What are applications for load line analysis

Simultaneous equations for circuit analysis

Simultaneous equations for circuit analysis

Mathematics for electronics

Question 1

Solve this equation for the value of x:

How many exact solutions does the above equation "all"> have

How many exact solutions does this equation have? Record the solutions to this equation in the following graph:

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If x + 5 = 8, then x = 3 (exactly one solution)
If x + y = 8 then there are an infinite number of solutions:

Follow-up question: Find the solution for x + 5 = 8 in the same graph.

Remarks:

This question starts with an extremely simple equation with one solution and goes on to another simple equation with an infinite number of solutions. While an infinity of correct answers seems impossible to reasonably handle, a graph handles the multitude of correct answer pairs represented as a line on a graph of infinite length quite well.

Question 2

Set the solutions on a graph to the equation y + x = 8:

On the same graph, draw the solutions to the equation y - x = 3. What is the meaning of the point where the two lines "# 2" cross? Show answer Hide answer

The intersection between the two lines represents the one set of solutions that satisfies both equations (where x = 2.5 and y = 5.5).

Remarks:

The purpose of this question is to gently introduce students to the concept of simultaneous systems of equations, in which a series of solutions satisfies more than one equation at the same time. However, it is important that students understand the basic concepts of graphing before attempting to answer this question.

Question 3

What does it actually mean to get a solution for a "simultaneous" system of equations? # 3 "> show answer hide answer

The solutions to a system of equations represent a unique combination of values ​​that satisfy all of the equations in that system. For a system with two variables, the solution is the intersection of two lines.

Remarks:

Many students have difficulty grasping the meaning of systems of equations. Discuss the meaning of equations and systems of equations with your students and make sure that the concept of simultaneity (solutions that satisfy all equations at the same time) is clarified.

Question 4

Put the equation y = x 2 in the following graphic:

On the same diagram, draw the equation y = x + 2. What is the meaning of the point where the two diagrams "# 4" cross? Show answer Hide answer

There are two points of intersection between the parabola (curve) and the straight line, which represent two different sets of solutions that satisfy both equations.

Challenge question: Solve this simultaneous system of equations without graphs, but through symbolic manipulation of the equations!

Remarks:

Here the solution through graphical representation can be a little easier than the symbolic solution. In principle, we can determine solutions for any pair of equations with roughly the same difficulty using graphs. The only real problem is precision: how exactly can we interpret about intersections? A practical example of a non-linear, simultaneous functional solution is load line analysis in semiconductor circuits.

Question 5

Load lines are useful tools for analyzing transistor amplifier circuits, but they can be difficult to understand at first glance. To understand what "load lines" are useful for and how they're determined, I'll apply one to this simple two-resistor circuit:

We will need to draw a load line for this simple two resistor circuit along with the "curve" for resistor R1 to see the utility of a load line. Load lines only have meaning when they are overlaid with other plots. First the characteristic curve for R 1 defined as the voltage / current relationship between the terminals A. and B.:

Next, I'll plot the load line as defined by the 1.5 kΩ load resistor. This "load line" expresses the voltage between the same two terminals (V FROM ) is available as a function of the load current to account for the voltage drop across the load:

At what value of current (I. R1 ) the two lines "# 5" intersect> show answer hide answer

I. R. = 8 mA is the same current value you would calculate if you had analyzed this circuit as a simple series resistor network.

Follow-up question: You may be wondering, "what is the point of drawing a" characteristic curve "and a" load line "in such a simple circuit when all we need to do to solve current is add the two resistances and divide this Total Resistance into Total Voltage? ”Well, to be honest, there's no point in analyzing such a simple circuit in this way other than illustrating how load lines work. My follow-up question to you is: Where would plotting a load line actually be helpful in analyzing circuit behavior? Can you think of any changes to this two resistor circuit that would require load line analysis to look for current?

Remarks:

While this approach to circuit analysis may seem silly - using load lines to calculate the current in a circuit with two resistors - it demonstrates the principle of load lines in a context that should be obvious to students at this point in their study. Discuss with your students how to get the two lines (one for resistance R 1 and the other that does the one for R 1 available voltage based on the total voltage of the source and the value of the load resistance).

