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I want to step down 5V to 3.3V at around 250mA.

As far as I see it, there are two options to consider:

  • Buck: more space, higher cost
  • LDO: less space, lower cost, more difficult to remove heat(?), less efficient(?)

What I am wondering is will the LDO be more efficient and better at doing this job? I've heard things like 6V to 5V solutions usually use LDO's instead of buck regulators because they are more efficient, but I'm wondering if this works for 5V to 3.3V?

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    \$\begingroup\$ A lot of common switching regulator designs do not do very well in terms of efficiency as their output voltage falls, because they tend to have a fixed-voltage loss through a schottky diode which becomes more and more important as the output voltage approaches the diode forward voltage. Just something to look out for. \$\endgroup\$ – user1844 Nov 8 '10 at 11:54
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Dropping 5 to 3.3 V at 250 mA will mean having to lose 0.425 Watt in the LDO, you will need a massive heat sink to make that work.

An LDO will never be more efficient than a buck converter, unless you need so little current that the power used by the regulator itself becomes an issue.

I have a mis-designed PCB right now where I tried doing exactly what you are proposing to turn 5 V into 3.3V at 200 mA and even though I have a large'ish copper plane as a heatsink the LDO still reaches 80 deg C in a few seconds.

I'm currently redesigning my power supply to use a MC34063A converter in stead.

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    \$\begingroup\$ .4W is not a massive heat sink. I have dissipated 1W with a ground plane without a problem. \$\endgroup\$ – Kortuk Nov 6 '10 at 17:05
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    \$\begingroup\$ Yeah, well, "massive heat sink" depends entirely on how large the PCB is, mine didn't have enough room for a full groundplane, so I lost. \$\endgroup\$ – dren.dk Nov 6 '10 at 17:37
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    \$\begingroup\$ Sorry to hear that Dren.dk. I guess we get by with what we can. You could always required chilled helium for the device. \$\endgroup\$ – Kortuk Nov 6 '10 at 17:42
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    \$\begingroup\$ At low voltage (e.g. 3V3) output, the freewheel diode in supplies like the MC34063A can be a significant cause of inefficiency too. If getting rid of 400mW in an LDO is a 'massive' task, then the 100mW in the diode might not be negligible either. \$\endgroup\$ – user1844 Nov 8 '10 at 11:49
  • \$\begingroup\$ Good point, there are tons of extra footprint needed for the 34063 as well, so it might not end up being the best possible solution, but for my application I don't have the room for the copperplane needed to get rid of the 425 mW, but the open area can hold the switcher and it simplifies my system to be able to go directly to 3.3V from m y input voltage, YMMV and all that. \$\endgroup\$ – dren.dk Nov 9 '10 at 10:11
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Many have already given you an opinion on the power efficiency, I would just like to bring up some of the reasons I have seen others do this.

  1. Noise immunity. buck/bost regulators, more broadly [SMPS][1], have very poor noise characteristics. They almost guarantee harmonics at the switching frequency. LDOs do not, they create very smooth power.

  2. Simplicity, you are only dropping a small voltage, keep your circuit clean and your components count low.

This noise immunity is normally one of the main reasons I see this. LDOs cannot be beat on this note, you pay power to get clean output power. The specific reason LDOs are so popular is related to the fact that you can use a buck/boost to get your voltage just barely above the operating voltage of your LDO. I have seen this often in 5V circuits, they boost power to 5.5V and then LDO it to the 5V rail. This gives very low noise high quality power while only suffering a 1/11 power loss, still getting about 90% power efficiency off of the LDO.

So, from this perspective, you could always drop the voltage to 4V with a buck and LDO it, but I would just LDO it and make sure you have it connected to a low resistance thermal path so that the heat is easily dissipated.

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  • \$\begingroup\$ Yeah... but most buck's I've seen have <5mVp-p output, which is very good, sure it's not as good as an LDO but does it really matter? \$\endgroup\$ – Thomas O Nov 6 '10 at 17:23
  • \$\begingroup\$ I agree with you on the simplicity factor though. \$\endgroup\$ – Thomas O Nov 6 '10 at 17:24
  • \$\begingroup\$ As we are talking I am currently reading EMC books. Yes, it does matter, and 5mVp-p may be what you see, but I have seen awful transients. You must remember they are switching pretty quickly for high efficiency. emissions will increase with a large load normally, especially a quickly time changing load, They also put significant conducted emissions on your power line. Both of these things can kill a device when it is put in front of the FCC testing. \$\endgroup\$ – Kortuk Nov 6 '10 at 17:41
  • \$\begingroup\$ Also, please note, as you get higher frequency transients you can start having a very hard time measuring the transients with an oscope. Any inductance can and will block the high frequency transients. \$\endgroup\$ – Kortuk Nov 6 '10 at 17:56
  • \$\begingroup\$ @Kortuk this is not always true, depending on boost frequency. In fact, if poor designed (boost + LDO) will have much worse results and cost. Most of linear regulators cannot reject properly buck/boost converter frequencies >1MHz (PSRR) \$\endgroup\$ – user1797147 Feb 3 '17 at 9:09
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LDOs will not be more efficient: (5 V - 3.3 V) * 250 mA = 0.425 W.

