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EE151_Exp1_V35

Experiment 1- Introduction to Electronic Equipment

Engineers test, measure, and verify…

PURPOSE

To become acquainted with the basic concepts and behavior of electrical voltages, currents, andresistances and to obtain a qualitative understanding of these concepts. To become familiar with basic

electronic measurements and test equipment including the ohmmeter, ammeter, and voltmeter. To

become proficient at proper measurement techniques including voltage across and current through

electronic components.

LAB EQUIPMENT

1 Agilent E3640A DC Power Supply1 Agilent 34401A Digital Multimeter

3 Decade Resistance Boxes (10k )

1 Decade Resistance Box with 0.25A rating (10-100)

STUDENT PROVIDED EQUIPMENT (Bring to Lab – From Student Project Lab Checkout Window)

8 Banana-to-banana leads1 Resistor Box

SECTIONSa. Background information: voltage, current, resistance measurements, voltage and current dividers

b. Instructor introduction/demonstration: power sources and measurement devices, example circuitconnections, spectrum analyzer, logic analyzer

c. Voltage Loop measurements

d. DC voltage supply measurements e. Current Division measurements f. Resistance measurements (optional)

PRE-LAB ASSIGNMENT1. Read the following background Information (Section 1) thoroughly, and review the lab

procedures that follow (Sections 2 and on). 2. See the course Blackboard site for additional preparations.

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Section 1: Background Information

Voltage, and Amperes, and Ohms,…

What is “voltage”, anyway?? What does “current” really represent physically?

You have probably encountered all of these basic electrical terms in your high school physics class.

But, do you really understand what each of these are? Before you dive into this major, it is importantthat you have a clear and intuitive sense of these fundamental electrical phenomena.

The simplest place to begin is with the concept of electrical charge. We know that there are threetypes of electrical charge that particles (like electrons and protons) can have: positive, negative, or

neutral (no net charge). We also know or have been shown enough times to believe that particles with

the same type of charge repel each other (that is they impose a repulsive electrical force on each other

that pushes them away in opposite directions), and that particles with opposite charges attract (imposean electrical force that draws them toward each other). Electrical charge is measured in units of

coulombs (C), where 1 coulomb represents the charge that must exist on each of two identical particles

in a vacuum so that they will repel each other with a force of 10-7 x c2 (c = speed of light = 2.997 x 108

meters/second) Newtons when spaced 1 meter from each other. One coulomb is a HUGE amount of charge. A fundamental quantity of charge that we more often encounter is the charge of an electron or

proton, which amounts to 1.602 x 10-19 C. We assign a negative value to the charge on the electron,

and a positive value to the charge of a proton.

If we set some charged particles in motion along the same path (like electrons flowing through a

copper wire), then we can observe and measure the amount of charge (q) flowing past a cross-

sectional slice of the wire at some point along its path in a given time span (t ). This rate of charge

flow is the electric current (denoted by “I”), which is measured in Amperes (or “amps”) where:

0lim 1 Amp 1 Coulomb moving past per second = 1 C/sec

t

charge dq q I

time dt t

A more familiar analogy to electric current is to think of water flowing through a pipe. In this case, wecould place a meter at the end of the pipe and measure the rate at which the water is flowing out of the

pipe, by measuring how much water (volume) goes through the meter in a given period of time:

3: = cm /secvolume Vol

F volume flowrate Unitstime t

The current in a wire (charge flow rate) is analogous to this measurement of water volume flow rate ina pipe.

Oh, My!!

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By convention, we consider positive charges flowing in a particular direction as representing a

positively-signed current in that direction. This is a little counterintuitive, since we know that many

types of currents we are familiar with are actually caused by the motion of electrons. However, bydefinition, the current caused by these moving electrons will have a negative sign if measured with

respect to the direction of physical motion of the electrons. Thus electrons moving to the right in a wire

cause a negative current in the direction pointing to the right (because negatively-signed charges are

moving toward the right), or a positive current in a leftward direction (because negative charges aremoving in a negative direction with respect to leftward motion). See Fig. 1. Confused yet??

-1912

1 6

1000( 1.602 10 C)= -80.1 10 / sec 80.1 picoamperes

5 10 sec

q I C

t

Fig. 1 Current Flow Direction Convention

Since both the sign and direction of current flow are needed to fully specify a current (and not just the

magnitude alone), we will often use arrows in our circuit diagrams to clearly indicate the assumed

direction of current flow in any wire or branch of an electric circuit. If we determine that the actualvalue of a current is negative when using these assumptions, then we simply have learned that ourassumption was incorrect, and that the actual direction of current flow was in the opposite direction

than our assumption in that particular branch.

