Feb 1, 2013 #1 cheech226 0 Joined Jun 25, 2011 Messages 1,481 Points 63 if i'm using a buck driver will the measured current across the battery neg and case be lower than the actual current drawn by the diode? seems this would be so.
if i'm using a buck driver will the measured current across the battery neg and case be lower than the actual current drawn by the diode? seems this would be so.
Feb 1, 2013 #2 awillis2589 0 Joined Jun 21, 2012 Messages 2,019 Points 48 I believe this is correct. With linear drivers you would be able to see current draw as current output.
I believe this is correct. With linear drivers you would be able to see current draw as current output.
Feb 1, 2013 #3 Blord 0 Joined Dec 24, 2007 Messages 5,356 Points 0 That is right, buck driver are more efficient than a linear drivers. The buck driver converts the excess voltage into current.
That is right, buck driver are more efficient than a linear drivers. The buck driver converts the excess voltage into current.
Feb 1, 2013 #4 cheech226 0 Joined Jun 25, 2011 Messages 1,481 Points 63 thanks, i know it works the other way with boosts. i just did a measurement of an m140 build and it was much less than i thought it should be.
thanks, i know it works the other way with boosts. i just did a measurement of an m140 build and it was much less than i thought it should be.