PHY 335 Unit 1 Instruments, measurements, DC circuits ...
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PHY 335 Unit 1 Instruments, measurements, DC circuits, Thevenin equivalents 09/18/2014 Michael Santana (Justin Thomas) Professor Michael Gurvitch
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Introduction: In Unit I, Instruments, measurements, DC circuits, Thevenin equivalents aims to better our understanding of electronics, and to highlight the techniques and instruments used in laboratory. Electricity, the engine behind electronics, is introduced by use of an analogy to water. Water dynamics or flow through a pipe is very similar to the flow characteristics of electricity. The flow of water through a cross-sectional area of the pipe (volumetric flow rate Volume/second) is analogous to current (electric charge flowing through a cross-sectional area of a wire per second) and is in units of Coulombs/second or Amps. Voltage (electrical potential) is the potential energy per charge in the surroundings of other charges divided by that charge and is in units of Joule/Coulomb or otherwise known as Volts. Voltage is analogous to the pressure difference between two sections of a pipe which give the potential to produce flow. Therefore, when talking of voltage and current, voltage is always measured across an object (resistor, etc.) whereas current is measured through an object (resistor, circuit, etc.) and not vise-versa. Due to the fact that water is approximately incompressible, it is assumed that like water electricity proceeds on its path at the same rate or current. Lastly, resistivity is a function of length and cross-sectional area and has the property of resisting flow. In the water analogy, a pressure differential is applied to opposite ends of the section of gravel (resistance), which induces flow from high pressure to low pressure. Furthermore, through the completion of assignments, we will come to better understand networks of resistors (series, parallel, and permutations of series and parallel), ways in which an ideal voltmeter and ammeter produce readings, Kirchhoff’s rules of algebraic sums of current in junctions and sums of voltage drops around closed loops, and how these methods can help us in better understanding circuit theory and analysis. In our assignments, we will learn the intricacies of a breadboard and its benefits in circuit design, how to measure resistances of standard resistors, variable resistors, and networks of resistors, how to account for instrumental uncertainties and random errors as well as how to mitigate systematic errors, techniques to measure internal resistances of the ammeter, DC voltage supply source, and voltmeter, how to build a simple current source and simple voltage divider, and the theory behind Thevenin’s equivalent circuits and their benefits.
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Equipment and Experimental Techniques: Keithley 177 MICROVOLT Digital Multi-Meter (DMM): Ω (350 V Peak), V (1 kV Max), A (2 A Max) It has six scales of precision for measurements of Ohms, Amps, and Volts (AC and DC). The DMM’s maximum instrumental accuracy is five digits. The DMM has internal resistance that needs to be accounted for when taking measurements as illustrated in assignment 5. It is also important to note that the internal resistance of the DMM on various settings (Ohms, Amps, and Volts) is dependent upon the scale of precision used. For example: when measuring internal resistance of an ammeter, it is necessary to choose a scale that is suitable for the desired function. If there will be high current supplied, we may want to choose a less sensitive scale. Keithley 195A Digital Multi-Meter (DMM): Ω (350 V Peak), V (1 kV Max), A (2 A Max). It has variable scales for precision, though it also has an auto adjust setting which automatically adjusts the scale for best precision and measurement. The DMM’s maximum accuracy is five digits.
Experimental: The Board Assignment 1: Get familiarized with the board layout using an ohmmeter (DMM on an ohms “Ω” function) to see the contact arrangement. Sketch the board layout in your lab book.
Displayed in the above image is a standard solderless breadboard similar to that used in PHY 335 Laboratory. The blue and red horizontal lines illustrate the power rails (bus strips) one to supply voltage and one ground. The vertical columns that are numbered (1-57) are connected vertically up until the middle divider, and not past that. Horizontal rows that are labeled by letters (A-J) are not connected. The middle divider is spaced out in such a way as to provide connectivity for
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integrated circuits. The colored circles on the right hand side (grey, blue, green, and red) are the power supply from 0 V (ground), +15 V, -15 V, and +5 V from top down.
Measuring Individual Resistors Assignment 2: measure three different resistors; measure a resistor in a MΩ range; measure your own resistance between your hands
≈10 Ω:
R1 = 151.17 Ω
≈1 kΩ:
R2 = 1.98 kΩ
≈100 kΩ:
R3 = 5.57 MΩ Resistance (wet fingers) = 0.75 MΩ Resistance (dry fingers) = 1.25 MΩ
Resistance of Body + Resistance of Resistor in Parallel: Resistor without contact: R = 5.57 kΩ Resistor with contact (theoretical): 1 1 1 = + 𝑅 𝑅1 𝑅2 1 1 1 = + 𝑅 1.25 𝑀Ω 5.57 𝑘Ω R = 1.02 MΩ
Resistor with contact (experimental): R = 0.35 MΩ
It was noted that when we measured the individual resistors and our hands made contact with the metal parts of the alligator clips, the resistance went down to R = 0.35 Ω. The experimental value was 0.63 Ω less than the theoretical value calculated, though the difference is believed to have come from our hands having still been damp and less resistive (dry hand resistance used in the calculation). When the effective resistance is calculated using Resistance (wet fingers) = 0.75
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MΩ, the theoretical value is significantly more accurate. It was also noted that effective resistance significantly degreases when holding the resistors tighter or by increases contact area.
Measurement Uncertainty Assignment 3: Evaluate and report measurement uncertainties for the three resistors you measured in #2.
Instrumental Error: R1 = 151.16 Ω
R2 = 1.9854 kΩ
R3 = 5.584 MΩ
R1 = 151.18 Ω
R2 = 1.9855 kΩ
R3 = 5.586 MΩ
R1 = 151.17 ± 0.01 Ω
R2 = 1.9855 ± 0.001 kΩ
R3 = 5.585 ± 0.001 MΩ
Random Uncertainty: R1 = 151.04 Ω
R2 = 1.9873 kΩ
R3 = 5.564 MΩ
R1 = 151.07 Ω
R2 = 1.9875 kΩ
R3 = 5.554 MΩ
R1 = 151.01 Ω
R2 = 1.9867 kΩ
R3 = 5.560 MΩ
R1 = 151.04 Ω
R2 = 1.9861 kΩ
R3 = 5.565 MΩ
R1 = 151.02 Ω
R2 = 1.9858 kΩ
R3 = 5.558 MΩ
R1 = 151.04 ± 0.03 Ω
R2 = 1.9867 ± 0.0009 kΩ
R3 = 5.560 ± 0.006 MΩ
Both instrumental and random uncertainties were accounted for and calculated for the three resistors used in our earlier analysis in Part II. The instrumental uncertainty is determined from fluctuating digits on our DMM. If no fluctuations are present, we approximate one unit of uncertainty on the last digit. As random uncertainty cannot be accounted for in one measurement, multiple identical measurements must be taken and averaged for the best number. The mean and standard deviation of the distribution are considered to be the best number and random error respectively, though we will not calculated either in this course. We do not account for systematic error. If there were to be some refinement necessary, it would have been taken care of prior to recording data and performing data analysis.
