Off topic question, I ask those of you who know your stuff..
If I wanted to use a 12v battery to power a something that requires 5V 1A power input (DC).. what would I need?
Offhand, (if it matters) I don't know recall how many amps my battery is - I can only say it's a 12v deep cycle battery I got at Walmart for about $120
You can't use a 12 volt battery to power a device that is designed to run on 5 volts! You will destroy it. Get a power supply that is 5 volts and rated at more than 1 amp. You would need a resistor in series that would drop the voltage to 5 volts to your device and a 2 watt resistor at least to use this battery. By the way, if your device actually is drawing an amp of current the battery won't last too long anyway.
Mark
The best way to handle this is to use a buck converter. These are very efficient, not too expensive, and won't waste power as would a resistor or a linear regulator. Do a web search on "12V to 5V converter" and you will find many options. Here are some, for example:
http://www.amazon.com/s?ie=UTF8&page=1&rh=i%3Aaps%2Ck%3A12v%20to%205v%20converter
Make sure you are getting a buck converter and not a linear type. Check the efficiency rating. It should be 90% or thereabouts.
You can figure your "run time" from the amp-hour rating (not the "amps") of your battery. Just divide the amp-hours by the current drawn at 12 Volts. For example, an 85 A-H battery supplying 2 amps at 12 Volts will run for 86/2 = 43 hours from a full charge condition.
Because of the high efficiency of the converter you can approximate the current drawn at 12 Volts by multiplying the current drawn at 5 Volts by 5/12. So for 1 Amp at 5 Volts the current drawn at 12 Volts is I = (5/12)x1A = 0.42 Amps. This is the number to use to calculate the "run time".
Neil
a LM7905 regulator circuit of sufficient power handling for your application in this case it appears to be 1 amp.
The 7905 is a negative 5 volt linear regulator as is its positive 5 volt cousin the 7805. Either of these would work but you will be losing more than half the battery energy as heat and the "run time" will be shortened.
Here are the approximate calculations. With a 1 Amp 5 Volt load from a 12 Volt battery the regulator will drop 12-5 = 7 Volts. The power lost will be 7 V x 1 A = 7 Watts which is wasted as heat and will require substantial heat sinking. The run time for an 85 A-H battery will be 85/1 or 85 hours.
A 90% efficient buck converter will waste about 0.56 Watt and the run time will be about 85/.42 = 202 hours and heat sinking will likely not be needed.
Though a linear regulator will work, the buck converter has advantages in both power lost and battery run time.
Neil
As switching supplies go, how much noise (if any) does a buck convertor cause?
The price is right and the power savings is very attractive.
I have used three converters, two boost and one buck, the difference being whether the voltage is increased or decreased, in sensitive analog applications in gamma radiometry and have had no issues with noise. One application uses a microcontroller and it works without problems.
In another application I used a boost converter to produce a regulated 1100 Volts DC from 8 VDC input and had some noise get into the very sensitive analog amplification circuitry (processing microvolt signals) but some metal shielding and filtering solved this problem.
These converters usually operate at between 30 and 50 KHz and produce waveforms with spikes so it would be expected that RF interference might happen but I have neither tested for nor observed this. Good shielding and filtering is the key for success in sensitive applications.
Neil