Hi all,
I recently bought some neat little LED pinspots that use 3 AG10 (3 X 1.5VDC) button cell batteries to power a single LED. The AG10's aren't going to last very long so I decided to mod the lights to use 9 volt batteries; longer-lasting and easier to change. Using an online LED resistor calculator, I installed 220 ohm resistors in the lights. I used the following values:

Source voltage = 9VDC
FV = 3.4VDC
FC = 30mA

These are nominal values - I don't have actual specs for the LEDs.

However, I notice that the LEDs are now running ~75% of the brightness that I get from using the AG10 batteries. Reducing the value of the resistor (150 ohm) doesn't change this, not surprisingly. There is no resistor used in the lights. If I use a straight 4.5VDC power source, such as 3AAA batteries, I get the full brightness again. I'd like to get the full brightness using the 9 volt battery. What am I missing here?

 

9v "transistor" batteries are usually made from six 1.5v cells; inside the rectangular case they are all strapped together. They almost always have a much higher internal resistance than a typical AA battery does, while having a far lower mAH capacity. But, you're using a series resistor to limit current, which means you are dissipating about half the power in the LEDs, most of the rest in the current limiting resistor, and some in the battery itself - depending on how much charge is left in the battery. If the 9v battery is alkaline, it will have a rating of around 500mAh, meaning it'll drop down to 7v at a 20-hour discharge rate of 25mA.

The AG10s also have a higher internal resistance than AA batteries do. It was their internal resistance that limited the current through the LEDs. You'll find this used frequently in cheap throwaway products.

You could use a "Joule Thief" type circuit to operate your LEDs on a single AA, C, or D cell. See this thread: http://forum.allaboutcircuits.com/showthread.php?t=42331

 

Understood, thanks! I did suspect the problem had to do with internal resistance of the 9 volt battery, but wasn't aware that the AG10's shared this feature. I'll likely change the power source to AA / AAA batteries to get the brightness I need.

 

Well, you can measure the approximate internal resistance by reading the short circuit current on a multimeter (use the 10A range.) Then it's a simple application of ohm's law to figure out the value of the resistance. Only do this for a few seconds though - the batteries *will* get hot and you will make them go flat if you do this for any significant period of time.

For example, shorting out a fully charged AA Duracell battery gives me: 4.32 amps. The nominal voltage is 1.5 volts, so the resistance (we can exclude the 0.01 ohm current shunt in the multimeter as that has a minimal effect): 347.2 milli-ohms.

 

A word of warning. Do not do this internal resistance check on a car battery as somebody I know did. It melted the leads to the meter (he kept blowing fuses so he replaced it with a nail). He was very luck the leads melted otherwise he would be up for a new meter.

 

A word of warning. Do not do this internal resistance check on a car battery as somebody I know did. It melted the leads to the meter (he kept blowing fuses so he replaced it with a nail). He was very luck the leads melted otherwise he would be up for a new meter.

Or worse. Car batteries can expand and explode if shorted under certain circumstances. It should go without saying, do not do this check on any big battery and any rechargeable batteries.

 

For your LED project, try using an LM317 adjustable voltage regulator. Experiment with different resistor values to get a 4.5v output from the 9 volt battery. When the battery weakens to let's say, 7 to 8 volts, the regulator will maintain a 4.5 volt output. I did something similar. I modified a lighted magnifier to use ten paralleled surface-mounted LEDs that operate on 3 volts.I used 4 AA batteries and a regulator. But, I don't know how it will work with a 9 volt.

 

If you're going to use an LM317, you might as well use it as a constant current regulator, instead of having a 4.5V output dropped down for the LEDs using resistors. It won't be much, if at all any more efficient, but it will work on a lower voltage (theoretically the LED's forward voltage + 1.5V.)