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Measuring Resistor Networks Assignment 4: Take the three resistors you have measured individually and connect them in series, then in parallel and finally in a series-parallel connection (you will need only three connections in total; you do not have to go through all possible permutations of the mixed series-parallel connection for the three resistors). Measure the total resistances of three-resistor combinations and compare measured values with the calculated ones (the latter based on known resistor-connection formulas (3) and (4)). Of course, use the measured rather than nominal individual values for these calculations. Save all this data for your report, in which you will show the circuits, the measured values, and the calculations.
Series:
Parallel:
Series - Parallel:
R1 = 151.04 Ω
R1 = 151.04 Ω
R1 = 151.04 Ω
R2 = 1.9867 kΩ
R2 = 1.9867 kΩ
R2 = 1.9867 kΩ
R3 = 5.560 MΩ
R3 = 5.560 MΩ
R3 = 5.560 MΩ
R (theory) = 5.562 MΩ
R (theory) = 140.36 Ω
R (theory) = 151.036 Ω
R (exp.) = 5.594 MΩ
R (exp.) = 140.72 Ω
R (exp.) = 151.399 Ω
𝑅 = 𝑅1 + 𝑅2 + 𝑅3
𝑅
1
=
1 𝑅1
1
1
+𝑅 +𝑅 2
3
1 𝑅
=
1 𝑅1
+𝑅
1
2 +𝑅3
Effective resistance was calculated for the three resistors in series, parallel, and series-parallel. Resistance for series was experimentally found to be 99.4% accurate with respect to the theoretical value, resistance for parallel 99.7% accurate, and resistance of series-parallel 99.8% accurate.
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How does the ohmmeter measure resistance? Internal resistance of an ammeter Assignment 5: Using two DMMs, figure out how a DMM measures resistance; specifically, find what currents flow through the measured resistor on different “Ω” measuring scales and on High and Low settings (if you have High and Low settings on your DMM). At the same time, estimate the internal resistance of an ammeter paying attention to the fact that it may change for different ammeter scales. As an exercise, calculate power dissipated in the measured resistors at the highest measuring currents (this power will be small).
How does the ohmmeter measure resistance?
Ω
2nd DMM in amps function
1-st DMM in ohms function
A Rx
Pair I: R = 2.66 kΩ
Pair II: R = 220 Ω
Pair III: R = 625 Ω
R = 1.99 kΩ
R = 152 Ω
R = 747 Ω
Pair I: I = 0.53 mA
Pair II: I = 0.50 mA
Pair III: I = 0.42 mA
R = 3.668 kΩ
R = 1.217 kΩ
R = 1.217 kΩ
I = 0.53 mA
I = 0.51 mA
I = 0.45 mA
R = 2.994 kΩ
R = 1.148 kΩ
R = 1.148 kΩ
The DMM is based on Ohm’s Law V = IR. To find the resistance of a resistor, the DMM must either apply a known voltage V0 across an unknown resistor and measure the resulting current or it may apply a known current I0 and measure the resulting voltage across the resistor. This was accomplished by connecting another DMM to the arrangement. While one DMM is measuring Rx, the other is measuring the current Ix through the resistor (the value measured for Rx includes the internal resistance of the ammeter Rx = R + rA). Three pairs of resistors were chosen, each
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pair with similar resistivity profiles. Having measured the resistance and current through each individual resistors in their respective pairs, I conclude that our DMM puts out a standard current I0 as the current does not change substantially between the measurements of any two resistors chosen to be paired together. Internal Resistance of an ammeter:
Ω
A
1-st DMM in ohms function
2nd DMM in amps function (scales can be changed)
Fig. 11. Direct measurement of ammeter’s internal resistance Resistance as a function of Ammeter Sensitivity Scale 1800
Internal Resistance (Ω)
1600 1400 1200
994.81
1000 800 600 400 100.551
200
10.45
1.4
0.44
0.02
0.2
2
0 0.0002
0.002
Ammeter Sensitivity (A)
The resistance is much larger at more sensitive scales, falling off quickly on less sensitive scales. The resistance falls off by about one order of magnitude (≈ 10n) at each boundary. Therefore, the ammeter is approximately ideal (zero resistance) only when it is used with less sensitive scales and high currents. As we will need to know the internal resistance of the ammeter at higher current scales later in the laboratory assignments, the value of rA is taken at the 200 mA scale, where rA=1.4 Ω. The highest power dissipated was resistor RX = 2,663 Ω
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𝑃 = 𝐼 2 𝑅 = (0.53𝑥10−6 𝐴)2 (3.668𝑥103 Ω) = 1.03𝑥10−9 𝑊 DC Power Supply Voltages Assignment 6: Connect power supply to the board, measure and record voltages which it produces.
Va (yellow) = 5.035 V Vb (green) = -12.134 V Vc (red) = 12.120 V V (ground) = 0 V
Internal Resistance of a DC Supply and of a Voltmeter Assignment 7: Experimentally find internal resistances of your DC power supply and of the voltmeter (DMM Keithley 179A on a “Volts” function). Explore all available settings and scales on each instrument: when investigating rDC of a DC supply, find it for all voltage settings: + 15 V, – 15 V, and + 5 V; when investigating rV of the voltmeter, explore all available (non-overloading) scales.
It is known that an ideal DC voltage source, such as a battery, should have zero resistance, but this is not the case experimentally. The voltage source does have an internal resistance rV, and it is necessary to find this unknown parameter. Similarly, we will want to establish the internal resistance of a voltmeter. I predict the resistance of the voltmeter will be quite large due to the fact that an ideal voltmeter has infinite resistance.
Internal resistance, r, of our DC voltage source:
A V0
R r
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In the image above, the dotted rectangular box encompasses the DC voltage supply with internal resistance r. An ammeter is also connected in series as well as a resistor R (load). R is chosen to be ≈ 100 Ω in order to be sensitive to r, as we expect r to be quite small. Furthermore, we worry about R being too small to handle the power dissipation as it will now be directly connected to the DC power source. Lastly, we take into account the internal resistance of the ammeter, rA, which was measured in assignment 5. As rA is dependent upon the ammeter sensitivity scale, we will chose a less sensitive higher current scale for 12.12 V/100 Ω ≈ 120 mA. Thus we take the reading of rA on the 200 mA scale which is measured to be rA = 1.4 Ω.
