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

Please allow more specifics. Like.

What are you needing out of the relay. (12V 1A)?

SPST or DPDT?

If so, here is an idea. Remove the led, replace it with a solid state relay, And replace the led off of the relay load side of the circuit.

Off the top of my head here is a relay that may work. But you will need to look around and see if this will fit your needs.

You may need a DPDT, or something handling more power, etc.

But let me know what you think of this solution.

It may also be possible to use a low voltage (On the input) solid state relay , this would provide isolation and an easy interface ?, optionally an opto isolator in series or replacing the LED in question with same result

Peter

Thankyou for your replies,

I have a fairly good understanding of simple electronics and some more technical things but IC's and this style circuit are still new to me.

I didn't understand this circuit and what the led was actually doing.

when i tested it with a multimeter and connecting between the pos or neg pins on the led i was always getting voltage even when the led was off just when the shift light was turned on i was getting more.

I only need a spst relay on it and about 1-2 amps would be max if i needed more i could run a larger relay off that one anyway.

i tried unsoldering the led and running a relay off the same pins but it didnt work because it wasnt turning it on or off and of course the voltage at the pins is 3.5v for the blue led so unless i can get a relay that has a 3-5v turn on then can handle 12v through?

hope this helps a little sorry for my lack of knowledge 

the LED drivers will not have anything close to the power needed to turn on a relay, you will need to use an opto isolator or at least a transistor to do the job

See this for a bit of a backgrounder http://www.element14.com/community/groups/internet-of-things/blog/2014/08/05/drive-big-things-with-added-safety-opto-isolators

I checked out your link Peter.

I have also tried to use a transistor on the led leg without any success. I wasnt sure what i exactly needed or what would work. When testing i know it always had voltage between both pins of the leds and their opposing supplies.

For example when i tested the voltage between the pos pin on the led and the ground i got a voltage and then the same thing happened between the neg pin on the led and the positive supply even when the led was turned off yet no voltage between the led's pins themselves.

I think from memory the negative pin was earthed straight to the negative so must have some form of positive change when reaching the desired rpm from the frequency- voltage chip.

so if it is a "switched positive" is there a particular type of npn transistor i need to use or is there a huge range that would work?

Thankyou.   

Greetings,
I have figured out little info about the circuit check the following images.

If I am guessing it right, looking at the led the internal right side(near capacitor) is larger which is negative part of led. So that pin is connected with resistor which is connected with 5th pin of IC. Now, the positive pin is connected with three resistor and the 1st pin of IC which I guess is the input to the led(this pins voltage is driving the led). Also if you look at the capacitor, you will find that the capacitor's positive pin is connected to the 1st pin of IC which also confirms that it is the input to the led. This part, I guess, you already knew.
Consider this circuit :

As you have mentioned that even when LED is off it is getting some voltage, I suggest you to add a Diode which will give a voltage drop of around ~0.7v (as shown in image). If again transistor stays on try adding another diode in series, total voltage drop will be around ~1.4v. I guess it will do the job. Don't forget to add resistor in between v+ and collector of transistor. For transistor consider TIP120, TIP31 ...
If this does not works, Please mention few things like:
1. IC number/Name
2. voltage across led when it is off.
3. Few more Images(without reflection, shadow).
4. Voltages across red & black wire, red & green wire.
5. Mention Circuit driving voltage(as I can see 1 watt or 2 watt resistors)

Good Luck

In the circuit you depict here I would have guessed the orange line is the positive power rail and therefore would not change much, the green line through the resistor I would have guessed to be the control line for the LED and the resistor between it and the pin on the IC is to limit the current through the LED.

The additional transistor in this case would not work as the supply rail for the chip is unlightly to be 12V, the transistor would have to be PNP and turn on then the base is pulled low, if it is NPN then the base would have to exced 12V in order to deliver 12V to the relay, if the chip is a 5V device then you will get about 4V only to the relay (Standard emitter follow). You indicate the transistor is NPN based on the description and the direction of the included DIODE. You have no current limiting in the circuit to the base so this could be an issue

I would suggest putting the transistor on the other side of the relay (12V to the relay, relay to the collector of the transistor and the emitter of the transistor to ground. put a few hundred ohms resistor between the base and the control signal

I would actually suggest measuring what is happening on the chip where you indicate the resistor is connected, this would be pin 5 by convention with the assumption the dot on the chip is pin 1 (Normal convention)

this pin I would expect to go from high volts when the LED is off, to close to ground when the LED is on, if this is the case you can use a NPN transistor to hold of a PNP transistor that in turn drives the relay (This provides a level of isolation from the 12 V to the chip which can not be avoided with only one transistor

Hope this helps

Peter

remind me again, what is the chip part number, if I am not clear from my waffiling above, I will draw up a diagram tonight for you

In the circuit you depict here I would have guessed the orange line is the positive power rail and therefore would not change much,

You might be right because this is another way that this led is powered, but conventional way to power a led is by giving voltage to the positive pin while keeping negative pin at ground. Your case is true too.

the green line through the resistor I would have guessed to be the control line for the LED and the resistor between it and the pin on the IC is to limit the current through the LED.

This might be also true, but current limiting resistor can be connected both ways i.e on positive pin as well as on negative pin of led to limit the current.

The additional transistor in this case would not work as the supply rail for the chip is unlightly to be 12V, the transistor would have to be PNP and turn on then the base is pulled low, if it is NPN then the base would have to exced 12V in order to deliver 12V to the relay, if the chip is a 5V device then you will get about 4V only to the relay (Standard emitter follow). You indicate the transistor is NPN based on the description and the direction of the included DIODE. You have no current limiting in the circuit to the base so this could be an issue

I have mentioned that the transistor is supplied with v+ but i forgot to mention it should be 12v as we can drive the 12v load directly, but this is the option 1(without relay). Transistors i have mentioned are of 3A/5A rating.

Current limiting resistor can be added to the base, The idea was to get the circuit up early and then do the minor corrections.

I would suggest putting the transistor on the other side of the relay (12V to the relay, relay to the collector of the transistor and the emitter of the transistor to ground. put a few hundred ohms resistor between the base and the control signal

Again another case and true but whenever there is a relay in the circuit I prefer to go with optocoupler/optoisolator. So this circuit can be used for this case too. You can use system supply at v+ and the where I have connected load that can be used to drive optocoupler which in turns drive the relay(connected with external 12v supply).

Here is the setup for the circuit.

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