AM 5 201 DC Circuit Analysis -1

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AM 5-201

BASIC ELECTRONICS

DC CIRCUIT ANALYSIS

December 2011

DISTRIBUTION RESTRICTION: Approved for public release. Distribution is unlimited.

DEPARTMENT OF THE ARMYMILITARY AUXILIARY RADIO SYSTEM

FORT HUACHUCA ARIZONA 85613-7070


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AM 5 201 Basic Electronics DC Circuits Analysis

ii Ver. 1.0

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AM 5 201 Basic Electronics - DC Circuit Analysis

Ver. 1.0 iii

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CONTENTS

1 DC CIRCUIT ANALYSIS..............................................................................1-11.1 REFERENCE................................................................................................ 1-11.2 OHM'SLAW .............................................................................................. 1-11.3 TEMPERATURE COEFFICIENT................................................................................ 1-5

2 HEAT AND POWER..................................................................................2-12.1 POWERDISSIPATIONINRESISTORS ................................................................... 2-1

3 THE SERIES CIRCUIT...............................................................................3-14 PARALLEL CIRCUITS................................................................................4-15 VOLTAGE DIVIDERS.................................................................................5-16 COMBINATION SERIES/PARALLEL CIRCUITS .....................................................6-1


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(Suggested corrections, or changes to this document, should be submitted through your StateDirector to the Regional Director. Any Changes will be made by the National documentation team.

DISTRIBUTIONDistribution is unlimited.

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The Versions are designated in the footer of each page if no version number is designated theversion is considered to be 1.0 or the original issue. Documents may have pages with differentversions designated; if so verify the versions on the Change Page at the beginning of eachdocument.

REFERENCES

The following references apply to this manual:

Allied Communications Publications (ACP):

ACP - 167 - Glossary of Communications Electronics Terms

US Army FM/TM Manuals

1. TM 5-811-3 - Electrical Design, Lightning and Static Electricity Protection2. TM 5-682 - Facilities Engineering Electrical Facilities Safety3. TM 5-690 - Grounding and Bonding in Command, Control, Communications,

Computer, Intelligence, Surveillance, and Reconnaissance (C4ISR) Facilities

4. TM 11-661 Electrical Fundamentals, Direct Current5. TM-664 Basic Theory and Use of Electronic Test Equipment

US Army Handbooks

1. MIL-HDBK 1857 - Grounding, Bonding and Shielding Design Practices

Commercial References

2. Basic Electronics, Components, Devices and Circuits; ISBN 0-02-81860-X, ByWilliam P Hand and Gerald Williams Glencoe/McGraw Hill Publishing Co.

3. Standard Handbook for Electrical Engineers - McGraw Hill Publishing Co.

CONTRIBUTORS

This document has been produced by the Army MARS Technical Writing Team under the authorityof Army MARS HQ, Ft Huachuca, AZ. The following individuals are subject matter experts whomade significant contributions to this document.

William P Hand


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1 DC CIRCUIT ANALYSIS

1.1 REFERENCE

Basic Electronics, Components, Devices and Circuits; ISBN 0-02-81860-X By William P Hand andGerald Williams.

1.2 OHM'S LAW

The German physicist George Simon Ohm conducted extensive experiments into the abilities ofvarious materials to conduct electric energy. He found that materials with the same physical sizehad different resistances. He was able to establish that, for a fixed voltage, the amount of currentflowing through any material depends upon type and physical size of material.

The ratio of voltage and current (resistance) for any given type of conductor is a constant that isdependent upon material and its dimensions. This ratio can be expressed mathematically asfollows:

E= IR

In this equation, Eisthe applied voltage in volts, Iis the current in amperes, and Ris the resistancein ohms, the unit of measure for resistance. The equation can be interpreted as:

1 voltwill force 1 ampereof current through a resistance of1 ohm,

When it is desired to specify how good a conductor is, rather than how much resistance it offers,the unit is the mho (ohm spelled backward),

Electric conductance/siemens The siemens is the practical unit of conductance, The siemens is thereciprocal of ohm since conductance is the reciprocal of resistance, The relationship between ohms

and siemens is given by G = R siemens,

A conductance of one siemens will permit a current flow of one ampere (A) under an electricalpressure of one volt (V), the siemens is a new unit honoring a pioneer in electricity,

NOTEFormerly, the unit of conductance was called the mho. The symbol forconductance is G, and the abbreviation for siemens is S,

The symbolforohm is the Greek letter (omega).

The symbolforthe mho is an upside-down omega, U.

The three forms ofOhm's law are as follows (Reference Fig. 1-1):


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Figure 1-1The Ohm's Law Memory Aid

The voltage (E)* that is required tomaintain a certain current in a circuit in which resistance isknown is equal to the product of the current (I) and the resistance (R).

E= IR

The current (I) in any circuit is equal to the voltage (E) applied, divided by resistance ofthe circuit.

The resistance (R) needed in a circuit so that a certain current (I) will flow when a certain voltage(E) is applied is equal to the voltage (E) divided by the current (I).

* Estands for electromotive force.

