The resistor connected to pin 2 (trigger) is shorted (both leads are in the same row).

 

OK, I moved one end down a few rows with nothing else, and also tapped in that green cap into that row, then the other end of the cap goes to +12V.

I touched something and the light came on for 1/2 second then turned off, but I don't know what did it. Light is still off and doesn't turn on if I unplug/replug the +9V Vcc

Here are more pictures.

Another angle

Another angle

 

The power rails now look good, but you have R1 and C1 wrongly connected. R1 should go from pin 2 to the +ve rail.
For now, remove C1 and connect your push-button between pin 2 and the ground rail. Does pressing the button flash the LED?

 

OK, I removed C1 and put the R1 directly to the +9V rail.

And yes, when I touch a jumper from ground to pin 2 and then let go, the LED stays on for 1/2 second. I'm not sure I understand how this will work when +12V pulses from the door lock to the positive line, the lights don't flash then?
Pic

 

Still seeing issues with the pin 2 connections. The side of capacitor which is not directly connected to pin 2 (the end that's in the left positive rail) needs to switch between 0 and 9 volts in order to generate the pulse which triggers the 555 to start timing. I think the way you had that rail being switched before would've worked fine. Providing constant power to that cap means no change on the trigger input. No change there equals no timing circuit.

Also the resistor is still connected wrong. You need one leg of that resistor to be where it's at and the other leg needs to connect to a steady, unswitched power source. I would suggest the positive rail on the right side, since the rail on the left might become a switched rail again. Alternately, you could leave both rails constant, which would allow you to move the resistor leg to the closer left positive rail. In that case you would have to move the leg of the capacitor that's currently in that rail to an available row and provide switched power to it so that you can generate start pulses.

The fundamental issue that keeps coming back here is in the nature of the breadboard. Each row of 5 adjacent holes are electrically connected. There's no reason to ever put both legs of a component in the same row, because then the internal breadboard connections of that row are short circuiting (bypassing) that component.

EDIT: I see others beat me to it!

 

OK, I removed C1 and put the R1 directly to the +9V rail.

And yes, when I touch a jumper from ground to pin 2 and then let go, the LED stays on for 1/2 second. I'm not sure I understand how this will work when +12V pulses from the door lock to the positive line, the lights don't flash then?
Pic

Glad you've got it working! When the trigger capacitor is reintroduced, it will be the key. The door lock signal turning back off after having turned on briefly will cause the capacitor to generate a negative pulse on the trigger input, which will start the timer. You're on the right track now.

 

Ok, so that means I should jump the R1 into an empty row, and put 1 wire of the green cap into that row and the other wire into the +9V rail? I tried that and tried to tap +9V to the green cap, but it doesn't turn on. It's only turning on from a negative tap.

Glad you've got it working! When the trigger capacitor is reintroduced, it will be the key. The door lock signal turning back off after having turned on briefly will cause the capacitor to generate a negative pulse on the trigger input, which will start the timer. You're on the right track now.

 

Ok edit - I put one leg of the green cap into the negative rail (not how the diagram shows it), and the other leg into the same row as pin 2. Then I cook the resistor and put it over to the other side of the board where it gets constant +9V from the battery.

I then touched +9V to the negative rail, and the light comes on and turns off, but isnt' that bad? I'm effectively short circuiting the positive and negative.

Ok, so that means I should jump the R1 into an empty row, and put 1 wire of the green cap into that row and the other wire into the +9V rail? I tried that and tried to tap +9V to the green cap, but it doesn't turn on. It's only turning on from a negative tap.

 

This is what it looks like now.

One pin of the resistor goes to pin 2, then is stretched across to the constant +9V battery.

In that same row as pin 2, one green cap leg is there, and then goes to negative. IF I put it positive and touch it with +, nothing happens. If I touch it while it's in the negative row with +9V the light turns on, but I'm short circuiting the negative and positive by doing that. But that's NOT how the diagram shows C1. It's supposed to be in the positive rail (I should say, getting a short + pulse from the door unlock +12V), not the negative.

 

This is what it looks like now.

One pin of the resistor goes to pin 2, then is stretched across to the constant +9V battery.

In that same row as pin 2, one green cap leg is there, and then goes to negative. IF I put it positive and touch it with +, nothing happens. If I touch it while it's in the negative row with +9V the light turns on, but I'm short circuiting the negative and positive by doing that. But that's NOT how the diagram shows C1. It's supposed to be in the positive rail (I should say, getting a short + pulse from the door unlock +12V), not the negative.

Ah, very interesting. I think (but am not at all certain) that I understand what's going on now. Alec is far more knowledgeable and experienced than I am, so we may want to wait for clarification from him, but here's what I think is happening:
Alec's simulation is based on the idea that the door unlock signal alternates between 12V and ground, but it's never floating. In that scenario, charging the cap up to 12V by turning on the door unlock signal, and then suddenly discharging the cap when the unlock signal drops to 0V will create a negative pulse on the 555 side of the cap, which is what is needed to start the timer.

On the breadboard test rig, we've been connecting the capacitor to 9V when the switch was on, but leaving it floating when the switch was off. With the leg left floating, there's no sudden discharge, so no pulse.

What you really need is to switch between high voltage and ground, but without short circuiting anything like your latest experiment did.

My best guess (emphasis on guess) is that you could use a resistor to ground (referred to as a pull down resistor) to discharge the cap and keep it low except when switch is closed to apply signal voltage to it. It would need to be a much smaller resistor than the pull up resistor on the 555 trigger input to have any chance of creating the necessary pulse.

