On Sun, 11 Mar 2007 10:13:06 -0500, the renowned John Popelish
wrote:

snip

Your schematic shows both LED and relay loads. The diodes
that all connect between the collectors and pin 10 really
have no practical use for the LED loads, since these produce
no voltage when their transistor switches off. They do have
a practical use for the relay loads, since these produce a
voltage that adds to their supply voltage, briefly, when
their transistor turns off. The diodes detour the inductive
current back to the positive supply, till the stored
magnetic field collapses. If you have several different
supply voltages feeding different loads, and some of the
loads are inductive, you could connect pin 10 to the most
positive voltage, so that none of the diodes can become
forward biased by any of the normal load swings, but can
still provide a path for the inductive current.

Note that there is a possible problem with this. If the inductive load
is switched and the stored energy from the inductance of the relay
coil goes into the power supply for the LEDs (suppose no LEDs are
turned on), then there may be nowhere for the energy to go.

With a lot of switching (or with a single switching but very light
capacitance on the LED power supply) the voltage could exceed the
voltage rating of the transistors.

If it is a possibility, a relatively large capacitor plus a bleed
resitor if there is no load, or a zener, will provide a simple
solution.