V (forvoltage) is also commonly used (V = I R, I = V/ R, R = V/ I).

In order tounderstand applications of the Ohm's law variations, we will work an example of eachtype.

(1) Example 3-1Problem: What voltage must be applied to a circuit containing a resistance of150 ohms and a required current of 3 amps in the circuit?


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Solution:

E= IR = (3) (150) = 450 volts

(2) Example 3-2Problem: If a lamp with a resistance of 75 ohms is connected across a voltagesource of 117 volts, what current will flow through the lamp?

Solution:


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(3) Example 3-3Problem: What resistance will cause a current of 1.2 amps to flow when avoltage of 150 volts is applied to the circuit?

Solution:

Up to this point we have not considered temperature in our calculations. We have considered only anormal room temperature of 20 degrees Celsius (20C). The resistance of any conductor dependsupon the number of free electrons in it. Therefore, as the temperature of a conductor is raised,energy is transferred to the atoms because of the temperature increase. This additional energy

given to the atoms causes them to move around more than they did, thus causing an increase indistance between atoms. The increase in distance will provide more space for free electrons tomove in without colliding with an atom.


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1.3 TEMPERATURE COEFFICIENT

When we heat most substances they will enlarge to some extent. As we increase emperature, it willcause an increasein resistance in mostsubstances. This is known as a positive temperaturecoefficient.Reference the graph shown in Figure 1-2 for copper. Note that it is approximately linear(a straight line graph) over the normal temperature range in which electronic equipment operates.

Figure 1-2Temperature Coefficient of Copper (Cu)

Scientists have measured the resistance of conductors at varying temperatures and havedetermined that most metals increase in resistance. The increase is approximately linear over awide temperature range.


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NOTES


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Ver. 1.0 2-1

2 HEAT AND POWERWhen current flows in a circuit there is some power (orJ2 R) loss in the circuit. This loss is in theform of heat.

(Power) Watts = I2R

Commercial resistors have two basic ratings, the resistance in ohms and a power rating expressed in

watts. A resistor must be able to dissipate the heat it produces. If the internal temperature of aresistor increases to a high level its resistance will change, and often it is a permanent change.

Most common lower wattage resistors, that is, to 2 watts, have their resistance values specifiedby a color code printed on their bodies. The wattage rating is simply a matter of the resistor'sphysical size, which, with a little practice, can easily be learned. Table 2-1 shows the resistor colorcode and how it works.

The color code should be memorized, as it will be used in almost all electronic equipment you willencounter.

1. Example 4-1

Problem: Figure the value of a resistor with the following color bands: red, firstband; green, second band; orange, third band; silver, fourth band. ReferenceResistor Color Code Chart in Table 1-1

Red Green Orange Silver

2 5 000 (3 zeros) 10%

Solution: 25,000 ohms 10%

2.1 POWER DISSIPATION IN RESISTORS

The power dissipated by a resistor is the product of current through it and voltage across it.

P= EI

This formula can also be stated in two other forms:

P=I2 R

1. Example 5-1Problem: Find power dissipation of a resistor with an applied voltage of 10 voltsand a current of 0.2 amps (the unit of measure is the watt).

Solution:

P = 10 x 0.2 = 2 watts


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Table

2-1

ResistorColorCode


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Ver. 1.0 2-3

The power dissipation can also be calculated when only current and resistance or voltage andresistance are known.

2. Example 5-2Problem: Find power dissipation of a 100 . resistor with a current through it of0.1 amp.

Solution:

P=I2 R

P=0.12 x 100

P= 1 watt

3. Example 5-3Problem: Find the power dissipation of a 100 ohm resistor when there is avoltage of 10 volts across it.

Solution:


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NOTES


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3 THE SERIES CIRCUITIn a series circuit the same current passes through EACH element in the circuit before completingits path to the source.

A simple series circuit is shown in Figure 3-1 (A). This circuit is made up of six lamps in series, like a

string of Christmas tree lamps. In order for the current to complete its path, it must flow througheach lamp in turn before flowing back to the battery. If anyone of the six lamps burns out, it willopen the circuit and current will cease the flow. This opening of the circuit is like a switch. A switchis a simple device to open a circuit whenever current flow is not wanted.

In part A of the figure, the circuit is redrawn showing resistors in place of the lamps. Many times acircuit such as that shown in part B of the figure can be more easily understood when symbols of anequivalent nature are used as shown in Figure 3-2.

Figure 3-1Series DC Circuit


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Figure 3-2Equivalent Series Circuit


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4 PARALLEL CIRCUITSIn Figure 4-1 we can see the differences in series and parallel circuits. In part A of the figure thereis only one path for current to follow. This single path through two resistors and back to the batterymakes it a series circuit. In part B of the figure, we have a much different situation. There are twodifferent paths that current can, and does, or follow. Currents II and h both flow from the battery

to the junction of the two resistors. At that point the current splits; current II will flow through RIand current will flow through R2. The two currents then recombine after flowing through theresistors and flow back to the battery, thus completing the circuit.

