On 3/5/2014 9:07 PM, doog wrote:
in hopes of explaining what is being talked around is this.
You want 5V from 11V so there's 6V difference. Those linear
regulators will give you the 5V at some current level(let's
say 1A for simplicity). So you get 5V at 1A and that's
5Watts(5V*1A) but that 1A of current is also involved in that
6V drop from 11V to get you 5V. The energy/power wasted in
that 6V drop is calculated by the current of your load( 1A )
times the drop( 6V ) which is 6Watts in the example.
This is why Gerald is correct in his statement that the linear
regulator is wasteful since in the example you only need 5W and have
6W of waste so have a total energy cost of 11W. A better solution
would be a DC->DC converter and you can get those on amazon with free
shipping from china(if you can wait) for less than $10. The take the
input power turn it into a AC signal which can then be chopped and
reassembled into a lower voltage with little energy loss.
for example:
http://www.amazon.com/gp/product/B008BHAOQO
Doug
The most important detail here is the 6W of waste heat. A linear
regulator regulates the Voltage by generating heat. The heat generated
in a linear regulator is measured in Watts and can be calculated as the
Voltage dropped across the regulator times the current through it. From
this discussion I understand that to be 6W (6V * 1A). The regulator will
have a specification for the temperature rise from the die to case (or
ambient) per Watt. A common LM340 in a TO-220 case, for example, has a
4C/W junction/case and 54C/W junction/ambient rating. In this example
that means with no heatsink the junction will be 324C over ambient
(54C/W * 6W). The 24V limitation is not the factor here, it's
dissipating the heat. You will need a heatsink capable of disipating at
least 17.6C/W to ambient just to keep the LM340 TO-220 from reaching its
rated 150C limitation with an ambient temperature of 20C (20C + (4 +
17.6C/W) * 6W = 150C). At elevated operating temperatures you will need
a better heatsink than that. I don't know what regulator you are using
as you have not stated that publicly yet, nor do I know the maximum
temperature for your application so I can't guess what size heatsink you
will actually need.
Short answer: Use a big heat sink or a switcher.
Artie
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