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I have a 24VDC supply that needs to be regulated down to 5VDC to power an Arduino Mega and a XBee Series 2 module.

How would you choose between a Switching Regulator (LM2576T) and a LDO Regulator (LM3480IM3)? I think 100mA is sufficient to drive the Arduino and XBee.

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  • \$\begingroup\$ Since this isn't a production-run project (otherwise you wouldn't be using Arduino Megas and XBee modules) just get one of the 5v buck regulators on eBay for $2 and call it done. \$\endgroup\$ – Bryan Boettcher Apr 16 '15 at 21:00
  • \$\begingroup\$ @insta For a production-run project, what would u use instead of XBee? \$\endgroup\$ – Nyxynyx Apr 17 '15 at 1:54
  • \$\begingroup\$ If you're only doing one, all the comments on efficiency are silly, unless you burn up the regulator or draw too much current from the supply. It is a big drop, and 2W in one package is a bunch, but it won't hurt your electric bill. Best solution is defined in terms of your objective-cheap once, easy once, cheap over many items, works with parts on hand, or what? \$\endgroup\$ – Ross Millikan Apr 17 '15 at 2:24
  • \$\begingroup\$ @Nyxynyx: whatever the XBee is made out of. \$\endgroup\$ – Bryan Boettcher Apr 23 '15 at 15:33
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Using a switcher instead of a linear regulator is a no-brainer without even doing the math. Also a LDO specifically is silly, since your problem is that you have a very large drop range. Any linear regulator would do, not just low-dropout ones.

In case some still need convincing, let's do the math. A linear regulator will drop 19 V, which times 100 mA is 1.9 W. That's more than any bare package in free air will be able to dissipate and stay within operating temperature. You'd need something like a TO-220 or TO-3 with a heatsink. That is going to be big and expensive.

A switcher will be smaller due to not having to get rid of so much heat. Let's say you get a buck regulator that is only 80% efficient (quite poor by today's standards). The output power is 5 V x 100 mA = 500 mW. 500 mW / 80% = 6.25 mW into the switcher, which means it will dissipate 125 mW. A SO-8 package or similar just normally mounted on the board can handle that kind of power. You would be able to feel it being warm, but you wouldn't even burn your finger.

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There is a huge difference between the input and output voltage, thats why I would go with the switching regulator.

The LDO dissipates the unwanted voltage, and it will heat your circuit a lot.

On the other hand, the switching regulator has a high efficiency (above 80%) and does not waste much power. Especially recommenceded when there is a significant difference between input and output voltages.

Example from a datasheet(GSM module): enter image description here

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First, you need to think about what you're asking. An LDO is a Low DropOut regulator. See, for instance. Now, "low dropout" isn't actually defined anywhere, but a good start is "functions normally with an input to output voltage drop of less than 2 volts". Since you are asking for advice on a 24 to 5 volt regulator, with an input-output difference of 19 volts, asking about the utility of an LDO is not remotely useful.

Assuming that you mean "linear regulator" as opposed to "switching regulator", the other answers do an excellent job of answering your question.

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    \$\begingroup\$ Nice work man, I too am tired of seeing LDO (a special and usually more expensive sub-class of the usual linear regulator) used in places where the design doesn't call for it. \$\endgroup\$ – KyranF Apr 16 '15 at 18:02
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In general a switching regulator is more power efficient than an LDO. Depending how much your circuit will consume I would prefer to use the switching regulator(LM2576T).

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