MOTM

This may be nothing but rehashing something many of you already know.  And,
I am not an expert on the subject.  But, I promised I would offer my
thoughts on this subject a while back.  Having said that, I am going to
offer a simple observation and apply ohms law to power supply DC wiring and
wire size in a simple exercise.

Please see the attached PDF file for the drawing I refer to in this
discussion.  I apologize for sending an attachment to list mail.  But, it is
only 5K, so I figured no one would mind.

In this drawing, I will assume that the power supply has a regulator which
is regulating to the power supply voltage at the power supply and not making
use of remote sensing.  Some modules (in particular that ones that have
LEDs) vary in the current that they draw from the power supply.  Other
modules are somewhat sensitive to fluctuations in power supply voltage.
Let's take one of each for this example.

Ohm's law tells us that anytime we have current flow across resistance, we
have voltage drop.  We know that all the components of our power supply
wiring have resistance.  We are going to look at ONLY the resistance of the
wire.  If we look at the drawing, we can assume that the voltage at the PCB
is lower than the voltage at the power supply by this amount:
E drop (wire A) = I (module left) + I (module right) x R (wire A).

The voltage at each module is also lower than the voltage at the PCB by this
amount
Module left:
E drop (wire B) = I (module left) x R (wire B)

Module right:
E drop (wire C) = I (module right) x R (wire C)

To find the voltage drop in the wiring to each module from the point of
regulation (the power supply), we have to add the voltage drops:

For module left total drop from the power supply:
E drop (wire A) + E drop (wire B)

For module right total drop from the power supply:
E drop (wire A) + E drop (wire C)

About now, you are probably cursing me for stating the obvious.  But, I
wanted it to be clear that current from one module effects the voltage
delivered to a different module.
See that voltage drop to module right = [I (right) + I (left) x R (wire A)]
+ [I (right) x R (wire C)]

And, varying current produces varying voltage drops.  So, in our example, we
see that varying load in module left is causing varying voltage to be
delivered to module right.  That is going to occur, no matter what we do
with wiring.  Our task is to minimize that to a negligible amount.

Since we have only 2 modules in the drawing, I am going to put some
out-of-proportion numbers in the formula to make a point.

module left I minimum = 20 mA;  maximum = 50mA
module right I = constantly around 30 mA
Voltage (+15) we will look at only one leg
resistance wire A, B and C all = 2 ohms

Voltage at PCB (I left min) = 15 - [(0.02 + 0.03) x 2] = 14.9 volts
Voltage at module left = 14.9 - (0.02 x 2) = 14.86
Voltage at module right = 14.9 - (0.03 x 2) = 14.84

Voltage at PCB (I left max) = 15 - [(0.05 + 0.03) x 2] = 14.84 volts
Voltage at module left = 14.84 - (0.05 x 2) = 14.74
Voltage at module right = 14.84 - (0.03 x 2) = 14.78

Delta E module left = .12 volts (.08%)
Delta E module right = .06 volts (.04%) 50% of delta E (left)

So, even though module right has a constant load, it is seeing a fluctuating
supply voltage because of the varying voltage drop across wire A caused by
the varying load of module left.  The problem is that wire A, B and C all
have the same resistance.  So, voltage drop from each module is imposed
equally across its own power cable and the power cable that supplies the PCB
(wire A in this case).

The solution, is to lower the resistance of wire A so that a larger
percentage of the varying voltage drops is found across the individual
module power supply wire.  Interaction between modules is then reduced.
Let's change the resistance of wire A from 2 ohms to 2/10 ohm and look at
these numbers again:

Voltage at PCB (I left min) = 15 - [(0.02 + 0.03) x 0.2] = 14.99 volts
Voltage at module left = 14.99 - (0.02 x 2) = 14.95
Voltage at module right = 14.99 - (0.03 x 2) = 14.93

Voltage at PCB (I left max) = 15 - [(0.05 + 0.03) x 0.2] = 14.984 volts
Voltage at module left = 14.984 - (0.05 x 2) = 14.884
Voltage at module right = 14.984 - (0.03 x 2) = 14.924

Delta E module left = .066 volts (.044%)
Delta E module right = .006 volts (.004%)  9% of delta E left

Notice how lowering the resistance of the wire A has reduced the delta E to
module right caused by the varying load of module left.  Or, to say it more
easily, module interaction due to power supply coupling has been reduced.
Now, these numbers are a little out of whack as the resistance of our power
wires was exaggerated some to make a point.

But, now we see the importance of using LARGE wire between the power supply
and the distribution PCBs.  This wire should represent the smallest portion
of resistance in the wiring between the power supply and the module.  I
always use # 12 or 14.

Sorry to have boar those of you to whom this discussion is elementary.
However, this question comes up frequently, and I hope this helps someone
understand why the wire to the distribution point has to be sized for
something other than it loadability.

Larry H
I'm tired., so forgive me if I have a slight math error along the way.  That
does not change the conclusion.