SparkFun Forums

[roach]

I'm trying to figure out power dissipation in voltage regulators. Basically, I want to know how much heat will be generated if I plug a 24V battery into a 5V regulator. Typically, the data sheet gives some figures (like 45deg. C/W, or something), but I'm not sure how to use this to figure out the actual heat that will be generated. I mean, do I calculate the power based on the voltage drop across the regulator, or what?

If so, this could be a problem. If I want to power a 5V, 5A circuit off a 9V battery, the power will be 20W. at 45deg.C/W, that gives 900deg. C. More than enough to reduce my PCB to a bubbling toxic mess. Is there something I'm doing wrong?

Please help this clueless newbie!

[amcfall]

To calculate heat generated figure out how much power it has to dump. If you're inpoutting 10V and outputting 5V, you're dumping 5V. If you are drawing 1A then you are dumping 1A/5V, so 5W.

I believe what you are looking at is it's thermal dissipation so you can figure out how hot it will get.

Using a linear reg to go from 24V to 5V is going to get very hot, dumping 19V as heat is a decent amount.

Avery

[leon_heller]

I'm trying to figure out power dissipation in voltage regulators. Basically, I want to know how much heat will be generated if I plug a 24V battery into a 5V regulator. Typically, the data sheet gives some figures (like 45deg. C/W, or something), but I'm not sure how to use this to figure out the actual heat that will be generated. I mean, do I calculate the power based on the voltage drop across the regulator, or what?

If so, this could be a problem. If I want to power a 5V, 5A circuit off a 9V battery, the power will be 20W. at 45deg.C/W, that gives 900deg. C. More than enough to reduce my PCB to a bubbling toxic mess. Is there something I'm doing wrong?

Please help this clueless newbie!

Yes, the power dissipated by the regulator will be the voltage drop across it multiplied by the current it is delivering. A 5V 5A load supplied from 9V will leave 4 V across the regulator giving 20 W. The 45 deg /W is probably the maximum dissipation of the regulator, which can be reduced considerably by using a suitable heat sink. A typical heatsink might be 7 C/W, giving 140C. Reducing the input voltage to 7V should get the figure down to something more manageable, or use a switching regulator.

Leon

[roach]

The regulator I'm looking at is the LT1083CP-5 from Linear. Claims to have 7.5A output current, and up to 20V operating input voltage, with 45deg.C/W power dissipation.

Operating at the MINIMUM input voltage of 6.5V, and drawing 5 Amps (say, to power twelve servos) will still give me 1.5*5*45= 337 deg.C of heat! I mean, seriously. That's frickin' ridiculous...

or use a switching regulator.

Hi Leon. What exactly is the difference between a regular LDO and a switching regulator?

All the datasheets I've seen for heatsinks give heat dissipation in terms of wattage, and talk about something called "Thermal resistivity". How can I take these metrics and apply them to the numbers I'm getting for power dissipation from the regulator's datasheet?

[Roko]

If you're going to be putting that much power through the LDO regulator, you'll need to slap a heatsink onto it to help dissipate the heat. When manufacturers quote maximums, often these values are not meant to be reached all at the same time, and especially not without proper heatsinking.

As for switching regulators, they work differently in that they chop the voltage (like PWM), as opposed to dissipating/wasting the excess voltage as heat. They are much more efficient (often around 90% efficient).

The drawback is that they are generally more complicated, as you often need external coils/capacitors. One exception is the LM2825 (Kind of a personal favorite ), which has all required circuitry inside of a 24 pin DIP package, so you don't need any external components. However it has a current limit of only 1 Amp...

[Roko]

Just found this page that will tell you how much of a heatsink you'll need;
http://www.ef-uk.net/data/heatsinking.htm

Scroll to the bottom for sample calculations, and it shows how to calculate the size of heat sink required. Although with my preliminary calculations, it seems to drop from 20v to 5v and sourcing 5 amps, you'll need one heckuva heatsink...

Another useful link on heatsinks is:
http://www.electronics-cooling.com/Reso ... n95_01.htm

[SOI_Sentinel]

A switching regulator basically PWMs a FET driver at speed (KHz to MHz) so you have to deal with switching losses (inefficienies in power when during the off/on or on/off transition period). However, you can get 80-95% power efficiency this way, depending on frequency, current, architecture, etc.

The power efficiency of your 6.5V to 5V conversion is 5/6.5 = 76%, not bad, but 5/24 = 21% efficient.

There's another problem with switching regulators than their complexity. Their switching frequency can interfere with sensitive electronics that may be on a harmonic (say 1MHz switching frequency, and your PIC is running at 4MHz nearby...). This can be avoided with careful filter design, though.

I've SEEN a homebuild 5A BEC before that would work for you, I just need to find it (was it on rc-groups or somewhere else...). It had a home-wound energy torroidial inductor and was about the size of a pack of cards IIRC. Don't know if they released plans, though.

[Philba]

what are the servos you are driving? do they need regulated power? I know a lot of people that don't bother with regulation on the power for their stepper motors. Since servos have feedback, any variation in voltage should be compensated for. just a thought.

alternatively, you could build several smaller power supplies for 3 or 4 servos. Might be cheaper than a big heatsink and expensive regulator.