Circuits &Devices Manual

8/6/2019 Circuits &Devices Manual

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Kirchhoff's Voltage Law (KVL)

Let's take another look at our example series circuit, this time numbering thepoints in the circuit for voltage reference:

If we were to connect a voltmeter between points 2 and 1, red test lead to point 2 andblack test lead to point 1, the meter would register +45 volts. Typically the "+" sign isnot shown, but rather implied, for positive readings in digital meter displays. However,for this lesson the polarity of the voltage reading is very important and so I will showpositive numbers explicitly:

When a voltage is specified with a double subscript (the characters "2-1" in the notation"E2-1"), it means the voltage at the first point (2) as measured in reference to the secondpoint (1). A voltage specified as "Ecd" would mean the voltage as indicated by a digitalmeter with the red test lead on point "c" and the black test lead on point "d": the voltageat "c" in reference to "d".


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If we were to take that same voltmeter and measure the voltage drop across eachresistor, stepping around the circuit in a clockwise direction with the red test lead of ourmeter on the point ahead and the black test lead on the point behind, we would obtainthe following readings:


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We should already be familiar with the general principle for series circuits stating thatindividual voltage drops add up to the total applied voltage, but measuring voltage dropsin this manner and paying attention to the polarity (mathematical sign) of the readingsreveals another facet of this principle: that the voltages measured as such all add up to

zero:

This principle is known as Kirchhoff's Voltage Law(discovered in 1847 by Gustav R.Kirchhoff, a German physicist), and it can be stated as such:

"The algebraic sum of all voltages in a loop must equal zero"

By algebraic, I mean accounting for signs (polarities) as well as magnitudes. By loop, Imean any path traced from one point in a circuit around to other points in that circuit,and finally back to the initial point. In the above example the loop was formed byfollowing points in this order: 1-2-3-4-1. It doesn't matter which point we start at or


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which direction we proceed in tracing the loop; the voltage sum will still equal zero. Todemonstrate, we can tally up the voltages in loop 3-2-1-4-3 of the same circuit:

This may make more sense if we re-draw our example series circuit so that allcomponents are represented in a straight line:

It's still the same series circuit, just with the components arranged in a different form.Notice the polarities of the resistor voltage drops with respect to the battery: thebattery's voltage is negative on the left and positive on the right, whereas all the resistorvoltage drops are oriented the other way: positive on the left and negative on the right.

This is because the resistors are resisting the flow of electrons being pushed by thebattery. In other words, the "push" exerted by the resistors againstthe flow of electronsmustbe in a direction opposite the source of electromotive force.

Here we see what a digital voltmeter would indicate across each component in thiscircuit, black lead on the left and red lead on the right, as laid out in horizontal fashion:


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If we were to take that same voltmeter and read voltage across combinations ofcomponents, starting with only R1 on the left and progressing across the whole string ofcomponents, we will see how the voltages add algebraically (to zero):


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The fact that series voltages add up should be no mystery, but we notice that thepolarityof these voltages makes a lot of difference in how the figures add. While reading voltageacross R1, R1--R2, and R1--R2--R3 (I'm using a "double-dash" symbol "--" to represent theseries connection between resistors R1, R2, and R3), we see how the voltages measuresuccessively larger (albeit negative) magnitudes, because the polarities of the individualvoltage drops are in the same orientation (positive left, negative right). The sum of thevoltage drops across R1, R2, and R3 equals 45 volts, which is the same as the battery'soutput, except that the battery's polarity is opposite that of the resistor voltage drops

(negative left, positive right), so we end up with 0 volts measured across the whole stringof components.

That we should end up with exactly 0 volts across the whole string should be no mystery,either. Looking at the circuit, we can see that the far left of the string (left side of R1:point number 2) is directly connected to the far right of the string (right side of battery:point number 2), as necessary to complete the circuit. Since these two points are directlyconnected, they are electrically common to each other. And, as such, the voltagebetween those two electrically common points mustbe zero.