Also discuss the meaning of the intersection of the two lines. What does the intersection of two graphs mean mathematically? What do the coordinate values ​​of the intersection represent in a system of simultaneous functions? How does this principle relate to an electronic circuit?

Question 6

Load lines are useful tools for analyzing transistor amplifier circuits, but they can be applied to other types of circuits as well. Take this diode-resistor circuit for example:

The characteristic of the diode is already plotted in the following diagram. Your job is to plot the load line for the circuit on the same graph and note where the two lines intersect:

What is the practical meaning of crossing these two plots? # 6 >> show answer hide answer

The two lines intersect at a current of approx. 1.72 mA:

Follow-up question: Explain why the use of a load line greatly simplifies the determination of the circuit current in such a diode-resistor circuit.

Challenge question: Assume the resistance value has been increased from 2.5 kΩ to 10 kΩ. What difference would this make in the load line plot and in the intersection between the two "notes hidden" plots?

While this approach to circuit analysis seems silly - using load lines to calculate the current in a diode-resistor circuit - it shows the principle of load lines in a context that should be obvious to students at this point in their study. Discuss with your students how the load line for this circuit is obtained and why it is straight while the characteristic curve of the diode is not.

Also discuss the meaning of the intersection of the two lines. What does the intersection of two graphs mean mathematically? What do the coordinate values ​​of the intersection represent in a system of simultaneous functions? How does this principle relate to an electronic circuit?

Question 7

For example, suppose you obtained the following two equations and looked for solutions for x and y that both satisfy at the same time:

If we manipulate the second equation to solve for y, we have a definition of y in x that we can use for the substitution in the first equation:

Show the substitution process in the first equation and how this leads to a single solution for x. Then use that value of x to solve for y, resulting in a set of solutions that is valid for both equations.

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If y + x = 8 and y = x + 3, then (x + 3) + x = 8. Hence

Remarks:

This question shows one of the many practical applications of algebraic substitution: solving simultaneous systems of equations.

Question 8

An interesting and useful property in mathematics is the transitive property:

If a = b and b = c, then a = c

Simply put, two variables must be the same if they both correspond to a common (third) variable. While this property isn't particularly profound or breathtaking, it is still useful for solving certain math problems.

For example, suppose you obtained the following two equations and looked for solutions for x and y that both satisfy at the same time:

Manipulate both equations to look for y, then explain how you could apply the transitive principle to solve for x.

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If 8 - x = y and 3 + x = y, then 8 - x must be 3 + x:

Solutions for x and y:

Remarks:

This method of solving a simultaneous two-variable equation is actually nothing more than a disguised substitution. However, some students find it easier to understand than a direct substitution.

Question 9

For example, suppose you obtained the following two equations and looked for solutions for x and y that both satisfy at the same time:

Well, you know that we can do whatever we want to do for either equation, as long as we do the same thing on both sides (on either side of the "equal sign"). This is the basic rule that we follow when manipulating an equation to solve a particular variable. For example, we can take the equation y + x = 8 and subtract x from both sides to get an equation expressed in y:

On the same principle, we can take two equations and combine them by adding or subtracting both sides. For example, we can take the equation y - x = 3 and add both sides of it to the respective sides of the first equation y + x = 8:

What useful result comes from this action "# 9"> Show answer Hide answer

We can use the result (2y = 11) to look for a value of y that, when plugged into one of the original equations, can be used to solve a value of x to satisfy both equations at the same time .

Remarks:

While it is not intuitively obvious to most people, the technique of adding two whole equations for the purpose of eliminating a variable is not only possible but very powerful when looking for solutions that satisfy both of the original equations. Discuss with your students why we can add y - x to y + x and add 3 to 8 to get the equation 2y = 11.

Question 10

Solve for values ​​of x and y that simultaneously satisfy both of the following equations:

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x = -3 y = 6

Next question: Solving this system of simultaneous equations using both substitutions (solving for one variable in one of the equations and substituting it for the other equation) and addition (adding the two equations together to produce a third equation with only one unknown).

