3
\$\begingroup\$

I'm building a car DVR using a GoPro. The GoPro can be turned on by shorting out two pins on the gopro bus and then turned off by shorting out the same two pins.

So when the car's ignition is turned on I need a 3 second pulse (555 timer probably) then when the cars ignition goes off I need another pulse - this is where I am struggling to come up with a solution. I need some type of edge triggering right? going from 0-12v fires off one pulse / triggers the 555 then dropping from 12 to 0v fires off another timer.

Any suggestions, I think I just need to set up two 555 to be positive edge and negative edge trigged right - how do I do that? Been a while since I dabbled in electronic so I'm a little rusty :)

\$\endgroup\$
4
\$\begingroup\$

I'm assuming you don't want to use a microcontroller for this, so here's a 555/logic based idea:

555 ignition

Disclaimer: I threw this together in 5 mins, and haven't used a 555 timer for ages, so I may have made a mistake (or three)
Anyway, the basic idea is simple, set the 555 up in monostable "one-shot" mode, and use an inverter gate and a couple of caps/diodes to produce negative pulses on ignition on/off. The inverter gate can be anything that has a reasonably low output impedance.
The inverter power pins are not shown, and ignore R3 (it's just there for SPICE simulation purposes)

Here's simulation of the circuit (on/off time of ignition shortened from "normal" use)
The ignition starts the simulation on, turns off at 1 second, and turns on again at 11 seconds, each time producing a 3 second pulse at the 555 output (this can be used to drive a transistor to switch your GoPro):

555 Ignition Sim

\$\endgroup\$
1
\$\begingroup\$

I'm guessing you could probably use an iTuner DCDC-USB power supply to accomplish this; it's available here and here.

Off the top of my head, I'm not sure what the on/off pulse timing is or if it can be adjusted, but I would think it could be done in the script mode: Manual; Advanced Manual

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.