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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
References: <firstname.lastname@example.org> <3DAC3026.email@example.com>
Date: Tue, 15 Oct 2002 15:42:41 GMT
NNTP-Posting-Date: Tue, 15 Oct 2002 08:42:41 PDT
Organization: EarthLink Inc. -- http://www.EarthLink.net
Fred Bloggs wrote:
> 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.
Ahh- 50 Hz means 100Hz ripple frequency- so that computes to 410uF- use
a 680uF if you want.
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