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You, my friend, are a genius! I just ordered a case of 36 from Dollar Store for a dollar each. Beets paying almost $3 for each input!

Have a simple circuit (LM317 , resistor , Pot) .

You can also use Pot to replace all resistor

Input Voltage 1.5 - 40v

In one situation, a user wanted to use a 5v DC power digital voltmeter or ammeter. His power source could be either a 9v dry cell or use in 12v vehicle battery. The quickest way is to use a serial resistor to bring down the voltage for 9v or 12v to around 5v. The actual voltage does not need to be exactly 5v. There is usually some tolerance. I guess around 4.75v to 5.25v is a good target range.

One could use a 1k Ohm potentiometer to dial in required resistor value. (VERY IMPORTANT) Set the potentiometer to the HIGHEST value. (VERY IMPORTANT) Put the pot in series with the meter and the voltage source. Turn the potentiometer down very very slowly till the voltage across the voltmeter is at 5v. Turn off the circuit. Isolate the potentiometer and use an Ohm meter to determine the resistance (Ohms). Now use the resistor and confirm final voltage.

If not already offered, ask seller if he could provide free resistor and shrink wrap to their 5v meters. Let seller know what is your source voltage.

- John

for example, so we use a voltage divider, say, a 5 ohm in series with a 7 ohm. now if we measure accross the 5 ohm resistor we get 5 volts. ace, now lets just connect it to my load. this load is also 5 ohms.

now are we applying 5 volts to the load.

NO.

now we have connected the 5 ohm load in paralel to the 5 ohm part of the potential divider, its total ohmage is now 2.5 ohms (product over sum) now if we use the potential divider sum, Vpd = 12 volts *2.5/(2.5+7) = 3.16 volts.

I hope that helped. another thing, assuming that we were only chargng one thing, and took that into account when calculating what resistors to use. (assuming we are still having a 5 ohm load.) that would give us a potential divider ratio of 2.5ohm:3.5ohm (with the 2.5 reprisenting a 5 ohm resistor in paralel with our load) we would get 5 volts. The problem with this is the wasted energy diserpated. The current drawn by this circuit is 12/6 = 2 amps. that is assuming the connecting wires have no resistance. now if we work out how much energy is disarpated at the 3.5 ohm resistor the formula is

Watts = I

^{2*R = 10.5 watts, wasted. a percentage of usefull energy used (efficiency)}total wattage = IV = 2*12 = 24 watts

the useful current is 1 amp, as the current is split between the 5 ohm resistor and load)

useful = I

^{2*R = 1*5 = 5 watts.}efficiency = 21%

i hope that this made sense, that i was polite, not teaching to suck eggs, not too advanced.

any questions message me, if i cant answer you il know someone who can.

pin 1 is input +, pin 2 is ground, and pin 3 is output +. this guy is also useful for a variety of voltage inputs and uses

There are similar regulators around for higher currents.