SparkFun Forums

[brainwav]

Hi, I have a 5VDC power supply and I need 3.3v, Is there a voltage regulator out there that can do this? 0.5Amps would be plenty. The LM7833 says it needs 5.3v, so I can't use that.

Brainwav

[upand_at_them]

Yes, there are 3.3V regulators. Check DigiKey or Jameco or Mouser or....

Or you could use an LM317 adjustable regulator. Google "LM317 calculator".

Mike

[wiml]

The key word is "low drop-out" or LDO regulator ... there are some specifically designed for generating a 3.3v supply from a 5v supply, can't remember any part numbers off hand.

[jayjay]

You can try the LP2981/5 or any of the LP298X series from National Semi, TI.com is a second source to these, you can choose the appropriate one based on your current requirements.

HTH
Jay

[JonChandler]

Remember to check power dissapation. At 0.5 amps, x a drop of (5 - 3.3), you may need a heat sink.

[Philba]

he will need a heat sink at 500 mA pull.

If heat or efficiency is an issue, check out some of the simple step-down switcher chips from linear or national. lots of design support from both of them (though national's web bench is a real PITA to use).

[upand_at_them]

Depends on how long the current is running. A 220 case can handle 1A intermittently and I doubt he'd even need a heatsink for 0.5A.

Mike

[Philba]

A case doesn't handle current, it dissipates heat. A TO220 has a thermal resistance of about 65C/W in air. 1/2 A gets you .85 W (1.7*.5). Ambient at 30C will give you 30 + 65*.85 or 85.25C. While this is not out of spec, its pretty darn hot. A closed case will drive ambient up and very likely will approach the operating spec max. Also, you'll need to derate for each degree above a certain temp (usually 30C) by some amount depending on the chosen device (see datasheet). A simple solution *could* be to leave a 1" sq are of copper on the PCB and bolt the tab to that. I don't recall the thermal resistance of that but it could be enough to avoid overheating.

Yes, you are right that if 500 mA is an intermittant max, it may be fine. You'll need to do a duty cycle analysis. If 500 mA is steady state, then I'd look a lot closer. I was, perhaps, rash in saying WILL but I would never run a device at that point with out a heatsink. Don't forget that you need to engineer for a broad range of ambient temperatures. Heat is the primary cause of device failure, I'll take the cheap insurance any day.

[upand_at_them]

Good points, Phil. Thanks for correcting me.

So, yes, he does need to know the duty cycle. I've run 2A through a TO220 case MOSFET with a low duty cycle, to switch current through nichrome wire, with no noticeable heat.

Mike

[Philba]

This whole area of thermal engineering is fairly confusing (at least it was to me) when you are just trying to get enough juice to your circuits. Too often people just ignore it. I've seen several articles that purport to explain it and they don't really help that much. It turns out to be quite simple when you understand that "thermal resistance" is a fancy term for the ability of a device (IC or heat sink) to conduct heat away. Lower resistance means less heat buildup, higher means more.

Mosfets are great examples of how thermal principles apply. If you turn a mosfet on with an Rds of .01 ohm and have 10V going through it, you get a voltage drop of .1V. At 2A, you are dissipating all of 200 mW and at 65C/W thermal resistance you will se a 13C temperature rise (65 *.2). At a 50% duty cycle, thats only 6.5C rise. No heat sink needed in either case. However, a mosfet that has an Rds of .1 in the same situation will see 10 times higher temp rise and will need a heatsink. A TO220 is generally rated up to a watt but at a watt, a heat sink is really needed. I generally use a heatsink above 1/2 watt.

Sorry for being a bit cranky in my posts. I sometimes forget that this is supposed to be fun.

Phil

[brainwav]

Thanks for the part # info! I will look into these. Another question: If I use a 500ma regulator and my circuit only uses 100ma, does this change the heat dissapation calcs?

[JonChandler]

The power dissapation is equal to the voltage drop across the regulator x the current draw.

For example, suppose the supply voltage is 5v, and you're regulating down to 3.3v. This is a 1.7 volt drop. Multiply this by the current, say 100 mA, which equals 170 mW.

If the supply voltage is higher, the dissapation will be greater. A 12v supply would increase the dissapation significantly:

(12 - 3.3) x 100 = 870 mW.

[Philba]

Thanks for the part # info! I will look into these. Another question: If I use a 500ma regulator and my circuit only uses 100ma, does this change the heat dissapation calcs?

Jon gave some good points. You do have to look at the package. What is the one you are considering? I suspect 100 mA will be no problem but you have to see what the thermal resistance is and then multiply it by the watts.

To recap:

• - calculate voltage drop (Vin - Vout)
  - multiply by average current to get watts (consider duty cycle)
  - multiply watts by thermal resistance to get temp rise
  - add ambient temp to get operating temperature
  - compare to package max from datasheet

[brainwav]

ok, I chose the LM317. I put in 5V and get out 3.28V (I've alculated the resistors needed). So 5-2.28 = 2.72v. Say I have max 200ma in the circuit.
2.72 x 0.2 = 0.544 Watts. This is were I got stuck. In the datasheet it says (theta)Jc = 3C/W and (theta)Ja = 19C/W. I guess I'll take the 19 one.
0.544 x 19C/W = 10.336C. 30 + 10.336 = 40.336C.
The MAx temp for this one is 125C, so I don't need a heat sink? Is that right?

[Philba]

which datasheet and which package? The national one http://cache.national.com/ds/LM/LM117.pdf
has lots of packages. K is the TO3 package and has a junction to ambient (Ja) of 35. T is the TO220, with a Thetaja of 50

By the way, your voltage drop is 1.72, not 2.72. not exactly good news for non thermal reasons

Your conclusion is correct but I'm suspicious of the 19C/W number. A TO3 should outperform the TO220 but the the national datasheet says the TO3 is 35. Using 1.72V, you'd get .344 watts. .344 * 50 is about 17.2C so you are looking at 42.2C. (they use ambient of 25C) No heat sink needed.

bad news, the LM317 is not a low dropout regulator. Look at the the chart for drop out voltage. I think it shows 200 mA as right on the line for around 1.7V. It's pretty close and you might get away with it but if you push the current up at all, you will have a problem. Interestingly enough, the dropout voltage goes down as the temp goes up.

There are several LDO linears that will work for you. In through hole there is the LD1117 from ST and national LM2937-3.3. both are fairly cheap. There are a lot of SMT parts that will do this as well.