What resistor value to drop 9v to 3-4v

[Ptownkid]

just as the topic states, how do I drop a 9v battery down to 3 or 4 volts?

 

[Test Point]

You could run it through a 3.5v 3 pin regulator or depending on current needs, you could cob together a simple zener regulator. It's not really advisable to drop a voltage through a resistor per se most of the time as the drop is dependant on current draw. if current swings around, so will the E drop.

TP

 

[Ptownkid]

I only need enough current to run 3 leds

 

[rodabod]

Search on google for a battery and LED calculator.

You need rough values for current draw for the LEDs.

I would recommend running them "cold" if you want to save battery power.

 

[AnalogPackrat]

V = IR and P = VI.

LEDs typically can take a maximum of 30mA or so. They will degrade in brightness if you run the current that high all the time. Better to use the superbright kind and run them at a lower current. You need to find out what the forward voltage drop of your LEDs is. It will vary with the color of the LED.

To find out, put a 2k or so resistor in series with the LED. Ground the cathode of the LED and run the other end of the resistor to 9V. The LED should light up. If not, its in backwards. Once it lights, measure the voltage across it (Vf).

Say it's 1.2V. That means that the resistor has 7.8V across it which means:

I = V/R = 7.8/2000 = 3.9mA

If you want more current recalc the resistor. Say you want to run at 10mA. The LED will still run at 1.2V (or whatever yours measure) and you'll have 7.8V across the resistor.

R = V/I = 7.8/0.01 = 780ohms

Make sense? Now, if you plan to run three LEDs, they will not all match in Vf and the one with the lowest will take the most current, so you should use a separate resistor for each (all the same value) with the top of each resistor connected to 9V and the other ends to their respective LEDs.

Analog Packrat

p.s. I suck at ASCII graphics, so no drawings...sorry.

 

[Ptownkid]

Awesome, thanks for the explaination, I was gonna ask, but I didn't want to be a bother. Teach a man to fish always works...

Cheers guys

 

[bcarso]

Agree so far...but if you have 3 LED's and they are all to be on at the same time, put them in series. Unless they are blue or white (blue with phosphor) and thus have high forward voltage (~>3V each) the 9V battery will be plenty of volts. Say they are 2.2V each nominally---then you will have 6.6V, 9 - 6.6 gives you 2.4V to work with, and for 10mA the series R will be 240 ohms. As the battery runs down of course the current will drop and with it the brightness, and this will happen as a function of voltage more quickly. But since the circuit is considerably more efficient than the one-R-per-LED approach your mileage will be higher.

Plus, the declining brightness is a built-in low battery indicator.

If constant brightness is critical an active current source will work until the dropout voltage is reached.