🚀 go-pugleaf

In article <1991Feb10.220719.17078@cbnewsl.att.com> wally@cbnewsl.att.com
(Brian S Walden) writes:
>I want to take my old Sony Dicsman D-5 in the car with me,
>the problem is that it requires 9 volts and 1/2 amp.

OK:  First, let's think 13.6 volts, not 12.  That's the actual (as opposed to
nominal) voltage when you're running the engine.

Second, you _can_ use a resistor/zener_diode/buffer_transistor circuit such as
the one you mention, but there are easier ways.

We want to carve about 4.6 volts off that 13.6.  If we call it 4.5, and you
can come up with a 4.5 volt zener, then just put the zener in series with the
load:

+ 12V o----------------|<-----------o + 9.1V
			4.5V
			zener
			diode

Negative is ground, of course.  Note that the zener diode is installed "back-
ward" to the way normal diodes are placed.  It's used in "reverse conduction"
mode.  A problem is that that diode is dropping 4.5 volts at half an amp...
that comes to 2.25 watts.  As a safety fudge, you'd want a diode that could
handle twice that...say, 5 watts.  5 watt zeners aren't cheap.  Also, note
that that drop assumes a significant current being drawn.  As the current gets
very small (<1mA, typically), the voltage will start to rise.  This _usually_
isn't a problem, but for a CYA it can't hurt to put a small load...say, an LED
with a current-limiting resistor (2.2 kilohms or so)...across the 9V side.
This will stabilize the current, and give you a pilot lamp to boot.

Stick a 3/4 amp or so fuse in the circuit, too.  Slo-blo is recommended.

Again, note the polarity of that diode.

Also, be warned:  If that zener should develop an internal short, the full
13.6 volts will be dumped into your load.  If you install it backward, you
will get just under 13 volts into your load. If you're lucky, the fuse will go
before your DiskMan.  _If_ you're lucky.... 

There's also a junk-box solution:  1 amp silicon rectifiers are widely avail-
able, cheap, and many of us have a bunch lying around.  The forward drop on a
silicon diode is nominally 0.7V; actually 0.626 is a _very_ good average fig-
gure.  Lesee...0.626 into 4.6 is...7.3bleep, so if we use seven rectifiers in
series, we lose about 4.4 volts leaving us with just under 9.2 volts.  The
circuit looks like this:


+12V o------->|-->|-->|-->|-->|-->|-->|-------------o +9V
	seven silicon rectifiers, 1A at 50V or so.

This time, note that the diodes all point downstream the way they're "sup-
posed" to.  If you need more current, it's easy to find 6A or even 10A
rectifiers.  Consider the rectifiers as all one component, then install the
fuse and pilot light as before.  Each diode dissapates about 0.626 times
the current.  Stay under 1 amp, and you should be safe.  The PIV of the
rectifiers isn't a worry, and 50V is about the lowest (and cheapest) type
you'll find.

You have some more safety here, too.  Failure of more than one of the diodes
(shorting-type failure, that is) is pretty unlikely in any situation that
doesn't involve a blown fuse.

Another advantage is that all the components pass the strongest test of
availability that I know.  Radio Schlock carries 'em....


					Best o' luck,
						d

PS -- A final caveat:  It's unlikely to happen, but make sure you have _sili-
con_ (not germanium) diodes, and make sure they're not Schottky or "hot car-
rier" silicon diodes.  All those diodes have a forward drop of about 0.3V, and
if you're using a rectifier chain, you're like to do unpleasant things to your
load.  Germanium rectifiers are rare, and Schottky rectifiers are both rare
and expensive.  If you're worried about it, you can measure the forward drop
very easily:  Just put the diode in series with a small load (say, a flash-
light lamp) powered with a battery, and measure the drop with a voltmeter.  It
should be between 0.6 and 0.7 volts.

						d


--
    "Got to slap these Goddamn Third World nations around, Flynn," he said,
    "until they learn some manners."     -- Gregory MacDonald, from _Flynn_
  Duke McMullan n5gax nss13429r phon505-255-4642 ee5391aa@triton.cirt.unm.edu