HIMNL9
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Re: The "Blue Ninja" - 445nm DIY Creation
Uhm, remember that the total heat that the linear drivers (ANY linear driver) need to dissipate, is also given from the difference between the load voltage and the power supply voltage.
I mean, if you have, as example, a LD that require 5V, and you build a linear driver with an LM317 for power it at 1A, then you have 5V (LD) + 2.5V (LM dropout) + 1.25V (Rsense dropout) = 8.75V of total required minimum voltage, for the assembly to work ..... say, 9V, for a bit of margin ..... now, the Rsense dissipate always the same power in heat, if the current don't change, so, 1.25W are from it ..... the regulator need to dissipate all the rest ..... there's a fixed 2.5W from the internal dropout voltage of the regulator itself, that don't change (still at 1A), plus all the rest of the voltage that you need to reduce for regulate the current at 1A ..... with 9V, your chip dissipate 2.75W for 1A, with 12V, your chip dissipate 5.75W, still for 1A .....
So, it's always better to choose the lowest voltage that you need (or the practical one that is the nearest one), for any assembly that uses linear drivers (just for not turn your chip in a frying plate ).
Leo:
In theory - would it be any more difficult for an LM317 to deal with a car battery, than with, say, 3x 10440 batteries ?
@rhd
if mean't the higher voltage.. hmm no. Easier to tell you the truth, as some 10440s would not withstand the 1A load and only supply around 600~700mA. The battery car was the only source of power I had :beer:
Uhm, remember that the total heat that the linear drivers (ANY linear driver) need to dissipate, is also given from the difference between the load voltage and the power supply voltage.
I mean, if you have, as example, a LD that require 5V, and you build a linear driver with an LM317 for power it at 1A, then you have 5V (LD) + 2.5V (LM dropout) + 1.25V (Rsense dropout) = 8.75V of total required minimum voltage, for the assembly to work ..... say, 9V, for a bit of margin ..... now, the Rsense dissipate always the same power in heat, if the current don't change, so, 1.25W are from it ..... the regulator need to dissipate all the rest ..... there's a fixed 2.5W from the internal dropout voltage of the regulator itself, that don't change (still at 1A), plus all the rest of the voltage that you need to reduce for regulate the current at 1A ..... with 9V, your chip dissipate 2.75W for 1A, with 12V, your chip dissipate 5.75W, still for 1A .....
So, it's always better to choose the lowest voltage that you need (or the practical one that is the nearest one), for any assembly that uses linear drivers (just for not turn your chip in a frying plate ).