[sdiy] Power Supply question
Happy Harry
paia2720 at hotmail.com
Fri Apr 6 17:37:09 CEST 2001
Hi Dave...
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<snip>
Can I simply create a voltage divider with a resistor and appropriate
sized zener diode before the regulators? I
>was planing on creating ~6V to feed the 5V regulators and ~17V to feed the
>15V regulators. Is this a good method to use?
It is possible. You should go at least 3V higher than the regulator
output... so 8V for the 7805... 18 for the 7815.
>I assume the resistor in the voltage divider will need to be 1/2W?
Woah !!! it could be a lot bigger than 1/2 watt... maybe several watts
depending on your expected load.
What is>the method for deriving the value of this resistor? I know how to
do it for>resistor-based voltage dividers, but unsure of what to do with a
>zener/resistor combination. I want to avoid drawing excess current from
the>unregulated 30V supply lines.
For a shunt regulator, you need to know (in advance) the current you will
draw. You want some voltage drop in the resistor, with some current to feed
the zener. With no load, the zener has to eat all the current.
An alternate trick it to put the zener in series with the regulator
input. A 12V zener will lower 30V to 18V... then you take the voltage
across the zener and multiply by the current through the zener... thats the
wattage required. If you use the whole 1A output, that will be 12 watts
dissipated in the zener (a BIG one).
Even using a resistor, you are looking at 12 watts... but it may be cheaper
to burn that power in a resistor.
IF you have a light load, this is a viable technique. With a heavy load...
its a loser. What current do you really need ?
BTW... this is a REAL BIG LOSER to get 5V unless your current draw is tiny.
This is why most linear supplies come in so many sizes... the losses are
just too high unless the transformer is designed for just the right voltage.
H^) harry
>
>Thanks in advance,
>Dave Magnuson
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