Making sense of this circuit?

[SimplyElectronics]

Hi Guys,

I am currently practicing my reverse engineering skills starting at creating a schematic from an existing circuit board.
I have been able to draw the schematic based on a simple led circuit powered from AC. (This is simplified, fuses and discharge resistor gone)

I know the 330nF cap is taking current down to 25mA before rectification.

After this, I am struggling to figure out what the electrolytic cap is doing (smoothing?) and why R1 is there.

I know what a zener diode does and how they work, but not sure what it is needed for here?


If someone could help me understand this I would be ever so grateful. I'm sure something will click with sudden realisation as soon as someone explains.
Thank You in advance

[Ian.M]

Your schematic looks borked and cant work as drawn.  Most likely the 100uF cap would explode if it was wired that way and it has less than a 350V rating!

 We'd need photos of the original board to see what you've done wrong.  One shot of each side of the board as square on and as near the same size as possible so we can flip one and overlay them to ease tracing the circuit, + shots at various angles to read markings on comonents.  Please make sure they are well lit and in focus.

[h.bashar]

Yeah that wont work and potentially very dangerous, maybe go back to the board you reverse engineered schematic from and carefully trace the wiring.

[SimplyElectronics]

Ahhh   

I really need more practice

Hopefully these images are good enough.

Thanks for your help. I really appreciate it.

[Ian.M]

After the bridge rectifier, from its +ve terminal there is a series circuit: R2 (grey 47R) - (D2 || R3) - (C2 (100uF)|| off_board LED) to its -ve terminal.

The *ONLY* good reason I can think of for the Zener D2 || R3 combo is to 'soften' the edges of the current pulse into C2 on each half cycle to reduce EMI due to snap recovery in the bridge rectifier diodes.  It has a minor effect on the voltage seen at the bridge rectifier input and thus the current through C1 and (rectified) through the LED, but changing the LED current would be more sensibly done by picking a different C1 or R2.

Edit: I'm wrong - see below
 

[SimplyElectronics]

I think I have it, your input really helped.

I got the images in photoshop and after much confusion and frustration finally got it right and replicated it in everycircuit.

The Zener diode seems to be there for overcurrent protection (at least up until a point until the LED is fried)
The Zener Diode (set at 3v breakdown) directs current away from the LED when voltage surpasses 3V thus protecting the LED during surge.

R1 (47ohm) doesn't really seem to do anything (perhaps someone could enlighten me?).
R2 (2kohm) has current flowing through it, but still not sure what it is achieving?

Obviously the first caps reactance limits the overall current to 25mA (240V/50Hz supply).

Let me know if I did a good job here, think I actually upskilled a little.

P.S is this really how difficult Reverse Engineering should be? This took me at least 2-3 hours (Partly due to much confusion)


Thank You All!

[Ian.M]

Looks like I got it wrong. You are right, the Zener and R3 are across the LED.

R1 is needed to limit the current during fast mains spikes.  They shoot straight through C1 as its impedance at frequencies >>50Hz is fairly low.

[MatthewEveritt]

I think I have it.

Yep, that's what I got too.

The Zener diode seems to be there for overcurrent protection (at least up until a point until the LED is fried)
The Zener Diode (set at 3v breakdown) directs current away from the LED when voltage surpasses 3V thus protecting the LED during surge.

It also keeps the voltage on the electrolytic within safe limits if the LED fails.

R1 (47ohm) doesn't really seem to do anything (perhaps someone could enlighten me?).
R2 (2kohm) has current flowing through it, but still not sure what it is achieving?

R1 also limits the current. Particularly important if the device is plugged in while the mains voltage is high (the sudden transition from unplugged (0V) to high voltage passes through the capacitor easily), or when it spikes.

R2 is probably there to 'steal' a bit of current. LEDs are visible at seriously low currents, so even the very small capacitance of a switch in the off position can allow enough current to allow it to glow slightly. R2 lets this small current flow without the voltage rising high enough to affect the LEDs.


Reverse engineering is a skill you gain through practice, getting it on your second try isn't bad going.

[SimplyElectronics]

I feel quite proud of myself, it's always a good feeling when you progress.

Thank you for your help. I can now sit back and relax. Perhaps another circuit tomorrow :p


Cheers!

[dentaku]

That reminds me of all the reverse engineering of simple circuits Big Clive does on his Youtube channel.
If you haven't watched it you should take a look.
https://www.youtube.com/user/bigclivedotcom

[SimplyElectronics]

Woo! 

I'm a long time subscriber of Big Clive. He has been a big inspiration to me and has inspired me to start my hobby in electronics.