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If we have a 12V DC input voltage and we drop that voltage using a switching regulator (LM2596) to 9V and then use a linear regulator (7805) to drop 9V to 5V, what is the output voltage going to look like?

Is the output going to be like a switching regulator output "noisy/non-linear" or linear?

voltages and part numbers are just for this example...

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    \$\begingroup\$ a switching regulator will output trash at 10MHz and 100MHz ringing, and spikes, because of parasitic resonances in the switching IC, the switching FET, and the enerty-storage L and C. The LDO will not handle those high frequencies. You must use R+C or L+C filters before the LDO. \$\endgroup\$ – analogsystemsrf Oct 13 at 4:14
  • \$\begingroup\$ I'm just gonna put This link here. \$\endgroup\$ – Hamed Oct 13 at 8:02
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The output from the linear regulator will be typical linear - smooth and flat.

That's not to say that your circuit as a whole will be clean. The switching frequency from the switching regulator can get into your circuit other ways. There can be ground bounce or ground loops where the switching noise gets into your analog signals. There can also be radiated interference from the switching regulator that your analog sections pickup and amplify.

Using a linear regulator to clean up after a switching regulator helps, but it doesn't get every thing.

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It's a common thing to do for noise-sensitive circuits. The linear regulator attenuates the noise on its input to a considerable degree over a wide bandwidth. This is part of what is called "line regulation" and "ripple rejection". The specific numbers or graphs are usually readily available in the datasheet.

For example, this TI datasheet covers the LM7805. Look at Figure 5 on page 9, which shows a ripple rejection figure of at least 50 dB all the way up to 100 kHz, and as high as 80 dB at lower frequencies.

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  • \$\begingroup\$ then it's a shame that the LM2596 runs at 150Khz, \$\endgroup\$ – Jasen Oct 12 at 21:57
  • \$\begingroup\$ @Jasen: Not really. Frequencies in that range and up are easily attenuated by other means. \$\endgroup\$ – Dave Tweed Oct 12 at 21:58
  • \$\begingroup\$ so why use a 7805? \$\endgroup\$ – Jasen Oct 12 at 22:00
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    \$\begingroup\$ Because it is very difficult to get >70 dB attenuation in the DC-2 kHz range by other means. Remember, the noise at the output of a switcher is not just the switching frequency! You also have to think about the transient response of its control loop. \$\endgroup\$ – Dave Tweed Oct 12 at 22:03
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    \$\begingroup\$ Ferrite beads and decoupling capacitors. \$\endgroup\$ – Dave Tweed Oct 12 at 22:44

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