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This question already has an answer here:

For a particular design, I need to take an unregulated voltage from a battery and regulate 5 different DC voltages from it. What's worse is that I need step-up and step-down converters. What's even worse is that some voltages need more current than others.

Note: To clarify the question, I've given the specific voltages that I need. However, keep in mind that I'm wondering what to do in general, not just this specific case.

The battery voltage is 11.1 V, which drops as it discharges. I need to create:

  • 15 V
  • 12 V
  • 5.5 V
  • 5.0 V
  • 3.3 V

The 12 V level needs to run at least 1 A, and the three lowest need to run about 200 mA each. The 15 V level doesn't use more than 50 mA.

So this is my question: would it be better to go from the battery to 15 V to 12 V to 5.5 V to 5.0 V to 3.3 V, or would it be better to just connect all 5 regulators directly to the battery?

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marked as duplicate by The Photon, Voltage Spike, Dmitry Grigoryev, uint128_t, winny Jul 5 '17 at 12:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ I would use some dual-output DC/DC converter with say, 15V and 5V (perhaps trimmed to 5.5V?) outputs. And regulate down from there. \$\endgroup\$ – Eugene Sh. Jun 26 '17 at 18:42
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    \$\begingroup\$ This has been asked several times before, although it might be hard to search out the exact best match from the old questions. In any case, the answer is, the efficiency doesn't change (much) and it comes down to where is it most convenient in your design to put the heat sinks. So there's no general procedure, just do your engineering design based on your requirements. \$\endgroup\$ – The Photon Jun 26 '17 at 18:47
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I have often wondered this myself, and each time what I come down to is a trade-off. The most obvious trade-off is the following:

Pro: Cascading them generally causes the lower-voltage regulators to stay cooler (you're not dropping as much voltage with them, so you're wasting less power in the form of heat).

Con: Your first regulator would need to be able to supply enough current for all of the circuitry running off each of the other regulators (and this is the case all the way down the chain). This means that your top-level regulator would have to be big and beefy compared to the regulators that are fed from it.

Sometimes the pro wins (for example, all of the circuitry on each of the power rails only draws milliamps of current, so you don't need a powerful regulator at the top), and sometimes the con wins (you can't find a top-level regulator that can supply enough current, so you opt for large heat sinks and extensive cooling systems).

You, as the designer, will need to analyze all potential cases and make sure the circuitry can handle any stress that it might see during normal operation.

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    \$\begingroup\$ Probably more important than efficiency is the order of the outputs on power up and power down. For many complex designs this is the most important factor. \$\endgroup\$ – Jack Creasey Jun 26 '17 at 22:02
  • \$\begingroup\$ @JackCreasey Excellent point. You don't want to backfeed certain circuitry. Good way to burn out protection diodes or the like \$\endgroup\$ – DerStrom8 Jun 27 '17 at 0:18
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You should also take care about making your system failsafe. I will go in a details by explaining what I've done in a recent project.

I've had battery pack and DC/DC to 5V/6A and in a need of 24V/2A supply. There were 2 options, use one DC/DC to to 24V/2A and the cascade that with 5V DC/DC supply or take Vpack->5V and Vpack->24V opton. Just to be clear, no LDO/Linear regulators were able to do that.

The end solution used another DC/DC from battery pack to make 24V/2A.

Pro's:

In case of hazard at 24V, only that part of circuit is being damaged.

Only one type of IC is avaiable.

Con's:

Increased PCB footprint of solution.

Now back to your question, everything is a tradeoff - without knowing anything about how much you are willing to pay for efficiency of power supply, no detailed answer is possible.

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