Next comes the trickier concept to grasp: voltage. Voltage is a measure of the amount of work thatmust be done to move a charge through some element in a circuit. This implies that it requires energy

to “force” charges through circuit elements. The voltage is a normalized measurement of this energyexpenditure. It is the amount of energy expended (or added, if we are talking about a voltage “source”)per unit of charge moving in an element. Thus, the units of voltage reflect this ratio of energy (Joules)

per unit of charge (Coulomb):

Joules

Coulomb

energyV voltage

charge 1 volt (V) = 1 Joule / Coulomb (J/C)

Voltage (also called electrical potential difference) is measured across an element in a circuit, fromone terminal to the other. Just as the sign and direction of current flow needed to be specified, we

likewise must also indicate the sign of a voltage change across an element. We indicate the terminal

with a higher electrical potential using a (+) sign, and the terminal having a lower potential using a (-)

sign. Using this convention, we can distinguish a passive, energy-absorbing circuit element (alsocalled a load ) as one having a positive current entering the (+) terminal and leaving the (-) terminal.

Conversely, an active circuit element (or power source) supplies energy to force positive charges (and

thus positive current) OUT of the positive (+), higher voltage terminal; and receives current back intoits negatively-signed terminal.

- - - - -

- - - - -I1= - 80 picoamps

Example:

1000 electrons move past you

going to the right

in 5 microseconds

-

I1= + 80 picoamps

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A simple plumbing analogy may again help you understand the concept of voltage and its relationship

to current in a simple circuit. Let’s say you have a closed loop of water pipe with no blockageanywhere in the pipe. (See Fig. 2) You fill the pipe with water, and connect a pump into the loop so

that water can flow out of the top (output) of the pump and return back to the bottom (intake) of the

pump. If the pump is turned off, no water flows. If the pump is turned on, the pump creates a higherpressure at its output than is present at its intake, which causes the water to begin to flow in the loop.

The electrical equivalents to this plumbing circuit would be a closed loop of wire with a power supply

or battery (a voltage source) connected in the loop. The pressure change between any two points in theplumbing circuit is analogous to the voltage or electric potential change between points in the electrical

circuit. The battery or power supply in the electrical circuit does work on the charged particles in theelectric circuit (imparts energy). The ratio of the amount of energy imparted (Joules) divided by the

amount of charge worked on (coulombs) by the power supply gives us the voltage that can bemeasured across its terminals. (1 volt = 1 Joule/coulomb).

3

Joules Joules

cm Coulomb

energy energyP pressure V voltage

volume charge

Again, the volume rate of water flow (cm3 /sec) in the pipes is analogous to the current (charge flow

rate) in the wires of the electrical circuit loop. As the pressure at the output of the water pump isincreased, increasing the pressure differential across the pump, the water flows faster in the pipes.

Likewise, increasing the electrical potential difference (voltage) supplied by the power source across

its terminals will increase the current (charge flow rate) in the circuit.

Fig. 2 Analogy of an Electrical Circuit to a Water (Hydraulic) Circuit

If the wire loop is severed somewhere, this is like cutting and capping off a pipe (not leaving it open toleak out water). In such a case, no current (or no water) can flow in either circuit, no matter how hardthe pump works or no matter how much voltage the source supplies. Thus, only “closed” circuit loops

with no breaks and branches with no dead ends can conduct current in an electrical circuit.

Electrical circuits are generally made up of a combination of circuit elements with at least one power

source (either a voltage source or a current source). Circuit elements can be connected in two basic

topologies – either in series, where they are connected sequentially, in-line, within a single circuit

branch; or in parallel, where they “straddle” each other, with the positive terminals of each connected

HigherPressure

Pump

LowerPressure

HigherVoltage

+

PowerSource

Lower - Voltage

Water

Flow

Current

Active Element(Source) Passive Element(Load)

+

-

I +

V

-

+

-

I +

V

-

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together, and likewise the negative terminals are connected together. Series-connected circuit elements

have the same current flowing through them, while different voltages may appear across each element.

Alternatively, parallel-connected components will have the exact same voltage across them, with thecurrent through each one generally being different (if the elements are different). More complicated

circuits can be formed by mixing multiple elements using both series and parallel combinations.