𝐼=
𝑉0 𝑟𝐷𝐶 + 𝑟𝐴 + 𝑅
𝑟𝐷𝐶 =
𝑉0 − 𝑅 − 𝑟𝐴 𝐼
+15 V: At 200 mΩ scale 𝑉 = 12.12 𝑉 𝑟𝐴 = 1.4 Ω 𝑅 = 100 Ω 𝐼 = 119.07 𝑚A
𝑟𝐷𝐶 =
12.12 𝑉 − 100 Ω − 1.4 Ω . 11907 𝐴 𝑟𝐷𝐶 = 0.3888 Ω
-15V: At 200 mΩ scale 𝑉 = −12.134 𝑉 𝑟𝐴 = 1.4 Ω 𝑅 = 100 Ω 𝐼 = −118.45 𝑚A
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𝑟𝐷𝐶 =
−12.134 𝑉 − 100 Ω − 1.4 Ω −.11845 𝐴 𝑟𝐷𝐶 = 1.0398 Ω
+15V: At 200 mΩ scale 𝑉 = 5.035 𝑉 𝑟𝐴 = 1.4 Ω 𝑅 = 100 Ω 𝐼 = 49.07 𝑚A 𝑟𝐷𝐶 =
5.035 𝑉 − 100 Ω − 1.4 Ω 0.04907 𝐴 𝑟𝐷𝐶 = 1.2085 Ω
Internal resistance of a voltmeter:
V
1-st DMM in volts function
2nd DMM in Ohms function
Ω
As discussed previously, the resistance of an ideal voltmeter is infinite. We will experimentally calculate the resistance of our voltmeter by connecting a DMM in Volts scale (V) to another DMM in Ohm’s scale (Ω). 𝑟𝑉 = 9.9370 𝑥 106 Ω 𝑟𝑉 ≈ ∞ Ω (Infinite resistance ideally)
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A Simple Current Source Assignment 8: Build and test a current source with I ≈ 15 μA in which you will be using V0 = 15 V, 0 < RV < 20 kΩ, and which will provide for current stability over the whole range of R V values of better than 3% (i.e. current will not change by more than 3%). Calculate power dissipated in (RV)max and in R0.
A R0
V0
RV
r
The above picture represents a circuit in which we aim to design a current source, or a circuit and/or device that has the ability to supply a constant current to a changing load (variable resistance). The larger dotted rounded square contains the DC power supply, its respective internal resistance, and a resistor R. R is chosen to be larger than RV in order to approximate RV to have negligible resistance in the current equivalence illustrated below. The variable resistor, indicated by a resistor with a diagonal arrow through it, can operate from 0 to RV(max) granted R0 ≫ RV. Although the constant current stability is better for greater values of R0, the maximum constant current is limited by R0 being so large (primary limitation of this design). 𝑅0 ≫ 𝑅𝑉 𝑚𝑎𝑥 𝐼=
𝑉0 𝑉0 ≈ 𝑟 + 𝑅0 + 𝑅𝑉 𝑅0 𝑅0 = 1𝑀Ω
𝑅𝑉 = 20.53 𝑘Ω 𝑉 = 12.12 𝑉 I = 13.209 mA 𝑅0 1,000,000 Ω = ≈ 2% 𝑅𝑉 20,530 Ω
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[Current will not change by more than 2% (≈constant)]
Power dissipated in RV(max) and R0:
Power dissipated in R0 𝑃 = 𝐼 2 𝑅 = 𝐼 2 (𝑟𝐷𝐶 + 𝑅0 ) = 12.908 𝑚𝐴2 (1,000,001 Ω) = 166.66 𝑊𝑎𝑡𝑡𝑠 Power dissipated in RV(max) 𝑃 = 𝐼 2 𝑅 = 𝐼 2 (𝑅𝑉 ) = 12.908 𝑚𝐴2 (20,530 Ω) = 3.42 𝑊𝑎𝑡𝑡𝑠 Simple Voltage Divider Assignment 9: Build a 10:1 voltage divider to divide 15 V from the power supply obtaining 1.5 V at the output to a ± 5% precision. Choose the resistors according to voltage divider formula, while at the same time guaranteeing low-enough power dissipation (absence of resistor heating).
Resistors we chosen in respect to the voltage divider formula, such that a 10:1 voltage divider ratio could be achieved guaranteeing a 12.12 V from the DC power supply to 1.212 V at the output within 5% accuracy. Similarly, resistors we chosen to guarantee low power dissipation across resistors. Resistors should not rise in temperature if low power dissipation is achieved. Vin R1
R1
=
Vin R2
R1 = 47.5 kΩ + 44.5 kΩ = 92.00 kΩ R2 = 9.94 kΩ
Vout
Vout R2
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𝑅1 92.00 𝑘Ω = = 9.25: 1 ≈ 10: 1 𝑅2 9.94 𝑘Ω 𝐼=
𝑉𝑖𝑛 𝑅1 + 𝑅2
𝑉𝑜𝑢𝑡 = 𝐼𝑅2 = 𝑉𝑖𝑛
𝑅2 𝑅1 + 𝑅2
Division Factor: k (≈10) 𝑘=
𝑉𝑖𝑛 𝑅1 = + 1 = 10.25 𝑉𝑜𝑢𝑡 𝑅2
Thevenin Equivalents 10. Thevenin equivalents Assignment 10: Build a circuit shown below; choose any resistors R1, R2, R3, R4 as long as they will not cause problems in terms of power dissipation (the easiest way to guarantee that is to choose them all in the kΩ range): A R1
R2 R3
5V R4
B
Measure individual resistors and the voltage (nominally 5 V) supplied from the power supply. Find Thevenin equivalents for this circuit; do this in two ways: a) experimentally, by using definitions of a Thevenin’s theorem, and b) by performing a calculation. Make sure the two results agree within the precision of your measurements and calculations. In your report draw and label a Thevenin equivalent circuit.
