Considering the dual rail linear power supply circuit diagram at the bottom of this page (https://sound-au.com/project05.htm), what might cause the +V rail to have ~30V AC as well as +15V DC, while the -V rail has 0V AC and -15V DC?
i.e. what could cause the +V rail to have an AC component?
I suspect bad rectifier diodes. But those have been replaced (not sure if replacements were good), and the issue persists. Anything else to check or replace, assuming a cheap DMM and no oscilloscope?
i.e. what could cause the +V rail to have an AC component?
I suspect bad rectifier diodes. But those have been replaced (not sure if replacements were good), and the issue persists. Anything else to check or replace, assuming a cheap DMM and no oscilloscope?
Probably a disconnected/incorrectly wired or missing ground on the +ve rail.
AC will not pass through a regulator with 4k7uF capacitors feeding it. The capacitors would explode!
AC will not pass through a regulator with 4k7uF capacitors feeding it. The capacitors would explode!
It could also be a DC-coupled meter with half-wave rectification, that is, no circuit issue at all.
Suppose you have a simple multimeter that does not insert a DC blocking capacitor when you switch to AC, and that just inserts one single diode. Depending on the polarity, the diode will or will not block the DC. When it doesn't block the DC, the meter reading will be pi/sqrt(2) ~= 2.22144 times the DC voltage. That's because meters are usually designed to indicate the RMS level when the input signal is a sine wave centred around 0. The average value coming out of a half-wave rectifier is sqrt(2)/pi times the RMS level when the input signal is a sine wave, so it has to indicate pi/sqrt(2) times whatever comes out of the rectifier.
To check my hypothesis, you can swap the meter wires. You should then see 33 V AC at the -15 V and none at the +15 V. A 100 nF capacitor between circuit and meter should then get rid of the "AC" on both outputs.
Suppose you have a simple multimeter that does not insert a DC blocking capacitor when you switch to AC, and that just inserts one single diode. Depending on the polarity, the diode will or will not block the DC. When it doesn't block the DC, the meter reading will be pi/sqrt(2) ~= 2.22144 times the DC voltage. That's because meters are usually designed to indicate the RMS level when the input signal is a sine wave centred around 0. The average value coming out of a half-wave rectifier is sqrt(2)/pi times the RMS level when the input signal is a sine wave, so it has to indicate pi/sqrt(2) times whatever comes out of the rectifier.
To check my hypothesis, you can swap the meter wires. You should then see 33 V AC at the -15 V and none at the +15 V. A 100 nF capacitor between circuit and meter should then get rid of the "AC" on both outputs.