Eventually (in EE 112) you will learn how to analyze and determine the current flows and voltages

throughout such complicated circuits. For now, we will stick with some simple circuit elementcombinations.

Fig. 3 Circuit Connection Types

Voltage and Current Relationship

Voltage, current, and resistance in an electrical circuit branch are related through Ohm’s Law

V I R (1)

in which V is voltage in units of volts (V), I is current in Amperes (A), and R is resistance in ohms ().

These parameters appear in the locations noted in the circuit schematic and physical diagram of Fig. 4below. The diagram on the left side of Fig. 4, which includes electronic components, is called a

schematic; the equivalent physical diagram appears to the right. The labels V S, I , R, and V R represent

the values of the source voltage, the current produced by the source, the resistor’s value (in ohms), andthe voltage drop across the resistor, respectively. It is important to note that current flows through a

circuit element, while voltage develops or is impressed across an element.

Fig. 4 Resistor Connected Across a Voltage Source

The size of a resistor (its resistance value) is analogous to the cross-sectional area of the pipe carrying

water. A low resistance presents less opposition to current flow; analogous to a large diameter pipe.Alternatively, a high resistance decreases current flow, just as a narrower diameter pipe reduces the

amount of water that can flow through it in a given period of time. Typical values of resistance range

from 1 – 1 M (106 ). A piece of wire should have a resistance value << 1 Sometimes inschematic circuit diagrams, the values of resistance are shown with only the order of magnitude

abbreviation and without the symbol: i.e., 2K for 2 kilohms (2x103 ), 2M for Megohm (2x106 ).

Series Connection

I

+

V2

-

+ V1 -

Parallel Connection

+

V

-

I1 I2

I

VS R

+

-

I

+

VR

-

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Voltage Measurements

To measure the voltage across an element like the resistor R shown above, a voltmeter is connected in

parallel with the element (i.e. the two voltmeter probe leads are connected with one touching each

terminal of the element being measured). It is important to note that all voltage measurements are

relative measurements; that is they indicate the change in electrical potential between the two probe

leads (across the circuit element being measured). Therefore, the measurement indicated on thevoltmeter is not an absolute quantity, but is referenced to whatever point the negative probe lead is

connected to. In order to make this clear, when we indicate a measured or expected voltage on a

circuit diagram, we will always indicate where it was measured – showing where the positivemeasurement probe was connected using a (+) sign, and showing where the negative (or reference)

measurement probe was connected using a (-) sign. This is illustrated in Figure 5, where the voltage

across the resister (VR) is shown with the + and – signs.

When measuring the voltage across an element that is in series with one or more other elements (Fig.

3), the positive lead should be attached to the node where a positive current enters the element and thenegative lead is attached to the other node where the current leaves the element. This will produce a

positive reading on the voltmeter.

Voltages can be measured across a single circuit element, or across combinations of circuit elements.Again, it is therefore imperative to always indicate exactly where the measurement was taken (where

both leads of the voltmeter were connected), using the + and – signs in any diagrams. For complicated

circuits, it is typical to measure the voltages at the circuit nodes of interest relative to the samereference voltage (same node in the circuit). Thus, the (-) lead of the voltmeter would remain touching

the same place in the circuit for all voltage measurements; usually at the negative (or lower voltage)

side of the power source for the circuit. This reference voltage point might be considered to be at 0volts or “ground” for the circuit. Voltage drops across circuit elements can be found from such

measurements by finding the difference between the voltages measured at the circuit nodes on eitherside of the circuit element. For example, looking at the series circuit shown in Figure 8 (ahead a few

pages), one could connect the (-) reference lead of the voltmeter to Node 3, and then measure the

voltage at Node 1 (relative to Node 3) and call it VNode 1, and the voltage at Node 2 (relative to Node 3)and call it VNode 2. Then, the voltage across the resistor R1 (called V1) would be:

V1 = VNode 2 - VNode 1

Note that ideally the voltage measured at a particular circuit node (like Node 1) should be the same nomatter where you put the (+) probe lead (as long as the other (–) lead is not moved). That is, there

should not be any voltage drops across any wires connecting circuit elements within a single circuit

node. Thus, if the (–) lead of the voltmeter is placed at Node 3 in Fig. 8, and the (+) lead is touched toeither the + side of Vs or the + side of resistor R1 (both points in Node 1), the same voltage reading

should occur.