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Experimental Method Thevenin Equivalents (black box): A V
A
B
𝑅𝑇𝐻 =
𝑉𝑇𝐻 𝐼𝑠𝑐
R1 = 44.67 kΩ R2 = 9.96 kΩ R3 = 2.03 kΩ R4 = 1.23 kΩ 𝐼𝑠𝑐 = 0.0252 𝑚𝐴 𝑉𝑇𝐻 = 0.3397 𝑉 𝑅𝑇𝐻 =
0.3397 𝑉 = 13.480 𝑘Ω 0.0000252 𝐴
Theoretical Method Thevenin’s Equivalents (4 resistors) 𝑉𝑇𝐻 =
𝑉1 (𝑅3 + 𝑅4 ) 5 𝑉(2.03 kΩ + 1.23 kΩ) = (𝑅3 + 𝑅4 ) + 𝑅1 (2.03 kΩ + 1.23 kΩ) + 44.67 kΩ 𝑉𝑇𝐻 = 0.3400 𝑉 𝑅𝑇𝐻 = 𝑅2 + [(𝑅3 + 𝑅4 ) ∥ 𝑅1 ]
𝑅𝑇𝐻
−1 1 1 = 9.96 kΩ + ( + ) = 12.990 kΩ 3.26 kΩ 44.67 kΩ
ISC
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Theoretical Method Thevenin’s Equivalents General (no black box):
R1
Vin
R1 A
A
Vin
R2 Vout = VTH
R2
B
B
𝐼=
𝑉𝑖𝑛 𝑅1 + 𝑅2
𝑉𝑜𝑢𝑡 = 𝐼𝑅2 = 𝑉𝑖𝑛
𝐼𝑠𝑐 =
𝑅2 𝑅1 + 𝑅2
𝑉𝑖𝑛 𝑅1
𝑉𝑇𝐻 = 𝑉𝑜𝑢𝑡 =
𝑅𝑇𝐻 =
ISC
A
𝑉𝑖𝑛 𝑅2 𝑅1 + 𝑅2
𝑉𝑇𝐻 𝑉𝑖𝑛 𝑅1 𝑅2 𝑅1 𝑅2 = = 𝐼𝑠𝑐 𝑉𝑖𝑛 (𝑅1 + 𝑅2 ) 𝑅1 + 𝑅2
1 1 1 𝑅1 + 𝑅2 = + = 𝑅𝑇𝐻 𝑅1 𝑅2 𝑅1 𝑅2
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Although simple circuits can be analyzed using Ohm’s law, more complex circuits require the use of Kirchhoff’s laws and Thevenin’s equivalents. Kirchhoff’s laws state that in any connection of multiple wires (junction) the sum of all currents is zero and in any closed loop of a circuit, the sum of all voltage drops is equal to zero. Thevenin’s Theorem postulates that any circuit of arbitrary complexity with two terminals containing resistors, voltage sources, and current sources can be substituted with a Thevenin’s equivalent circuit with one equivalent resistor RTH and one equivalent voltage source VTH. If one wishes to find the equivalent resistance RTH of a black box schematic (intricacies of circuit design are unknown other than the few requirements for Thevenin’s Theorem previously mentioned), he/she can use a voltmeter to measure the voltage across the two terminals represented by VTH. Similarly, he/she can use an ammeter to measure short-circuit current through the circuit represented by Isc. RTH = VTH / Isc. A theoretical method to approaching non-black box Thevenin’s equivalents is represented above. Thevenin’s Analysis of a Voltage Divider with a load Assignment 11: Make a simple 1:2 voltage divider with two 10 k resistors, apply 15 V to it (measure the actual resistor values and input and output voltages). Calculate VTH and RTH for this circuit. Then, as you have done above in #10, experimentally find VTH and RTH by performing measurements according to Thevenin's definitions. Compare calculated and measured values. They should agree to a precision of your measurements and calculations. Record them for the Report. Now attach three different load resistors RL to the output of this divider; first, RL 10 RTH, then RL RTH (comparable), and finally RL 0.1 RTH (measure all resistors, including the load). Show these three cases on labeled Thevenin equivalent circuits. Measure voltages across these loads and calculate expected voltages based on Thevenin equivalent circuits. You should achieve an agreement between measured and theoretical voltages.
The first voltage divider (1:2) created using two 10 Ω resistors allowed the calculation of VTH and RTH, a Thevenin’s equivalent circuit. If a load resistor is then attached to the Thevenin’s equivalent circuit, VTH of the original equivalent circuit acts as Vin and RTH acts as R1 and the load resistor acts as R2. We can then calculate the VTH’ or Vout’ of the new Thevenin’s equivalent circuit. Using this formulation, we derived an equation to reconcile expected voltages based on the new Thevenin’s equivalent circuit. An agreement between experimental and theoretical voltages was achieved for RL ≈ 10RTH, RL ≈ RTH, and RL ≈ 0.1RTH.
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VTH
RTH = R1R2 /(R1 + R2)
Vout′ RL
R1 = 9.947 kΩ R2 = 9.944 kΩ Vin = 12.12 V Vout = 6.069 V
Theoretical:
𝑉𝑇𝐻 =
𝑉𝑖𝑛 𝑅2 = 6.059 𝑉 𝑅1 + 𝑅2
𝑅𝑇𝐻 =
𝑅1 𝑅2 = 4.974 𝑘Ω 𝑅1 + 𝑅2
𝐼𝑠𝑐 =
𝑉𝑇𝐻 = 1.21 𝑚𝐴 𝑅𝑇𝐻
Experimental: 𝑉𝑇𝐻 = 6.069 𝑉 𝐼𝑠𝑐 = 1.19 𝑚𝐴
𝑅𝑇𝐻 =
𝑉𝑇𝐻 = 5.087 𝑘Ω 𝐼𝑠𝑐
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Thevenin Analysis of a voltage divider with a load: RL ≈ 10RTH ≈ 50 kΩ = 51.65 kΩ RL ≈ RTH ≈ 5 kΩ = 4.74 kΩ RL ≈ 0.1RTH ≈ 500 Ω = 486.7 Ω
𝑉𝑇𝐻′ =
𝑉𝑇𝐻 𝑅𝐿 𝑅𝑇𝐻 + 𝑅𝐿
Experimental: VTH (RL1) = 5.537 V VTH (RL2) = 2.961 V VTH (RL1) = 0.541 V
Theoretical: VTH’ (RL1) = 5.524 V VTH’ (RL2) = 2.927 V VTH’ (RL3) = 0.529 V
Conclusion: Beyond the layout of a breadboard, measurements of resistance of individual resistors and networks of resistors in series and parallel, and measurements of internal resistances of a DC power supply, voltmeter, and ammeter, things start to get interesting when we start building simple current sources (device that provides constant current to a variable resistor or changing load to high precision) and simple voltage dividers (device which steps down voltage to a fraction of the input voltage) which sets the basis for Thevenin’s equivalence circuits. When designing a simple current source it is important to remember that the current stability will be better for larger values of R0, though the size of R0 limits the maximum constant current available to the circuit. Similarly, most of the power is dissipated in the larger resistor and only a small portion is used by RV. In many circumstances, a voltage divider is necessary to step down
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some voltage to a fraction so that it can be connected to a resistor (load) further in the circuit. This introduces Thevenin’s Theorem, which postulates that any circuit consisting of resistances, voltage sources, and current sources can be replaced using Ohm’s Law and some algebra to obtain a circuit containing only a single equivalent resistance RTH and single equivalent voltage source VTH. This can be approached experimentally using the black box Isc method or theoretically. This method of equivalence is furthered when one introduces a load after a voltage divider where the VTH for the previous circuit becomes Vin for the new circuit, and RTH for the previous circuit becomes R1 of the new circuit with the load resistor now being R2 in the new circuit. If the voltage of the new circuit is divided, the same rules for RTH and VTH hold as RTH old circuit and R2 new circuit would be connected in parallel. Our measurements of voltage across the load resistors using the previous Thevenin’s equivalent circuit was verified both experimentally and theoretically to high accuracy (≈ 99%).
