The Cyber-Spy.Com Usenet Archive Feeds Directly
From The Open And Publicly Available Newsgroup
This Group And Thousands Of Others Are Available
On Most IS NNTP News Servers On Port 119.
Cyber-Spy.Com Is NOT Responsible For Any Topic,
Opinions Or Content Posted To This Or Any Other
Newsgroup. This Web Archive Of The Newsgroup And
Posts Are For Informational Purposes Only.
From: Fred Bloggs
User-Agent: Mozilla/5.0 (Windows; U; Win 9x 4.90; en-US; rv:1.0.1) Gecko/20020823 Netscape/7.0
X-Accept-Language: en-us, en
Subject: Re: Power Supply Basic Design Req
Date: Tue, 15 Oct 2002 15:11:04 GMT
NNTP-Posting-Date: Tue, 15 Oct 2002 08:11:04 PDT
Organization: EarthLink Inc. -- http://www.EarthLink.net
George Shaw wrote:
> Can anyone suguest a design and the values or fomular needed for a
> simple DC power supply.
> Input voltage is 230V @ 50Hz
> using a full wave rectifier (which one?) to give 20V out @ 20mA
> current with less than 0.5V peak-to-peak ripple.
> I would like to model this in MultiSim so need the values etc.
> Please email me as I dont often get to check this group. As many
> suguestions, formula would be helpful.
The 0.5Vp-p ripple translates to 0.5/(srt(3)*2)=0.14 V,rms for a ripple
factor of 0.14/20= 0.7%. Your equivalent load resistance is 20/20m=1K
ohm so that the required filter capacitor for full wave rectification at
50Hz is about 1/[ 2 sqrt(3) F 1K 0.007]=820uF -use 1000uF.
Go Back To The Cyber-Spy.Com
Usenet Web Archive Index Of
The sci.electronics.design Newsgroup