The internal circuit of the voltmeter contains the equivalent circuit configuration shown inside the

dashed blue box in Fig. 5 below:

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Fig. 5 Voltmeter Connected Across Resistor R

in which R int is the internal resistance of the voltmeter, while V meter is the voltage sensing element

which provides the measurement. The internal resistance R int should be very large to minimize the

amount of current drawn from the voltage source by the meter. A significant amount of current drawnfrom the circuit under test can contribute to measurement errors by creating an undesired voltage drop

across probe leads. This is because real physical wires actually have a very small resistance, even

though we usually ignore this in our circuit diagrams or measurements, since these resistances are

usually very small compared to the circuit element resistors purposely placed in a circuit (usually byseveral orders of magnitude). When making precise measurements, however, we will need to keep in

mind that these small resistances exist.

Current Measurements

‘Conventional’ current flow ( I in Fig. 4) refers to the direction of positive charge flow in a circuit. Thisdirection is opposite to electron (negative charge) flow. In the circuit above, positive charges are

imagined to emanate from the positive terminal of the voltage source (Vs), travel through the resistor

from the top lead to the bottom lead, then return to the negative terminal of the source.

To measure the current in the loop using an ammeter , the circuit must be opened (broken) and the

ammeter inserted in series with the loop. This is shown in Fig. 6 below in both schematic form and aphysical layout. In this case, a low value for the internal resistance of the meter Rint is desired to

minimize the disturbance to the circuit and consequently maximize the accuracy of the measurement.An Rint value comparable to R will cause the current to decrease and introduce significant measurement

error.

Fig. 6 Ammeter Connected in Series with Resistor R

I

VS R

+

-

Voltmeter

Ammeter

ILO

LO

Ammeter

(+) Probe

Lead

(-) Probe

Lead

+

VR

-

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Resistance Measurements

To measure the resistance of a circuit element using an ohmmeter , the element must be disconnectedfrom all other elements and power sources. Other resistive elements connected across the element

being measured will cause you to measure the combined resistance of both resistors together, rather

than the resistance of the one element you were trying to measure. Also, since the ohmmeter actuallyuses its own, internally-generated current to measure the resistance, any external power sources

supplying current to the element being measured with “fool” the ohmmeter and result in an inaccurate

measurement.

When measuring most resistances (>>10), the simplified method using only 2 wires connected acrossthe device being tested (resistor R in Fig. 7) is adequate for getting an acceptably accurate

measurement. For low-value resistance measurements, i.e. resistance values similar to the resistance

of the wires being used to make the measurement, the four-wire measurement feature on ohmmeters ispreferred (see Fig. 7). In this figure, all components within the dotted line boxes are internal to the

meter; and the Rwire lines represent the probe leads. The primary difference between the two methods

is that the test current supply and voltage sense lines are separate leads in the four-wire configuration,while they are one and the same in the two-wire arrangement.

Two-Wire Resistance Measurement

Four-Wire Resistance Measurement

Fig. 7 Two-Wire vs. Four-Wire Resistance Measurement of Resistor R

An ohmmeter measures resistance by forcing a known current I source through the element to bemeasured (in this case the resistor R) and sensing the resulting voltage V meter developed across the

element. The resistance is calculated using Ohm’s Law ( R = V meter / I source). Since all wires havenonzero resistances Rwire, a nonzero voltage drop develops across the test/sense line wires in the two-

wire configuration. The size of this voltage drop depends on how big the resistor R is compared to the

resistance of the wires. The extra voltage drop across the two probe leads is measured by the sensing

voltmeter; and therefore a falsely large measurement will be displayed (a measurement of the unknown

resistor R plus the resistance in the probe wires (2 x Rwire)).

LO

4W

Sense

LO LO

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In the four-wire configuration, there is only a very small amount of current in the voltage sense lines

(on the right side of the circuit diagram); substantially less than in the test current (on the left). This

reduced current also reduces the voltage drop across the small wire resistances in the sense lines, andthus allows the voltmeter to obtain a more accurate voltage measurement of the unknown resistor R

alone.

Kirchhoff’s Voltage and Current Laws (KVL and KCL); Voltage and Current Dividers

Kirchhoff’s Voltage Law (KVL) states that the net voltage changes around any complete circuit loop

must sum to zero. Stated another way, the sum of the voltage drops across all resistors in a circuit loop

must exactly equal the total amount of voltage that is supplied by any sources in the loop. Thus, inFigure 8 below, the voltage drops across resistors R1 and R2 must exactly equal the voltage increase

created by the power supply Vs. This law comes from the principle of conservation of energy, as it

implies that no energy is created or lost in a closed loop circuit. Any energy added by a power sourcemust be dissipated, converted, or stored by the other elements in the circuit.

