Krutz
0
- Joined
- Nov 21, 2007
- Messages
- 1,733
- Points
- 48
why do you think the flexdrive isnt efficient?
for example, if your diode would need 6.0v and be driven at 200mA and your battery is a 3.0v singlecell. if the currentdraw measured at the tailcap (at the input side of the driver) was 400mA, the driver would be 100% effective! the input current is double the output current, but the driver raises the voltage to the double too.
your numbers:
18650 cell, i assume 3.9v input (from a 3.7v nominal cell)
215mA "input" draw
that means: input: 3.9v x 215mA = 0,84w
149mA output
diode voltage-drop: 5.42v
that means: output: 5.42v x 149mA = 0,81w
that calculates to more than 96% of the input electricity is stepped up and put out! I call that pretty efficient! ;-)
manuel
for example, if your diode would need 6.0v and be driven at 200mA and your battery is a 3.0v singlecell. if the currentdraw measured at the tailcap (at the input side of the driver) was 400mA, the driver would be 100% effective! the input current is double the output current, but the driver raises the voltage to the double too.
your numbers:
18650 cell, i assume 3.9v input (from a 3.7v nominal cell)
215mA "input" draw
that means: input: 3.9v x 215mA = 0,84w
149mA output
diode voltage-drop: 5.42v
that means: output: 5.42v x 149mA = 0,81w
that calculates to more than 96% of the input electricity is stepped up and put out! I call that pretty efficient! ;-)
manuel
