# Simple circuit help...

Discussion in 'Electronic Basics' started by MacGyver, Oct 4, 2005.

1. ### MacGyverGuest

At least I think it is simple...

I have 2 12v wingtip lights that I need to be able to control with
one switch and one wire running through a wing in an airplane.
This is for just one wing. That's the gist of it.

I just bought piggyback strobes for the wingtips, but I don't want to
have them on all the time that I have the navigation lights on.

I was thinking of a relay that has a remote control, or has a signal
run through the positive wire or could possibly have a low & high
setting and if we send 12 volts to the relay it trips, but it won't
trip at say 10 volts.

Also was thinking about if I turn the switch on once just the light
comes on. Flip the switch off & on again in under 2 seconds and the
lights & strobes come on. Leave the switch off & everything turns off.

I was thinking about relays with a capacitor to keep the line live for
2 seconds to be able to flip another relay... but now I am confusing
myself again.

Any Ideas???

Seems like it should not be too difficult.
Thanks so much for any advise!!!

Sean Gallagher
Pheonix, AZ

2. ### Bob MonsenGuest

Ah, the easiest and most reliable thing is to run another wire. You can
use that wire to either switch a transistor, or to pass the current to the
light.

If you can't do that, you can build a little circuit that will 'remember'
how long ago the power was shut off, and turn on a circuit if it
wasn't too long ago. Thus, you can flip the switch for a second, flip
it off, then flip it on again, and the strobe will start.

The basic memroy circuit is an RC node, A below. It charges slowly,
and discharges even more slowly. If Vcc is dragged to ground (through
the position light) the diode will keep it from draining.

The voltage on this node can be compared against another node that
charges up and drains quickly, B below. If both are charged up shortly
after the power comes up, the circuit latches on. If either are low,
the circuit does not latch.

The latching circuit is a simple "SCR" configuration, using mosfets.

A is pulled somewhat slowly up to 12V while the power is on, but
decays very slowly back to ground when the power is off.

10k
Vcc-----|>|--/\/\/\/-----o-------.------- A
| |
| \
| /
| \
22uF--- / 1 Meg
--- \
| /
| \
| |
| |
Gnd----------------------o-------'

B creates a quick spike, which decays back to Gnd in few ms.

||
Vcc------||---o------o---------B
|| | |
| \
1uF | /
| \
- / 10k
^ \
- /
| \
| |
| |
Gnd-----------o------'

Here is the main switch. When the power comes on, A will start to
charge up with an RC constant of about 1/5 of a second. B will spike
up immediately, then drain in about 1/20 second. So, the first time,
one will be low, the other will be high. Thus, one of the N-MOSFETs
below will be off, and no current will flow through them. Because of
this, 2N7000 will never be turned on, and the P-MOSFET will also never
be turned on.

However, after a second, A will be charged up to nearly 12V. If the
power is then turned off, it'll slowly drain down to ground, taking
about 100 seconds to get all the way down. Thus, if within about 10
seconds, the power is restored, both mosfets will be on briefly,
causing the gate of the 2N7000 to be pulled up to 12V. This will turn
it on, causing current through the 100k resistor to bring the gate
voltage of the P-MOSFET down, turning it on. Once it's on, it'll keep
the gate of the 2N27000 high, so the load will have power until the
12V supply goes away.

Vcc -----o-----------------o-----------------------------.
| | |
||-| \ |
||<' N-MOSFET / |
A----||-. \ 100k |
| / |
| \ |
||-| | |
||<' N-MOSFET | |
B----||-. o--------------------------||-|
| | ||>' P-MOSFET
| | ||-. Low Rds(on)
'-----------------)-------. |
| | |
| | |
'-|| | |
2N7000 .>|| | |
|-||----o-------\/\/\/\-------o----- to 12V load
| 10k
|
|
Gnd -----------------------o----------------------------------- load gnd

Use a big P-MOSFET with low Rds(on), because that will dissipate
almost no power, so it won't get hot, or lower the voltage too
much.

You may want to put a resistor from the load output to ground, perhaps
10k. That will ensure that it won't turn on incorrectly when the power
comes up if there isn't a load pulling the drain of the P-MOSFET to ground.

There may be gotchas in this circuit, and I haven't built it, so ymmv.

3. ### Guest

Am I getting this right?

With 1 switch, you want to set 4 different scenarios:

1 - Off
2 - Nav Lights
3 - Strobe Lights
4 - Nav lIghts + Strobe Lights
Back to Off

Please let me know if this is correct.
--