gwbaltzell
Enlightened
It would make no sense to tear down a 28A to simply wind up with 4 LR44 (A76) cells unless there was a significant cost advantage. The advantage to other owners, if the keymate could take the 28A directly, would have been to be able to use a single premade battery. Now the PX28L is made up of two cells and you might want to try two of these cells. I doubt if you'll have room for three. The cells are 2L76 (2LR76, CR1/3N, CR11108, DL1/3N, K58L). Since these are lithium they can deliver a higher current and have a flatter discharge voltage than akaline. But with two you will be starting out at the original design voltage!
Now you need to understand that 7 V figure is meaningless. This is not how LEDs operate. It is power that produces light. And power is measured in watts. Light output is measured in watts or lumens. And directional light falling on a surface is measured in candela and is not useful unless you also know the angle.
To measure watts going into the LED you must measure, with the LED connected, both the voltage across (called the forward voltage or Vf) and the current through it (measured with the meter in series). LEDs are rated by their nominal, maximum and absolute maximum current. The type used in your keymate are likely rated 20 mA (0.020 A), 30 mA and absolute max. of 50 mA. Some flashight makers ignore the ratings and will go up to 80 mA. But in all cases something must regulate the current to the LED, because it will draw enough to self destruct. What regulates (actually limits) the current in the simplier Photons (I, II) is the internal resistance of the cells that drive it. If your keymate is being driven at 80 mA and the Vf is in the typical range, lets say 3.4 V, then the watts in would be 0.272 and lumens out would likely be no more than 4 tops. It is two factors that cause you to percieve it as "brighter" than an E1E. The first is that more of the output is concentrated in a tighter area, this is called directionality and has an angle associated with it. The second, as pointed out before, is the color. The white LED has more energy in the scotopic (night vision) color of blue-green and therefore appears "brighter". This is why I suggested replacing the white LED with a blue-green if seeing colors was not important to you.
George
Now you need to understand that 7 V figure is meaningless. This is not how LEDs operate. It is power that produces light. And power is measured in watts. Light output is measured in watts or lumens. And directional light falling on a surface is measured in candela and is not useful unless you also know the angle.
To measure watts going into the LED you must measure, with the LED connected, both the voltage across (called the forward voltage or Vf) and the current through it (measured with the meter in series). LEDs are rated by their nominal, maximum and absolute maximum current. The type used in your keymate are likely rated 20 mA (0.020 A), 30 mA and absolute max. of 50 mA. Some flashight makers ignore the ratings and will go up to 80 mA. But in all cases something must regulate the current to the LED, because it will draw enough to self destruct. What regulates (actually limits) the current in the simplier Photons (I, II) is the internal resistance of the cells that drive it. If your keymate is being driven at 80 mA and the Vf is in the typical range, lets say 3.4 V, then the watts in would be 0.272 and lumens out would likely be no more than 4 tops. It is two factors that cause you to percieve it as "brighter" than an E1E. The first is that more of the output is concentrated in a tighter area, this is called directionality and has an angle associated with it. The second, as pointed out before, is the color. The white LED has more energy in the scotopic (night vision) color of blue-green and therefore appears "brighter". This is why I suggested replacing the white LED with a blue-green if seeing colors was not important to you.
George
