The 1.414 factor is the maximum assuming zero source resistance. It will be
less with series resistance.
73
Bill wa4lav
At 01:07 PM 9/5/02 +0100, Ian White, G3SEK wrote:
>Steve Thompson wrote:
>>
>>>For an inductive input filter the RMS ripple current is equal to the output
>>current if you have a >single phase supply. For a capacitive input
>>filter the max ripple current will be equal to the >square root of 2
>>(1.414) times the DC output current.
>>
>>
>>Doesn't it vary with source resistance and C/R values? I use a rule of
>>thumb of 3x for capacitive input - it was very instructive putting a .1
>>ohm resistor in the bottom of the capacitor stack and hanging a 'scope
>>across it. The example circuit in the Duncan software has a cap input PSU
>>putting 450V across 5k. DC current is 89mA, rms current in the cap is
>>226mA and 248mA in the transformer winding.
>
>I haven't checked that example, but one "gotcha" in the Duncan program is
>that the default results in the tables include the power-on cycles where
>current can be very high.
>
>The solution is to set the "Report from" time a few cycles in.
>
>Also the original PSU Designer did not include high-voltage rectifiers
>like we use. If you go to the PSUD site via
>www.ifwtech.co.uk/g3sek/in-prac/index.htm (scroll down to August 2001)
>then you can see more details and also download a modified RECTIFIERS.TXT
>file which includes a typical high-voltage string of 6*1N5408
>
>PSU Designer is very highly recommended - it can answer just about all
>your power supply questions.
>
>
>--
>73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
> Editor, 'The VHF/UHF DX Book'
>http://www.ifwtech.co.uk/g3sek
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