Near Space
acts like a 13-foot-thick slab of
concrete. This is a good thing for us,
since we don’t handle extreme
exposures to radiation very well. A
cosmic ray diving into the Earth’s
atmosphere is called a primary
cosmic ray. The flux of primary
cosmic rays at the top of the atmosphere is about one cosmic ray per
square centimeter per second — or
about the same flux of raindrops
during a rain shower.
Secondary Cosmic Ray
Production
When cosmic rays slam into
Earth’s atmosphere, they are primarily colliding with nitrogen and oxygen molecules. Upon impact, the primary cosmic ray shatters the mole-
cule, creating a shower of lower
energy particles from the subatomic
zoo — subatomic particles, like
neutral and charged pions, neutrons,
and more protons. Neutral pions
decay into gamma rays, which later
create electron-positron pairs.
Charged pions decay into muons, a
heavier relative of the electron. This
shower of particles created by the
Solution to the Voltage Divider Question
F
o
r
E
l
e
c
t
r
o
n
i
c
s
NUTS & VOLTS
E
v
e
r
y
t
h
i
n
g
As you will recall, in the July column, I
asked the question, “How can we prove
that using a fixed resistor of a value equal
to the geometric mean of the range of a
variable resistor yields the greatest range in
values in the voltage divider circuit?” Well,
the readers of this column replied with the
answer.
I will send Treasure Valley Near Space
Program patches to the first two responses because, together, they showed me the
way to mathematically prove my observation. By the time this column goes out,
Erick McAfee of Texas and Colorado
College Physics Professor Val Veir should
have their patches. The patches were carried to an altitude of 104,571 feet onboard
my near spacecraft launches at the Great
Plains Super Launch on July 3, 2004.
From Erick and Val’s suggestions, plus
A. A. Klaf’s book Calculus Refresher (great
book, by the way) and some fooling around
on my part, I have put together the following proof.
The Plan of Attack
The voltage range generated by a
voltage divider is equal to the maximum
voltage generated by the divider, minus the
minimum voltage generated by the voltage
divider. If the output voltage of the voltage
divider is graphed with respect to the
possible values of the fixed resistor, we get
a curve that is low on its two extreme ends
and has a peak value somewhere near the
middle. That peak value of the voltage range
will occur at a point over the fixed resistor
value that is equal to the geometric mean
of the maximum and minimum resistance
values of the variable resistor.
By using calculus, we can make this
into a max-min problem and find the point
at which the slope of the output voltage
curve goes to zero (this is at its peak
value). At this point, we will find that the
value of the fixed resistor is equal to the
square root of the product of the minimum
and maximum resistances of the variable
resistor or, in other words, the geometric
mean of the maximum and minimum values
of the variable resistor.
The Proof
We begin with:
V = Vmax - Vmin
Note that:
Vmax = VA · [Rfixed / (Rfixed + Rmin)]
and
Vmin = VA · [Rfixed / (Rfixed + Rmax)]
I’m going to drop VA (the voltage applied to
the voltage divider circuit) from the math,
since it’s just a constant and doesn’t affect
the best value of the fixed resistor.
Making my substitution, I get the following:
V = [Rfixed / (Rfixed + Rmin)] - [Rfixed / (Rfixed
+ Rmax)]
Now, take the derivative with respect to
the fixed resistor value and set everything
equal to zero:
V / d Rfixed =
d[Rfixed / (Rfixed + Rmin)] / d Rfixed - d[Rfixed
/ (Rfixed + Rmax)] / d Rfixed = 0
According to A.A. Klaf, the derivative of the
equation S = R / (R + A) with respect to R
is equal to A / (R + A)2
Making this substitution, I get the following:
Rmin / (Rfixed + Rmin)2 - Rmax / (Rfixed +
Rmax)2 = 0
Flip the ratios (by moving them to the
opposite side of the equality), multiply out
the squared numerator, and we get:
(Rfixed2 + 2RfixedRmax + Rmax2) / Rmax -
(Rfixed2 + 2Rfixed Rmin + Rmin2) / Rmin = 0
Divide the terms by either Rmin or Rmax (the
value in the denominator) and we’ll have:
Rfixed2 / Rmax + 2Rfixed + Rmax - (Rfixed2 /
Rmin + 2Rfixed + Rmin ) = 0
Note that we can move the minus sign into
the second half of the equation and subtract the 2Rfixed to end up with:
Rfixed2 / Rmax + Rmax - Rfixed2 / Rmin - Rmin
= 0
Combining like terms and moving them to
opposites of the equation gives us:
Rfixed2 / Rmax - Rfixed2 / Rmin = Rmax - Rmin
Factor out the Rfixed2 and move the
remainder to the other side of the equation and we get:
Rfixed2 = (Rmax - Rmin) / (1/ Rmax -1 / Rmin )
Now, at this point, I ran into a brick wall,
but, knowing what I had and what I wanted,
I discovered the following equation by playing around with some algebra:
Rmin Rmax (1 / Rmin - 1 / Rmax) = Rmax - Rmin
I substitute in the left side of the equation
for the Rmax - Rmin in the previous equation
to get the following:
Rfixed2 = Rmin Rmax (1 / Rmin - 1 / Rmax) / (1
/ Rmax - 1 / Rmin )
Divide out the like terms and we get:
Rfixed2 = Rmin Rmax
So, the condition of having the maximum
voltage range occurs when:
Rfixed2 = Rmin Rmax
or when the value of the fixed resistance is
equal to the geometric mean of the maximum and minimum resistances of the variable resistor.
98
SEPTEMBER 2004
