Hi,all I am designing a power supply circuit(12V-3.3V, 12V-1.2V)for a board which include two fpga,a/d,dsp and the total power dissipation is about
10W+. Now I am a little confused about the function of regulator and DC-DC converter. In my first opinion, the DC/DC convert had more efficiency. But the FPGA EVM board from altera use LM2678 and LT1085 to provide the power. Does I really need a DC-DC converter?
In the context of the original post 'wording', the LM2678, _is_ a DC-DC converter. Hence the prototype board has got a DC-DC converter. However such units only using a simple inductor, or a PWM without an inductor, are commonly referred to as 'switching regulators', with the phrase 'DC-DC converter' tending to be used more commonly for devices that involve a transformer, or some other form of voltage 'shift'. Hence the prototype board is using a switching regulator to keep the power losses to a reasonable level (I'd guess generating the 3.3v with this), and then using the LDO linear regulator, to give the 1.2v. The advantage of the switching regulator, is the high efficiency. It's disadvantage is noise. As said above, the device needs a switching converter, or a large heatsink. Remember also that using a device like this, lowers the delivery requirements. The poster talks about power dissipation, if he actually means that the final circuit 'consumes' this much, and making a guess that most of the current is drawn from the 3.3v rail, with (guessing), perhaps
1A at 1.2v, and 2.6A at 3.3v, then using linear regulators to deliver this would give a power dissipation in the regulators of (12-3.3)*2.6 = 22.6W, and (12-1.2)*1 = 10.8W. The system would be using 9.8W itself, bringing a total required from the incoming supply of 43.2W. 3.6A at 12v... However using the 2678, and the LT1058, would give instead 3.3v at 3.6A (because the linear regulator is running from this rail), with an efficiency of (say) 90% (the switcher can do slightly better than this), so (3.3*3.6)/0.9 = 13.2W from the incoming supply, and only about 1.32W needing to be dissipated in the switcher. The LT1085, would dissipate (3.3-1.2)*1 = 2.1W. The current drawn on the 12v rail becomes only 13.2/12 = 1.1A.Best Wishes
While a regulator dissipates the unwanted voltage of 12-3.3 = 8.7V or 12-1.2 = 10.8V times the nominal current, the DC/DC converts at 90% or so efficiency.
You're not going to dissipate 10W easily and noiseless.
10W require quite a heatsink.Rene
-- Ing.Buero R.Tschaggelar - http://www.ibrtses.com & commercial newsgroups - http://www.talkto.net
schreef in bericht news: snipped-for-privacy@g14g2000cwa.googlegroups.com...
Well, the LM2678 *is* a switcher - not an ordinairy regulator. So it is much more efficient than a linear regulator. If you need one, you have to decide yourself.
Infact they are all DC/DC converters, but when speaking about DC/DC converters, it's often switched mode supplies, often with galvanic separation too. I suppose you could even call a resistor divider a DC-DC converter ;)
-- Thanks, Frank. (remove \'q\' and \'invalid\' when replying by email)
You seem to have gotten some good answers to most of your questions.
I will just point out that switching supplies should have input and output filtering on them unless you don't care about putting a lot of noise on the input power supply for some reason.
I overlooked this once.
--Mac
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.