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ECE 304: Two-Mesh Design Project
Problem^1
In the circuit of Figure 1, a sensing device provides a current I 1 = 10 mA and a receiving device demands a current of I 2 = 5 mA. The matching T-section of resistors is subject to the conditions that it keeps the input voltage V 1 in the range –20V ≤ V 1 ≤ 20V and the output voltage V 2 in the same range. Find values for the resistors R 1 , R 2 and R 3.
V_
{R_3}
PARAMETERS:
R_1 = 1500
R_2 = 1000
R_3 = 3000
I_1 = 10mA
I_2 = 5mA
V_
0
V_M
{R_2}
{R_1}
{I_1} I_1 I_2 {I_2}
FIGURE 1
Schematic for the problem
Procedure
Our procedure is to analyze the circuit to find the constraints on the resistors imposed by the current limitations and node voltage requirements. We will find formulas expressing the range of resistance values that meet these conditions. Then we will plot these constraints using a spreadsheet. Then we will verify the spreadsheet using PS PICE.
Analysis
Using Kirchhoff's laws we find EQ. 1 V 1 =I 1 (R 1 +R 2 )−I 2 R 2
EQ. 2 V 2 = I 1 R 2 −I 2 (R 2 +R 3 ).
Approach to the design problem
We introduce an extra variable into the problem, namely, the voltage on the middle node (top of the center resistor). Let’s call this voltage VM (M = middle node). The reason for introducing this variable is an anticipation that its behavior will be easy to understand and make solutions to the design problem easy to understand. We will see later whether this idea works. From Ohm’s law we find EQ. 3
VM = ( I 1 −I 2 ) R 2.
Using VM, we can rewrite EQ. 1 and EQ. 2 as follows: EQ. 4 V 1 =VM+I 1 R 1
EQ. 5 V 2 = VM−I 2 R 3.
(^1) This example is based on a suggestion by Larry Huelsman
Next we solve for the resistor values using EQ. 3, EQ. 4 and EQ. 5. From EQ. 4, R 1 is found as EQ. 6 below. EQ. 6
1
1 M (^1) I
V V R − =
From EQ. 5, R 3 is found as EQ. 7
2
M 2 (^3) I
V V R − = ,
and rearranging EQ. 3, R 2 is found as EQ. 8
1 2
M (^2) I I R V −
=.
Putting in the values given for I 1 =10 mA and I 2 = 5 mA, we find EQ. 9
k Ω 10
V V R 1 1 M − =
EQ. 10
k Ω 5
V R 2 = M
EQ. 11
k Ω 5
V V R 3 M^2 − =.
Intuitive behavior of equations
The expected advantage of VM as a variable is that we know intuitively how it should behave. For example, because I 1 flows from V 1 to VM, VM ≤ V 1. Likewise, because I 2 flows from VM to V 2 , we know VM ≥ V 2. Or, putting these two facts together, we know intuitively that VM lies between V 1 and V 2 , that is, EQ. 12 V 1 ≥ VM ≥ V 2.
Using our understanding of VM it is easy to understand EQ. 9: R 1 → 0 as VM → V 1 because the current in R 1 is fixed at I 1 = 10 mA. As VM → V 1 , the voltage drop across R 1 is reduced, so R 1 has to be smaller to allow the same current to flow. A similar understanding applies to EQ. 11.
Bounds on resistor values due to voltage limits
Looking at EQ. 9 assuming VM is given, the value of R 1 increases linearly as V 1 increases, which means the maximum value of R 1 occurs when V 1 is its maximum value of V 1 = 20V. The reason R 1 is a maximum when V 1 = 20 V is that V 1 = 20V will place the largest voltage drop across R 1 , so to keep the current at I 1 = 10 mA, R 1 will be its biggest. In other words, the maximum value R1MAX is given by EQ. 13
k Ω 10
20 V R (^1) MAX M
− =.