Remarks:

Nothing special here - just practice solving for a two-variable system of equations.

Question 11

Solve for values ​​of x and y that simultaneously satisfy both of the following equations:

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x = 5 y = -2

Remarks:

Nothing but "drilling" (practicing) here.

Question 12

Solve for values ​​of x and y that simultaneously satisfy both of the following equations:

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x = -2 y = 3

Remarks:

Nothing but "drilling" (practicing) here.

Question 13

If we want to solve for the value of three related variables (ie x + y + z = 0), how many equations do we need in our "system" of simultaneous equations?

What does the solution (x, y, z) represent graphically for a system of equations with three variables?

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Three variables require three equations to solve. Graphically, the solution set represents the point at which three planes of infinite surface intersect.

Remarks:

Ask your students to graph the scenario of three variables and three equations versus two variables and two equations. Where is the solution theorem shown in a system with two variables and two equations? How many do we extrapolate from this situation to one involving three variables and three equations?

Question 14

Many circuit analysis techniques require the solution of "systems of linear equations", sometimes referred to as "simultaneous equations". This question is a set of practical problems in solving simultaneous linear equations that will give you plenty of practice with various solving techniques (including what your calculator can do for solving).

Systems of two variables:

x + y = 5x - y = -62x + y = 7
x - y = 12x - y = 4x - y = 2
3x - 2y = -1-10x + 2y = 03x - 5y = -13
5x + y = -6-3x - 5y = -28-X + 2y = 5
1000x - 500y = 0-15000x + 2200y = -662009100x - 5000y = 24
550x + 2500y = 55507900x - 2800y = 28300-5200x - 2700y = -6.5

Systems of three variables:

x - y + z = 13x + 2y - 5z = -21x + y + z = 0
-X - y + z = -1x - 3y + z = 82x - y - 4z = -9
x + y + z = 3-X - y - z = -12-2x + 2y - z = 12
x + y - 2z = -12-4x - 3y + 2z = -3219x - 6y + 20z = -33
3x - 2y + z = 19x - 2y + 3z = -14x + 5y - 3z = -17
-4x + 3y - 5z = -45-2x + 7y - z = 3-7x + 2y - 8z = 9
890x - 1000y + 2500z = -15002750x - 6200y + 4500z = 17500
3300x + 7200y - 5100z = 21500-10000x + 5300y - 1000z = 8100
-X + y - z = 06x - 2y - 3z = 5
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Systems of two variables:

x + y = 5x - y = -62x + y = 7
x - y = 12x - y = 4x - y = 2
x = 3 ; y = 2x = 10 ; y = 16x = 3 ; y = 1
3x - 2y = -1-10x + 2y = 03x - 5y = -13
5x + y = -6-3x - 5y = -28-X + 2y = 5
x = - 1 ; y = - 1x = 1 ; y = 5x = - 1 ; y = 2
1000x - 500y = 0-15000x + 2200y = -662009100x - 5000y = 24
550x + 2500y = 55507900x - 2800y = 28300-5200x - 2700y = -6.5
x = 1 ; y = 2x = 5 ; y = 4x = 0. 001924 ; y = - 0 . 001298

Systems of three variables:

x - y + z = 13x + 2y - 5z = -21x + y + z = 0
-X - y + z = -1x - 3y + z = 82x - y - 4z = -9
x + y + z = 3-X - y - z = -12-2x + 2y - z = 12
x = 1 ; y = 1 ; z = 1x = 4 ; y = 1 ; z = 7x = - 3 ; y = 3 ; z = 0
x + y - 2z = -12-4x - 3y + 2z = -3219x - 6y + 20z = -33
3x - 2y + z = 19x - 2y + 3z = -14x + 5y - 3z = -17
-4x + 3y - 5z = -45-2x + 7y - z = 3-7x + 2y - 8z = 9
x = 2 ; y = - 4 ; z = 5x = 6 ; y = 2 ; z = -1x = - 5 ; y = 3 ; z = 4
890x - 1000y + 2500z = -15002750x - 6200y + 4500z = 17500
3300x + 7200y - 5100z = 21500-10000x + 5300y - 1000z = 8100
-X + y - z = 06x - 2y - 3z = 5
x = 2. 215 ; y = 1. 378 ; z = - 0 . 8376x = - 5 . 171 ; y = - 9 . 322 ; z = - 5 . 794

