[sdiy] Best way to get a +3.3V rail from +15V?

[Josh]

The same total power is dissipated when using two devices, but they
are sharing the load.  Like two skinny dudes picking up a big box
instead of one bodybuilder.  In some cases it's easier to find the
skinny dudes.

Inserting a zener diode in series before the 3V3 regulator and
dropping the voltage there is also an option.  I have a dsPIC design
similar to yours where the 3V3 rail is stepped down from 15V.  I drop
8.7V across the zener and the rest across a linear regulator.  The
1N5345BRLG zener diode can dissipate a couple watts no problem.  I'm
also only using the PCB for heatsinking the 3V3 regulator.

And this is much simpler than a switching solution.

-Josh

On Thu, Mar 7, 2013 at 11:49 AM, Neil Johnson
<neil.johnson97 at ntlworld.com> wrote:
> Scott Gravenhorst wrote:
>>
>> I'm wondering about using another regulator between +15 and +3.3.  Maybe a +8v.  I would think
>> that the heat dissapation should be less for the +3.3 regulator.  Some heat might also be expected
>> on the +8v reg.  It's either that or a power resistor as previously mentioned.
>
> How would that help?
>
> Single regulator:
>   Vdrop x Iload = (15 - 3.3) * 0.2 = 11.7 * 0.2 = 2.34W
>
> Dual regulator:
>   Vdrop1 x Iload = (15 - 8) * 0.2 = 7 * 0.2 = 1.4W
>   Vdrop2 x Iload = (8 - 3.3) * 0.2 = 4.7 * 0.2 = 0.94W
>
>   Ptot = 1.4 + 0.94 = 2.34W
>
> The same power is dissipated, this time shared between two heatsinks.
> You'd get the same effect using a single regulator and a larger
> heatsink.
>
> Neil
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