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- Newsgroups: sci.electronics
- Path: sparky!uunet!gumby!destroyer!news.iastate.edu!class1.iastate.edu!sehari
- From: sehari@iastate.edu (Babak Sehari)
- Subject: Re: 12 V -> 3V, 0.6 A
- Message-ID: <sehari.721325974@class1.iastate.edu>
- Originator: sehari@class1.iastate.edu
- Sender: news@news.iastate.edu (USENET News System)
- Organization: Iowa State University of Science and Technology, Ames, Iowa.
- References: <sehari.721280522@class1.iastate.edu> <1992Nov9.114833.26094@bernina.ethz.ch>
- Date: Mon, 9 Nov 1992 16:19:34 GMT
- Lines: 37
-
- In <1992Nov9.114833.26094@bernina.ethz.ch> schaerer@isi.ethz.ch (Thomas Schaerer) writes:
-
- >Babak Sehari (sehari@iastate.edu) wrote:
- >:
- >: There are many ways to do this the cheapest way is to put:
- ^^^^^^^^^
-
- >Sorry but all this is nonsense bcause of the high current. If you realise
- >a linear regulator you will generate very high lostpower. In this example
- >will be 0.6A * 9V = 5.4W. The better way is to build a switched regulator.
- >The advandage is the high efficiency ang therefore less powerdissipation.
- >There are many ic which can do this with a few additional components.
- >Take the Switching-Regulator L296 from SGS-THOMSON. There is a kit with
- >pcb-board and the needed components. Ask a SGS-THOMSON-Distributor.
-
- >So I hope I can help you. Greeting from Thomas
-
- You can spend 1000s of $s on a voltage transformer to make it as efficient as
- possible. However, 5.4W in a car is not really that important. You
- lose much more, just by not having a new battery in your car, or just
- by turning on your lights. Here, the question is Engineering compromise.
-
- I guess you could use gold wire to wire up your car, too! -:) Even though,
- it has less resistance than regular wire; the costs does not justify the
- savings in this case.
-
- However, If you want to do this most efficient way, you should change it
- to AC, using a Class B Amp, then send it through step down transformer
- , with super conducting wires a ferrite core -:), than convert it back
- to DC. The efficiency would be %85*%98*%85=%70. That is 0.7W Vs. 5.4W
- loss. However, it costs more than $500 Vs. $5 of my design. This does not
- justify 4.7W power saving in a car, which more power is consumed for
- ornamental reasons. For example an extra stop light consumes 30W alone. -:)
-
- With highest regards,
- Babak Sehari.
- --
-