1. Lecture #24 Review for Exam #2
ECE 410 – Spring 2012
Lecture #24
Review for Exam #2

2. Circuit Terminology

3. Practice How Many Nodes? How Many Essential Nodes?
How Many Branches? How Many Essential Branches?
How Many Meshes?

4. Node-Voltage Method 1st Steps in Node-Voltage Method
Make a neat layout of your circuit with no branches crossing over
Mark the essential nodes
Pick one of the essential nodes to be your reference node and mark it. (typically the node with the most branches)
Define the node voltages (voltages from reference node to other essential nodes)

5. Node-Voltage Method Continued
With the node voltages labeled in your circuit you are ready to generate your node-voltage equations
This is done by writing each of the essential branch currents leaving the essential nodes in terms of the node voltages and then summing them to zero in accordance with Kirchoff’s current law.

6. Write Node Voltage Equations
(3)
(1)
(2)
Write branch currents exiting each essential node in terms of the node voltages and sum them to zero in accordance with KCL.

7. Using N-V Method with dependent sources
In order to solve a circuit with dependent sources, the node-voltage equations must also include the constraint equation that governs the dependent source written in terms of the node voltage.

8. Supernode Technique
Sometimes a current flowing out of a node cannot be expressed in terms of the node voltages due to a voltage source between essential nodes
These nodes can then be treated as a single ‘supernode’.

9. Mesh Current Method Only applicable to planar circuits.
Mesh current is the current that exists around the loop of a mesh.. It is not necessarily a branch current.
Apply KVL around each mesh loop, expressing all voltages in terms of the mesh current.
This will create enough equations to solve the circuit
Once mesh currents are determined, the branch currents (and voltages) can be determined.

10. Mesh-Current Setup Example
4 equations, 4 Unknowns (i1, i2, i3, and i4)

11. Solve equations for four mesh currents
Example Continued
With mesh currents found, group current terms in each equation
Solve equations for four mesh currents

12. Mesh Current Technique with Dependent Sources
Much like the Node Voltage technique, solving circuits with a dependent source with the mesh current technique requires the inclusion of a constraint equation for the dependent source.
These equations should be expressed in terms of the mesh currents.

13. Special Cases with Mesh Current
When using the mesh current technique, current sources in your circuit can have similar effects as voltages sources in the node voltage technique
Current sources can at times directly indicate a mesh current, simplifying your solution
Sometimes a current source can prevent you from writing an expression for the voltage drop across it in terms of the mesh currents or anything else
This situation leads to the concept of the supermesh (analogous to the supernode)

14. Supermesh
If a current source lies in a branch of your circuit, you can ‘remove’ the current source when creating your mesh currents.. Avoiding the branch that contained the source

15. Source Transforms
Another method like Delta-Wye and Parallel/Series combinations to simplify circuits
Can switch between equivalent sources: a voltage source with series resistor or current source with parallel resistor

16. How to Transform
Each of the two equivalent sources must produce the same current through any load connected across the source. RL RL
(current divider) OR

17. Example What is Power of 6V source? i i = (19.2V-6V)/16 = 0.825A
P = i*v = 0.825A*6V = 4.95W

18. Extraneous Resistors
Resistors in parallel with the voltage source, or in series with the current source have no effect on the load.

19. Terminal Behavior
We are often interested in looking at the behavior of an electrical circuit at output terminals.
If a load is hooked to a circuit, how does the circuit behave relative to the load?
If the circuit is made of linear elements, the entire circuit can be simplified into a Thevenin (or Norton) Equivalent circuit.

20. Justification
We can consider the entire circuit a black box with two output terminals. The behavior at these terminals can be modeled by a voltage source and a series resistor.

21. Finding the Thevenin Voltage
If we look at the terminal with no load attached (open circuit output), the voltage between the output terminals is VTh.
With an open circuit, no current flows
This means no voltage drop across RTh
Hence Vab = VTh

22. Finding Thevinen Resistance
Now lets look at what happens when the output is short circuited
If we can find the short circuit current and have already found VTh, then we can easily calculate RTh
isc

23. Example
Find Vab (open circuit voltage) in the following circuit. v1
Use node voltage:
Vab = VTh = 32V

24. Example Continued Find closed circuit current and calculate RTh v1
Again… Use node voltage:

25. Final Equivalent Circuits
The resulting Thevinin Equivalent Circuit is shown below:
Using a source transform the Thevenin circuit can be represented as a current source in parallel with the same RTh. This is the Norton Equivalent Circuit.

26. Alternate Methods
Other methods exist for finding the Thevenin equivalent. The simplest is simply using source transforms and parallel/series simplifications.

