Hello,
When I opened an optic smoke dedector,I saw the pcb layout of photodiode was different.
there was no pcb ink,I attached the photo.Why pcb ink has not been used?

 

Well, I don't have an answer. But having worked in the electronics industry for 30 years I've seen incorrect silkscreen as well as missing silkscreen (the ink you're talking about). It may just be something that was missed either by the designing engineer or by the company that produced the raw board.

I was working for a company named Teleco Oilfield Services back in the 80's. They bought a lot of boards. One morning I walked out the front door to go to work and I found several bare boards laying on the grass. Lo and behold, they were Teleco boards. But where did they come from? I also recognized the boards as being a new project, one that was considered "Secret" (not military secret, company propriety secret). I picked them up and took them to work. At work, upon closer inspection they were factory defects. Someone at the factory must have taken them home then decided there was nothing practical they could do with the boards. Part of the defect was the silkscreen was incorrectly placed, running ink over some of the via's. (through hole) As production boards they were useless. Still, I found it odd that they ended up outside my door. I wondered if someone was up to something.

Anyway, your PCB - - - it's probably just an error.

 

Thanks for reply, but I think they've made it on purpose.(I dont know why )
Because it looks shapely and under the photodiode.
I thought maybe there was a special reason.

 

The purpose is to lower stray leakage on the surface of the pcb, because the photodiode produces only a tiny current. A photo of the bottom would be nice to see, I would expect it being right next to an opamp with another bit of unmasked pcb and maybe even a guard ring around the sensitive node of the opamp.

Also, what is missing is not only the white silkskreen, but also the green soldermask.

 

Thanks for reply.
Here is the IC(A5366) that connected to photodiode,there is an opamp in it.
But there is no guard thing on pcb.



And second question:
The photodiode has 3 pins, I've search on the internet but I could not see the photodiode that has 3 pins.
One of these pins looks like metal case:What is this pin used for?

 

You're calling it a photodiode. It could be an IR Phototransistor. Could be UV phototransistor, same thing, just the opposite end of the light spectrum.

Same thing as what's found in TV's with remotes. Well, somewhat similar in that it has three leads.

 

yes,you are right but explanation of the IC (part number:5366) has confused my mind
It says for the detect pin "This is the input to the photoamplifier and is connected to the cathode of the photodiode. The photodiode is operated at zero bias and should have low dark leakage current and low capacitance. A shunt resistor must be installed in parallel with the photodiode."

 

Perhaps the 'missing' parts are for a different feature that this detector does not have. It would make sense to make a 'universal' pwb art work, and provide different (upgrade) features using the same pwb.

Some smoke detectors allow for remote testing with a handheld device. Could be what the IR (36khz) three leaded part is for.

Did you notice the chamber with the 2 leds ? One is an emitter and the other is a detector.

They are designed such that smoke will cause the detector IR led to sense photons coming from the emitter IR led when there is smoke in the chamber, but will not detect when there is no smoke. Smoke causes photons to scatter.

The 200k shunt resistor provides the detecting circuit with a known input value and allows a current to flow as the photo diode is operating with zero bias.

The cover is to minimize (or eliminate) other phontoic pollution.

Some detectors have the ability to sense when the chamber is getting dirty and prompt the user to clean the chamber as dirty chambers cause false alarms.

 

Hi again,
I measured "the cover" and it is connected to +9V.
What could the purpose of this cover be?To minimize photonic pollution?
Actually I did not understand why +9V?