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- Newsgroups: sci.electronics
- Path: sparky!uunet!spool.mu.edu!yale.edu!ira.uka.de!chx400!bernina!schaerer
- From: schaerer@isi.ethz.ch (Thomas Schaerer)
- Subject: Re: 12 V -> 3V, 0.6 A
- Message-ID: <1992Nov9.114833.26094@bernina.ethz.ch>
- Sender: news@bernina.ethz.ch (USENET News System)
- Organization: Swiss Federal Institute of Technology (ETH), Zurich, CH
- X-Newsreader: TIN [version 1.1 PL6]
- References: <sehari.721280522@class1.iastate.edu>
- Date: Mon, 9 Nov 1992 11:48:33 GMT
- Lines: 45
-
- Babak Sehari (sehari@iastate.edu) wrote:
- :
- : There are many ways to do this the cheapest way is to put:
- :
- : 12-3
- : --------- = 13 diodes in series with the input voltage of your CD player,
- : 0.7
- :
- : Like this:
- :
- : +12V -|>|-|>|-|>|-|>|-|>|-|>|-|>|-|>|-|>|-|>|-|>|-|>|- +3V
- : GRD --------------------------------------------------
- :
- : The problem might be generated if the input voltage swings. To avoid
- : this problem try putting a 2N3055 transistor as follows:
- :
- : +12V |>|-|>|-|>|-|>|-|>|-|>|-|>|-|>|-|>|- ---+3V
- : | \_/ Emitter
- : R |
- : |--|
- : |
- : ||>|-|>|-|>|-|>|-|>|-|
- : |
- : GRD ---------------------------------------------------------+-----
- :
- :
- : R=1000 ohm
- : Trans=2n3055
- : Diodes= 1N4001 or 1N4005
- :
- : With highest regards,
- : Babak Sehari.
- : --
-
- 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
-
-
-