Re: Need help building a split voltage supply
From: JoJo (N_at_NE.nothing)
Date: 08/27/04
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Date: Fri, 27 Aug 2004 11:30:44 -0400
"Robert C Monsen" <rcsurname@comcast.net> wrote in message
news:OOAXc.52818$9d6.20002@attbi_s54...
>
> "JoJo" <N@NE.nothing> wrote in message
> news:6mwXc.111166$4h7.15647526@twister.nyc.rr.com...
> >
> >
> > I tried 3 4001 in series and a 1K both as a load across the 5V and
> > in series
> > with the diodes.
> >
> > 3 diodes in series - 1K as load = 4.36V (same reading using 4 diodes
> > too)
> > 3 diodes in series - 1K in series =4.75V
> >
> >
>
> Here is some information that might be helpful, some of which you
> might already know.
>
> (please view in courier font)
>
> On real schematics, the symbol for a diode is a filled in black arrow
> with a line on the point. The line on the diagram corresponds to the
> line on the actual diode. This is an ascii equivalent:
>
> ->|-
>
> Current flows in the direction of the arrow, from left to right, when
> you put voltage across it.
>
> This is the ascii symbol for a resistor:
>
> --/\/\/---
>
> Series means end to end, whereas parallel means all ends tied
> together.
>
> Series:
> 1k
> 5V ---->|---->|---->|----/\/\/--- GND
>
> Parallel:
>
> +--->|----+
> | | 1k
> 5V-+--->|----+--/\/\/--- GND
> | |
> +--->|----+
>
> If you put them in series, then the voltage across the resistor shown
> will be about 2.9V. Replacing the resistor with your circuit that
> requires 3V is what the original responder was saying.
>
> If you put them in parallel, then the voltage across the resistor will
> be about 4.3V. This is probably not what you want, cause it doesn't
> drop the voltage to near 3 volts for the remainder of the series
> elements.
>
> If you don't have a series resistor, like this
>
> Series:
>
> 5V ---->|---->|---->|--- GND
>
> or
>
> Parallel:
>
> +--->|----+
> | |
> 5V-+--->|----+---- GND
> | |
> +--->|----+
>
> then you are shorting out your 5V supply, and probably pulling the
> supply down to 4.75 by trying to suck far too much current through it.
> Don't do this, you will burn out your power supply and fry your
> diodes. You should always have a resistor, or something that provides
> resistance between the 5V and GND terminals.
>
> One other thing, a Zener diode is a special diode, which is used
> reversed from the usual orientation (the line points to the positive
> side.) It works by preventing current from flowing until a voltage
> threshold is reached, and then allowing allowing a virtual short above
> that. If you put it in series with a resistor, and put more voltage
> across it than the voltage rating of the zener, the voltage across the
> zener will be approximately the voltage limit, and the voltage across
> the resistor will be whatever is left over, that is, V+ - V- - Vzener.
>
> Hope this helps!
>
> Regards,
> Bob Monsen
>
>
Thanks for clarifying, worked like a charm :-)
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