Already quite a lot for smallish (SOT-23) LDOs, at least a DPAK is likely necessary. The design (not the efficiency) could be improved with series resistors at the LDO's input to take heat away from the IC and into the resistors, but make sure the voltage drop across the resistors Rser × Imax doesn't get too big for the highest current that is required. At Imax and at the low end of the available input voltage Vin, min, you still need to meet the LDO's minimum input voltage, i.e.

Vout, max + Vdrop, LDO, max ≥ Vin, min - Rser × Imax.

This trick sometimes help if you can't dissipate all the heat within the LDO's own package and want to spread it across more components. Also, the series resistors in front of the LDO sometimes act as a poor man's short circuit protection, given they can handle the full input voltage for a while.

All this is cheap and dirty, so yes: Might well be worth using a buck.

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It depends on your requirements:

  • For a high-efficiency digital circuit: buck.
  • For precision, low-noise analog circuits: LDO!
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    \$\begingroup\$ I said it, but you kept it short and sweet! +1! \$\endgroup\$ – Kortuk Nov 6 '10 at 21:27
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It is not quite true than an LDO will never be more efficient, as at some point the switching losses and supply current for the switcher will outweigh the benefits.

Oh, and 34063A is a pretty lousy converter as switchers go - for 5 V to 3.3 V it wouldn't surprise me if the benefit is minimal. There are much better converters for this voltage range.

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For digital signals, use a buck converter. Often times you will find a solution that is smaller than LDO solutions, given that inductors has gotten quite a small footprint and the number of external components needed is low.

If you need both digital and analog, you want to clean the signal by using an LDO. In your example, you might use dual DC/DC converts to get both digital and analog voltage out of a single chip. For instance you can get a chip that converts 5V to 3.3V digital, and then connect that output to get 3.0V analog voltage.

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I think you have a misconception about LDO.

LDO means low-drop-out or when you need a very few difference from Vin to Vout. What you're trying to do doesn't require a LDO, a regular 7805, LM317 or another crap will perform the same (read poor).

You can think about linear regulator's efficiency as Vout/Vin so in your example, it's clearly that 3.3/5 = 66% is a poor number. This means that at anytime, your regulator will heat atmosphere with the rest of 34%.

Even with so poor efficiency, a linear can work very well as long as power dissipated on it (that is, make the difference Pin and Pout) will be adequate for regulator's package + natural cooling or PCB plane (read rising package temp at 50 deg for example). This can be easily calculated from datasheets.

But if you're trying to convert 3 from 3.3 will achieve 90.9%, much better (and cheaper) than most of buck regulators. In this case you will need a LDO (and a good one), since 300mV can't be handled by LM317.

So in your case, buck will be far better in terms of efficiency.

Cheers,

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Buck converters usually perform poorly at 'standby' currents of just a few microAmps.

I've actually used battery powered designs combining both an ldo and a buck converter together, where the uC runs of an ldo, and switches on a buck converter powered circuit which consumed ~300mA for a few minutes at a time.

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Well I think I know of a simpler solution. You can use LM117/LM317 IC to do your job and since your current limit is 250mA this should be the best option and you don't have to worry about the heat as these can go up to 1.5A. The requirement here is the input voltage should be at least 1.5V more than the output voltage.

I have used these even without any heat sink for such small currents and they go perfectly fine. Here's the data sheet hope this helps you and the circuit is is not that complex. For a safer side you can find out if you need the heat sink or not by using the formula provided in the data sheet.

http://www.national.com/ds/LM/LM117.pdf

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  • \$\begingroup\$ LM317 in a TO-220 package - too big for my application. And the smaller packages have issues with heat dissipation. Thanks for your suggestion. Heat produced in a regulator is a function of power dissipation, not necessarily load current, though - a regulator dropping 12V to 3.3V will dissipate more than one dropping 5V to 3.3V and hence get hotter. \$\endgroup\$ – Thomas O Nov 6 '10 at 16:11
  • \$\begingroup\$ I'm using an LM317 on my breadboard prototypes though, to do the same thing, but those don't have to be small. \$\endgroup\$ – Thomas O Nov 6 '10 at 16:11
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    \$\begingroup\$ Just becasue it is rated for 1.5A doesn't mean it won't melt if you drop too much power in it. An LM317 will get as hot as any LDO or other linear reg in the same package for the same current and voltage drop. \$\endgroup\$ – mikeselectricstuff Nov 6 '10 at 16:15

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