ANSWERS : Technology

okay, so i'm making a 9 volt battery usb charger, but the voltage regulator takes in the 9 volts and turns it into 5 by burning off  volts as heat. i was wondering if you could put a resistor before the regulator to limit the amount of power going in, thus preventing wastage? and if so, what resistance would it be? any help would be greatly appreciated and i will subscribe to you : D  

Comments

[lemonie]

The problem you would have with a resistor, is that the output voltage would be current-dependent. Also, you would have little current-control, beyond the resistor and battery.
This is not a good idea for a charger.

L

[jeff-o]

Use a switching DC-DC converter. There are dozens and probably hundreds of different devices that will drop 9V down to 5. And, they'll do it with over 90% efficiency, unlike a regulator.

Visit digikey or mouser and do a search for DC-DC converter and scroll through the multiple options. Each chip will have a datasheet, and that datasheet will probably have one or more example circuits that you can copy outright and use in your design.

You could also switch to using four AA batteries, with a total output of 6V, so you're only wasting 1V instead of 4...

[orksecurity]

Investigate a switching-mode regulator. They're a bit more complicated, but they are more efficient.

[Re-design]

The best way to prevent wastage is to start with the proper voltage so you don't HAVE to reduce it.

[steveastrouk]

You can't prevent wastage- that unused energy has to go somewhere ! and adding a series resistor can be a bad idea.Just heat sink the regulator

[Jack A Lopez]

You could put a resistor in series with the input to the regulator, but it won't save you any power losses.  All these components are in series (the load, the regulator, the resistor), so therefore they share the same current, I.

Consider the example with just the regulator and the load.  And suppose the current through both of them is I=100 mA. The power dissipated by the load is 5V*100mA = 500mW=0.5W.  The power dissipated by the regulator is (9V-5V)*100mA= 400mW=0.4W.  Total power losses are 0.4W +0.5W = 0.9W = 9V*100mA

Now suppose you put a 10 ohm resistor in series with the regulator. The voltage  drop across this resistor is 10 ohm * 100 mA = 1.0 V.  The drop across the regulator is now 8V-5V=3V.  And the drop across the load is 5V (because it's regulated).  The power dissipated by the resistor is 0.1W, by the regulator is 0.3W, by the load is 0.5W.  Total power losses are 0.1W + 0.3W +0.5W = 0.9W = 9V*100mA, same as before.