July 1999 Mixed Signal Products

ApplicationReport

SLOA025

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iii Understanding Basic AnalogCircuit Equations

ContentsIntroduction 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Laws of Physics 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Voltage Divider Rule 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Current Divider Rule 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Thevenins Theorem 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Superposition 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Calculation of a Saturated Transistor Circuit 8. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Transistor Amplifier 9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Conclusions 10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

List of Figures1 Ohms Law Applied to the Total Circuit 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Ohms Law Applied to a Component 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Kirchoffs Voltage Law 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Kirchoffs Current Law 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Voltage Divider Rule 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Current Divider Rule 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Original Circuit 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Thevenins Equivalent Circuit for Figure 7 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Example of Thevenins Equivalent Circuit 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Analysis Done the Hard Way 6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Superposition Example 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 When V1 is Grounded 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 When V2 is Grounded 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Saturated Transistor Circuit 8. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Transistor Amplifier 9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Thevenin Equivalent of the Base Circuit 9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

iv SLOA025

1Understanding Basic AnalogCircuit Equations

By Ron Mancini

ABSTRACTThis application report provides a basic understanding of analog circuit equations. Onlysufficient math and physics are presented in this application report to enableunderstanding the concepts.

IntroductionAlthough this application note tries to minimize math, some algebra is germaneto the understanding of analog electronics. Math and physics are presented inthis application note in the manner in which they are used later, so no practiceexercises are given. For example, after the voltage divider rule is explained, it isused several times in the development of other concepts, and this usageconstitutes the practice. This application note builds on each concept after it hasbeen explained, thus, if you want to get familiar with the concepts, read it frombeginning to end.

Circuits are a mix of passive and active components. The components arearranged in a manner that enables them to perform some desired function. Theresulting arrangement of components is called a circuit or sometime a circuitconfiguration. The art portion of analog design is designing the circuitconfiguration. There are many published circuit configurations for almost anycircuit task, thus all circuit designers need not be artists.

When the design has progressed to the point that a circuit exists, equations mustbe written to predict and analyze circuit performance. Textbooks are filled withrigorous methods for equation writing, and this application note does not supplantthose textbooks. But, a few equations are used so often that they should bememorized, and these equations are considered here.

There are almost as many ways to analyze a circuit as there are electronicengineers, and if the equations are written correctly, all methods yield the sameanswer. There are some simple ways to analyze the circuit without completingunnecessary calculations, and these methods are illustrated here.

Laws of PhysicsOhms law is stated as V=IR, and it is fundamental to all electronics. Ohms lawcan be applied to a single component, to any group of components, or to acomplete circuit. When the current flowing through any portion of a circuit isknown, the voltage dropped across that portion of the circuit is obtained bymultiplying the current times the resistance.

2 SLOA025

V R

I

Figure 1. Ohms Law Applied to the Total Circuit

V IRIn Figure 1, Ohms law is applied to the total circuit. The current, (I) flows throughthe total resistance (R), and the voltage (V) is dropped across R. In Figure 2,Ohms law is applied to a single component. The current (IR) flows through theresistor (R) and the voltage (VR) is dropped across R. Notice, the same formulais used to calculate the voltage drop regardless of what portion of the circuit thecalculation is made on.

V RIR

VR

Figure 2. Ohms Law Applied to a ComponentKirchoffs voltage law states that the sum of the voltage drops in a series circuitequals the sum of the voltage sources. Otherwise, the source (or sources)voltage must be dropped across the passive components. When taking sumskeep in mind that the sum is an algebraic quantity.

V R2

R1

VR1VR2

Figure 3. Kirchoffs Voltage Law

VSOURCESVDROPS

V VR1 VR2Kirchoffs current law states; the sum of the currents entering a junction equalsthe sum of the currents leaving a junction. It makes no difference if a current flowsfrom a current source, through a component, or through a wire, because allcurrents are equal. Kirchoffs law is illustrated in Figure 4.

I4 I3

I1 I2

Figure 4. Kirchoffs Current Law

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3 Understanding Basic AnalogCircuit Equations

IIN IOUT

I1 I2 I3 I4

Voltage Divider RuleWhen the output of a circuit is not loaded, the voltage divider rule can be usedto calculate the circuits output voltage. Assume that the same current flowsthrough all circuit elements. Equation 6 is written using Ohms law as V = I(R1 + R2). Equation 7 is written as Ohms law across the output resistor.

V R2

IVO

R1

I

Figure 5. Voltage Divider Rule

I VR1R2

VD IR2Substituting equation 6 into equation 7, and using algebraic manipulation yieldsequation 8.

VD VR2

R1R2A simple way to remember the voltage divider rule is that the output resistor isdivided by the total circuit resistance. This fraction is then multiplied by the inputvoltage to obtain the output voltage. Remember that the voltage divider rulealways assumes that the output resistor is not loaded; the equation is not validwhen the output resistor is loaded by parallel component. Fortunately, mostcircuits following a voltage divider are input circuits, and input circuits are usuallyhigh resistance. When a fixed load is in parallel with the output resistor, theequivalent parallel value comprised of the output resistor and loading resistor canbe used in the voltage divider calculations with no error. Many people ignore theload resistor if it is ten times greater than the output resistor value, but thiscalculation can lead to a 10% error.