PHY 335 Unit 1 Instruments, measurements, DC circuits, Thevenin equivalents 09/18/2014 Michael Santana (Justin Thomas) Professor Michael Gurvitch
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Introduction: In Unit I, Instruments, measurements, DC circuits, Thevenin equivalents aims to better our understanding of electronics, and to highlight the techniques and instruments used in laboratory. Electricity, the engine behind electronics, is introduced by use of an analogy to water. Water dynamics or flow through a pipe is very similar to the flow characteristics of electricity. The flow of water through a cross-sectional area of the pipe (volumetric flow rate Volume/second) is analogous to current (electric charge flowing through a cross-sectional area of a wire per second) and is in units of Coulombs/second or Amps. Voltage (electrical potential) is the potential energy per charge in the surroundings of other charges divided by that charge and is in units of Joule/Coulomb or otherwise known as Volts. Voltage is analogous to the pressure difference between two sections of a pipe which give the potential to produce flow. Therefore, when talking of voltage and current, voltage is always measured across an object (resistor, etc.) whereas current is measured through an object (resistor, circuit, etc.) and not vise-versa. Due to the fact that water is approximately incompressible, it is assumed that like water electricity proceeds on its path at the same rate or current. Lastly, resistivity is a function of length and cross-sectional area and has the property of resisting flow. In the water analogy, a pressure differential is applied to opposite ends of the section of gravel (resistance), which induces flow from high pressure to low pressure. Furthermore, through the completion of assignments, we will come to better understand networks of resistors (series, parallel, and permutations of series and parallel), ways in which an ideal voltmeter and ammeter produce readings, Kirchhoff’s rules of algebraic sums of current in junctions and sums of voltage drops around closed loops, and how these methods can help us in better understanding circuit theory and analysis. In our assignments, we will learn the intricacies of a breadboard and its benefits in circuit design, how to measure resistances of standard resistors, variable resistors, and networks of resistors, how to account for instrumental uncertainties and random errors as well as how to mitigate systematic errors, techniques to measure internal resistances of the ammeter, DC voltage supply source, and voltmeter, how to build a simple current source and simple voltage divider, and the theory behind Thevenin’s equivalent circuits and their benefits.
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Equipment and Experimental Techniques: Keithley 177 MICROVOLT Digital Multi-Meter (DMM): Ω (350 V Peak), V (1 kV Max), A (2 A Max) It has six scales of precision for measurements of Ohms, Amps, and Volts (AC and DC). The DMM’s maximum instrumental accuracy is five digits. The DMM has internal resistance that needs to be accounted for when taking measurements as illustrated in assignment 5. It is also important to note that the internal resistance of the DMM on various settings (Ohms, Amps, and Volts) is dependent upon the scale of precision used. For example: when measuring internal resistance of an ammeter, it is necessary to choose a scale that is suitable for the desired function. If there will be high current supplied, we may want to choose a less sensitive scale. Keithley 195A Digital Multi-Meter (DMM): Ω (350 V Peak), V (1 kV Max), A (2 A Max). It has variable scales for precision, though it also has an auto adjust setting which automatically adjusts the scale for best precision and measurement. The DMM’s maximum accuracy is five digits.
Experimental: The Board Assignment 1: Get familiarized with the board layout using an ohmmeter (DMM on an ohms “Ω” function) to see the contact arrangement. Sketch the board layout in your lab book.
Displayed in the above image is a standard solderless breadboard similar to that used in PHY 335 Laboratory. The blue and red horizontal lines illustrate the power rails (bus strips) one to supply voltage and one ground. The vertical columns that are numbered (1-57) are connected vertically up until the middle divider, and not past that. Horizontal rows that are labeled by letters (A-J) are not connected. The middle divider is spaced out in such a way as to provide connectivity for
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integrated circuits. The colored circles on the right hand side (grey, blue, green, and red) are the power supply from 0 V (ground), +15 V, -15 V, and +5 V from top down.
Measuring Individual Resistors Assignment 2: measure three different resistors; measure a resistor in a MΩ range; measure your own resistance between your hands
≈10 Ω:
R1 = 151.17 Ω
≈1 kΩ:
R2 = 1.98 kΩ
≈100 kΩ:
R3 = 5.57 MΩ Resistance (wet fingers) = 0.75 MΩ Resistance (dry fingers) = 1.25 MΩ
Resistance of Body + Resistance of Resistor in Parallel: Resistor without contact: R = 5.57 kΩ Resistor with contact (theoretical): 1 1 1 = + 𝑅 𝑅1 𝑅2 1 1 1 = + 𝑅 1.25 𝑀Ω 5.57 𝑘Ω R = 1.02 MΩ
Resistor with contact (experimental): R = 0.35 MΩ
It was noted that when we measured the individual resistors and our hands made contact with the metal parts of the alligator clips, the resistance went down to R = 0.35 Ω. The experimental value was 0.63 Ω less than the theoretical value calculated, though the difference is believed to have come from our hands having still been damp and less resistive (dry hand resistance used in the calculation). When the effective resistance is calculated using Resistance (wet fingers) = 0.75
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MΩ, the theoretical value is significantly more accurate. It was also noted that effective resistance significantly degreases when holding the resistors tighter or by increases contact area.
Measurement Uncertainty Assignment 3: Evaluate and report measurement uncertainties for the three resistors you measured in #2.
Instrumental Error: R1 = 151.16 Ω
R2 = 1.9854 kΩ
R3 = 5.584 MΩ
R1 = 151.18 Ω
R2 = 1.9855 kΩ
R3 = 5.586 MΩ
R1 = 151.17 ± 0.01 Ω
R2 = 1.9855 ± 0.001 kΩ
R3 = 5.585 ± 0.001 MΩ
Random Uncertainty: R1 = 151.04 Ω
R2 = 1.9873 kΩ
R3 = 5.564 MΩ
R1 = 151.07 Ω
R2 = 1.9875 kΩ
R3 = 5.554 MΩ
R1 = 151.01 Ω
R2 = 1.9867 kΩ
R3 = 5.560 MΩ
R1 = 151.04 Ω
R2 = 1.9861 kΩ
R3 = 5.565 MΩ
R1 = 151.02 Ω
R2 = 1.9858 kΩ
R3 = 5.558 MΩ
R1 = 151.04 ± 0.03 Ω
R2 = 1.9867 ± 0.0009 kΩ
R3 = 5.560 ± 0.006 MΩ
Both instrumental and random uncertainties were accounted for and calculated for the three resistors used in our earlier analysis in Part II. The instrumental uncertainty is determined from fluctuating digits on our DMM. If no fluctuations are present, we approximate one unit of uncertainty on the last digit. As random uncertainty cannot be accounted for in one measurement, multiple identical measurements must be taken and averaged for the best number. The mean and standard deviation of the distribution are considered to be the best number and random error respectively, though we will not calculated either in this course. We do not account for systematic error. If there were to be some refinement necessary, it would have been taken care of prior to recording data and performing data analysis.
