Hello
I just need to understand the logical flow in this PCB,.


12V AC goes to the input terminal and then 12V AC signal goes to the input of bridge rectifier. then it should be connected to one of the capacitors but I don't understand which capacitor is this ?

Any help or suggestion much appreciated

 

We need a photo of the back of the board to help.

I can guess, with some confidence, that the 1000mf capacitor next to the bridge is across it.

 

We need a photo of the back of the board to help.

I can guess, with some confidence, that the 1000mf capacitor next to the bridge is across it.

I understand without looking layout it's is difficult to say anything. I don't have a photo of the back of the board.
I also think the output of bridge goes to input of 1000mf capacitor but what would be the next component ?

 

Are you trying to reproduce the circuit? If so, there's no reason to reverse engineer this board. It is a common application for the 7805 and there are many sources for schematics.

 

I don't have a photo of the back of the board.

How did you get a picture of the front of the board? Do you have a cell phone? Snap a shot then e-mail it to yourself. Then upload the picture here.

The negative lead from the BR goes to the neg of C1000. It also goes on to the base of the 7805 and to the neg of the 5V header. I see a power indicator, likely an LED. It's anode is connected to the 2K2 resistor which is connected to the 12V+ from the BR. The cathode from the LED goes to neg. The C10 should be connected between neg and 5V+ going to the header. The C104, I would expect that to be on the input side of the 7805, between +12 and neg.

Very basic stuff.

 

Hello
I just need to understand the logical flow in this PCB,.

View attachment 173768
12V AC goes to the input terminal and then 12V AC signal goes to the input of bridge rectifier. then it should be connected to one of the capacitors but I don't understand which capacitor is this ?

Any help or suggestion much appreciated

Based on the silkscreen, the C1000/35 is a 1000uF capacitor at 35V to filter the rectified AC from the diodes. The white side is usually related to the white stripe on the cap (negative). However, without seeing the traces, it is possible that the C10/63 is used for the filtering, since it has a higher voltage rating. Check the "+" trace from the rectifier to see which cap it goes to.
The C104 is a bypass cap on the output, to suppress noise/RF. The 2.2K resistor may be part of the 7805 regulator circuit or simply a bleeder resistor for one of the caps
This is a basic 7805 regulator circuit, and schematics can be found on the Internet. This board just adds the AC input and bridge rectifier to the standard "7805" regulator schematic like this one:
http://www.twovolt.com/2016/08/22/lm7805-tiny-module-schematic-and-pcb-layout/
Most of these "clone" types of boards skip a few components to save money, but the basic schematic is standard...

EDIT: Tony is right, the 2.2k is for a LED, I didn't see the LED at first...

 

Assuming a green LED, likely forward voltage is probably close to 3 volts.
12v - 3v = 9v
9v ÷ 2200Ω = 0.004A (a.k.a 4 mA)
4mA is a soft glowing LED current. Not intended to be super bright.

IF the LED is red then likely it has a 2fV (forward voltage).
12 - 2 = 10
10 ÷ 2200 = 0.0045 (4.5 mA)
Again, a very low current, not bright but definitely visible.

One thing for sure, the LED is NOT on the output side of the 7805. If your 7805 fails open the LED will still light, and you may thing it's working when it isn't.

 

Download the datasheet for the 7805 regulator and read it. The 1000 uF cap is the bulk filter capacitor, first thing after the rectifiers. The two smaller caps are required by the 7805 for noise reduction, transient response, and stability. The recommended values vary among manufacturers. Because the regulator input pin is so close to the bulk filter, there is no explicit need for an additional input capacitor. For that reason, both of the smaller caps might be across the output connector. A bottom view of the board will confirm this (or not).

The resistor is the current limiter for the LED.

ak

 

Assuming a green LED, likely forward voltage is probably close to 3 volts.

Small, garden variety green LEDs typically come in at 2.0 V - 2.1 V.

ak

 

Tony, if it is 12V AC input, the DC voltages will be much higher on the input, more like 16-17VDC
At 16V, red LED would be 16-2 = 14V drop
14/2200 = about 6ma.
Not much difference in brightness I would agree.... It may be worth changing the resistor, depending on the type of LED being used.

Also, that circuit can work with lower voltages, down to about 7V. The lower the AC voltage (and DC), the less heat on the 7805 at 1 Amp

 

Small, garden variety green LEDs typically come in at 2.0 V - 2.1 V. ak

Thank you.

Tony, if it is 12V AC input, the DC voltages will be much higher on the input, more like 16-17VDC
At 16V, red LED would be 16-2 = 14V drop
14/2200 = about 6ma.
Not much difference in brightness I would agree.... It may be worth changing the resistor, depending on the type of LED being used.

Also, that circuit can work with lower voltages, down to about 7V. The lower the AC voltage (and DC), the less heat on the 7805 at 1 Amp

Yes, you're right. That UPs the current to the LED some but not significantly. Thanks for bringing up the RMS true value.