I have merged your other topic with this one as it seems to be just a continuation of the same thing.
It's an electronic ballast for fluorescent lights.
The neon screwdriver will lead you up the garden path, it's not reliable.
A while ago, I bought a modern LED and button cell version of the neon screwdriver.
It's worse than useless, lighting-up like a fluorescent tube under a powerline just about anywhere.
The red block is a capacitor to lower the mains voltage.
The 220 or 330 kΩ resistor is to discharge the capacitor (the red block).
Then the 4 diodes are a full bridge rectifier.
The DC current is smoothed by the electrolytic capacitor.
The resistor behind the electrolytic capacitor is to discharge it.
The resistor on the left has burned out. It is probably a low value to act as a fuse or to limit the current a little.
The burnt resistor shows value in multimeter if I change it is it becomes ok?
I'm sticking with #3 and #18.
you HAVE a multimeter!
I think its to provide the current needed by the load without ohmic heating. A 220nF has a reactance of 15k at 50Hz, so limiting the current to about 16mA rms
The burnt out component IS a low value resistor - the value is printed on the board ???R.
I think its there to prevent high voltage at the output if the load fails
I'm mystified why the electrolytic hasnt blown its top when the circuit is being tested with no load.
I figured out the schematic by the photo of the back side, because I watch bigclivedotcom.
The electrolytic capacitor is a 400V type or so.
The capacitor (red block) limits the current to the led. When the circuit is on with a load, the most voltage is over that capacitor, that was what I was thinking of
The burnt out resistor is not to prevent a high voltage, its value is too low for that.
The question is, should sunil125 repair it or get rid of it ?
- Get rid of it, it is not safe, it is dangerous.
- Just put 10Ω in it and see what happens.
- It is not worth it, and it is not safe and dangerous.
- Did I mention that it is not safe ?
Total 2 resistors burnt out left side one 470 ohm another one replaced near the electrolyte capacitor that is 470k ohm. I replaced 470 k ohm with 47 k ohm is it a problem?
It is a transformerless capacitive power supply. As mentioned before it is DANGEROUS! Do not touch it when powered and wait a moment until the caps discharge after power is removed. Interesting app note: Transformerless Power Supply Design.pdf (561.9 KB)
Yes, it is the discharge of the electrolyte capacitor.
If your mains voltage is 240V, then there is 330V over the 47k resistor, which is a current of 7mA, which is 2.3 Watt. That 47k resistor will go up in flames.
You better stop doing this.
I will replace 470 k ohm with 470 k ohm. Then I will check what happens.
Very dangerous, ALL components at mains potential.
Probe screw driver hopeless.
The unit has been short circuited, check the LEDs it supplies.
Two burnt resistors are 470R and 470k ohm. I by mistake replaced 470k ohm with 47k ohm resistor . See attached image.
I have just noticed the device from OP in the heat shrink looks different than the "naked" device. The electrolyte cap is "standing" on the naked board and have so short leads I don't think it can be laid down.
Unless OP desoldered it and then soldered back, otherwise looks a match
What is the output dc voltage of this circuit?
It is more constant current source than a constant voltage. Output voltage depends on the load.