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Kirchhoff's Voltage Law (sometimes denoted as KVL for short) will work for anycircuitconfiguration at all, not just simple series. Note how it works for this parallel circuit:

Being a parallel circuit, the voltage across every resistor is the same as the supplyvoltage: 6 volts. Tallying up voltages around loop 2-3-4-5-6-7-2, we get:

Note how I label the final (sum) voltage as E2-2. Since we began our loop-steppingsequence at point 2 and ended at point 2, the algebraic sum of those voltages will be thesame as the voltage measured between the same point (E2-2), which of course must bezero.

The fact that this circuit is parallel instead of series has nothing to do with the validity ofKirchhoff's Voltage Law. For that matter, the circuit could be a "black box" -- itscomponent configuration completely hidden from our view, with only a set of exposedterminals for us to measure voltage between -- and KVL would still hold true:


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Try any order of steps from any terminal in the above diagram, stepping around back tothe original terminal, and you'll find that the algebraic sum of the voltages always equals

zero.

Furthermore, the "loop" we trace for KVL doesn't even have to be a real current path inthe closed-circuit sense of the word. All we have to do to comply with KVL is to begin andend at the same point in the circuit, tallying voltage drops and polarities as we gobetween the next and the last point. Consider this absurd example, tracing "loop" 2-3-6-3-2 in the same parallel resistor circuit:


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KVL can be used to determine an unknown voltage in a complex circuit, where all othervoltages around a particular "loop" are known. Take the following complex circuit(actually two series circuits joined by a single wire at the bottom) as an example:

To make the problem simpler, I've omitted resistance values and simply given voltagedrops across each resistor. The two series circuits share a common wire between them(wire 7-8-9-10), making voltage measurements between the two circuits possible. If wewanted to determine the voltage between points 4 and 3, we could set up a KVL equationwith the voltage between those points as the unknown:


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Stepping around the loop 3-4-9-8-3, we write the voltage drop figures as a digitalvoltmeter would register them, measuring with the red test lead on the point ahead andblack test lead on the point behind as we progress around the loop. Therefore, thevoltage from point 9 to point 4 is a positive (+) 12 volts because the "red lead" is onpoint 9 and the "black lead" is on point 4. The voltage from point 3 to point 8 is a positive


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It is important to realize that neither approach is "wrong." In both cases, we arrive at thecorrect assessment of voltage between the two points, 3 and 4: point 3 is positive withrespect to point 4, and the voltage between them is 32 volts.

REVIEW:

Kirchhoff's Voltage Law (KVL): "The algebraic sum of all voltages in a loopmust equal zero"

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MAXIMUM POWER TRANSFER THEOREM

NORTONS THEOREM SUPERPOSITION THEOREM

Sometimes in engineering we are asked to design a circuit that will transfer themaximum power to a load from a given source. According to the maximum powertransfer theorem, a load will receive maximum power from a source when itsresistance (RL) is equal to the internal resistance (RI) of the source. If the sourcecircuit is already in the form of a Thevenin or Norton equivalent circuit (a voltageor current source with an internal resistance), then the solution is simple. If the

circuit is not in the form of a Thevenin or Norton equivalent circuit, we must firstuse Thevenins orNortons theorem to obtain the equivalent circuit.

Heres how to arrange for the maximum power transfer.

1. Find the internal resistance, RI. This is the resistance one finds by looking backinto the two load terminals of the source with no load connected. As we haveshown in the Thevenins Theorem and Nortons Theorem chapters, the easiestmethod is to replace voltage sources by short circuits and current sources byopen circuits, then find the total resistance between the two load terminals.

2. Find the open circuit voltage (UT) or the short circuit current (IN) of the sourcebetween the two load terminals, with no load connected.

