Help designing a circuit to automatically push a dash switch for approximately 15 minutes

Thread Starter

ee4060

Joined May 11, 2018
5
I am looking to design an automotive (12v) based circuit that will close a switch when the circuit is powered up for approximately 15 minutes (+/-1 minute) before opening the switch until being reset (by having the car turned off and losing power). I considered using an RC circuit but I believe the time delay is to long to reliably use just RC. I am also looking into using a 4060 timer IC but do not have much experience with IC's and was wondering if a simpler device/circuit existed for such a long time frame.

The objective is to fool the ECU into thinking the user has the button depressed for up to approximately 15 minutes. (The long duration is because the time between the circuit being powered up and the ECU actually recognizing the input can be up to 15 minutes.) The device has to allow the sensing wire to go high after the specified time frame, otherwise the ECU will see the low input all the time and set a logic fault code ignoring future inputs.

If someone with a better understanding of long delay timer circuits could help me out I would be incredibly grateful!!
 

AnalogKid

Joined Aug 1, 2013
8,538
Is your intent to have a transistor connected in parallel with the dash switch, and sink approx. 10 mA for 15 minutes? If so, then where does the power for the circuit come from? A connection to the left side of R1 in your drawing, or somewhere else?

CD4060
2 resistors and 1 capacitor for the oscillator
1 resistor and 1 capacitor for the power-on reset
1 diode to inhibit the oscillator after 15 minutes
2 2N7000 MOSFETs to invert the 4060 Q14 signal and drive the switch line
1 decoupling capacitor
1 drain pull up resistor

ak
 

Thread Starter

ee4060

Joined May 11, 2018
5
Is your intent to have a transistor connected in parallel with the dash switch, and sink approx. 10 mA for 15 minutes? If so, then where does the power for the circuit come from? A connection to the left side of R1 in your drawing, or somewhere else?

CD4060
2 resistors and 1 capacitor for the oscillator
1 resistor and 1 capacitor for the power-on reset
1 diode to inhibit the oscillator after 15 minutes
2 2N7000 MOSFETs to invert the 4060 Q14 signal and drive the switch line
1 decoupling capacitor
1 drain pull up resistor

ak

Thank you for the response! I was thinking of using a transistor in parallel with the switch to sink the 10mA. My other thought was using the 4060 to just run current through a resistor in parallel with the switch while powering it off of the sensing circuit, dragging the sensing circuit down to about 5V (anything less than about 6v the sensing circuit detects as a low state, or button press). But i think that would cause stability/reliability issues in the chip with the with the chips source voltage variation.. Also, I'm not sure about the idle power consumption of the CD4060, and if i can even have it on that circuit at all without the sensing circuit detecting the voltage drop. If i need to use the CD4060 and a transistor in parallel I can power it from the source to the left of R1, it would just be tapped into a key controlled B+ circuit. In the parts list above how exactly are you using the MOSFET's? I just started researching CD4060's so this isn't exactly my strongest skill set. Thanks again for the help!!

P.S. Is the circuit diagram broken in the original post? I cant see it in my browser anymore for some reason
Circuit.PNG
 

AnalogKid

Joined Aug 1, 2013
8,538
The 4060 circuit cannot be powered from the line it is switching to ground, because then its own power source is switched to ground. The circuit must be powered from somewhere that comes on and goes off when you want because per your first post, its power source also is the control signal to the circuit.

4060 idle current is about 1 uA for the chip, plus static current through the oscillator resistors, approx 120 uA. This forma a voltage divider with R1, reducing the voltage at the sense point by approx. 1.2%.

ak
 
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