Remarks:

I suggest that you show your students how to use the equation functions of their scientific calculators on their own. In my experience, both young and older students willingly accept this challenge as they learn how to do an enormous amount of hand calculations with their calculators!

Question 15

Suppose you had to choose a fixed resistance value (R) to form a voltage divider circuit, with a known resistance value of the potentiometer, the value of the source voltage, and the desired adjustment range:

Solve for R and show the equation you set up to do it.

Tip: think about the series resistance-voltage divider formula. . .

VR = V total  R.

R. total

 
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R = 20.588 kΩ

Remarks:

Make sure your students do their equations in front of the class so everyone can see how they did it. Some students may choose to apply Ohm's law to solving R, which is good, but for the purpose of developing equations that fit problems it may not be the best solution. Challenge your students to find a single equation to solve for R with all known quantities on the other side of the "equal sign".

Question 16

Suppose you had to choose a potentiometer value (R) to form a voltage divider circuit, with a known fixed resistance value, the source voltage value, and the desired adjustment range:

Solve for R and show the equation you set up to do it.

Tip: think about the series resistance-voltage divider formula. . .

VR = V total  R.

R. total

 
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R = 121.43 kΩ

Follow-up question: You won't find a potentiometer with a full-range resistance of exactly 121.43 kΩ. Describe how you can take a standard potentiometer and connect it to one or more fixed value resistors to get the full range you want.

Remarks:

Make sure your students do their equations in front of the class so everyone can see how they did it. Some students may choose to apply Ohm's law to solving R, which is good, but for the purpose of developing equations that fit problems it may not be the best solution. Challenge your students to find a single equation to solve for R with all known quantities on the other side of the "equal sign".

The follow-up question is very practical as it is impossible to find potentiometers prepared for any value of full resistance. Instead, you have to work with what you can find. These are usually nominal values ​​like 10 kΩ, 100 kΩ, 1 MΩ, etc.

Question 17

An engineer needs to calculate the values ​​of two resistors to set the minimum and maximum resistance ratios for the following potentiometer circuit:

First, write an equation for each circuit and show how the resistances R 1, R 2 and assemble the 10 kΩ of the potentiometer to form the ratio (a / b). Then use simultaneous equation solving techniques to get the actual resistance values ​​for R 1 and R 2 to calculate.

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a

b

(Minimum) = R. 1

R. 2 + 10000

a

b

(maximum) = R. 1 + 10000

R. 2

R. 1 = 15.77 kΩ

R. 2 = 515.5 Ω

Remarks:

This very hands-on application of simultaneous equations was actually used by one of my students to determine the lower and upper bounds for the voltage gain setting of an inverting op-amp circuit!

Question 18

The voltage gain of an emitter transistor with an emitter circuit is roughly equal to the collector resistance divided by the emitter resistance:

For this reason, calculate the required resistance values ​​for the following fixed-value resistor (R 2 ) and potentiometer (R. 1 ) to give this common mode amplifier an adjustable voltage gain range from 2 to 8:

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R. 1 (Pot) = 9 kΩ

R. 2 (fixed) = 3 kΩ

Remarks:

Ask your students how to create a standard potentiometer like 10 kΩ with a maximum (maximum) resistance of only 9 kΩ.

Question 19

The voltage gain of an emitter transistor with an emitter circuit is roughly equal to the collector resistance divided by the emitter resistance:

To do this, calculate the required resistance values ​​for the following fixed-value resistors (R 1 and R 2 ) to give this emitter an adjustable voltage gain range from 4 to 7:

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R. 1 = 13.33 kΩ

R. 2 = 3.333 kΩ

Remarks:

Have your students show how to set up the system of equations for the two resistance values. This is good practice in front of the class so that everyone can (possibly) see different possible solutions.

