Help with Making a RPM tacho Shift Light circuit open a 12v Relay

Here is the setup for the circuit.

The positive, and negative orientation of the diode is the key.

Ah yes so where i have the +ve leg on the led coming off the relay i need to put the cathode on the diode and bridge it straight across the relay solenoid in parallel if that makes sense?

you got it.

you got it.

Although i cant remember if i have the relay coming off the negative pin on the led or the positive... i have a feeling it may be switched negative and constant pos.... cant remember in that case would i flip the diode around?

nope. here is what happens in a relay, that requires the diode.:

Since an inductor (the relay coil) cannot change it's current instantly, the fly-back diode provides a path for the current when the coil is switched off. Otherwise, a voltage spike will occur causing arcing on switch contacts or possibly destroying switching transistors.

But, if you place the diode the wrong way, it will act like a short, and there will not be enough voltage to the relay to activate it.

Done properly, When the current through a coil is switched off, the coil (being an inductor) will try to maintain the current. When there is no path for this current the voltage across the coil will increase rapidly, and the current will find a path, right through the isolation of a chip or transistor, destroying that component. The diode provides a path for this current, so the energy stored in the coil can be dissipated safely.

Yes i understand that but what i mean is that the switch (SW1) in your diagram is actually negative on mine from memory (will clarify later) so in that case my positive lead is constant so then if i didnt switch the diode around wouldnt it just short and create high current to run and not activate the relay?

DIODE will always point to the positive supply side, no matter where you place the switch, it is to protect against back EMF when the switch opens as stated

the switch does not dictate the current flow or the polarity of the supply, thats all about how the battery is wired up in the circuit

Peter

Yeah thats what i thought so If mine is a switched negative then i need to face the diode in the opposing way to the diagram. I will draw up a bit of a circuit later to confirm and also see if mine is switched negative like i think cheers