If you trust me enough to experiment some more (and I totally understand if you'd rather wait for Alec's reply) you could move the capacitor leg which you currently have grounded to a new, empty row, then connect a small-ish resistor (maybe 1k to 2k) from that row to ground. That will keep the capacitor leg at ground voltage most of the time like your current test rig does, but it won't create a total short circuit when you hit the capacitor with 9V. Just to be clear here, the pull up resistor on the other side of the cap would stay as is, and this new pull down resistor would be on the other side of the cap - the pull up and pull down resistors would not be directly connected.

 

OK, I was on the right track if you are correct.

From what I can tell with my limited knowledge, the 12V pulse from the door just goes +12, then trails down quickly. There is another wire in the harness for ground if that matters. The door lock voltage is not a binary on / off thing.

I'll try the small resistor to pull it down right now.

Edit, smallest resistor I have is 100ohm, and I can't figure out if it's supposed to go from the positive rail to an empty row with the other resistor? I don't understand.

Edit #2 - 100ohm resistor works and pulls it down. When I touch the capacitor with +9V the light comes on for 1/2 second and turns off.

I think this is good!

 

Ok, first, everything that is on the 555 trigger input (pin 2) should stay exactly as it is.

Take the leg of the cap that is currently in the ground rail and move it to an empty row. Take the new resistor (100ohm should work for testing this, although it's lower than you'd want to use in the real circuit) and connect one side of it to the same row that you just put the free cap leg in. Connect the other end of the new resistor to the ground (negative/common) rail.

With the new setup you should be able to touch 9V to the capacitor (on the pull down resistor side, not on the 555 pin 2 side) without short circuiting anything. When you release the 9V connection there, the pull down resistor should drop that side of the capacitor back to ground pretty quickly, hopefully quickly enough to generate the trigger pulse needed on the other side of the cap!

 

OK as you said, we have one 555 working (I think) as intended.

If 100ohm is too low, what should I use? I read something about using a 10 time difference in value between the input resistance and the pull down resistance. I think it's a 100K ohm resistor, so is 1K ohm OK? It might not work, let me try 1K ohm.

 

OK as you said, we have one 555 working (I think) as intended.

If 100ohm is too low, what should I use? I read something about using a 10 time difference in value between the input resistance and the pull down resistance. I think it's a 100K ohm resistor, so is 1K ohm OK? It might not work, let me try 1K ohm.

Congratulations! Now that you've got one working, adding the second should be much easier. You shouldn't need the additional pull down resistor for the second 555 like you do for the first, because the output from the first 555 will swing high and low, never floating. Keep up the good work, and good luck with phase two!

 

P.S. I would expect 1k to work just fine, but that's just a hunch. Definitely worth trying. Even higher would be good if it works... 2.2k, 3.3k, 4.7k. Try anything you've got. The higher you can go and still have it work, the less power you're wasting when the door lock signal is on.

 

Thank you! Getting there. Thanks for sticking it out.

I have two chained together now, but the first 555 only blinks once, but the second 555 blinks twice as intended. (By tapping +9v to the cap going to pin 2.)

The diagram shows the first 555 being the trigger source right? But he second one's trigger makes it work as intended.

I have the recommended diodes now also but couldn't find a TVS at Radio Shack.

 

I may be slightly oversimplifying, but I believe this is essentially true: Any given 555 circuit can't run for a specific count. Depending upon how the circuit is designed, a 555 will either cycle on and off forever, or just run one time when triggered. In order to achieve specific counts, additional circuits have to be added. This can be done by chaining lots of 555s together or by using separate counting circuits to trigger, stop, and reset repeating 555s.

In this circuit, if I understand it correctly, neither 555 blinks twice. Each 555 runs once. The first "blink" will eventually be controlled directly by the incoming unlock signal. The first 555 timer starts timing when the initial unlock signal ends, and it is timing the delay between blinks, not providing a blink. The second 555 starts timing when the first 555's output signal ends, and it provides the second blink.

So, if you've got the two timer circuits wired together properly, but presumably haven't yet added the MOSFET, relay, etc, then the proper test would be to connect your LED only to the output of the second timer. What should happen is that if you touch power to the first 555's capacitor like before, then release, there should be a short delay, then a short pulse from the LED. For now you just have to imagine that while you're touching the power to that capacitor, the LED is lighting for the first blink.

If the two timers seem to be working properly together, providing a delay followed by a blink from the LED, then it's time to think about adding some diodes and maybe the MOSFET to achieve double blink action!

If you add D1 (at the input capacitor to pull down resistor junction) and D4 (on the output of the second 555) then you can join their two cathodes and feed the LED from that. With that connection in place, you'll see the full effect: while you're applying power the LED will be lit, then when you release power there will be a brief delay followed by a blink.

If any of this isn't making sense or isn't working, post pics of the new setup with both 555s connected and I'll take a look in the morning.

 

I'm 99% sure that the TVS isn't needed for breadboard testing off a 9V battery. It'd be a good idea to get it before trying vehicle installation though.

 

So I just realized that my advice above to add the two diodes and drive the LED through them could backfire. Depending on the total forward voltage of the LEDs, plus those of the new diodes and the voltage drop coming out of the 555, it's possible that there won't be enough voltage to make anything happen with the diodes driving the LED directly. Incorporating the MOSFET would solve that problem, so if simply testing the diodes doesn't work, we can try skipping ahead to the adding the MOSFET too.

 

Glad you're making progress!
Ebeowolf17 has got it sussed and describes the intended circuit operation correctly. The 100 Ohm pull-down resistor you added on the first 555 won't be needed when the circuit is finally installed in the car, since for now it's acting in place of the unlocking motor resistance.