The minimum value of R 1 is zero and occurs when V 1 = VM. According to EQ. 10, for a non-negative value of R 2 we must have EQ. 14 VM ≥ 0
From EQ. 11 for a given value of VM, the value of R 3 decreases linearly as V 2 increases, reaching zero when V 2 = VM. Again, the reason is that when VM is fixed and V 2 is increased, the
automatically update when the data input on C HARTS is changed, allowing immediate visual appraisal of the benefit of the change. The worksheets shown in the row under C HARTS are supplementary, serving to remin the user how the spreadsheet works, w
d hat problem it has solved, what the notation means, and what PSPICE verifications have been done to insure everything is working. The worksheets are separately introduced next.
Worksheet CHARTS
FIGURE 3
Worksheet CHARTS showing USER I NPUT field for data and design point values from ; the maximum voltage has been labeled as V (^) CC = 20 V
Figure 3 values are plotted. this area of the worksheet is
DESIGN shows the C HARTS worksheet where data is input and charts of the resistor To avoid accidental overwriting of the F ROM D ESIGN area, protected. Instructions to unprotect it are provided in case we forget how later on.
Worksheet DESIGN
FIGURE 4
DESIGN worksheet that contains formulas for the values of the bounds and of R_
The D ES e all the ll use
M _V ARIES
IGN worksheet contains the formulas used in the design. It is useful to hav formulas in one place. Then the other worksheets are made by copying these formulas, so a the same formulas. If a formula is changed, it is changed here, and then copied to the other worksheets.
Worksheet V
FIGURE 5
orksheet V M _VARIES showing the formula used to calculate R (^) 1MAX
Figure 5 rmed using the D ESIGN
the
culations. V (^) M is a column variable, and for each
Worksheet SCHEMATIC
W shows the worksheet VM _VARIES. This worksheet is easily fo worksheet: we simply highlight the cells F15:G18 in Figure 4, copy them, and go to a blank worksheet where we use E DIT/PASTE SPECIAL with the T RANSPOSE box checked. That pastes columns as rows on the new worksheet. The implementation of EQ. 13 using an I F-statement as shown in Figure 5 and Figure 4 serves to check for negative values of R 1 , and inserts #N/A (not available) if a negative value would result. Negative values should not happen if we keep VM ≤ 20 V, a requirement of EQ. 12 already discussed. Worksheet VM _VARIES does all the graphical cal value in the VM column the resistor value R 2 is found, and the limiting values on R 1 and R 3.
FIGURE 6
CHEMATIC worksheet
Figure 6 orksheet, which is useful to remind you what the variables in the spreadsheet mean.
The S shows the SCHEMATIC w
V_M
{R_2}
5.000mA
0
1 10.00V
{I_1} 10.00mA
{R_1}
V_
R_1 = 100 R_2 = 2000 R_3 = 5000
I_1 = 10mA I_2 = 5mA
{I_2}
5.000mA
{R_3}
0V
PARAMETERS: 11.00V V_^ -15.00V
FIGURE 9
PSPICE verification of spreadsheet results in Figure 8
Verification of design boundaries
A useful test of the spreadsheet (and also the analysis behind the spreadsheet) is to check points on the boundary. On the boundary V 1 or V 2 or both should be at their limiting values because their t set the boundaries in the first place. An example of a design point on the boundary is show
limits are wha n in Figure 10 below.
FIGURE 10
An example design point where R 1 and R 3 both are at their limiting boundaries, and the spreadsheet predicts V 1 = 20 V and V 2 = –20 V, which are limiting values for the voltages
V_M
{R_2}
5.000mA
0
V_1^ 13.00V
{I_1} 10.00mA
20.00V
{R_1}
PARAMETERS:^ V_ R_1 = 700 R_2 = 2600 R_3 = 6600
I_1 = 10mA I_2 = 5mA
{I_2}
5.000mA
{R_3}
0V
-20.00V
FIGURE 11
PS PICE verification of the design point of Figure 10
Figure 11 shows that PSPICE agrees with the spreadsheet predictions for V 1 and V 2. Of course, PSPICE input was R 2 , not VM , so another check is that V (^) M from PS PICE is the value V (^) M = 13V used in the spreadsheet.