A very important distinction should be noted here about circuits. It is not totally correct to statethat two components connected across each other make a parallel circuit. As you can see in Figure6-1, both series and parallel circuits across the power source, in this case, the battery. In part A,the series circuit, as we already know, the same current is flowing through both resistors and thebattery. In part B of the figure, RI and R2 are connected across each other and the combination ofthe two resistors is connected across the battery. The two resistors RI and R2 cannot be in seriesbecause different currents flow through each resistor.

It can be seen that the way in which the current flows in a circuit determines whether it is a seriesor a parallel circuit; and that the total current or combined current, is the sum of the currentsflowing through each branch, as shown by

It = I1 + I2 + i3 ----

In parallel circuits. the total current flowing through any parallel combination can always be foundby applying Ohm's law to each branch in turn and then adding the resulting currents.

Figure 4-1Series and Parallel Circuits

Example 4-1Problem: What is the total current flowing, when a 200 ohm, a 150 ohm, and


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a 300 ohm resistor are connected in parallel across a 100 volt source?

Solution:

100II =

200= 0.5

10012=

150= 0.66

10013=

300= 0.33

It = 0.5 + 0.66 + 0.33 = 1.49amps

From this example it can be seen that the total current is much greater than the current throughany one of the individual parallel branches.

We now have enough data to state Kirchhoffsfirst law:

At any junction point in an electrical circuit, the algebraic sum of the currents entering the pointmust be equal to the algebraic sum of the currents leaving the point.Let us look at an example of this law.

Example 4-2Problem: What is the current through the resistor R2 in the circuit?

Diagram for example 4-2.

Solution:

Therefore, from Kirchhoffs Law


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In the examples it was shown that the branches is the lighting circuit in a house. Each light in thehouse is connected in parallel with all the others to the 117 volt source, and switching one of thelights on or off has no effect on the others in the house.


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NOTES


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5 VOLTAGE DIVIDERSThe basic voltage divider normally consists of a series of resistors having two input terminals, acrosswhich the input voltage from the power source is applied. The outputs are across the variousresistors depending upon what part of the input voltage is desired. The simple voltage divider is aspecial series circuit.

Shown in Figure 5-1 is an example of a voltage divider.

Example 5-1Problem: Figure 5-1 shows a simple voltage divider. Find the voltage between eachterminal and ground.

Solution:a. Calculate total current

It = E/(Rl + R2 + R3) (Ohm's law) It = 100/1000 = 0.1 amp

Using the resistance between terminal A and ground and the total current, calculatethe voltage

E= IRE= 0.1 X 100E= 10 volts

Using the total resistance between terminal B and ground and the total current,calculate the voltage

E= IRE= 0.1 X 300 + 100

E= 0.1 X 400

Figure 5-1 Simple Voltage Divider.

E= 40 volts


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There is no need to calculate the voltage between terminal C and ground because it isdirectly across the battery.

Suppose we increase the value of resistors RI, R2, and R3 to 6k, 3k, and lk,respectively,and solve the problem again.

a. Calculate the total current. Keep in mind that k means 1,000.

b. Using the resistance between terminal A and ground and the total current,calculate the voltage.

E= IRE= 0.01 X 1000 E = 10 volts

c. Using the total resistance between terminal B and ground and the totalcurrent, calculate the voltage.

E= IRE = 0.01 X 3000 + 1000 E = 0.01 X 4000E = 40 volts

Voltage betweenterminal and first second

ground example example

A 10 volts 10 volts

B 40 volts 40 volts

C 100 volts 100 volts

The two examples make the important point that the terminal voltages in a voltage divider dependonly upon the resistance ratios and not on the absolute values of the resistors.


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6 COMBINATION SERIES/PARALLEL CIRCUITSAs the name implies, the series-parallel circuit is a combination of series and parallel circuits.Figure 6-1 shows the two basic types of series-parallel circuits and their simplified equivalentcircuits. As can be seen from Figure 6-1, a series-parallel circuit can be defined as one that has thecombined characteristics of both series circuits and parallel circuits in the total network. If thereare two or more components in a parallel branch in a complex network, all of the characteristics ofa parallel network apply to this part of the complex network. If there are two or more components

in series in the complex network, all of the characteristics of a series network apply to this part ofthe circuit.

The best way to solve a series-parallel network problem is to break it down into separate paralleland series networks. When this separation is done, each of the sections can be simplified in turn.

In example 6-1 we will show a simple series-parallel network problem along with each step of thesimplification process.

Example 6-1Problem: A 12-ohm resistor is placed in series with a parallel network of 10 ohms and40 ohms. A 100 volt supply is connected to this series-parallel combination. What is

the current through each resistor? What is the total current?

Circuit for example 6-1 Solution:

Step 1: Isolate R2 and R3. Find the equivalent resistance:

(10) (40) 400R eq = 10 + 40

=50

= 8 ohms


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Now the circuit is reduced to:

Circuit equivalent for example 6-1

Step 2: Find the series equivalent resistance.

Rt= 12 + 8 = 20 ohms

AM 5 201 DC Circuit Analysis -1

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