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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:

Schematic

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:

enter image description here

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

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    \$\begingroup\$ Have you tried using a linear regulator or a battery? If so, did the 'very high noise level' then go away? \$\endgroup\$ Jul 29, 2021 at 21:13
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    \$\begingroup\$ 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. \$\endgroup\$
    – rdtsc
    Jul 29, 2021 at 21:55
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    \$\begingroup\$ Or impose adequate filtering (may involve L and C) between the PSU and amplifier, taking care to keep the audio grounds clean. \$\endgroup\$ Jul 30, 2021 at 12:55
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    \$\begingroup\$ 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.) \$\endgroup\$
    – rdtsc
    Aug 2, 2021 at 12:18
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    \$\begingroup\$ 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. \$\endgroup\$
    – rdtsc
    Aug 2, 2021 at 21:53

1 Answer 1

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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.

schematic

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.

enter image description here

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  • \$\begingroup\$ The 12 VDC → 9 VDC regulator gets its current from an SMPS (230 VAC → 12 VDC), and the switching regulator is mounted together with two capacitors (one on the 12 V and one on the 9 V side) and a diode on a small board: m.media-amazon.com/images/I/61a2WgTRxhL._AC_SL1170_.jpg \$\endgroup\$
    – Neppomuk
    Nov 6, 2021 at 23:40
  • \$\begingroup\$ Your photo shows a tangled mess of wires all over a breadboard. Each one of those messy wires picks-up interference and spreads its signals to nearby other wires. Many ICs cannot work like that, they need neat and tidy wiring, usually on a compact pcb.. \$\endgroup\$
    – Audioguru
    Nov 6, 2021 at 23:46
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    \$\begingroup\$ @Neppomuk. Do you have a oscilloscope? Can you add a scope shot of the noise to your question? Knowing the frequency components would be useful if the circuit above does not help. Also, how much power are you drawing from your supply? Can you add that to your question? \$\endgroup\$ Nov 6, 2021 at 23:51
  • \$\begingroup\$ @Audioguru My breadboard is already filled up with various components, so that I simply don't have anymore space left. Of course I will make a PCB once I've got my testing work done, but this will still take some time. :-( \$\endgroup\$
    – Neppomuk
    Nov 7, 2021 at 0:06
  • \$\begingroup\$ @Neppomuk I'm not sure why you showed my the photo of your DC-DC converter and it's caps. But I think it is worth mentioning that DC-DC converters are designed for efficiency. The circuit I am offering gives up some of that efficiency in exchange for quietness. Despite it's name, a "capacitance multiplier" doesn't simply act like a large capacitor. \$\endgroup\$ Nov 7, 2021 at 0:32

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