Can Someone explain the working of this Circuit (MODIFIED ONE)?

Hi,
here is a circuit that i found in a device .. all i know is that this circuit provides 5.6v constant output at the "output" labelled in the image and takes AC voltage (220V r.m.s. and 50 Hz ). The labels on the diodes were not visible so i used random diodes in the simulation just one thing that the diode labeled "D3" is a zener diode. further that the capacitor is polypropylene film it reads as (1.1k 250v). i don't know weather it is 1.1kF or what because i think its not possible so i have attached an image of the capacitor too.
can some one explain how this thing works ??

IN Simulation when i set the value of the AC to 6Vpeak and 50Hz the result is as it is in real life . but when i change the value to 320VPeak(220V r.m.s) the result varies and goes up-to very high values ..

it is awesome because it converts ac to dc without any transformer or something so kindly take a look and do tell if u find out something .

Picture of Can Someone explain the working of this Circuit (MODIFIED ONE)?
IMG00215.jpg
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It might help to know what kind of device the circuit came from?
usbg3rd (author)  mpilchfamily5 years ago
A Disco light application.. supplies constant 5V to an IC on the circuit and that's it.
Now Got any idea about the circuit working ?
Your schematic is missing allot of the circuit.

Is the circuit getting power directly from the wall or is it going through a power brick? Where is the power connector?
usbg3rd (author)  mpilchfamily5 years ago
Its complete and on simulation its working fine. the ac source signal shows 320V peak (220V r.m.s) @ 50Hz . i don't know how its working and that's what i am asking. its practicable and is implemented to . so now do u have a clue or something how its working ?
I'm not familiar with the simulator your using. But it looks like you have an AC and a DC voltage source on the side you have labeled as output. Why did you choose those to positions as test points? Why do you have a signal generator in the schematic?

Having a signal generator in the simulator in that position and a scope where you have it will just be reading whatever the signal generator is outputting.
usbg3rd (author)  mpilchfamily5 years ago
first i am using Proteus7.8
second its mot generating signal its providing 220vrms @50Hz to the circuit ans there is no other power-supply ac or dc besides this ..
third i chose those terminals because they can provide 5V dc which i am interested in but don't know how the 5v is produced without any rectifier /step down transformer ??
The circuit as you have it cannot take 220VAC and covert it to 5VDC. At best the diodes are rectifying the voltage to around 220VDC but it will be a very dirty DC output. The circuit needs allot more to it to take the input voltage down to 5VDC,

usbg3rd (author)  mpilchfamily5 years ago
The Circuit is working fine on simulation and its working practically too so there is nothing wrong with it here is the image chek this out its giving 5.6- 6V DC output
honda supply test.png
Like i said before the circuit as you have it in the simulator won't work in the real world. If the function generator symbol is your input side then you have a wire going directly from the input to the output, bypassing all the components. Rendering half the cycles of the input unaffected by the components.

Let me put it more simply. I look at the picture of the device and i see right away that your simulation is missing 2/3 of the components on the board. There is more going on to convert the 220VRMS to 5VDC then 3 zener diodes, 2 capacitors and 1 resistor. All the components on that board are likely connected and effect each other. A voltage reading at a single point on the board is being affected by all the components not just the few you think are the ones responsible. So your real world reading are being skewed unless you have physically isolated the remaining components on the board. (i.e. cut the traces or removed the components) For now forget about the simulation.

Lets start at the beginning. You want to know how the circuit works. Before you can figure that out or ask for help you need to reverse engineer the entire board. Figure out what components are what and what each component is connected to. You need to do this for the whole board not just a small section of it. Lay out all the components into a ruff schematic program and start linking the traced as you find them on the board. When your done you can start reorienting them into a more organized and easier to read layout. Then you'll start to have a better idea of what is going on in the circuit.

To find what traces link what you can use a continuity setting on your DMM. Follow the visible traces from component to component and verify they are linked with your DMM. I'm working on a project now where i'm reverse engineering an old crank powered radio. I've considered doing an instructable on reverse engineering a circuit but it gets pretty involved and often requires special equipment.
usbg3rd (author)  mpilchfamily5 years ago
Its Direct ac from the wall no adapter etc in between i have measured with DMM. simple wires.
Like I said previously, the cap is probably 1.1kilo pico farads, or 1.1nF.

The circuit is better, but still wrong, because the resistor is NOT in series with the supply, which it has to be.


This is STILL a dangerous circuit, even when you get it right BECAUSE THERE IS NO MAINS ISOLATION.

Do NOT use it if there is ANY chance of you touching ANYTHING associated with it.


Steve