# From 12V DC to 5+ V DC How to?

Discussion in 'Electronic Basics' started by [email protected], Aug 24, 2006.

1. ### Guest

Hi all,

I should probably know this but I am not sure.
I have a motorcycle that has a 12V output jack. I have a device that I
need to connect to it which is rated for 5V(+). I also have another
device which is a 12V device.
I need to connect both devices to the Jack but I need to reduce the
voltage for one of them from 12 to 5.
I am guessing this is done with a voltage regulator but I haven't used
one so I was hoping to find out if this is the right solution and if so
how to do it so its nice and clean. Is there voltage regulators that
are nice and modular that I can put in between the source and the
device?
Do I need to consider anything else?

Thanks

2. ### Guest

I should also point out that the power supply for the device shows the
following:
OUTPUT: +5V 1.5A
Output Power: 8W MAX

3. ### Roger DewhurstGuest

Use diodes in series. Each will drop the voltage by about 0.6 volts.

R

4. ### ehsjrGuest

You have to use a voltage regulator, because the input
voltage will increase while the bike is running. A simple
series resistor or a group of series diodes won't work.
Here's a circuit:

10 W -----
+12 ---[2.5R]---+---Vin|LM350|Vout---+---> +5V
| ----- |
[.1uF] | |
| +--[270R]--+
| | | +
| [820R] [1uF]
| | |
Gnd ------------+---------+----------+

You will need a good heatsink on the LM350, something
rated 7 degrees C per watt or better. Digikey HS276-ND
is rated 2.5 degrees C per watt and costs \$1.44 - it
will keep the LM350 under 60 degrees C, which is great.
The 2.5 ohm resistor needs to be 10 watts. The 270 and
820 ohm resistors can be 1/2 watt. The 2.5 ohm resistor
will dissipate 5.625 watts, and the LM350 will dissipate
close to 9 watts, worst case.

For a smaller and non-heat producing solution, you could
use 445-2435-ND from Digikey. It is a DC-DC converter
module that accepts 9 to 18 volts in, and produces
5 volts out at up to 2 amps. But it is more expensive.

Ed

6. ### defaultGuest

Probably no

That is a good "cheap and dirty" way to reduce the power. They don't
cut down the voltage they just switch it on and off rapidly - the
average power is less and efficiency is high - but it won't work for
devices that are sensitive to voltage or anything with a large filter
capacitor - the cap will charge to the peaks and you'd have 12V going
into the 5V electronics.

It can be made useful - with an inductor, pulse width modulator
circuit, and a degree in electronics - and cheaper to design from
scratch.

That particular gizmo is a controller for a brushless motor. It is
designed to supply both a variable voltage and variable frequency to a
small three phase motor to allow it to run from batteries - very
efficient way to control a motor speed - but not what you say you
want.

If you have a 5 volt motor and want to run it on 12 V just build a
simple single phase pulse width modulator to lower the average power
to the motor (a brush type motor doesn't care that it's getting 12
volt pulses for 40% "ON TIME" versus the 5 volts it is rated at)

7. ### defaultGuest

Are you sure you need 1.5 amps? A lot of simple there terminal
regulators will fit the bill - and a lot of them only put out 1 amp.
You can get up to 5 amp three terminal regulators but cost and
complexity increases.

Lot of places on a bike would make a good heatsink - the old Triumphs
used to mount a big 50 watt zener diode to the air intakes for the
carbs - that was the "regulator" for the electrical system.

8. ### HKJGuest

Your can use this program to design a power supply with a voltage regulator:
http://www.miscel.dk/MiscEl/miscelPowerSupply.html

It might be a good idea to add some overvoltage protection before the
regulator. A power resistor in series with the supply and a 20 volt
transient suppressor diode to ground will probably be enough.

9. ### Guest

about is that I am going from 12 to 5 which is a 7V drop. I am told
that will generate lots of heat. But I would think this device would
take care of that

10. ### ehsjrGuest

That's a *nice* looking device. Don't know if one will
work for you, tho - you might need 2. Your original
power supply is capable of 1.5 amps - maximum for the
device at the url is 1 amp. If your current requirement
is < 1 amp, one should work fine. The url says they can
be put in parallel, so if you use 2 in parallel you can
exceed 1.5 amps.

Heat is not a problem with switching regulators like it is
with linear regulators, and DE-SWADJ is a switcher. Sounds
like you would be in good shape with one (or perhaps 2)
of them.

Ed