Similarly, Kirchhoff’s Current Law (KCL) requires the net sum of all currents entering a circuit node

(a connection point where multiple circuit branches meet – see Fig. 8) must exactly equal the amountof current leaving the node. This law has its basis in the principle of conservation of charge, as no net

charge is produced or lost at a circuit node. (There’s no place to “store” the electrons that make up the

current (charge flow) in a wire connection; and we can’t create, transform, or destroy any of them at aconnection, either.)

A voltage divider is formed when two or more resistive elements are connected in series with a voltagesource (see Fig. 8 below). In this case, the voltage supplied by the source (Vs) is divided into smaller

voltage drops across each of the resistors in the loop, in proportion to the resistance values of these

elements.

Fig. 8 Voltage Divider: Two Series Elements

This principle can be derived mathematically using Ohm’s Law and the Kirchoff’s Laws stated above.

The currents in all branches of the single loop are identical since the current entering each node (Node

1 between +V s and R1, Node 2 between R1 and R2, and Node 3 between R2 and -V s) must equal thecurrent exiting that node, as per KCL. Voltages along the loop in the direction of the current I include

voltage drops (decreases) developed across resistors R1 and R2 (V 1= I R1 and V 2= I R2, respectively), anda voltage rise (increase) across the source V s. According to KVL, these voltages sum to zero,

1 2 1 2 1 2 1 20 ( )

s s sV V V V IR IR V I R R V V (2)

Therefore, the total resistance Rtotal driven by voltage source V s is the sum of R1 and R2,

21 R R I IRV totals and I = Vs / ( R1 + R2) (3)

Solving equation (3) for I and using the relation V 1 = I*R1 yields,

I

+

+

-

-

Node 2 Node 1

Node 3

+

VNode 2

-

+

VNode 1

-

+ V1 -

+

V2

-

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21

11

21

11 R R

RV R

R R

V IRV s

s

(4)

Using the expression for I derived from (3) and the corresponding relation for R2 (V 2 = I*R2) yields

21

22

R R

RV V s

(5)

Equations (4) and (5) are known as voltage divider relations. They can be interpreted as indicating thatproportionally more voltage is dropped across the larger of two resistances when they are connected inseries.

A current divider is formed when two or more parallel elements are connected with a current source(see Fig. 9 below).

Fig. 9 Current Divider: Two Parallel Elements

As per KCL, the current I s supplied to the node that includes the upper terminals of R1 and R2 (circled)must equal the current exiting the node. This is analogous to a single inlet water pipe connected to two

branch pipes. The total amount of water entering by the single inlet pipe must be carried away by one

or the other of the two branch pipes. Thus the water flow (current) splits between the two pipes.Therefore, for the electrical equivalent, using Ohm’s Law again:

21

21 R

V

R

V I I I totaltotal

s (6)

From Ohm’s Law (1), the voltage developed across R1 and R2 is I 1 R2 and I 2 R2, respectively. Since the

voltage developed across the R1 and R2 combination is V total (Fig. 9), the source current can beexpressed through Ohm’s Law as

total

total

s R

V I (7)

Substituting equation (7) into (6) and solving for Rtotal yields

21

21

R R

R R Rtotal

(8)

Substituting (8) into (7) yields

21

21

R R R R I V stotal

(9)

Since I 1 = V total / R1 and I 2 = V total / R2,

21

12

21

21

R R

R I I and

R R

R I I ss

(10)

The expressions in equation (10) are known as current divider relations. It states that for two parallelbranches, the larger the resistance is in one branch, the larger the current is in the other branch. Here

I1 I2

Vtotal

+

-

Circuit Node

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again, the water analogy agrees with this idea. If water is forced to branch off to two different pipes, it

makes intuitive sense that more water (more current) would flow through the pipe that presents less

resistance (the wider pipe) than an alternative narrower pipe with more resistance. Electrical currentfollows this same principle. Electric current will divide in proportion to the relative resistances of the

paths ahead of it; choosing the “path of least resistance” (sending more current through the lower

resistance path and less through the higher resistance path).

Closed Loops in Electrical Circuits

In order for current to flow in any particular branch of a circuit, it must be part of a complete loop that

also passes through some voltage or current source. Thus, if a circuit branch leaving a node does not

connect back in to the circuit at some other point (at some other node), but instead dead ends with theunconnected end hanging out in the air, then no current can flow in that branch. The branch with a

connection to the rest of the circuit at only one end would be considered an “open circuit” at the

unconnected end.

