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

Question

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:

Oscilloscope output of the linear converter Oscilloscope output of the switched-mode converter

Answer 1

SMPS's generate noise and linear regulators will often pass noise through. One way to quiet a noisy power supply is to add a "capacitance multiplier" between the PS and the load. You lose some voltage, but for audio circuits that is often acceptable.

^{simulate this circuit – Schematic created using CircuitLab}

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.