EXP. 1.2 - DC CIRCUITS - THEVENIN EQUIVALENT CIRCUIT
Before you come to class
1. Read carefully the Horowitz/Bill references for this experiment.
2. Read carefully these experiment inssssttions.
Introduction
Linear circuits are those that contain only linear elements. Active
elements are those having a source of power within them, as opposed to
passive elements that contain no sources of power. Common linear pas
sive elements are resistors, capacitors and inductors. The common
flashlight battery or a regulated power supply are linear active ele
ments. Circuits containing linear elements can be put into simpler form
for numerical circuit analysis by using well-known theorems for series
and parallel passive elements, and by using Thevenin's theorem and/or
Norton's theorem for active circuits. Electronic power supplies and
function generators are examples of linear active circuits.
In this experiment, we shall study the function generator as an ex
ample of an active circuit.
A linear active circuit (assumed to have only 2 terminals) is defined
as any circuit whose terminal voltage, V, is a linear function of the
current, I, that flows out of one terminal and into the other. The cir
cuit may be either a DC, or an AC, circuit. In the next experiment we
study AC circuits in detail. The function generator delivers AC voltage
and current (at fairly low power), but it has been designed not to dis
play any internal capacitive or inductive effects in its output charac
teristics, usually found in AC circuits.
Thevenin's Theorem
A simple circuit consisting of a constant amplitude voltage, of magni-
tude VTh (the Thevenin voltage), in series with a resistance RTh (the
Thevenin resistance) is the electrical equivalent of any linear active
circuit, provided that:
Voc = the open circuit terminal voltage (output current I
the short circuit terminal current (terminal voltage V
0)
[This current is also called the Norton current in the Norton theorem.]
RTh = /Voc / Iscl = the magnitude of the (-) slope of the V-I graph.
R(Thevenin) Vout
v
LOAD;Slope = m = L1V/L1I=~
10
Current (I)
Figure: Thevenin equivalent circuit and Voltage (V) vs. Current (I) graph.
EXP. 1.2 Thevenin equivalent circuit for function generator
1. Connect the signal generator output to one DO channel input, using a
coaxial cable with dual banana plug terminals. Be sure to connect
the GROUND (cable shield) banana plugs to the COMMON output terminal
of the function generator, and to the GROUND input of the DO.
2. Display a I-kHz sine waveform on the DO. Use 1 V/DIV, on the DO. Use
AC input so that the function generator OFFSET control can be ig-
nored. Display a large number of cycles to permit easy peak-to-peak
voltage measurements.
3. Set up a suitable table for measured, and for calculated, data. Re
cord ALL pertinent data - name, date, station, equipment informa
tion, etc., and measured data.
4. Adj ust the function generator for an 8. 0 V peak-to-peak vol tage.
This is your value of Voc. (The DO input resistance is 1M ohm, which
draws negligible current, for our purposes.) Keep the 8.0 V peak-to
peak output fixed as the frequency is varied during the complete
course of this experiment, by making adjustments of the frequency
generator output.
5. Set the decade resistor to 2,000 ohm. Do not at any time decrease
the decade box resistance below 100 ohm. Then connect it to the
function generator, using a second dual banana plug coaxial cable.
This places the resistor in series wi th the Thevenin equivalent
vol tage and the Thevenin resistance of the function generator. It
keeps the DO as a voltage measuring device, measuring the terminal
voltage of the function generator.
6. Adjust the decade resistor to a value that makes V = 4.0 V, equal to
one-half of the open-circuit voltage. Justify in mathematical form
that this value of the decade resistor equals the Thevenin resis
tance. [This measurement method is a quick way to measure the Theve
nin resistance provided that the circuit is capable of delivering
the current that is required at this load resistance.]
7. To prove that the function generator is a LINEAR device, take data
for a terminal voltage-versus-current graph.
7.1 Suggestion: Take data for peak-to-peak voltages equal to 7.0 to 1.0
volts, in steps of 0.5 V adjusting the decade (load) resistance RLoM
to give each of these values. Use the data to calculate respective
currents I with I=V/RLoM •
7.2 Plot the voltage-current graph, current on the horizontal axis. By
drawing two estimated l-standard deviation separated slopes on your
data, calculate the slope for this graph and the error on the slope.
Determine the Thevenin resistance from the slope.
7.3 Include the Thevenin circuit with your data.
8. Write a summary of your experiment. See end of this section for
guidelines.
Final comment: This experiment shows you the important fact that the
output voltage amplitude of your function generator depends upon how
much current your circuit draws from the generator. For DC circuits, the
current will change only when the load resistance is changed. For AC
circuits, the load impedance often depends upon frequency, and so fre
quency changes, alone, will cause the function generator output current
and output voltage to change. You are, therefore, cautioned to monitor
the function generator output voltage in all experiments.