This is like the case of the wall sockets in your house. These are “dead end” wires connected back to a

serious power source. No current can flow out of these until a device is connected using two wires –one to take in the current from the higher voltage side, and one to return the current completing thecircuit loop. If no device is plugged in to the socket, a voltage would still be measured across the two

wires of the socket (the voltage is available for a connected device); however no current is flowing. So

it is possible to have a voltage difference between wires with no current running between them. Fromthe viewpoint of Ohm’s Law, an open circuit is like having an infinite resistance between the two open

ends of the wire.

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LABORATORY ACTIVITIES

Section 2: Instructor Guided Activities: Proper Use of the Equipmenta. Decade resistor boxes (setting, connecting)

1) Practice series and parallel element connections (Fig. 11).

b. Digital Multimeter1) Resistance measurement (inputs (2W / 4W), 2W / 4W selection), Not in circuit!

i. Measure the resistance across a wire ( 0 ) using both the two-wire and four-wire

techniques (Fig. 7).

c. DC voltage supply introduction:

1) Adjusting voltage (High/Low range, Voltage/Current button, > < digit selection)

2) Setting current limit & overvoltage protection

3) Enable / disable output

4) Constant current vs. constant voltage operation

d. Digital Multimeter

1) Voltage measurement (input locations, DC V selection)

i. Measure the voltage across Power Supply terminals.

ii. Measure the voltage across resistor using Multimeter.

2) Current measurement (input locations, Shift DC V = DC I selection)

i. Measure the current through resistor using Multimeter

ii. Compare to current measurements on front panel of Power Supply

1K

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Section 3: Laboratory Exercises, Voltage Loop MeasurementsPurpose: In this section, you will practice constructing simple circuits, making differential and

referenced voltage measurements, and implementing proper circuit modifications for in-line current

measurements. You will also study how changing resistances in different circuit branches affect the

voltage across and current through those branches. Finally, you will investigate how an “open

circuit” affects the voltages and current in a circuit loop.

a. Construct the circuit defined in Fig. 10 (below) with the Agilent E3640A Power Supply set to 6V

with a 0.1A current limit and three decade resistance boxes each set to 10k . Draw a copy of theschematic diagram for the circuit (left drawing in Fig. 10) in your notebook.

Fig. 10 Series Connected Components

b. Measure the voltage across each of the resistors R1, R2, and R3 individually (3 DC voltagemeasurements); and across the Power Supply terminals (Vs), with the circuit connected and the

power supply output ON so that current is flowing in the circuit. Record these voltages on the

diagram in your notebook, at the locations where they were measured. Be sure to indicate thepolarity of each voltage measurement, using the “+” sign to indicate where in the circuit the

voltage probe wire connected to the red (V) terminal in the multimeter was touching, and using the“-“ sign to show where the wire from the black (LO) terminal on the meter was touching.

c. Measure the two currents (I1 and I2) shown at two different places in the circuit (the current IN to

R1 and the current OUT of R2). This requires you to place the multimeter in series with your

circuit at two different places (two separate measurements). Show these currents using arrows andadd their measured values to your circuit diagram.

d. Change the power supply voltage to 12V and repeat the measurements (voltages across eachelement, including the power supply, and the current through at least one element in the circuit).

Record these on a new circuit diagram in your notebook.

e. Maintain a 12V supply voltage, but change resistors R1 and R3 to 5k each and repeat the voltagemeasurements and at least one current measurement. Draw a diagram of the circuit you are testing

in your lab notebook, and record these measured values on your diagram in the appropriatelocations.

f. To get a more unified view of how a circuit is working, it is typical for one point in the circuit(usually the (-) side of the power supply) to be used as the common reference point for all other

voltage measurements in the circuit. This way, all voltage readings are relative to the same point in

I1

VS

R1

R2

R3

+

-

I1 +

+

+

-

-

-

10k

10k

10k

6V

I2

I2

ab c

d

e

f g

Ref

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the circuit (the voltage reference point-which we would usually call the “common” or “ground”

reference). Connect the (LO) terminal of the multimeter to the (-) side of the power supply so that

this is your common reference point. Then measure the voltages of each of the circuit connectionpoints (a through g) relative to this one reference by touching the probe wire from the (V) terminal

of the meter to each of these points. Show these measured values on the circuit diagram in your

notebook at the locations they were measured. Indicate the reference point used for these

measurements using the ground reference symbol:

on your diagram.

g. Remove the wire connecting the (-) side of resistor R2 and the (+) side of resistor R3. Draw this

circuit in your notebook. Measure the voltage across each circuit element (R1, R2, R3, Vs), and

the current in to R1 (+ side), to see how the break in the circuit is manifested in yourmeasurements. Record these values on your diagram. Measure the voltages at each circuit node

relative to the power supply (-) terminal again (as in part (f)) for this “broken” circuit, and record

these also on your diagram.