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Measuring Resistor Networks Assignment 4: Take the three resistors you have measured individually and connect them in series, then in parallel and finally in a series-parallel connection (you will need only three connections in total; you do not have to go through all possible permutations of the mixed series-parallel connection for the three resistors). Measure the total resistances of three-resistor combinations and compare measured values with the calculated ones (the latter based on known resistor-connection formulas (3) and (4)). Of course, use the measured rather than nominal individual values for these calculations. Save all this data for your report, in which you will show the circuits, the measured values, and the calculations.
Series:
Parallel:
Series - Parallel:
R1 = 151.04 Ω
R1 = 151.04 Ω
R1 = 151.04 Ω
R2 = 1.9867 kΩ
R2 = 1.9867 kΩ
R2 = 1.9867 kΩ
R3 = 5.560 MΩ
R3 = 5.560 MΩ
R3 = 5.560 MΩ
R (theory) = 5.562 MΩ
R (theory) = 140.36 Ω
R (theory) = 151.036 Ω
R (exp.) = 5.594 MΩ
R (exp.) = 140.72 Ω
R (exp.) = 151.399 Ω
𝑅 = 𝑅1 + 𝑅2 + 𝑅3
𝑅
1
=
1 𝑅1
1
1
+𝑅 +𝑅 2
3
1 𝑅
=
1 𝑅1
+𝑅
1
2 +𝑅3
Effective resistance was calculated for the three resistors in series, parallel, and series-parallel. Resistance for series was experimentally found to be 99.4% accurate with respect to the theoretical value, resistance for parallel 99.7% accurate, and resistance of series-parallel 99.8% accurate.
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How does the ohmmeter measure resistance? Internal resistance of an ammeter Assignment 5: Using two DMMs, figure out how a DMM measures resistance; specifically, find what currents flow through the measured resistor on different “Ω” measuring scales and on High and Low settings (if you have High and Low settings on your DMM). At the same time, estimate the internal resistance of an ammeter paying attention to the fact that it may change for different ammeter scales. As an exercise, calculate power dissipated in the measured resistors at the highest measuring currents (this power will be small).
How does the ohmmeter measure resistance?
Ω
2nd DMM in amps function
1-st DMM in ohms function
A Rx
Pair I: R = 2.66 kΩ
Pair II: R = 220 Ω
Pair III: R = 625 Ω
R = 1.99 kΩ
R = 152 Ω
R = 747 Ω
Pair I: I = 0.53 mA
Pair II: I = 0.50 mA
Pair III: I = 0.42 mA
R = 3.668 kΩ
R = 1.217 kΩ
R = 1.217 kΩ
I = 0.53 mA
I = 0.51 mA
I = 0.45 mA
R = 2.994 kΩ
R = 1.148 kΩ
R = 1.148 kΩ
The DMM is based on Ohm’s Law V = IR. To find the resistance of a resistor, the DMM must either apply a known voltage V0 across an unknown resistor and measure the resulting current or it may apply a known current I0 and measure the resulting voltage across the resistor. This was accomplished by connecting another DMM to the arrangement. While one DMM is measuring Rx, the other is measuring the current Ix through the resistor (the value measured for Rx includes the internal resistance of the ammeter Rx = R + rA). Three pairs of resistors were chosen, each
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pair with similar resistivity profiles. Having measured the resistance and current through each individual resistors in their respective pairs, I conclude that our DMM puts out a standard current I0 as the current does not change substantially between the measurements of any two resistors chosen to be paired together. Internal Resistance of an ammeter:
Ω
A
1-st DMM in ohms function
2nd DMM in amps function (scales can be changed)
Fig. 11. Direct measurement of ammeter’s internal resistance Resistance as a function of Ammeter Sensitivity Scale 1800
Internal Resistance (Ω)
1600 1400 1200
994.81
1000 800 600 400 100.551
200
10.45
1.4
0.44
0.02
0.2
2
0 0.0002
0.002
Ammeter Sensitivity (A)
The resistance is much larger at more sensitive scales, falling off quickly on less sensitive scales. The resistance falls off by about one order of magnitude (≈ 10n) at each boundary. Therefore, the ammeter is approximately ideal (zero resistance) only when it is used with less sensitive scales and high currents. As we will need to know the internal resistance of the ammeter at higher current scales later in the laboratory assignments, the value of rA is taken at the 200 mA scale, where rA=1.4 Ω. The highest power dissipated was resistor RX = 2,663 Ω
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𝑃 = 𝐼 2 𝑅 = (0.53𝑥10−6 𝐴)2 (3.668𝑥103 Ω) = 1.03𝑥10−9 𝑊 DC Power Supply Voltages Assignment 6: Connect power supply to the board, measure and record voltages which it produces.
Va (yellow) = 5.035 V Vb (green) = -12.134 V Vc (red) = 12.120 V V (ground) = 0 V
Internal Resistance of a DC Supply and of a Voltmeter Assignment 7: Experimentally find internal resistances of your DC power supply and of the voltmeter (DMM Keithley 179A on a “Volts” function). Explore all available settings and scales on each instrument: when investigating rDC of a DC supply, find it for all voltage settings: + 15 V, – 15 V, and + 5 V; when investigating rV of the voltmeter, explore all available (non-overloading) scales.
It is known that an ideal DC voltage source, such as a battery, should have zero resistance, but this is not the case experimentally. The voltage source does have an internal resistance rV, and it is necessary to find this unknown parameter. Similarly, we will want to establish the internal resistance of a voltmeter. I predict the resistance of the voltmeter will be quite large due to the fact that an ideal voltmeter has infinite resistance.
Internal resistance, r, of our DC voltage source:
A V0
R r
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In the image above, the dotted rectangular box encompasses the DC voltage supply with internal resistance r. An ammeter is also connected in series as well as a resistor R (load). R is chosen to be ≈ 100 Ω in order to be sensitive to r, as we expect r to be quite small. Furthermore, we worry about R being too small to handle the power dissipation as it will now be directly connected to the DC power source. Lastly, we take into account the internal resistance of the ammeter, rA, which was measured in assignment 5. As rA is dependent upon the ammeter sensitivity scale, we will chose a less sensitive higher current scale for 12.12 V/100 Ω ≈ 120 mA. Thus we take the reading of rA on the 200 mA scale which is measured to be rA = 1.4 Ω.
