# Project help (probably simple solution)

Discussion in 'The Projects Forum' started by alexfloyd21, Feb 12, 2013.

1. ### alexfloyd21 Thread Starter New Member

Jan 28, 2012
9
0
I'm just learning circuits, so this'll probably be a simple project.

Project
I have 2 power sources: P1(12V) and P2(6V)

When P1 and P2 are switched on, the circuit needs to activate.
When P1 is switched off (while P2 is still on) I need it to power an led for 5-10 seconds and then go back to checking for both P1 & P2 to start the circuit over. If P2 is never switched on... it needs to do nothing.

P2 is going to power the light.

Thanks!

2. ### praondevou AAC Fanatic!

Jul 9, 2011
2,936
488
Yes this is easy. If you give us more details. What circuit is being powered by these two sources? How is current divided? Or does P2 have priority over P1?

3. ### alexfloyd21 Thread Starter New Member

Jan 28, 2012
9
0
I haven't started on the circuit yet... just trying to come up when a design first.

P2 needs to provide the current to the led. P1 is just acting as a trigger. I think I can design a circuit that will turn on the led when P1 is off and P2 is on, but I don't need it to automatically perform this without P1 being on first.

4. ### praondevou AAC Fanatic!

Jul 9, 2011
2,936
488
Sorry I still don't get it.

Please confirm or correct the following assumptions:

1.
P2 (6V) is your main power supply that powers the LED (when P1 is turned off) AND some other circuit you didn't reveal yet?

2.
With P2 off the additionnal circuit and LED remain OFF, regardless of what happens with P1?

3.
What happens to the additional circuit when P1 goes off and the LED turns on? Will it still be powered from P2 (6V)?

4. To turn on the additional circuit, is there a sequence P1 and P2 have to follow?

If you want some help, give more details. Many people watch your thread but don't know what you want.

Otherwise you get this:

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5. ### alexfloyd21 Thread Starter New Member

Jan 28, 2012
9
0
Sorry I still don't get it.

Please confirm or correct the following assumptions:

1.
P2 (6V) is your main power supply that powers the LED (when P1 is turned off) AND some other circuit you didn't reveal yet? Yes... and a small motor.

2.
With P2 off the additionnal circuit and LED remain OFF, regardless of what happens with P1?yes

3.
What happens to the additional circuit when P1 goes off and the LED turns on? Will it still be powered from P2 (6V)?yes

4. To turn on the additional circuit, is there a sequence P1 and P2 have to follow?The only sequence is that the both have to be powered initially and when P1 switches off and P2 remains on, the led and motor turn on.

I apologize for being unclear... my son and I are playing around with circuits and I have next to no experience with them, so I'm trying to find some answers to questions he's had while we were looking at the different logic gates.

6. ### praondevou AAC Fanatic!

Jul 9, 2011
2,936
488
Could be something like this:

This will turn on the motor and LED once every time the 12V turns off (for the time set by the 100k and 100uF cap. This may not work if the 12V turns off too slowly. The two schottky diodes are needed , otherwise you damage the 555.

The FET needs to be one that turns already fully on at 6V gate-source voltage. Shouldn't be a problem as there are thousands on the market.

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7. ### alexfloyd21 Thread Starter New Member

Jan 28, 2012
9
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Thank you. Again to point out my lack of skills... could you explain what's taking place in the circled area? Also... I couldn't read the size of that capacitor. I need to get more familiar with the 555 timer IC.

Thanks again!

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8. ### praondevou AAC Fanatic!

Jul 9, 2011
2,936
488
The monostable 555 is being triggered with a negative going pulse at the trigger input (pin2). The output timer only starts when the trigger input pulse ended, i.e went up again.
Here the trigger input is normally at 6V level through the resistor.

When the input signal goes from 0V to 12V nothing happens, the pulse appearing on the right side of the capacitor is shorted to 6V through the upper schottky diode.
When the signal goes from 12V to 0V then the right side of the capacitor is pulled low for a short amount of time, until this side charges up to 6V again. The timer starts.

However, I have a question: When the input is not 12V does that mean it is open? Or is it actively pulled down to 0V level? If the latter this circuit should do it. If the input is either 12V or floating it will not work, something else need to be added.