leigh
Well-known member
I read the tacked-on blue wire incorrectly - although it looked like it connected to the fuse, it actually doesn't - a PCB trace on the board has been (I assume, intentionally) severed, so appearances were deceiving. See photo.
Instead, the blue wire is part of the scheme that powers the lamp for the VU meter. The two leads to the lamp connect to the + and - sides of the power supply, (after two stages of RC filtering, but before the final regulators), where the DC voltage is about 23 volts per side (so it's 46 V total being put across the lamp circuit). The lamp is in series with a series pair of 640 ohm, high-wattage resistors, and when the circuit is fired up, there's a 5v drop across the lamp.
Anyways, that doesn't point to a solution of any kind, because even if the lamp had shorted (which it hasn't), those two resistors would still limit the current through the lamp circuit to about 36mA.
Which really only leaves the first two capacitors (the ceramic and the 1st electrolytic, which is 1000uF/35V) as the possible shorting points (with those 51 ohm resistors in series with the power rails after that point) - so maybe with that 50 ohm power resistor I'm using instead of the fuse, it's limiting current just enough not to have one of those borderline components fail.
Either that, or, it's an incorrectly labeled fuse, as PRR has mentioned. For what it's worth, putting the fuse in series with the 50 ohm power resistor does NOT blow the fuse.
Thank you all for your invaluable input. This seems like a whole lot of head-scratching for a lousy cheap power board, but I'm learning something, so that makes it worth it for me... glad you are all willing to participate in that process too!
Instead, the blue wire is part of the scheme that powers the lamp for the VU meter. The two leads to the lamp connect to the + and - sides of the power supply, (after two stages of RC filtering, but before the final regulators), where the DC voltage is about 23 volts per side (so it's 46 V total being put across the lamp circuit). The lamp is in series with a series pair of 640 ohm, high-wattage resistors, and when the circuit is fired up, there's a 5v drop across the lamp.
Anyways, that doesn't point to a solution of any kind, because even if the lamp had shorted (which it hasn't), those two resistors would still limit the current through the lamp circuit to about 36mA.
Which really only leaves the first two capacitors (the ceramic and the 1st electrolytic, which is 1000uF/35V) as the possible shorting points (with those 51 ohm resistors in series with the power rails after that point) - so maybe with that 50 ohm power resistor I'm using instead of the fuse, it's limiting current just enough not to have one of those borderline components fail.
Either that, or, it's an incorrectly labeled fuse, as PRR has mentioned. For what it's worth, putting the fuse in series with the 50 ohm power resistor does NOT blow the fuse.
Thank you all for your invaluable input. This seems like a whole lot of head-scratching for a lousy cheap power board, but I'm learning something, so that makes it worth it for me... glad you are all willing to participate in that process too!