27. Source Deletion
Here is another method that works well for circuits containing only independent sources.
Find RTh by ‘deactivating’ all the independent sources in the circuit and finding the resulting equivalent resistance from terminal a to terminal b.
‘Deactivating’ a votage source turns it into a short circuit, ‘deactivating’ a current source turns it into an open circuit.

28. Modification for Dependent Sources
If dependent sources exist in the circuit, you can deactivate the independent sources and connect a test voltage source or test current source to the output terminal.
The voltage delivered by the test source divided by the current supplied by the test source is equal to RTh.

29. Basics of Power Transfer
We have some network of sources and linear elements and will connect this network to a load RL. We want to maximize power in the load.
This process is easiest to analyze by using the Thevenin equivalent circuit.

30. Analyzing Thevenin Circuit with Load
The power dissipated in the load can be written as P=iL2*RL
By using the Thevenin Equivalent circuit, finding an expression for this power is straight-forward.

31. Finding condition for Max Power Transfer
To find the load value that give maximum power dissipated in the load, you must take the derivative of the Power expression with respect to the load resistor and set it equal to zero:
Max Transfer when Load and Thevenin Resistance are the same

32. How much power is transferred?
Using our relation for max power transfer (RL=RTh) we can use our relation for power dissipated in the load to find the total power transferred.
Power transferred increases as RTh decreases.
Since RTh=RL

33. Superposition All linear systems obey the principle of superposition.
This means that whenever a linear system is driven by multiple sources, the total response of the system is the sum of the responses to each individual source
This principle is used in all areas of engineering and physics and is of course useful in Circuit Theory

34. What does this mean for solving circuits?
The value of voltage or current relating to any element in a circuit can be thought of as the sum of the voltages and currents resulting on that element from each of the independent sources in the circuit individually. v1 v2 =
v1’
v2’
v1’’
v2’’ +

35. Uses for superposition
Superposition is highly useful if you have a circuit that you have already solved that is altered by adding a new source of some type.
All you need to do is deactivate all the original sources and find the values resulting from your new source
The true altered values in your circuit are just the original values plus what results from the new source

36. Example
Problem 4.9
With only 24V source:
With only 40mA source:

37. Sensitivity Analysis
When creating a circuit you must use real components with inexact values
Resistors vary from their nominal value by up to 20% depending on the type specified
Resistors with more accuracy cost more money.
You need a way to be able to know what tolerance in resistor values your circuit can accept and still manage to meet specifications
Spending extra money only on resistors that create the biggest change in system performance is highly desired.

38. Example Sensitivity Analysis
Express output value in terms of the system components: + i
vout -

39. Continued
To find the variation caused by a specific component, take the derivative of the output variable with respect to that component.

40. Numerical solution
Put variable parameters into the equation to find out how quickly vout changes with respect to the resistance of R1.
This can be done for every component in the circuit

41. Output Data for our circuit
DC SENSITIVITIES OF OUTPUT V(N00465)
ELEMENT ELEMENT ELEMENT NORMALIZED
NAME VALUE SENSITIVITY SENSITIVITY
(VOLTS/UNIT) (VOLTS/PERCENT)
R_R E E E-02
R_R E E E-03
R_R E E E-02
V_V E E E-02
This agrees with our calculated value
Nominal Output Voltage = 2.941V

42. Calculating Effects due to Component Change
If we want to know how much the output will change due to the change in a component value we can just multiply the sensitivity by the amount of change.
EXAMPLE
If we increase the value of resistor R3 by 100Ω

43. Effects of multiple changes
Using the superposition principle, we can see that the total change in output variable due to changes in multiple elements is just the sum of the changes due to each element change.
EXAMPLE
R3 increases by 100Ω, R1 and R2 increase by 100Ω each.

44. Comparison to Exact Value
Sensitivity analysis only gives an estimate based on a linear assumed change and is less accurate the larger the variation you use. Here is the exact PSpice simulation for the changes we made:
Sensitivity analysis shows Vout=2.955V
Match is exact in our case

45. Resistor Tolerances
Real resistors have tolerance values that define the amount of variation in their resistance value that can be expected.
Resistors typically come with tolerances of 20% (rare), 10%, 5%, 2% or 1%.
A 1k resistor with 10% tolerance can have a value anywhere between 900Ω and 1100Ω.
A 1k resistor with 1% tolerance may have a value anywhere between 990Ω and 1010Ω

46. Example Circuit Here’s a circuit simulated in PSpice:
The output voltage is nominally 34.27V
What range would the output voltage have if we used 10% resistors?
What if we used 5%
What about 1%

47. PSpice Sensitivity Output
The normalized sensitivity for each element can be multiplied by the % tolerance of the resistor to determine what he maximum shift in output voltage could be.
To find Vout max, choose the sign of your percentage tolerance to result in a positive voltage shift, to find Vout min, choose the sign of your percentage tolerance to result in a negative voltage shift.