𝐼=
𝑉0 𝑟𝐷𝐶 + 𝑟𝐴 + 𝑅
𝑟𝐷𝐶 =
𝑉0 − 𝑅 − 𝑟𝐴 𝐼
+15 V: At 200 mΩ scale 𝑉 = 12.12 𝑉 𝑟𝐴 = 1.4 Ω 𝑅 = 100 Ω 𝐼 = 119.07 𝑚A
𝑟𝐷𝐶 =
12.12 𝑉 − 100 Ω − 1.4 Ω . 11907 𝐴 𝑟𝐷𝐶 = 0.3888 Ω
-15V: At 200 mΩ scale 𝑉 = −12.134 𝑉 𝑟𝐴 = 1.4 Ω 𝑅 = 100 Ω 𝐼 = −118.45 𝑚A
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𝑟𝐷𝐶 =
−12.134 𝑉 − 100 Ω − 1.4 Ω −.11845 𝐴 𝑟𝐷𝐶 = 1.0398 Ω
+15V: At 200 mΩ scale 𝑉 = 5.035 𝑉 𝑟𝐴 = 1.4 Ω 𝑅 = 100 Ω 𝐼 = 49.07 𝑚A 𝑟𝐷𝐶 =
5.035 𝑉 − 100 Ω − 1.4 Ω 0.04907 𝐴 𝑟𝐷𝐶 = 1.2085 Ω
Internal resistance of a voltmeter:
V
1-st DMM in volts function
2nd DMM in Ohms function
Ω
As discussed previously, the resistance of an ideal voltmeter is infinite. We will experimentally calculate the resistance of our voltmeter by connecting a DMM in Volts scale (V) to another DMM in Ohm’s scale (Ω). 𝑟𝑉 = 9.9370 𝑥 106 Ω 𝑟𝑉 ≈ ∞ Ω (Infinite resistance ideally)
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A Simple Current Source Assignment 8: Build and test a current source with I ≈ 15 μA in which you will be using V0 = 15 V, 0 < RV < 20 kΩ, and which will provide for current stability over the whole range of R V values of better than 3% (i.e. current will not change by more than 3%). Calculate power dissipated in (RV)max and in R0.
A R0
V0
RV
r
The above picture represents a circuit in which we aim to design a current source, or a circuit and/or device that has the ability to supply a constant current to a changing load (variable resistance). The larger dotted rounded square contains the DC power supply, its respective internal resistance, and a resistor R. R is chosen to be larger than RV in order to approximate RV to have negligible resistance in the current equivalence illustrated below. The variable resistor, indicated by a resistor with a diagonal arrow through it, can operate from 0 to RV(max) granted R0 ≫ RV. Although the constant current stability is better for greater values of R0, the maximum constant current is limited by R0 being so large (primary limitation of this design). 𝑅0 ≫ 𝑅𝑉 𝑚𝑎𝑥 𝐼=
𝑉0 𝑉0 ≈ 𝑟 + 𝑅0 + 𝑅𝑉 𝑅0 𝑅0 = 1𝑀Ω
𝑅𝑉 = 20.53 𝑘Ω 𝑉 = 12.12 𝑉 I = 13.209 mA 𝑅0 1,000,000 Ω = ≈ 2% 𝑅𝑉 20,530 Ω
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[Current will not change by more than 2% (≈constant)]
Power dissipated in RV(max) and R0:
Power dissipated in R0 𝑃 = 𝐼 2 𝑅 = 𝐼 2 (𝑟𝐷𝐶 + 𝑅0 ) = 12.908 𝑚𝐴2 (1,000,001 Ω) = 166.66 𝑊𝑎𝑡𝑡𝑠 Power dissipated in RV(max) 𝑃 = 𝐼 2 𝑅 = 𝐼 2 (𝑅𝑉 ) = 12.908 𝑚𝐴2 (20,530 Ω) = 3.42 𝑊𝑎𝑡𝑡𝑠 Simple Voltage Divider Assignment 9: Build a 10:1 voltage divider to divide 15 V from the power supply obtaining 1.5 V at the output to a ± 5% precision. Choose the resistors according to voltage divider formula, while at the same time guaranteeing low-enough power dissipation (absence of resistor heating).
Resistors we chosen in respect to the voltage divider formula, such that a 10:1 voltage divider ratio could be achieved guaranteeing a 12.12 V from the DC power supply to 1.212 V at the output within 5% accuracy. Similarly, resistors we chosen to guarantee low power dissipation across resistors. Resistors should not rise in temperature if low power dissipation is achieved. Vin R1
R1
=
Vin R2
R1 = 47.5 kΩ + 44.5 kΩ = 92.00 kΩ R2 = 9.94 kΩ
Vout
Vout R2
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𝑅1 92.00 𝑘Ω = = 9.25: 1 ≈ 10: 1 𝑅2 9.94 𝑘Ω 𝐼=
𝑉𝑖𝑛 𝑅1 + 𝑅2
𝑉𝑜𝑢𝑡 = 𝐼𝑅2 = 𝑉𝑖𝑛
𝑅2 𝑅1 + 𝑅2
Division Factor: k (≈10) 𝑘=
𝑉𝑖𝑛 𝑅1 = + 1 = 10.25 𝑉𝑜𝑢𝑡 𝑅2
Thevenin Equivalents 10. Thevenin equivalents Assignment 10: Build a circuit shown below; choose any resistors R1, R2, R3, R4 as long as they will not cause problems in terms of power dissipation (the easiest way to guarantee that is to choose them all in the kΩ range): A R1
R2 R3
5V R4
B
Measure individual resistors and the voltage (nominally 5 V) supplied from the power supply. Find Thevenin equivalents for this circuit; do this in two ways: a) experimentally, by using definitions of a Thevenin’s theorem, and b) by performing a calculation. Make sure the two results agree within the precision of your measurements and calculations. In your report draw and label a Thevenin equivalent circuit.
