voltage conversion puzzle.

Hello

How would I convert 13.2 volts dc from my motorcycle down to 9 volts dc for
my helmetcam?
Something in-line in the wire?
A resistor?
The camera uses 50 milliamps during operation.
Please reply to



Thanks for any replies.

 

A 3-terminal voltage regulator and a couple of capacitors.

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KotPT -- "for stupidity above and beyond the call of duty".

 

Quick, cheap and dirty: a 50 ohm 1/2 watt resister in series with a 9
volt zener diode to ground.

 

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.

 

You obviously only put the circuit in series with the one component that
requires the 9V.

On the bike side of the resistor will still be 13.2V. The load side of
the resistor would be 9V. The zener diode will only draw enough current
to drop 4.2V on the resistor at idle (84 ma using a 50 ohm series
resitor proposed) and would use 352 milliwatts (.084^2 x 50)

But to properly design this circuit you need to know what the current
demand of the device is so that you do not drop more than 4.2V when it
is drawing current though the series resistor. Therefore, it cannot
draw more than 84 ma.

 

I don't think you understand how zener diodes work. They are normally
set-up in reverse bias. Putting it in backwards would be forward
biasing the diode and you would sink the full 12V to ground until the
resistor or the Diode fried opened.

 

There... I fixed it for you.

 

I don't think they make a 3 pole 7809 regulator, rather 7808, 7812,
7805... & the negative voltage 7900 series. what you want for exactly
9 volts is an adjustable 3 pole LM317 regulator with a voltage divider
(75 ohms) on the ground leg to set up the 9 volts & (25 ohms) on the
top rail to 12Volts to set up the 9 volt output Also use a 1 uf cap
from the output to ground to reduce any noise generated by the
regulator and more importantly to keep the curcuit from oscillating,

SEE reference below:
http://www.kitsrus.com/projects/k68.pdf

An alternative (cheap and dirty method) is to pick up a 9 volt zener
diode (reverse bias to ground) and a series 50 ohm resistor in front
in series with the 12 volt line (supply side) and the 1 uf cap on the
output of the zener diode both in parellel from 12 volts to ground.

Bob Nixon RZ-350

 

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.

 

Holly Crap 1 message turned on the 'light' for me
Use an inverter from 13 volts to AC.
Plug in the factory supplied AC. TO DC. 9 Volt adaptor.
Plug the camera into the 9 volt supply plug.
A bit of a pile of junk to transport in my backpack but it should work
I'll use a meter to see the result before plugging in the camera.

Thanks for the techinical feedbacks and electronics lessons .

Brian

 

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

 

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.

 

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

The above is correct, I was in too big a hurry. Mine would be fine for
and unloaded voltage divider but that's not what an LM 317 is. BTW, we
all should have allowed for a working running bike regulated voltage
od 13 volts instead of 12 volts.

 

A couple of bucks and an hour's work.
I picked up one at West Marine a couple years back.
A reasonable addition to any bike.

 

there are some real smart replies in the stack here... however a
person could run the camera off of a small 9v battery for 5 dollars
and skip tying to power it from the bike... a battery might run the
thing for hours at a time.

Phil Scott

 

50 ma at 9 vdc semms like 180 ohm r. To drop 9vdc on this load and 4.2
(9+4.2=13.2) on a series load , use around an 85 ohm resister.....doest this
sound right?....not sure I try with my bike and camera though.

 

Clueless.

 

1st off, give me a little credit. I knew the divider was on the output
side (between top rail third leg & ground. However my main concern
with the 13 or even 14 regulated volage is the heat dissapation of
that 4-5 volts across the LM317. It should handle 5 volts @100 mills
easily if the properly mounted epoxy package LM 317 has the metal
backside flat on unpainted metal.Otherwise that half watt (100mills X
5 volts) could eventually fail from continuous over heating. Just
playing on the safe side

 

you can usa a Zener diode to drop it, or use the voltage
drop in a series of common doides to do the same,
or purchase a dc-dc convertor which costs quite a lot. ~$75.
the cheap diodes have a forward drop of about 2/10 of a volt drop each.
sam

 

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