PCB needs repairs

A schematic would seriously help here.

edit: the 18V is just the rectified 12Vac. i would be looking for the state of the 5v rail and then wondering if the led is powered directly off that rail or from a pin on whatever flavour of uController that is

 

I commented on that 'PCB' already, in it's original thread and you never replied?

As i said before, a schematic would be very helpful here - especially as that control device appears to have lost it's descriptive printing. the rest of the circuit 'looks' pretty straight forward.

on the Right there are a pair of transistor arrays that are driven by what ever the big IC is. bottom there is a 555 so on one pin of that you ought to be able to find some pulses - google ne555 for a pin-out description.
data sheets are readily available and have examples of most configurations so you can even work out what it's supposed to be doing too

below the two large caps looks like a linear reg be it 7815/7812/7805 i can't read round corners need schematic!

the led however, and i'm guessing here, appears to be connected to the input pin on the reg, so if it's not lit there is a fairly fundamental issue at the front end of the supply (or the LED's popped). though i could be wrong, it's not easy to tell with a picture of a populated double sided PCB what's connected to what.

If you can get a schematic for this board i might be able to help a bit more.
As this is off-topic for this thread and you already have a thread for this, please continue it there - I have posted this reply there also

QC

 

OK no schematic, but if you swapped out all the IC's and they worked then frankly theirs not much left on that board to fix!
I don't mind helping you fix this but your going to have to at least try & learn a few things along the way - it'll all help for your fixing of future things i promise.

A 'linear regulator' or sometimes called a 'three terminal regulator' is a basic building block of older power supplies. It's got 3 pins, IN, GND (or 0V if you prefer) and OUT - looking at it from the front and reading left to right (put "7805 datasheet" into google and you'll see what i mean). It's function is to turn in incoming DC voltage into a lower DC voltage , in this case 5V (that's what the 5 stands for in 7805), it then wastes the remainder in heat. other common variants are 7812 & 7815 - can you guess what their outputs are
so in this circuit it takes the rectified 12V ac input (rectified by the bridge rectifier (black four legs top of board)) that's now 18V dc and turns it into 5V. The two large capacitors will sit either side of it so with your meter you should be able to read 18V across one of them and 5V across the other - anything else & the 7805 is dead.

Briefly on the 555, it's a very common IC, been around since the arc - just download a data sheet and scan through it, you may not want understand how it works but it'll show you how to use it (and therefore fault find it). I don't think your Fluke 77 does frequency measurements so to truly know if the 555's OK you'll need an oscilloscope (or a meter that does Freq)

Check out those measurements on the reg & let me know how you get on.

 

Yes, that's basically it. pin 1 is the input and pin 3 is the output.measuring between these pins and ground (the middle pin (and also the tab)) should give you the voltages your expecting,

QC

 

Sorry, I missed you extra reply,

the input cap smooths the rectification from the, err rectifier diodes (otherwise it would just be essentially dc but with a BIG wobble on it!). the second cap on the output smooths the output of the reg.

 

hmm, i'm not really surprised about the caps, they are both 35V jobs.
"On the bad pcb I'm getting 0V on both legs of the 7805" so the 7805 is not getting a supply.
I'm not sure, as the components are covering a lot of the upper board tracks, but it looks a bit like the relay controls the 18V going to one of the caps. have you checked the function of the relay or just assumed "as it clicks it's working"?

 

Hi partyanimallighting, I've been following this thread and I think it's time to put my two cents in.

You've actually given plenty of information, with your photos of the top and bottom sides of the board. Here's what I've figured out about the design.

As you and QuantumCheese have already figured out, your 12V AC is rectified by the bridge rectifier (between the big electrolytics and the edge of the board) and smoothed by one of the electrolytics (the one closest to the fan connector). You should measure about 18VDC across it with power applied.

This voltage then passes through the contact of the relay (bottom left corner) and to the second electrolytic (the one closest to the big IC, which is a microcontroller), and to the two ULN2803 driver ICs on the right side, and to the three connectors for colour, shutter and iris.

The relay is driven by the transistor in the bottom left corner, which is driven by the 555. I would normally expect this circuit to be a "watchdog" circuit. While the firmware inside the microcontroller is running correctly, it toggles (i.e. generates a continuous stream of pulses on) one of its pins, and these pulses are detected by the 555. While the pulses are present, the 555 keeps the transistor and the relay energised, so the outputs will work. If firmware fails, the pulses will (or should) stop, and the relay drops out. This is a safety mechanism to prevent damage to the equipment being controlled, if the microcontroller goes haywire (due to program corruption or a bug).

In this case, I can't see the connection from the micro to the 555, so I can't be sure that the 555 is being used in this way. There are some signals from the microcontroller that go to connectors J11 and J12, which are close to the 555, and I would like to know what those connectors are for.

As for the 5V power rail, it is derived from the first electrolytic by the 7805 regulator. The middle pin is 0V. The right hand pin (on the top side view in your first post) should have +18V from the first electrolytic, and the other end pin should have +5V on it. This +5V rail powers the microcontroller, and drives the LED.

Since you say the LED isn't lighting up, perhaps the 5V rail is not present. This could be due to a failed 7805, but you say you've replaced that. It could also be caused by a short circuit on the 5V rail. There are several capacitors across the 5V rail, which could have failed. You need to find out for sure whether the 5V rail is present or not. If it isn't, then you could try replacing the two mustard-coloured capacitors that are on either side of the 7805.

If the 5V rail is present but the LED isn't lighting, the fault is either the LED itself, or the current limiting resistor, which is connected to the LED with a track that's visible on the top side. It looks like the colour bands are green, brown, brown, which is 510 ohms, which is about right.

I'd be interested to know what J11 and J12 are for, and the results of your testing now that you know more about how the circuit is supposed to work.

 

OK. Can you confirm for me that you're measuring all your voltages with the black lead of your multimeter connected to the "0V" rail of the circuit board? That's the tab at the top of the 7805. You need to measure all your voltages with the black wire of the multimeter connected to that tab. Are you doing that?

So you have 18V on one side of the 7805 and nothing on the other side, and you've replaced the 7805 so you know it's not faulty. That seems to indicate there's a short on the 5V rail. The 7805 should be getting hot, because its output is being shorted out.

The next step is to check for a shorted 5V rail. Set your multimeter to resistance range, leave the black lead connected to the 7805 tab, and put the red lead to the left side pin of the 7805. If there's a short, the multimeter should read only a few ohms, or less.

If you get that result, I would suspect the two decoupling capacitors on the 5V rail. These are the little mustard-coloured components on either side of the 7805. To start with, just remove them, and re-measure for a short.

If you need to replace them, check the markings. They're probably marked "104". If so, replace them with "0.1 microfarad ceramic capacitors", rated for 10V or more (they'll probably be rated for 50V). These are very common components and should be available from Radio Shack or any local electronics shop.

 

Since all the ICs are socketed remove them all and test the 7805 again... If the short still percsist you have then narrowed it down to only a few components... If the short is gone you have narrowed it down to an IC...