I want to step down 5V to 3.3V at around 250mA.
As far as I see it, there are two options to consider:
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?
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.
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.
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.
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.
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.
It depends on your requirements:
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.
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.
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.
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,
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.
