 3.5Amps.
– Abel
May 22 at 2:35
• @Abel Thanks, would any junction diode that exceeds 3.5 Amps work?
– 190E
May 22 at 2:48
• Use schottky diode type, they have lower power losses here. Expect around 1W heat in the diode.
– Jens
May 22 at 3:11
• Whats wrong with you car switches? May 22 at 4:51
• You don't need the resistor. The diodes will work fine without it. May 22 at 6:34

You can maintain your circuit using a “Dual schottky diode common cathode 10A”. It does what you need, in a single package (TO-220 or TO-247).
You can repurpose one from a used/scrap of a PC power supply or buy a new piece as MBR2045C (10A/diode x 45V) or SBR1060CT (5A/diode x 60V).

This shown Dual diode device would be better than the second listed one because a higher current rated drops less voltage at same current level (lower % of max rating) and a lower maximum reverse voltage also presents a lower voltage drop for the same current level. The total influence may show a small difference in heat dissipation, maybe changing from 4W to 5W - both devices will need a small heatsink.
The above rationale can be used to comparatively check hundreds of other dual diode components that could perform similarly; Mouser and other component distributors have selection tools that can help.

The right diode and the resistor are not needed.

Just omit the resistor and connect the right lamp (L2) directly to the upper switch (S2).

The diode prevents the left lamp being powered when the right lamp is lit.

As others have stated the diode needs to be rated for current taken by the lamp. The diode may get warm in use as it will dissipate some power. A silicon diode would drop about 0.7V so with a 21W lamp taking just under 2A it will create just over 1W of heat. An overrated diode with 5A or 10A capability may be appropriate. A schottky diode would reduce the voltage drop and so produce less heat.

simulate this circuit – Schematic created using CircuitLab

PS. It makes it easier to answer the question if you put reference designators (eg D1, D2, S1, S2) on the various components so the answers can be unambiguous.

• A "shortly" diode? or schottky? Like this though, simple. May 22 at 21:17
• @SolarMike - I've never heard of those. Thanks. !@#\$% autocorrect. May 22 at 23:18

You may use one SPST and one DPST switch.

• As much as the op may love making logic circuits and using diodes, it's best when the user interface is already the appropriate signal.
– Abel
May 23 at 0:17
• @Abel - Thank you very much for your observation, Abel. May 23 at 3:19

When using semiconductors you need consider the temperature of the filament in the bulb. Many times this is referred to as inrush when turning the lamp on. I have seen 15x published as a rule of thumb. My experience is that 15x is a very conservative number. As you get colder the resistance becomes much lower. You would see inrush decaying surge well over 20 amps. The current follows Ohm's law and as the filament gets warmer the resistance gets higher. You need to determine this inrush pulse and check if the diode you pick will support that. I would suggest the best solution would be a mechanical switch and or relay combination such as suggested by vu2nam.