Questions: Section 3

1) Were the voltages across each of the resistors in part (b) equal? Why or why not?

2) How did the sum of three voltages compare to the power supply voltage (Vs)?

3) Were the currents in part (c) equal? Why or why not?

4) When the power supply voltage was increased from 6V to 12V in part (d), what was theproportional change in voltage across each of the three resistors?

5) When the power supply voltage was doubled from 6V to 12V in part (d), what was the proportionalchange in current through each of the three resistors?

6) With the changes in resistors R1 and R3 in part (e), what fraction of the power supply voltage was

dropped across each of the resistors?

7) How did the current in the circuit loop change when you altered the two resistances?

8) Show that “Ohm’s Law” agrees with all of your measurements in part (e), by computing the value

of each resistor R1, R2, R3 based on your actual voltage and current measurements. (Show thevoltage and current measurements, the computed resistance, and the actual resistance you set for

each in a table.)

9) Turn in a drawing of your circuit from part (f) that includes all the measurements that you made in

parts (e) and (f). Show the voltage across and the current through each resistor. Show the circuit

connection point (a through g) voltage values relative to the common reference point that youmeasured, and label these points (a – g) in the diagram. Be sure to also include the resistor

numbers (R1, R2, R3) and their values on the diagram; as well as the power supply voltage Vs and

its value. (Hand-drawn diagrams are fine, as long as they are neat.) Label the diagram “Section 3f

Circuit Diagram”.

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10) Explain (justify) your measurement results from part (g). Why was the current at the value you

measured? Explain why you got the measurements you did for the voltages across each resistor?

Can you account for where the voltage from the power supply was dropped in the circuit (unlesswe have proven Dr. Kirchhoff wrong!!)?

Section 4: Laboratory Exercises, DC Voltage Supply CharacteristicsPurpose: In this section, you will investigate how a DC power supply responds to changing resistive

loads, including those that exceed its preset safe operating limits.

a. Set the Agilent E3640A Power Supply to 5V with a 0.25A current limit.

b. Connect one decade resistor box across the power supply terminals with the resistance set to 100 .

c. Place a voltmeter (multimeter set for voltage measurement) across the power supply terminals toaccurately measure its actual voltage output for this procedure. You will use the power supply’s

own current display readout (on the front panel of the power supply) for simultaneous

measurements of the actual current output of the power supply.

d. Record both the voltage and current output of the power supply as you decrease the decade box

resistance from 100 down to 10 in 10 steps. Try to do this quickly, particularly for the lowerresistance settings, as the higher currents will lead to more power dissipation, which will heat up

the resistor boxes internally.

e. Adjust the power supply’s current limit down to 0.10 A. Repeat your voltage and current

measurements while again decreasing the decade resistor box from 100 to 10 in 10 steps.

f. Plot the data taken in steps (d) and (e) above on the same axes (same plot), with voltage in units of

volts on the horizontal (x) axis and the current (in amps) on the vertical (y) axis. Turn in a copy of this plot with your report, and label it ‘Section 4, Power Supply I-V Characteristics, Voc = 5V.’

Label each of the two sets of plotted data with the appropriate current limit setting (‘I sc = 0.25A.’ or

‘Isc = 0.10A.’). Note that the ‘oc’ and ‘sc’ subscripts represent ‘open circuit’ and ‘short circuit,’respectively. The ‘open circuit voltage’ is the voltage output expected when there is nothing

connected to the power supply output terminals. The ‘short circuit current’ is the maximum currentthat would be provided if the two output terminals of the power supply were connected together

with a wire (short circuited). [PLEASE DO NOT TRY TO PROVE THIS!!]


a. Did the current obey Ohm’s Law (V = IR) for ALL points on your graph? V is voltage in volts (V), I is current in amps (A), and R is resistance in ohms (). Show this using a table as in Section 3.

b. On the plot of section 4, step e, identify the “constant voltage” and “constant current” operatingregions, by adding these labels to the appropriate part of the plotted data.

c. Explain in your own words why the two plots appear as they do? At what point do they eachchange abruptly? Why? How are they behaving differently on each side of the abrupt change?