Once we have found RI, we know the optimal load resistance(RLopt = RI). Finally, the maximum power can be found:

In addition to the maximum power, we might want to know another important

quantity: the efficiency. Efficiency is defined by the ratio of the power received bythe load to the total power supplied by the source. For the Thevenin equivalent:
http://www.tina.com/English/tina/course/10norton/nortonhttp://www.tina.com/English/tina/course/12super/superhttp://www.tina.com/English/tina/course/9thevenin/theveninhttp://www.tina.com/English/tina/course/10norton/nortonhttp://www.tina.com/English/tina/course/10norton/nortonhttp://www.tina.com/English/tina/course/9thevenin/theveninhttp://www.tina.com/English/tina/course/10norton/nortonhttp://www.tina.com/English/tina/course/coursehttp://www.tina.com/English/tina/course/10norton/nortonhttp://www.tina.com/English/tina/course/12super/superhttp://www.tina.com/English/tina/course/9thevenin/theveninhttp://www.tina.com/English/tina/course/10norton/nortonhttp://www.tina.com/English/tina/course/9thevenin/theveninhttp://www.tina.com/English/tina/course/10norton/norton


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and for the Norton equivalent:

Using TINAs Interpreter, it is easy to draw P, P/Pmax, and as a function ofRL.The next graph shows P/Pmax, the power on RL divided by the maximum power,Pmax, as a function ofRL (for a circuit with internal resistance RI=50).

Now lets see theefficiency as a function ofRL.


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The circuit and the TINA Interpreter program to draw the diagrams above areshown below. Note that we we also used the editing tools of TINAs Diagramwindow to add some text and the dotted line.


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Now lets explore the efficiency ( ) for the case of maximum power transfer,where RL = RTh.

The efficiency is:

which when given as a percentage is only 50%. This is acceptable for someapplications in electronics and telecommunication, such as amplifiers, radioreceivers or transmitters However, 50% efficiency is not acceptable for batteries,


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power supplies, and certainly not for power plants.

Another undesirable consequence of arranging a load to achieve maximum

power transfer is the 50% voltage drop on the internal resistance. A 50% drop insource voltage can be a real problem. What is needed, in fact, is a nearlyconstant load voltage. This calls for systems where the internal resistance of thesource is much lower than the load resistance. Imagine a 10 GW power plantoperating at or close to maximum power transfer. This would mean that half of theenergy generated by the plant would be dissipated in the transmission lines andin the generators (which would probably burn out). It would also result in loadvoltages that would randomly fluctuate between 100% and 200% of the nominalvalue as consumer power usage varied.

To illustrate the application of the maximum power transfer theorem, lets find the

optimum value of the resistor RL to receive maximum power in the circuit below.

Click here to load orsave this circuit

We get the maximum power if RL= R1, so RL = 1 kohm. The maximum power:

A similar problem, but with a current source:

Click here to load orsave this circuitFind the maximum power of the resistor RL.


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We get the maximum power if RL = R1 = 8 ohm. The maximum power:

The following problem is more complex, so first we must reduce it to a simplercircuit.

Find RI to achieve maximum power transfer, and calculate this maximum power.

Click here to load orsave this circuit First find the Norton equivalent usingTINA.

Click here to load orsave this circuit

Finally the maximum power:


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{Solution by TINA's Interpreter}O1:=Replus(R4,(R1+Replus(R2,R3)))/(R+Replus(R4,(R1+Replus(R2,R3))));

IN:=Vs*O1*Replus(R2,R3)/(R1+Replus(R2,R3))/R3;RN:=R3+Replus(R2,(R1+Replus(R,R4)));Pmax:=sqr(IN)/4*RN;IN=[250u]RN=[80k]Pmax=[1.25m]

We can also solve this problem using one of TINAs most interesting features, theOptimization analysis mode.

To set up for an Optimization, use the Analysis menu or the icons at the top rightof the screen and select Optimization Target. Click on the Power meter to open itsdialog box and select Maximum. Next, select Control Object, click on RI, and setthe limits within which the optimum value should be searched.

To carry out the optimization in TINA v6 and above, simply use theAnalysis/Optimization/DC Optimization command from the Analysis menu.

In older versions of TINA, you can set this modefrom the menu,Analysis/Mode/Optimization , and then execute a DC Analysis.

After running Optimization for the problem above, the following screen appears:


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After Optimization, the value of RI is automatically updated to the value found. Ifwe next run an interactive DC analysis by pressing the DC button, the maximumpower is displayed as shown in the following figure.


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NORTONS THEOREM SUPERPOSITION THEOREM

Copyright 2011 DesignSoft, Inc. All rights reserved.

