[UFO Chicago] Short Post-Meeting: The voltage regulator

[Neil R. Ormos]

jordanb at hafd.org wrote:

> The voltage regulator in my lamp is a L4940V12.

> To answer nate's question, it is connected to
> ground and power dissipation is "internally
> regulated." As long as the input voltage does
> not exceed 30V, the device will not have trouble
> dissipating power.

I don't think that's the correct reading of the
device datasheet.  In the "Absolute Maximum
Ratings" section (Table 1), the datasheet states
that output current and power dissipation are
"internally limited."  I believe this means that
there is no short-term absolute maximum rating for
Io and Pd that the user need worry about, because
the device will internally limit the output
current to avoid instantaneous damage.

However, that doesn't mean you can ignore
long-term power dissipation and its effect on the
junction temperature of the device.  This is a
series-pass regulator, so the entire load (output)
current passes through the device.  The power
wasted in the device is calculated as the product
of output current multiplied by the voltage drop
across the device:

  Pw=(Vi-Vo)*Io

and that wasted power integrated over time shows
up as heat which needs to be dissipated. (This is
a simplification but it's good enough for this
discussion.)  The rate at which power is
dissipated depends on whether there is a heat
sink, how good it is, and whether there is any
other cooling.  If the heat is not removed, the
junction temperature will rise above the absolute
maximum rating, and the device may fail.

Table 1 specs a maximum junction temperature of
150 C, and Table 2 specifies the
Junction-to-Ambient thermal resistance at 62.5
C/W.  The interaction between these parameters is
shown in Figure 26 (incorrectly labelled "Load
Transient Response"; should be labelled "Total
Power Dissipation").  At an ambient temperature of
25 C, with no heat sink, the device can dissipate
2.5 watts.  Exceeding that power will overheat the
device.

Jordan, I don't think you'll have a power
dissipation problem with your headlight (example:
Vi=22 V; Vo=12 V; Io=5*0.020=0.1 A; Pw=1 W).  If
other people want to build something similar, but
with more LEDs or higher-current LED's, they might
need a heat sink.

--Neil