It might have cooked U2 if it applied >12.5V to pin 12. Before fixing it, fire it up and measure that voltage. Also when you pull U4 manually jumper u2-12 low and high to see if the output changes (assuming pin 13 is low - OR gate..).
The rest should be OK but you don't know until you get U4 gone.

The voltage reg might not be happy now that I think about it. It won't mind short circuits / overloads but won't like being back-driven by a seriously failed U2. Just have to see.

See the edit above for a suggestion to fix the circuit until you rev the board.

Have fun.

I've read through your edit...it makes sense, but I'm not exactly clear on how to execute that given the config of the board. Should I scratch out the board line between R13 and Pin12, replacing this by a resistor from the R13 / D5 junction and Pin12?

Is it actually possible to scratch out a copper trace?

 

Yes, you can scratch through a copper circuit board trace. You can repair it later, too. Very versatile, these circuit boards.

 

You can cut copper traces with an X-ACTO knife. Use firm pressure and hack away at the trace in a thin spot -OR- Press the blade into the trace at 2 close-together places and scrape or lift the trace between the cuts with the tip of the blade. Be careful and don't slice yourself, please. A Dremel tool with a small ball cutter tip is a must-have to do much PCB prototyping though. However you do it, cut the thin trace going to U4-12. Just make a gap, it doesn't have to be big. Safety glasses recommended here.

Remove R13 (1.8K LED resistor). Replace it with 2 resistors inserted vertically, one leg of each R in the R13 holes. Twist the remaining leads together and solder a small wire from this junction to U4-12 that you just cut free. 30AWG Kynar wire-wrap wire is ideal for this purpose. This will make a voltage divider off the 18V supply that won't crush your CMOS input.

Add some big resistor across the LED (33K-68K would work fine). This is to pull the CMOS input down when the LED is OFF. Tack solder it to the back of the board. You could also put it from U4-12 to ground if that's easier.

As for values for the 2 resistors replacing R13:
Assuming LED Vf=1.5V you have 16.5V nom. across R13 giving about 9ma of current. You want to limit the voltage at U4-12 to about 85% of Vdd or 10.2 - 10.5V. The voltage at the wired junction of the 2 resistors is Vf LED+ Vresistor so we want ~9V across the resistor closest to the LED and (18V-9V-1.5Vf)= 7.5V across the resistor closest to the 18V supply. At ~9ma this gives about 820 ohms for the one closest to the 18V supply and 1Kohms for the one closest to the LED. You should double check the values to fit your exact input volts and Vf LED.

Good luck.

EDIT: #12 posted while I was typing.. right again as usual
EDIT EDIT: I sometimes remove dead DIP ICs by clipping the leads at the package with flush cutters then removing the legs individually - easier.
EDIT3: I guess as long as you are scabbing resistors on the back of the PCB, you could just make a dedicated divider with 2 Rs across the DC input jack. The ratio will change but you can use high value Rs

 

I don't think you can power the 555 with the cmos ic.

 

I don't think you can power the 555 with the cmos ic.

Maybe if it were a CMOS 555 but yeah.. why?
Good catch.​

 

I don't think you can power the 555 with the cmos ic.

What issue do you see? It is a cmos 555. I breadboard this and it seemed to work.

 

I noticed the collector of Q1 has no current limiting resistor for the LED.

 

I breadboard this and it seemed to work.

You may be able to get away with it, but it is not the proper way to do it.

 

There seem to be a lot of unconnected pins on those chips. You haven't left any inputs floating, have you? Clearly not the main problem, but maybe another source of heat for the hex inverter?

 

dledge,

Exercise caution when posting images. We now know who you are and where you live. Most (all?) of us here are good guys, but everything here is accessible via a simple Google search, and there are wackos in the world who may use such information for nefarious purposes. This is the internet, for Pete's sake!

 

There seem to be a lot of unconnected pins on those chips. You haven't left any inputs floating, have you? Clearly not the main problem, but maybe another source of heat for the hex inverter?

No pins are floating. This is a four layer board, so unused inputs are tied directlyto ground.

 

You may be able to get away with it, but it is not the proper way to do it.

What would be the proper way?

 

I noticed the collector of Q1 has no current limiting resistor for the LED.

The Q1 transistor connects to an external LED circuit that is expecting 12 vdc.

 

This is a four layer board...

Some thirty posts into the thread you finally happen to mention this?

 

Some thirty posts into the thread you finally happen to mention this?

Sorry...I assumed the '+' and '-' symbols on the pinouts within the board schematic made this clear.

 

Brevor said: ↑
You may be able to get away with it, but it is not the proper way to do it.

What would be the proper way?

Offhand i'm not sure, I have never used a 555 in a circuit. But it is usually done using an enable or reset pin.

 

If it is a LMC555 you can do it.

 

If it is a LMC555 you can do it.

How about a SE555P. This chip appeared to have the same voltage rating as the other CMOS chips.

 

How about a SE555P. This chip appeared to have the same voltage rating as the other CMOS chips.

Looks like the pin out on the two chips are the same. Hopefully the all I'd have to do is adjust the resistor and capacitor values.

 

No, It needs to be the CMOS version. The regular one needs more current (7ma.) than you can supply with the CD chip (<1ma.).

http://www.ti.com/lit/ds/symlink/lmc555.pdf

edit: add datasheet