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Section 5: Laboratory Exercises, Current Divison MeasurementsPurpose: In this section, you will investigate how current divides in different circuit branches

according to the resistances it encounters. You will also measure the voltage characteristics of series

and parallel resistor combinations.

a. Construct the circuit defined in Fig. 11 below with the Agilent E3640A Power Supply providing

6V with a current limit of 0.1A. Set all three resistances equal to 10k

as indicated in the figure.Draw a copy of this circuit diagram in your lab notebook.

Fig. 11 Current Measurement Circuit

b. Measure the current flowing through each of the three resistors. Record these on your drawing.Then, measure the voltages across each resistor and add them to your diagram. Finally, using the

black (-) terminal of the power supply as the common reference for your measurements, measure

the voltage at each node of the circuit. Record these values on your diagram as well, in theappropriate locations.

c. Change resistor R3 to 5k . Draw this circuit in your notebook. Repeat all your current and

voltage measurements for this circuit to see what has changed (including voltage across eachelement and the voltage of each node relative to your common reference point).


1. Turn in a copy of the drawing of your circuit from part (c) that includes all the measurements that

you made. Show the voltage across and the current through each resistor, and the connection nodevoltages relative to the common reference. Be sure to also include the resistor numbers (R1, R2,

R3) and their values on the diagram; as well as the power supply voltage Vs and its value. (Neat,

hand-drawn diagrams are fine.) Label the diagram “Section 5c Circuit Diagram”.

2. What was the relation between the current flowing through R1 and the sum of the currents flowing

through R2 and R3 for both circuits? Explain any noticeable observations.

3. Based on your measurements already taken, how much current would you expect there is flowing

into the (-) terminal of the power supply in step (c) above? [Do not measure this current. Baseyour answer on what you have learned so far.]

10k

10k

10k

6V

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4. Explain the similarities and differences in the voltages measured across each resistor in part (b).

Since all the resistors were the same value (10k ), shouldn’t they all have the same voltage acrossthem?

5. For part (c), which of the two resistors R2 (10k ) or R3 (5k ) had more current flowing throughit? What fraction of the current leaving resistor R1 flowed through this resistor? Why was there

more current in this resistor than in the alternate current path through the other resistor?

Section 6: Laboratory Exercises, Resistance MeasurementsPurpose: In this section, you will again practice 2-wire and 4-wire resistance measurements, and

determine what situations they are each best suited for.

a. Measure the resistance of a decade box set to 0 using both the two-wire and four-wire methods.

First, connect one pair of banana-to-banana leads between the V Input terminals on themultimeter (Agilent 34401A) and the decade box to measure resistance using the two-wire method

( 2W button). Record the resistance value.

b. To use the four-wire method, add a second pair of banana-to-banana leads between the 4W sense

terminals on the multimeter and the decade box, then press the 4W button. Record the resistancevalue and compare to the value obtained using the two-wire method.

c. Set a decade resistance box to 1k and repeat the above measurement procedure, using both 2W

and 4W methods.


1. For the 0resistance setting on the decade box measured in part (a), did the two different

resistance measurement methods (2-wire or 4-wire) yield the same results? Why or why not? If different values were obtained, which value was larger and why?

2. For the 1k resistor, did the resistance measurements result in the same values? If different values

were obtained, which value was larger and why?

LEARNING OBJECTIVES FROM EXPERIMENT #1:(This is what you are responsible for being able to do after this experiment. If you can’t, you should go over

this with your lab partners before the next lab session, and make additional notes in your lab notebook.)

1. Define voltage, current, and resistance.

2. Measure voltage, current, and resistance (2-wire and 4-wire) using a digital multimeter

3. Properly wire multimeter leads for all measurements (series, parallel, or out of circuit)4. Properly wire both series and parallel combinations of resistors with a power supply.

5. Apply Ohm’s Law to predict the current, voltage, or resistance in a circuit branch.

6. Predict the effects of a broken or missing connection in a circuit on the voltages and currents

measured in the affected branch7. Describe the effects of setting a current limit on a power supply output, and how it affects the

voltage and current supplied when the limit is exceeded.

8. Predict how current will divide (relative amounts) at circuit nodes connecting to branches withdifferent amounts of resistance.

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