Current Divider RuleWhen the output of a circuit is not loaded, the current divider rule can be used tocalculate the current flow in the output branch circuit (R2). The currents I1 and I2in Figure 6 are assumed to be flowing in the branch circuits. Equation 9 is writtenwith the aid of Kirchoffs current law. The circuit voltage is written in equation 10with the aid of Ohms law. Combining equations 9 and 10 yields equation 11.

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4 SLOA025

I R2 VI2I1

R1

Figure 6. Current Divider Rule

I I1 I2

V I1R1 I2R2

I I1 I2 I2R2R1 I2 I2

R1R2R1

Rearranging the terms in equation 11 yields equation 12.

I2 IR1

R1R2

The total circuit current divides into two parts, and the resistance (R1) divided bythe total resistance determines how much current flows through R2. An easymethod of remembering the current divider rule is to remember the voltagedivider rule. Then modify the voltage divider rule such that the opposite resistoris divided by the total resistance, and the fraction is multiplied by the input currentto get the branch current.

Thevenins TheoremThere are times when it is advantageous to isolate a part of the circuit, andanalyze just the isolated part of the circuit. Rather than write loop or nodeequations for the complete circuit, and solving them simultaneously, Theveninstheorem enables us to isolate the part of the circuit we are interested in. We thenreplace the remaining circuit with a simple series equivalent circuit, thusThevenins theorem simplifies the analysis.There are two theorems that do the similar functions, and the second theorem iscalled Nortons theorem. Thevenins theorem is used when the input driver is avoltage source, and Nortons theorem is used when the input drive is a currentsource. Nortons theorem is rarely used, so its explanation is left for the readerto dig out of a textbook if it is ever required.

V

R3C

R1

R2

X

X

Figure 7. Original Circuit

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5 Understanding Basic AnalogCircuit Equations

The rules for Thevenins theorem start with the component or part of the circuitbeing replaced. Referring to Figure 7, look into the terminals (point XX in thefigure) of the circuit being replaced. Calculate the no load voltage (VTH) as seenfrom these terminals (use the voltage divider rule). Look into the terminals of thecircuit being replaced, short independent voltage sources, and calculate theimpedance between these terminals. The final step is to substitute the Theveninequivalent circuit for the part you wanted to replace.

VTH

R3C

RTH

Figure 8. Thevenins Equivalent Circuit for Figure 7The Thevenin equivalent circuit is a simple series circuit, thus further calculationsare simplified. The simplification of circuit calculations is often sufficient reasonto use Thevenins theorem because it eliminates the need for solving severalsimultaneous equations. The detailed information about what happens in thecircuit that was replaced is not available when using Thevenins theorem, but thatis no consequence because you had no interest in it.As an example of Thevenins theorem, lets calculate the output voltage (VO)shown in Figure 9A. The first step is to stand on the terminals XY with your backto the output circuit, and calculate the open circuit voltage seen. This is a perfectopportunity to use the voltage divider rule to obtain equation 13.

V VOUTR2

R1 RS

XY

(a) The Original Circuit

Vth VOUT

Zth R3

XY

(b) The Thevenin Equivalent Circuit

R4

R4

Figure 9. Example of Thevenins Equivalent Circuit

VTH VR2

R1RStill standing on the terminals X-Y, step two is to calculate the impedance seenlooking into these terminals (short the voltage sources). The Theveninimpedance is the parallel impedance of R1 and R2 as calculated in equation 14.Now get off the terminals X-Y before you damage them with your big feet. Stepthree replaces the circuit to the left of X-Y with the Thevenin equivalent circuit VTHand RTH.

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6 SLOA025

RTHR1R2

R1R2The final step is to calculate the output voltage. Notice the voltage divider rule isused again. Equation 15 describes the output voltage, and it comes out naturallyin the form of a series of voltage dividers, which makes sense. Thats anotheradvantage of the voltage divider rule; the answers normally come out in arecognizable form rather than a jumble of coefficients and parameters.

VOUT VTHR4

RTHR3R4 V

R2R1R2

R4R1R2

R1R2R3R4

The circuit analysis is done the hard way in Figure 10, so you can see theadvantage of using Thevenins Theorem. Two loop currents, I1 and I2, areassigned to the circuit. Then the loop equations 16 and 17 are written.

V

R3

VOUTI1

R1

I2R4R2

Figure 10. Analysis Done the Hard Way

V I1R1R2 I2R2

I2R2R3R4 I1R2Equation 17 is rewritten as equation 18 and substituted into equation 16 to obtainequation 19.

I1 I2R2R3R4

R2

V I2 R2R3R4

R2

R1R2 I2R2

The terms are rearranged in equation 20. Ohms law is used to write equation 21,and the final substitutions are made in equation 22.