48. Analysis in Excel
ELEMENT
NORMALIZED
% tolerance
NAME
VALUE
SENSITIVITY 10 5 1
(VOLTS/UNIT)
(VOLTS/PERCENT)
voltage shift
R_R1
1.00E+02
7.19E-02
0.72
0.36
0.07
R_R2
2.00E+01
-2.64E-01
-5.28E-02
0.53
0.26
0.05
R_R4
1.50E+01
1.38E+00
2.06E-01
2.06
1.03
0.21
R_R5
2.50E+01
5.42E-01
1.35E-01
1.35
0.68
0.14
R_R3
3.00E+00
-8.29E-01
-2.49E-02
sum
4.67
2.33
0.47
Vout max
38.94
36.60
34.74
Vout min
29.60
31.94
33.80
What if we want to keep the max output voltage under 37 volts,
Which resistor(s) should we attack first in terms of reducing tolerances?

49. Analysis Continued
ELEMENT
NORMALIZED
Tolerance
NAME
SENSITIVITY %
(VOLTS/PERCENT)
R_R1
7.19E-02 10
0.72
R_R2
-5.28E-02
-10
0.53
R_R4
2.06E-01 5
1.03
R_R5
1.35E-01
1.35
R_R3
-2.49E-02
-0.25
sum
3.38
Vout max
37.65
ELEMENT
NORMALIZED
Tolerance
NAME
SENSITIVITY %
(VOLTS/PERCENT)
R_R1
7.19E-02 10
0.72
R_R2
-5.28E-02
-10
0.53
R_R4
2.06E-01 5
1.03
R_R5
1.35E-01
0.68
R_R3
-2.49E-02
-0.25
sum
2.71
Vout max
36.98
We can reduce the maximum output variation from +/- 4.67V, to only +/-2.71V, by only improving the tolerance on two most sensitive resistors.
A complete change of all the resistors would have only reduced the output variation to +/- 2.33V

50. Op Amps
Operational Amplifiers (Op Amps) are a basic building block of many electronic circuit designs.

51. OpAmp Terminals
Op amps have at minimum five basic terminals that are uses
An inverting input (marked with -)
An non-inverting input (marked with +)
A positive power supply
A negative power supply
An output terminal

52. Output behavior
In an ideal op amp, the open loop gain is infinite, so if the amplifier is to remain in the linear region Vn=Vp this is called the virtual short.
In addition, an ideal op amp has no current flowing into the inverting and non inverting terminals. ip=in=0
How does the amp remain in this condition?
Negative feedback is used to keep the input voltages equal
This means that there is a circuit path from the output of the op amp to it’s inverting terminal

53. Example Negative Feedback Control System #1 - HVAC
Lets look at other negative feedback systems to help with understanding
One of the most ubiquitous is the thermostat for your AC/furnace
It compares two inputs (the desired temperature Tset and the actual ambient temperature Troom)
Feedback loop
Troom - +
AC/Furnace control
If Troom > Tset output is negative and AC cools room
If Troom < Tset output is high and furnace heats the room
Tset
Troom is forced to Tset
What happens on a very cold or hot day when HVAC can’t keep up?

54. Method for solving circuits with negative feedback op amp
Assume that it is operating in linear regime (i.e. that the output isn’t saturated)
Thus assume vn=vp (remember in = ip = 0)
Find vp it’s value will not depend on the output of the op amp
Find an expression for vn this will include the output voltage of the op amp
Set vn = vp… you now have a governing equation for the op amp circuit
Use this equation to solve for whatever parameter is of interest
If you find the output is saturated at the supply limit, use this limiting value for the output voltage in your final equation.

55. The Inverting Amplifier
The type of circuit shown in the previous assessment problem is the general layout of an inverting amplifier
In the linear region:
The inverting gain is:

56. Summing Amplifier
An extension of the inverting amplifier is the summing amplifier
Superposition principle can be used to show that the output of the summing amp is the sum of outputs of each of the individual inputs.

57. Non-Inverting Amplifier
The non-inverting amplifier amplifies the input voltage buy a positive gain factor.
In the linear region:
The gain is:

58. Differential Amplifier
The differential amplifier is used to amplify the difference between two input voltages
If:
In the linear region:
The gain is: