Hello everyone!
I'm looking for a circuit that amplifies the current for a 12v input so it's strong enough to trigger a relay coil.
Here's the scenario I've got going:
Recently i purchased a smoke detector and was planning on wiring it to my security system but the detector made the system malfunction. It turns out the detector outputs 0.5 to 1v when idle and around 12v when in alarm mode and the securty system is based on closed circuit operation rather than voltage monitoring so i got a 12v relay only to find out the 12v current from the detector is too weak to trigger the coil. What can i do to amplify this current so the relay works?
Any points in right direction are much appreciated! Thanks in advance

 

Welcome to AAC!

Use an NPN transistor or N channel MOSFET to energize the relay coil. Any general purpose NPN transistor could be used, but be mindful of the 1V detector output when it isn't alarming. A MOSFET like 2N7000 would avoid that problem because it's treshold voltage is 0.8-3V and it only conducts 1mA at that voltage.

EDIT: assuming the coil resistance isn't less than 100 ohms or so.

 

Below is the sim of dl324's suggested circuit with a MOSFET driver:
Note the added diode, which is needed to protect the MOSFET from any voltage spike caused by the relay coil inductance when the relay is turned off.
It shows the relay coil current (yellow trace) versus the alarm signal (green trace).

 

Thanks to you both for the time and the diagram. I have ordered the parts and I'll be giving this a try!

 

Would it help to add a resistor between the detector and the mosfet to eliminate the idle voltage causing a false alarm? If yes, at what resistance? Because it apparently goes above 1v sometimes and the 2n7000 datasheet mentions the minimum gate threshold as 1 volt.

 

Would it help to add a resistor between the detector and the mosfet to eliminate the idle voltage causing a false alarm?

No.

What is the worst-case voltage when the detector isn't alarming?

 

I've seen it reach 1.3v at max

 

No.

What is the worst-case voltage when the detector isn't alarming?

i tested again and left it on for an hour monitoring the idle voltage and it stayed within 0.47 to 0.59 volt.
apart from that i just got the parts and assembled the circuit but as soon as i touch the gate of the mosfet with bare hand the relay starts going crazy (flipping very fast). the relay is a 12 volt HKP JQC-3FP(T73).

 

it did that a few times i powered it on and then stopped. i replaced the mosfet and now when i power it on by connecting a 12v power supply + to the coil pin 1 and - to the source of mosfet the coil is triggered and stays that way instead of waiting for the 12v alarm.

 

What is the nature of the detector output? Is it being driven (i.e. not floating), when the alarm isn't alarming? How much current can it source?

 

Would it help to add a resistor between the detector and the mosfet to eliminate the idle voltage causing a false alarm?

Yes. Good thought, right idea, but - wrong solution.

It will take two resistors. The input impedance to the FET is so high (megohms) that it would take a huge series resistor value to reduce the voltage seen by the gate. The FET's internal path from the gate to the source, which would be the shunt leg of an attenuator, is basically a very thin sheet of glass. As resistors go, that's *very* large.

The answer is an external 2-resistor voltage divider that will attenuate both the high and low signal values without depending on the FET characteristics. For example, if you attenuate the sensor signal by, say, 50%, then the 12 V part is attenuated to 6 V, which still is plenty high enough for the 2N7000, and the 1 V part down to 0.5 V, which should be low enough to eliminate false triggers. The good news is that because the FET needs so little current to change state, the resistors in the attenuator can be a high value that doesn't load down the sensor signal. Two 10K resistors should be OK. If not, increase to two 47K or 100 K parts.

There is a trade-off here. The higher the resistor values, the less loading of the sensor signal (good). BUT, the greater the chance that induced electrical noise will false-trigger the FET (bad). The solution is to add a small capacitor across the shunt (lower) leg of the attenuator, from the FET gate to GND. This forms a simple noise filter. There is a chance that hanging a capacitor on the sensor output will upset it somehow, so I would start big with something like 1.0 uF just to see. If the sensor doesn't mind, fine. If it does, drop down to 0.1 uF, until you find a value it can tolerate.

And, just for grins, there is another way: In the two-resistor attenuator, replace the series resistor with a small value zener diode, such as 3 V, and keep the shunt resistor at 10K or 100K. Now, the sensor high output voltage will be dropped by the zener to a big, fat 9 V, but the sensor low voltage will not reach the FET, and the shunt 10K resistor will pull the gate down to 0 V. This is because any sensor output voltage below 3 V will not be high enough to bias the zener diode into conduction.

If you don't have access to low voltage zener diodes, two small signal diodes (1N914, 1N4148, etc.) in series will add enough voltage drop to work. If you don't have those, don't worry about it. The two-resistor attenuator should solve everything.

Where are you located?

ak

 

idk how to measure that but it has 4 pins.
two of them take 12v power and the other 2 are out with 0.5v average on idle and 12v on alarm. when i connect the output to an led strip (that draws 220 milli amps normally when connected to a proper psu) the output voltage of the detector drops to 7.92v. when connected to the relay coil it drops to 3.3v.
Currently the issue is not the detector. Shouldn't the mosfet not allow current flow when the gate pin is open and not connected to anything else? simply applying the V1 on the diagram to the circuit activates the relay with the gate pin open. also the 2nd mosfet stopped working as well and now it doew nothing. i have 2 more mosfets at hand but also i noticed the notes on the mosfet say STK 7000 H41. Is it different than the 2N7000? i believe i have been sent the wrong one. here's some pics, maybe i wired it wrong. dont mind the solder mess, i'll use a new board when i figure it out

 

It seems to be that STK7000 is not interchangeable with 2N7000. I'll be having a word with the seller :|

 

it did that a few times i powered it on and then stopped. i replaced the mosfet and now when i power it on by connecting a 12v power supply + to the coil pin 1 and - to the source of mosfet the coil is triggered and stays that way instead of waiting for the 12v alarm.

Minus side of the relay coil should connect to the drain and the source to ground or negative 12 volts.
From the second photo I see the mosfet is in backwards.
The STK7000 and 2N7000 should be interchangeable as they have the same pinout. The relay circuit should look like this:

 

Minus side of the relay coil should connect to the drain and the source to ground or negative 12 volts.
From the second photo I see the mosfet is in backwards.
The STK7000 and 2N7000 should be interchangeable as they have the same pinout. The relay circuit should look like this:
View attachment 300149

ok i made this but as soon as i connect the +12v of the smoke detector the mosfet's temperature skyrockets.

 

Show a photo of the circuit please. Check the diode it may be shorted.

 

Show a photo of the circuit please. Check the diode it may be shorted.

Yes! the diode was dead. replaced and when i connected the detector, tick!
but it also gets triggered with the idle voltage of the detector. do i need to connect the detector's minus output to ground? it's currently not connected to anywhere.
i'm drawing the traces on the pic, be ready in 2 mins

 

do i need to connect the detector's minus output to ground?

YES!

 

YES!

ok even with the detector's minus output grounded it still triggers in idle.
The resistors show 98.5k ohms of resistance.

 

I can't see the detector connected to the perf board. Expand the photo on the left side please.
Without the opto the circuit should look like this: