Question

The resistances of commercially-available discrete resistors are restricted to particular sets. For example, the available values of resistors with 10% tolerance are selections from the E12 set multiplied by a power of ten from 10^0 through 10^5. The E12 set is
E12={10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82}

Answer 1

Vout = R2/(R2+R1) * Vin, Thus Ratio=Vout
/Vin= R2/(R2+R1)
With given voltage values here Ratio=1/4
To find Rout
, one needs to suppress the independent source, so we short out the voltage
source. Therefore Rout=R1||R2
So the requirements of the question are:
• Rout min < Rout = R1*R2/(R1+R2) < Rout max (1)
• Ratio*(1-ε) < Vout
/Vin = R2/(R1+R2) < Ratio*(1+ε) (2)
By inspecting inequalities (1) and (2) closely, one can notice if we divide the two, a bound
for R1 can be found, so R1 can be selected independent of R2, which is helpful in simplifying
the design.
However, there is a subtle point in dividing ranges. Note that
• If a < x < b and c < y < d
• Then a/d < x/y < b/c (Not the simple division of bounds of a/c < x/y < b/d)
With this in mind, the bounds for R1 can be found by dividing inequalities of (1) with (2):
Rout min / (Ratio*(1+ε)) < R1 < Rout max
/ (Ratio*(1-ε))
With the values of this problem (Ratio=0.25, ε=0.1, Rout max = 30k and Rout min =10k), R1
bounds are as follows: 10k/0.275 < R1 < 30k/0.225, which results in:
36.4k < R1 < 133.3k
With such a bound for R1, there are many options in E12 set that could work (39k, 47k, 56k,
68k, 82k, 100k, 120k are all possible). However for a conservative design, I'll choose a value
mid-range, for instance 68k.
Therefore R1=68kΩ.
With the value of R1 selected, now we can choose R2 such that Vout
/Vin ratio satisfies the
nominal value:
Vout
/Vin = R2/(R1+R2) = 1/4 in this case. Therefore 3*R2=R1 and thus R2=22.67k. However
we are limited to E12 possible resistor values. Thus I choose the closest value of 22k for R2.
However since this is not the exact value derived from Vout
/Vin ratio, let's check to see if our
design satisfies the 10% tolerance limit on Vout
/Vin .
With
• R1=68 kΩ
• R2=22 kΩ
Vout
/Vin = 0.244 which is well within 10% deviation of 0.25 nominal value of the desired
ratio. So our design is confirmed to satisfy all requirements! J
In order to find Vout min and Vout max with 10% deviation of resistor values, note that
maximizing the numerator and minimizing the denominator, maximizes a ratio.
• Vout max = 1.1R2 / (0.9R1+1.1R2) * Vin = 1.1*22k / (0.9*68k + 1.1*22k) * 80
= 0.283 * 80 = 22.67 V
• Vout min = 0.9R2 / (1.1R1+0.9R2) * Vin = 0.9*22k / (1.1*68k + 0.9*22k) * 80
= 0.209 * 80 = 16.74 V

:))))))

Answer 2

for different values!!!!!