Bells and whistles
EXCEL allows us to implement bells and whistles. These conveniences make sense only if the spreadsheet is used a lot, and they aren’t worthwhile for a single design. Here’s an example, which introduces features more valuable in complex designs. The spreadsheet has a check feature to warn the user whether the design is in spec. This warning is in cell F8 on C HARTS , and is implemented using an IF-statement as shown below:
=IF(OR(ABS(V_1)>20,ABS(V_2)>20),"OUT OF SPEC","IN SPEC")
The IF s er the voltages are out of spec. If they therwise it prints the value IN SPEC. To emphasize the
tatement evaluates the OR condition to see wheth are, it prints the value O UT OF SPEC. O violation of specifications, C ONDITIONAL FORMATTING is used to change the font color depending on whether the design is in or out of specification. C ONDITIONAL FORMATTING is implemented by highlighting the cell F8 and selecting the menu F ORMAT/C ONDITIONAL FORMATTING. The menu is filled out as shown in Figure 12.
FIGURE 12
Conditional formatting menu for making the font red for a violation of specification, and for within specification; this is how the menu is filled out: once OK is clicked, E XCEL
To add s shown Figure 13 below.
blue will implement the entries as text strings with quotation marks to the warning, a label is added to the chart when a violation occurs. An example i in
Design variables are inventions related to concept. There is no cut-and-dried approach to concept, but the spreadsheet can help. Spreadsheet organization is tied to concept. For example, spreadsheet layout depends on what variables are selected, and which are elected independent, which dependent. Also, the spreadsheet displays trade-off curves. Explaining these curves forces us to think, learn more, and perhaps change our approach. The spreadsheet is a partner in our thinking, and the design can take new turns as our minds and our spreadsheet evolves.
Summary of solution to design problem
Finally, we summarize the solution to the design problem:
The specifications that I 1 = 10mA and I 2 = 5mA for input and output voltages less than 20V in magnitude are satisfied for values of R 1 and R 3 , that satisfy inequalities EQ. 13 for R 1 and EQ. 16 for R 3 at the selected value of V (^) M. That is, there is a range of allowed values for R 1 found below the lowest downward sloping curve in Figure 10, and for R 3 found below the highest upward sloping curve in Figure 10. For a solution, R 2 must satisfy EQ. 10, also shown in Figure
- For any selected R 1 , R 2 and R 3 , the voltages are reported in the C ALCULATED box, as determined by EQ. 1 and EQ. 2.
Exercises
Below are other spreadsheets related to this problem. You’ll see spreadsheets affect your vice versa. In these problems, the “presented” spreadsheet means the spreadsheet just disc
- Red 2_V ARIES as
the presented spreadsheet so it uses R 1 and R 3 as variables and R1_V ARIES and as calculation worksheets, and does not use the variable V (^) M or worksheet ce of
selection of R 1 , R 2 and R 3 to meet a spe es
l solve the design problem for any values for I 1 and
thinking and ussed. esign the presented spreadsheet so it uses R 2 as a variable and R calculation worksheet, and does not use the variable V (^) M or worksheet V M _VARIES at all. C HARTS should show a chart for the allowed values for R 1 and R 3 similar to the presented spreadsheet, but using R 2 as x- axis.
- Redesign R3_VARIES VM _VARIES at all. C HARTS should show a chart for the allowed values for R 2 for any choi R 1 and R 3.
- Redesign the presented spreadsheet to guide cification for the power consumed by the resistors. The voltages V 1 and V 2 and the valu of I 1 and I 2 still must satisfy the original problem requirements.
- Redesign the presented spreadsheet so it will solve the design problem for any positive values for I 1 and I 2 input on C HARTS , regardless of which of I 1 and I 2 is larger. Assume the currents are in the range from 0 to 10 mA.
- Redesign the presented spreadsheet so it wil I 2 input on C HARTS , positive or negative. Assume the currents are in the range from – mA to 10 mA.
- Redesign the presented spreadsheet so it will explore the design problem as a function of positive values for I 1 and I 2 , regardless of which of I 1 and I 2 is larger. The new spreadsheet will contain additional worksheets I1_VARIES and I2_VARIES , and place charts on CHART show how R 1 , R 2 , and R 3 behave as these currents are changed. Assume the currents are in the range from 0 to 10 mA.
S to