Question 20

Suppose you need to choose two resistor values ​​to get a voltage divider with a limited range of adjustment. One of these resistances is in the value (R 1 ) while the other is variable (a potentiometer known as the rheostat R 2 is connected ):

Establish a system of simultaneous equations to calculate for both R 1 as well as for R 2 and show how to get to the solutions for each one.

Tip: think about the series resistance-voltage divider formula. . .

VR = V total  R.

R. total

 
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R. 1 (fixed) = 4.286 kΩ

R. 2 (Pot) = 19.048 kΩ

Follow-up question: You cannot find a potentiometer with a full-range resistance of exactly 19.048 kΩ. Describe how you can take a standard potentiometer and connect it to one or more fixed value resistors to get the full range you want.

Remarks:

Make sure your students do their equations in front of the class so everyone can see how they did it. Some students may choose to apply Ohm's law to solving both resistances, which is good, but for the purpose of developing equations that will fit problems it may not be the best solution. Ask your students to develop a series of equations that can be used according to R 1 and R 2 solve, and then use simultaneous equation solving techniques to find solutions to each solution.

The follow-up question is very practical as it is impossible to find potentiometers prepared for any value of full resistance. Instead, you have to work with what you can find. These are usually nominal values ​​like 10 kΩ, 50 kΩ, 100 kΩ, etc.

Question 21

Suppose you need to choose two resistor values ​​to create a voltage divider with a limited range of adjustment:

Establish a system of simultaneous equations to calculate for both R 1 as well as for R 2 and show how to get to the solutions for each one.

Tip: think about the series resistance-voltage divider formula. . .

VR = V total  R.

R. total

 
Reveal answer hide answer

R. 1 = 5.25 kΩ

R. 2 = 2.25 kΩ

Remarks:

Make sure your students do their equations in front of the class so everyone can see how they did it. Some students may choose to apply Ohm's law to solving both resistances, which is good, but for the purpose of developing equations that will fit problems it may not be the best solution. Ask your students to develop a set of equations that can be used according to R 1 and R 2 solve, and then use simultaneous equation solving techniques to find solutions to each solution.

Question 22

Use simultaneous equations to find the values ​​of R 1 and R 2 to calculate that are necessary to give this voltage divider the specified setting range:

V out (Minimum) = 3 volts V. out (Maximum) = 8 volts

R. 1 = R 2 =

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R. 1 = 2 kΩ

R. 2 = 3 kΩ

Remarks:

Have your students demonstrate their solving methods in class so you can observe their problem solving skills and they can see multiple solving methods.

Question 23

Use simultaneous equations to find the values ​​of R 1 and R 2 to calculate that are necessary to give this voltage divider the specified setting range:

Vout (minimum) = 5 volts Vout (maximum) = 12 volts

R. 1 = R 2 =

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R. 1 = 4.2857 kΩ

R. 2 = 71429 kΩ

Remarks:

Have your students demonstrate their solving methods in class so you can observe their problem solving skills and they can see multiple solving methods.

Question 24

The voltage gain of an inverting op-amp circuit is defined by the ratio of feedback to input resistance:

Calculate the required values ​​of R 1 and R 2to limit the minimum and maximum voltage gain of this op-amp circuit to 5 and 30, respectively, when using a potentiometer in the middle with a full-range resistance of 5 kΩ:

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R. 1 = 1.2 kΩ

R. 2 = 31 kΩ

Remarks:

This is a very practical example of using simultaneous equations in analog circuit design.

Question 25

Calculate the required values ​​of R 1 and R 2to limit the minimum and maximum voltage gain of this op-amp circuit to 10 and 85, respectively:

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R. 1 = 2 kΩ

R. 2 = 153 kΩ

Remarks:

This is a very practical example of using simultaneous equations in analog circuit design. A common mistake students make while setting up the equations is forgetting that the gain of a non-inverting amplifier is the ratio of the feedback and ground resistances plus one!

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