Am really a noob in this regards, therefore, need help to build an easy 12V circuit which when circuit "A" is powered off enables "B" circuit to switch off for 2-3 seconds and then turns on. Hopw I have clarified myself, I have searched it but not found the answer. Would really appreciate the help.
Thanks so much everyone for the prompt suggestions, will get the material and build both the circuits tomorrow and will let you know the results.
It is a required for cluster of my vehicle and circuit 'B' is basically always open circuit as it supplies power to retain clock/date memory in the cluster. Here I will add that if this circuit 'B' is disconnected for longer duration the memory of clock/date is reset.
As far as circuit 'A' is concerned it is basically the ignition switch and when it is ON nothng is required to be done to circuit 'B', however, when 'A' is switched OFF then 'B' has to be disconnected for 2-3 seconds and then connected again to overcome a problem of the dials illumination (i.e the power remains ON in the cluster when the vehicle is switched OFF if not cutoff). Hope I have clarified myself 🙂
Best Answer
The following circuit should accomplish what you want:
Normally pin 3 of the 555 timer is off, which keeps the P-channel MOSFET Q1 (high side switch) on, which powers circuit B (output on the left).
When circuit A (input on the left) is powered off, the line going into the trigger input (pin 2) of the 555 is monetarily brought low (C1 and R1 act as a differentiator) and triggers the 555 configured as a monostable. The output pin 3 goes high for 2.5 seconds (determined by the combination of R1 and C2), then pin 3 goes back to 0, turning the MOSFET back on.
The MOSFET is rated at 5A. If for some reason you can't drive circuit B from the MOSFET directly, a lower-amperage MOSFET could be used to operate a relay.