I2 V

R2R3R4R2R1R2 R2

VOUT I2R4

VOUT VR4R2R3R4 R1R2

R2R2

This is a lot of extra work for no gain. Also, the answer is not in a usable formbecause the voltage dividers are not recognizable, thus more algebra is requiredto get the answer into usable form.

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7 Understanding Basic AnalogCircuit Equations

SuperpositionSuperposition is a theorem that can be applied to any linear circuit. Essentially,when there are independent sources, the voltages and currents resulting fromeach source can be calculated separately, and the results are addedalgebraically. This simplifies the calculations because it prevents writing a seriesof loop or node equations.

V1 VOUT

R1

R2R3

V2

Figure 11. Superposition ExampleWhen V1 is grounded, V2 forms a voltage divider with R3 and the parallelcombination of R2 and R1. The output voltage for this circuit (VOUT1) is calculatedwith the aid of the voltage divider equation. The circuit is shown in Figure 12. Thevoltage divider theorem yields the answer quickly.

V2 VOUT2

R3

R2R1

Figure 12. When V1 is Grounded

VOUT2 V2R1 R2

R3R1 R2Likewise, when V2 is grounded, V1 forms a voltage divider with R1 and the parallelcombination of R3 and R2, and the voltage divider theorem is applied again tocalculate VOUT1.

V1 VOUT1

R1

R2R3

Figure 13. When V2 is Grounded

VOUT1 V1R2 R3

R1R2 R3After the calculations for each source are made the components are added toobtain the final solution.

VO V1R2 R3

R1R2 R3 V2

R1 R2R3R1 R2

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8 SLOA025

The reader should analyze this circuit with loop or node equations to gain anappreciation for superposition. Again, the superposition results come out as asimple arrangement that is easy to understand. One looks at the final equationand it is obvious that if the sources are equal and opposite polarity, and R1 = R3,then the output voltage is zero. Conclusions such as this are hard to make afterthe results of a loop or node analysis unless considerable effort is made tomanipulate the final equation into symmetrical form.

Calculation of a Saturated Transistor CircuitThe circuit specifications are: when VIN = 12 V, VOUT

9 Understanding Basic AnalogCircuit Equations

Transistor AmplifierThe amplifier is an analog circuit, and the calculations, plus the points that mustbe considered during the design, are more complicated than for a saturatedcircuit. This extra complication leads people to say that analog design is harderthan digital design (the saturated transistor is digital i.e.; on or off). Analog designis harder than digital design because the designer must account for all states inanalog, whereas in digital only two states must be accounted for. Thespecifications for the amplifier are an ac gain of four and a peak-to-peak signalswing of 4 volts.

12 V

VOUT

VIN

RC

CERE2

RE1

R1

R2

12 V

CIN

Figure 15. Transistor AmplifierIC is selected as 10 mA because the transistor has a current gain (b ) of 100 at thatpoint. The collector voltage is arbitrarily set at 8 V; when the collector voltageswings positive 2 V (from 8 V to 10 V) there is still enough voltage dropped acrossRC to keep the transistor on. Set the collector-emitter voltage at 4 V; when thecollector voltage swings negative 2 V (from 8 V to 6 V) the transistor still has 2 Vacross it, so it stays linear. This sets the emitter voltage (VE) at 4 V.

RCV12 VC

IC

12 810 400

RE RE1RE2VEIE

VEIB IC

VEIC

410 400

Use Thevenins equivalent circuit to calculate R1 and R2 as shown in Figure 16.

IB

R1 || R2

VB = 4.6 VR2

R1 + R212

Figure 16. Thevenin Equivalent of the Base Circuit

IBIC

10 mA100 0.1 mA

VTH12R2

R1R2

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10 SLOA025

RTHR1R2

R1R2We want the base voltage to be 4.6 V because the emitter voltage is then 4 V.Assume a voltage drop of 0.4 V across RTH, so equation 35 can be written. Thedrop across RTH may not be exactly 0.4 V because of beta variations, but a fewhundred mV does not matter is this design. Now, calculate the ratio of R1 and R2using the voltage divider rule (the load current has been accounted for).

RTH0.40.1 K 4 K

VTh IBRTh VB 0.4 4.6 5 12R1

R1 R2

R175 R2

R1 is almost equal to R2, thus selecting R2 as twice the Thevenin resistance yieldsapproximately 4 K as shown in equation 35. Hence, R2 = 11.2 K and R1 = 8 K.The ac gain is approximately RC/RE1 so we can write equation 38.

RE1RCG

4004 100

RE2 RERE1 400 100 300 The capacitor selection depends on the frequency response required for theamplifier, but 10 m F for CIN and 1000 m F for CE suffice for a starting point.

ConclusionsThe application note presents the minimum number of physics laws andequations required for beginning analog analysis. The laws and equations aresimple, but when applied correctly, they are powerful. As you proceed further intothe realm of analog analysis or into analog design, the physics laws andequations get more complicated, but they are understandable.

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