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I'm working on a dual rail power supply that will provide DC current for 4 power amplifiers.

Two of them require 20V and the other two 25 - 50V. All 4 draw a maximum of 2.5A of current, hence my transformer is 24+24V, 10A. Currently I have the output of the transformer to a rectifier bridge and on each rail 10.000µf capacitors, resulting in 35V.

My question is: how I drop this voltage to 20V, allowing a maximum of 5A of current to flow through? I've read about shunt regulators but it seems I would need a very large resistor and zener for this amount of current.

Are there other options? Which is the most suited for this case?

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  • \$\begingroup\$ If the noise of switching reg is fine with you, do what John said. Otherwise you need to reconsider your transformer. Hifi linear sources use just the right transformer output voltage[s] so they don't need to drop huge amounts of voltage thereafter. \$\endgroup\$
    – got trolled too much this week
    Commented Nov 4, 2015 at 20:05
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    \$\begingroup\$ A good power amp may not even need its rails to be regulated at all. The gain is set by its internal components, not the supply, and it does whatever is needed to produce an output that is X * input, where X is the gain (gross simplification). Therefore, supply rails that move based on the input voltage or the instantaneous load are not necessarily a problem. This allows the regulator to be omitted entirely and the circuit to be powered directly from the first set of filter caps. \$\endgroup\$
    – AaronD
    Commented Nov 4, 2015 at 20:19
  • \$\begingroup\$ That's true @AaronD, actually I just found out that both of my amps work with unregulated voltage, so I will send the 24V straight from the transformer to the less power amplifier (it supports up to 28V and has rectification/filter bank) and for the other I have PCB for the rectification and filter banks. Thanks for your answers and comments. \$\endgroup\$
    – Carlos Eduardo da Fonseca
    Commented Nov 4, 2015 at 22:19

2 Answers 2

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A switching regulator like John suggests would work in principle, and any experienced electronic engineer would probably find it trivial, but keep in mind that that chip is effectively funneling all of that power through a 4mm square plastic package with no leads and that pretty much all the heatsinking is on the bottom (board side) of the chip. This makes the PCB design a bit tricky in that area for the uninitiated. See section 10 of its datasheet.

A less demanding solution in terms of heat removal would be to separate out some of the functions that that chip does, thus providing more surface area for the heat to dissipate, but it only shifts the PCB difficulties into avoiding parasitic interactions between parts that should not interact.

To keep it simple, I'd be tempted to go with a separate, lower voltage transformer/rectifier/filter for the smaller amps. If you happen to find a transformer that has a single primary and multiple secondaries that collectively meet your requirements, you can make it that much more compact, but those are comparatively rare.

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  • \$\begingroup\$ Also a good approach. However, the switching regulator above is not a synchronous design so not all of the power will be dissipated in the device, much of it will be in the external diode. Still, a proper layout both electrically and thermally is critical and not trivial for someone who has never done it. \$\endgroup\$
    – John D
    Commented Nov 4, 2015 at 21:28
  • \$\begingroup\$ I'm accepting @AaronD's answer as it complements John's and gives other options. See my conclusion on the comment above. Thanks everyone. \$\endgroup\$
    – Carlos Eduardo da Fonseca
    Commented Nov 4, 2015 at 22:22
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Use a switching regulator like the TPS54541 from TI or the equivalent from any of the other power semiconductor vendors:

TPS54541

TI has an online tool called WEBENCH that can help you with the design, as do some of the other suppliers.

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