How do I get a 5-Volt Output?

Hi,

The needed input voltage in a PCB is 5 Volts. The source for power are batteries. How do I get the 5 Volts from the batteries to power the PCB?

Thanks!

Here are the simplest solutions:
(Image attached)

If load current is constant,
resistor value is 1V/load current in Amps.

Resistor power rating:
1V x load current in Amps = power in Watts

What you need is a DC to DCan converter. You don't state the battery voltage or your desired current so this answer is pretty vague.. Simplistically, If your battery voltage is less than 5V you need a boost converter and if it's more than 5V you need a buck converter. If you go on the National or TI website (National and TI are now the same company.) there is design software where you input your input values and output requirements and the software returns bunches of possible designs with parts lists, schematics and costs. Both national and TI have simulation software. One is webench designer, which I see as a sidebar ad.I think I like the TI software better, but I may be confused (It's not unusual).

Of course it's WAY easier to just buy what you need over the internet.

Another possibility is a Low Dropout Regulator IC (LDO), such as the:

* MCP1702-5002E/TO [Digi-Key part#: MCP1702-5002E/TO-ND]
* LP2950CZ-5.0 [Digi-Key part#: LP2950CZ-5.0/NOPB-ND]

Make sure to include the required capacitor on the output [and in some cases, input] -- check the datasheet!

This is less efficient than a DC-to-DC converter [Stephan's suggestion], though, but it's a simpler/less expensive solution.

And, also, as Stephan pointed out, you neglected to divulge the battery type/voltage, which can make a world of difference. Also, if the battery will drain down to below 5 Volts + the dropout voltage of the regulator, while the battery still has power to give, then this is a poor solution -- the DC-to-DC converter would be better in that case.

A low dropout linear regulator would work if the battery voltage is above 5v. However it has a couple of problems, particularly if the battery voltage is way above 5v and when the output current is large. A linear regulator basically converts the lost energy (the voltage drop times the current) into heat. This makes it inefficient and the heat that's generated has to be dealt with. If you drop auto voltage (the battery's 12.6v and when the alternator's on the voltage is about 14.5v) at 1A then worst case you generate 9.5W of heat, which is way beyond a simple heatsink, and your efficiency is about 35%.

If you are using batteries such as D cells you can get the battery voltage closer to 5v, but that also has a problem. Battery voltage varies with the discharge rate and charge state. Even if the battery voltage is 6.3v at a 1A discharge you have to start by dissipating 1.3W, which is not that hard, but must be considered.

If your battery voltage is close to 5v then a buck/boost switching power supply would work for battery voltages above and below 5v, allowing you to get more out of your batteries.

A linear regulator is a MUCH simpler solution. I would only consider building a switching regulator if you need the efficiency, there's a heat problem or the battery voltage will be close to or below 5v during discharge.

Some of the switching power supply designs may have design evaluation kits including circuit boards, which would make things much easier.

I just finished a real sweet design for auto voltage to an adjustable voltage of 2 to 5v at up to 2 A using a TPS54325. It's small and cheap and hardly gets warm. The board's about 0.4" x 2". It has reversed voltage and overcurrent protection, which made it a little bigger.

What _he_ said!

For an example, here are some step-up and step down units from a seller on Ebay.
Just put in your terms in the top box, such as "power supply 5V step up" (or down).

http://stores.ebay.co.uk/amart2010/_i.html?rt...1011298929&_trksid=p4634.c0.m14&_sop=15&_sc=1

If you want a circuit to build your own, try this link:

http://lmgtfy.com/?q=5V+power+supply+circuit+diagram