Is an LM2596 sufficient as a 9 VDC power source for hi-fi audio applications?

I'm currently using an LM2596 DC-DC converter (as part of a break-out board) to power my TDA7318 mixer chip, which in turn gets audio signals from a sound HAT and an FM radio module. Input voltage is 12 V, output is 9 V. As I have experienced some trouble with a very high noise level on the TDA's outputs (in fact, you can hardly hear the sound itself), I've got the following question: Is the LM2596 good enough (in terms of noise / ripple / other impurities in output voltage) for feeding power into audio circuitry like the TDA?

UPDATE: Here is a hand-drawn schematic of the mixer and its connections. The amp isn't depicted here as I haven't implanted it yet. Instead I'm using headphones to listen to the mixer's output:

UPDATE #2: Replacing the switching regulator with a linear LM317T one did not help: The noise remained as it used to! The only cause I can't exclude is the wiring. The orange wire on the top right delivers +9VDC, its green neighbor is the ground wire. The black thing named Visaton is one of the coupling capacitors. See this picture:

UPDATE #3: I've run my oscilloscope over both DC–DC converters and made one 1-second video of each output:

• Have you tried using a linear regulator or a battery? If so, did the 'very high noise level' then go away? Jul 29, 2021 at 21:13
• Hi Neppomuk, the LM2596 is a switch-mode supply, which by definition means that it will definitely create "noise" or "ripple" on it's output. This is usually not good for audio circuits. Two solutions: use a linear regulator supply only (lossy and inefficient, but virtually noise-free), or bump the LM2596 output up a volt or two, and follow that with a low-dropout linear regulator. Research these terms to get a better idea what they are and how they work. Jul 29, 2021 at 21:55
• Or impose adequate filtering (may involve L and C) between the PSU and amplifier, taking care to keep the audio grounds clean. Jul 30, 2021 at 12:55
• The L78S09 is a linear regulator, yes. Note page 11, that the minimum input voltage for this L78S09 is 12V, 12V-9V=3V, thus it is not a "low-dropout" regulator. (You will get the most efficiency by dropping the least voltage across the linear regulator -- some can be found with a dropout voltage of 0.6V or so, if efficiency is very important.) Aug 2, 2021 at 12:18
• If you have an oscilloscope, view the +12V at all possible conditions (assuming it is an automobile: stopped, starting, running, revving, stopping, etc.) You might find that "12V" is more like 9V-44V very short spikes. Always use a fuse in automotive circuits. Aug 2, 2021 at 21:53

The "capacitance multiplier" is essentially a low pass filter feeding a voltage/emitter follower. With the value of R1 and C1 chosen, there is about a 1V loss but a 0.6 $$\V_{p-p}\$$ 1 kHz ripple is significantly attenuated. Obviously, if you use a switching supply, it will have a much higher frequency ripple, and C1 can be reduced accordingly. (Also, the ripple will not be a nice sinusoid like in this example). In this configuration, R1 needs to be kept low because the output is reduced by the voltage drop across R1 plus $$\V_{be}\$$. Alternatively, one could use a Darlington pair to reduce the needed base current through R1 (allowing R1 to be larger), but losing an extra diode drop due to the Darlington pair.