Motorcycle Forum / General / Sportbikes / May 2007

The other solutions provide voltage regulation/safety. If for some
reason your voltage at the battery goes to 20V they will protect the
camera.

If you want to keep it real simple you can just use a couple of diodes
in series from the 13.2 volts to the cam. I would use ~5 1N4001 diodes
wired end to end.

Switched 12V from cycle ->|->|->|->|->|----camera

They only cost about 8¢ each, are good in this design for ~12 Watts, and
should drop the voltage about .8V to 1.2V per diode. 5 is a good
starting place and can be adjusted with the addition or removal of
diodes later.

Rick

Checking my math, make that a 60 ohm 1 watt, a 1 watt zener, and if
you hook the zener up backwards you'll have zero volts out and will be
smoking the resister.

    Huh ?

    That's going to result in the ENTIRE system of the bike being
pulled down to 9 V, at the trickle rate of the resistor, ~ .25 A,
which means ~ 2 W.

    Given that the diode is a switch, as long as you are above 9V,
you get :

12 V / 50 Ohm = .24 A drain to chassis

.24 A * 12 V = 2.88 Watts

Once the system is drained below V-f ( 9 ), the switch ( diode )
opens.

Once you turn the bike off, it will drain dwon to 9V and you'll never
get it started again.  When it's running, you'll STILL be at 12 V in
the system, with a wasted drain load of ~ 1 % of the output of the
regulator ( until the resitor burns out, if you don't puyt in a 5W
nominal ).

You are much better off to go with the 3 terminal regulator and 3
caps, 1 small (.01uf) , two large (100uf or so).... They are designed
to be abused. Zener's don't like it and tend to fry at very
inopportune moments.  
put one big cap between input and ground, another between output and
ground, and the small ones goes from input to output

I can't decipher your instructions-- if you are trying to get 9V
using the circuit in the k68.pdf file referenced above, setting
R2 to 75 Ohms and R1 to 25 Ohms will give you 5V not 9V.  There's
no need to use such small value resistors, BTW, something in the
hundreds or even thousands of Ohms would work just as well.

If you want 9V output, R2 = R1 * 6.2

Vo = 9V
Vr = 1.25
Vo * R1 / (R1 + R2) = Vr
Vo/Vr = (R1 + R2)/R1
Vo/Vr = R1/R1 + R2/R1
Vo/Vr - 1 = R2/R1, plugging in for Vo and Vr gives R2/R1 = 6.2

Pick something sane for R1 and R2 such as R1 = 330 & R2 = 2k, gives
about 2% less than 9V, close enough.

Forget the zener diode method, LM317 is the way to go.

Gotta love this thread. Talk about a geekfest.  ;-)

How about you  buy a $19.95 plug in 12v -> 9V 100ma
converter from Radio Schlock.

http://www.radioshack.com/sm-9-volt-vehicle-dc-to-dc-adapter--pi-2102592.html

I'd be astonished if you had to spend $30 for an
over-the-counter DC -> DC adapter.

Forget the 12V -> 120AC -> 9VDC stuff.

Also forget cobbling one together with diodes
and resistors.

That's the second person that has recomended using a bunch of series
diodes.  Why on earth would you use diodes instead of a single series
resistor?  Either way (diodes or resistors) there would not be any
regulation of the 9v with variations of battery (supply) voltage and the
output 9V would vary under varying load (current).

A 9.1V zener diode and a series resistor selected for the appropriate
nominal voltage drop at peak device load would be a more elegant
solution.  The zener would only sink current (and consume power) when
the input voltage gets too high or under less than peak load.

If this is too simple, you could build the regulator circuit shown on
this web page for a couple of bucks:
http://www.uoguelph.ca/~antoon/circ/car912.htm

Here is a web page that shows all of the solutions people have been
talking about:
http://www.cpemma.co.uk/diodes.html