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Experimental Method Thevenin Equivalents (black box): A V
A
B
𝑅𝑇𝐻 =
𝑉𝑇𝐻 𝐼𝑠𝑐
R1 = 44.67 kΩ R2 = 9.96 kΩ R3 = 2.03 kΩ R4 = 1.23 kΩ 𝐼𝑠𝑐 = 0.0252 𝑚𝐴 𝑉𝑇𝐻 = 0.3397 𝑉 𝑅𝑇𝐻 =
0.3397 𝑉 = 13.480 𝑘Ω 0.0000252 𝐴
Theoretical Method Thevenin’s Equivalents (4 resistors) 𝑉𝑇𝐻 =
𝑉1 (𝑅3 + 𝑅4 ) 5 𝑉(2.03 kΩ + 1.23 kΩ) = (𝑅3 + 𝑅4 ) + 𝑅1 (2.03 kΩ + 1.23 kΩ) + 44.67 kΩ 𝑉𝑇𝐻 = 0.3400 𝑉 𝑅𝑇𝐻 = 𝑅2 + [(𝑅3 + 𝑅4 ) ∥ 𝑅1 ]
𝑅𝑇𝐻
−1 1 1 = 9.96 kΩ + ( + ) = 12.990 kΩ 3.26 kΩ 44.67 kΩ
ISC
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Theoretical Method Thevenin’s Equivalents General (no black box):
R1
Vin
R1 A
A
Vin
R2 Vout = VTH
R2
B
B
𝐼=
𝑉𝑖𝑛 𝑅1 + 𝑅2
𝑉𝑜𝑢𝑡 = 𝐼𝑅2 = 𝑉𝑖𝑛
𝐼𝑠𝑐 =
𝑅2 𝑅1 + 𝑅2
𝑉𝑖𝑛 𝑅1
𝑉𝑇𝐻 = 𝑉𝑜𝑢𝑡 =
𝑅𝑇𝐻 =
ISC
A
𝑉𝑖𝑛 𝑅2 𝑅1 + 𝑅2
𝑉𝑇𝐻 𝑉𝑖𝑛 𝑅1 𝑅2 𝑅1 𝑅2 = = 𝐼𝑠𝑐 𝑉𝑖𝑛 (𝑅1 + 𝑅2 ) 𝑅1 + 𝑅2
1 1 1 𝑅1 + 𝑅2 = + = 𝑅𝑇𝐻 𝑅1 𝑅2 𝑅1 𝑅2
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Although simple circuits can be analyzed using Ohm’s law, more complex circuits require the use of Kirchhoff’s laws and Thevenin’s equivalents. Kirchhoff’s laws state that in any connection of multiple wires (junction) the sum of all currents is zero and in any closed loop of a circuit, the sum of all voltage drops is equal to zero. Thevenin’s Theorem postulates that any circuit of arbitrary complexity with two terminals containing resistors, voltage sources, and current sources can be substituted with a Thevenin’s equivalent circuit with one equivalent resistor RTH and one equivalent voltage source VTH. If one wishes to find the equivalent resistance RTH of a black box schematic (intricacies of circuit design are unknown other than the few requirements for Thevenin’s Theorem previously mentioned), he/she can use a voltmeter to measure the voltage across the two terminals represented by VTH. Similarly, he/she can use an ammeter to measure short-circuit current through the circuit represented by Isc. RTH = VTH / Isc. A theoretical method to approaching non-black box Thevenin’s equivalents is represented above. Thevenin’s Analysis of a Voltage Divider with a load Assignment 11: Make a simple 1:2 voltage divider with two 10 k resistors, apply 15 V to it (measure the actual resistor values and input and output voltages). Calculate VTH and RTH for this circuit. Then, as you have done above in #10, experimentally find VTH and RTH by performing measurements according to Thevenin's definitions. Compare calculated and measured values. They should agree to a precision of your measurements and calculations. Record them for the Report. Now attach three different load resistors RL to the output of this divider; first, RL 10 RTH, then RL RTH (comparable), and finally RL 0.1 RTH (measure all resistors, including the load). Show these three cases on labeled Thevenin equivalent circuits. Measure voltages across these loads and calculate expected voltages based on Thevenin equivalent circuits. You should achieve an agreement between measured and theoretical voltages.
The first voltage divider (1:2) created using two 10 Ω resistors allowed the calculation of VTH and RTH, a Thevenin’s equivalent circuit. If a load resistor is then attached to the Thevenin’s equivalent circuit, VTH of the original equivalent circuit acts as Vin and RTH acts as R1 and the load resistor acts as R2. We can then calculate the VTH’ or Vout’ of the new Thevenin’s equivalent circuit. Using this formulation, we derived an equation to reconcile expected voltages based on the new Thevenin’s equivalent circuit. An agreement between experimental and theoretical voltages was achieved for RL ≈ 10RTH, RL ≈ RTH, and RL ≈ 0.1RTH.
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VTH
RTH = R1R2 /(R1 + R2)
Vout′ RL
R1 = 9.947 kΩ R2 = 9.944 kΩ Vin = 12.12 V Vout = 6.069 V
Theoretical:
𝑉𝑇𝐻 =
𝑉𝑖𝑛 𝑅2 = 6.059 𝑉 𝑅1 + 𝑅2
𝑅𝑇𝐻 =
𝑅1 𝑅2 = 4.974 𝑘Ω 𝑅1 + 𝑅2
𝐼𝑠𝑐 =
𝑉𝑇𝐻 = 1.21 𝑚𝐴 𝑅𝑇𝐻
Experimental: 𝑉𝑇𝐻 = 6.069 𝑉 𝐼𝑠𝑐 = 1.19 𝑚𝐴
𝑅𝑇𝐻 =
𝑉𝑇𝐻 = 5.087 𝑘Ω 𝐼𝑠𝑐
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Thevenin Analysis of a voltage divider with a load: RL ≈ 10RTH ≈ 50 kΩ = 51.65 kΩ RL ≈ RTH ≈ 5 kΩ = 4.74 kΩ RL ≈ 0.1RTH ≈ 500 Ω = 486.7 Ω
𝑉𝑇𝐻′ =
𝑉𝑇𝐻 𝑅𝐿 𝑅𝑇𝐻 + 𝑅𝐿
Experimental: VTH (RL1) = 5.537 V VTH (RL2) = 2.961 V VTH (RL1) = 0.541 V
Theoretical: VTH’ (RL1) = 5.524 V VTH’ (RL2) = 2.927 V VTH’ (RL3) = 0.529 V
Conclusion: Beyond the layout of a breadboard, measurements of resistance of individual resistors and networks of resistors in series and parallel, and measurements of internal resistances of a DC power supply, voltmeter, and ammeter, things start to get interesting when we start building simple current sources (device that provides constant current to a variable resistor or changing load to high precision) and simple voltage dividers (device which steps down voltage to a fraction of the input voltage) which sets the basis for Thevenin’s equivalence circuits. When designing a simple current source it is important to remember that the current stability will be better for larger values of R0, though the size of R0 limits the maximum constant current available to the circuit. Similarly, most of the power is dissipated in the larger resistor and only a small portion is used by RV. In many circumstances, a voltage divider is necessary to step down
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some voltage to a fraction so that it can be connected to a resistor (load) further in the circuit. This introduces Thevenin’s Theorem, which postulates that any circuit consisting of resistances, voltage sources, and current sources can be replaced using Ohm’s Law and some algebra to obtain a circuit containing only a single equivalent resistance RTH and single equivalent voltage source VTH. This can be approached experimentally using the black box Isc method or theoretically. This method of equivalence is furthered when one introduces a load after a voltage divider where the VTH for the previous circuit becomes Vin for the new circuit, and RTH for the previous circuit becomes R1 of the new circuit with the load resistor now being R2 in the new circuit. If the voltage of the new circuit is divided, the same rules for RTH and VTH hold as RTH old circuit and R2 new circuit would be connected in parallel. Our measurements of voltage across the load resistors using the previous Thevenin’s equivalent circuit was verified both experimentally and theoretically to high accuracy (≈ 99%).
