here you go.

 

Does the relay activate if the + lead from the detector is not connected.

 

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).

That's a perfectly normal response of a MOSFET, since its gate is a very high impedance, so it is likely responding to the stray AC main's voltage your body is picking up.
The gate of a MOSFET should never be left floating.

 

Most smoke detectors are powered from a 9V alkaline battery and their output is a loud BEEP. Some smoke detectors also light a bright white LED. Which output are you using?

 

I would increase the resistor values to at least 1MΩ each to minimize idle current.

 

Does the relay activate if the + lead from the detector is not connected.

no it does not. i tried putting a 10k resistor, multiple diodes between the + lead and the gate input but it still activated with the idle volt. connecting a piece of wire with 0 volts does not activate the relay anymore. also here's a pic of the sticker on the detector if that helps.

 

Read the voltage from + signal to minus 12 volts with the detector idle again please.

 

Most smoke detectors are powered from a 9V alkaline battery and their output is a loud BEEP. Some smoke detectors also light a bright white LED. Which output are you using?

It takes 12 to 32v input, has two leds that flash green on idle every 4-5 seconds and stay bright red when on alarm. No beeps. i'll attach the datasheet.

 

Read the voltage from + signal to minus 12 volts with the detector idle again please.

[idle]
with the - lead of detector connected to psu ground (it's set to 12v), it reads 12.26v.
with the - lead open, it reads 0.5-0.7v

 

[idle]
with the - lead of detector connected to psu ground (it's set to 12v), it reads 12.26v.
with the - lead open, it reads 0.5-0.7v

There's the problem, the output in idle is 12 volts not .5 volts. You don't measure voltages by disconnecting the - lead.
Have you tried this configuration?

 

Your "Fire Alarm" product from Iran uses an (old fashioned?) photoelectric detector and an external power supply. ALL household smoke alarms here in Canada use a radioactive ionization detector and a battery.

 

Your "Fire Alarm" product from Iran uses an (old fashioned?) photoelectric detector and an external power supply. ALL household smoke alarms here in Canada use a radioactive ionization detector and a battery.

This is for a security system not household. Please back off while I'm in the process of trying to help the TS.

 

Your "Fire Alarm" product from Iran uses an (old fashioned?) photoelectric detector and an external power supply. ALL household smoke alarms here in Canada use a radioactive ionization detector and a battery.

Which are banned in my state, Massachusetts. All smoke detectors here are photoelectric.

 

Well unfortuntely this is iran we're talking about. Things of this sort don't follow these global regulatory mandations and that's outside the topic of discussion here. I'm just trying to see if i can by any means utilize this detector (hence i'm here seeking expert help) or i'll have to return it.

 

There's the problem, the output in idle is 12 volts not .5 volts. You don't measure voltages by disconnecting the - lead.
Have you tried this configuration?
View attachment 300161

this was my initial attempt and the voltage from the detector is too weak to energize the coil. drops to 3.3v.
the 12.26v i read earlier i believe originates from the power supply, not the detector itself.

 

this was my innitial attempt and the voltage from the detector is too weak to energize the coil. drops to 3.3v

Then you need to add an inverter to the relay circuit like this:

 

the 12.26v i read earlier i believe originates from the power supply, not the detector itself.

Confusing. Let's try this again.
Read the voltage from minus 12 volts to the + signal from the detector when idle please all connected to the relay circuit.

 

Then you need to add an inverter to the relay circuit like this:
View attachment 300163

well! this is going to take a bit. I'll try this as a last effort and then just go look for a detector that has the relay included. Thank you a lot for spending your valuable time helping a clueless soul. God bless

 

See my post #37.

 

Confusing. Let's try this again.
Read the voltage from minus 12 volts to the + signal from the detector when idle please all connected to the relay circuit.

sorry for the confusion! lemme clear things up. I have a variable power supply set to 12v that powers the detector.
the other two pins on the detector are DC output. right now it's in idle and the voltage between the output pins is 0.6 to 0.8v.