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

Aug 24, 2006 9 Replies

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



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

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

R >

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 | ----- | | Adj | [.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

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)

----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----

formatting link
The #1 Newsgroup Service in the World! 120,000+ Newsgroups

----= East and West-Coast Server Farms - Total Privacy via Encryption =----

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.

----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----

formatting link
The #1 Newsgroup Service in the World! 120,000+ Newsgroups

----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Your can use this program to design a power supply with a voltage regulator:

formatting link

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.

okay what about this thing will this work? The only thing I am worried 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

formatting link

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

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required