SUPERPOSITION THEOREM

MAXIMUM POWER TRANSFER THEOREM KIRCHHOFF'S LAWS

The superposition theorem states that in a linear circuit with several sources, thecurrent and voltage for any element in the circuit is the sum of the currents andvoltages produced by each source acting independently.

To calculate the contribution of each source independently, all the other sourcesmust be removed and replaced without affecting the final result. When removing avoltage source, its voltage must be set to zero, which is equivalent to replacingthe voltage source with a short circuit. When removing a current source, itscurrent must be set to zero, which is equivalent to replacing the current sourcewith an open circuit.

When you sum the contributions from the sources, you should be careful to taketheir signs into account. It is best to assign a reference direction to each unknownquantity, if it is not already given. The total voltage or current is calculated as thealgebraic sum of the contributions from the sources. If a contribution from asource has the same direction as the reference direction, it has a positive sign in
http://www.tina.com/English/tina/course/10norton/nortonhttp://www.tina.com/English/tina/course/12super/superhttp://www.tina.com/English/tina/course/11maxim/maximhttp://www.tina.com/English/tina/course/13kirch/kirchhttp://www.tina.com/English/tina/course/coursehttp://www.tina.com/news.rsshttp://www.tina.com/English/tina/course/coursehttp://www.tina.com/English/tina/course/10norton/nortonhttp://www.tina.com/English/tina/course/12super/superhttp://www.tina.com/English/tina/course/11maxim/maximhttp://www.tina.com/English/tina/course/13kirch/kirch


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V = -IS * R*R1/(R+R1) = -2*10*10/(10+10) = -10 V.

Finally,

the unknown voltage is the sum of V and V: V = V + V =5 + (-10) = -5 V.

Note that the signs of the partial answers V and V had an important role in thesolution. Be careful to determine and use the correct signs.

{Solution by TINA's Interpreter}

{Using the superposition theorem}

V1:=-Is*R*R1/(R+R1);

V1=[-10]

V2:=Vs*R/(R+R1);

V2=[5]

V:=V1+V2;

V=[-5]

Example 1

Find the currents shown by the ammeters.

Clickhere to load or save thiscircuit


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The following figure shows the steps of the superposition method for the solution.

In the first step (left side of the figure above), we calculate the contributions I1 andI2 produced by the source V2. In the second step (right side of the figure), we

calculate the contributions I1 and I2 produced by the source V1.

Finding I1 first, we should calculate R13 (the total resistance of parallel connectedR1 and R3) and then use the voltage division rule to calculate V13, the commonvoltage across these two resistors. Finally, to calculate I1 (the current through R1),we should use Ohms law and divide V13 by R1.


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With a similar consideration for all the quantities:

And

Finally, the result:


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You can check the correctness of the steps using TINA as shown in the figuresabove.


{Use the superposition method!}

{We use doubled subscript because

the Interpreter does not allow the' and " as an index.

the second subscript means the first or second measuring}

I11:=V2*R1*R3/(R1+R3)/(R2+R1*R3/(R1+R3))/R1;

I21:=V2*R1*R3/(R1+R3)/(R2+R1*R3/(R1+R3))/R3;

I31:=-V2/(R2+R1*R3/(R1+R3));I12:=-V1/(R1+R2*R3/(R2+R3));

I22:=V1*R2/(R2+R3)/(R1+R2*R3/(R2+R3));

I32:=V1*R3/(R2+R3)/(R1+R2*R3/(R2+R3));

I1:=I11+I12;

I1=[50m]

I2:=I21+I22;

I2=[250m]

I3:=I31+I32;

I3=[-300m]

Example 2

Find the voltage V and the current I.

Clickhere to load or save thiscircuit

The figure shows how can you use the superposition theorem:


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{Solution by TINA's Interpreter!}

{Using the superposition method !}

I1:=Is*R1/(R1+R1);

I2:=-Vs/(R1+R1)

I:=I1+I2;

I=[0]

V1:=0;

V2:=Vs;

V:=V1+V2;

V=[2]

Example 3

Find the voltage V.

Click here to load or save thiscircuit

And the superposition:
http://www.tina.com/English/tina/course/12super/Super%2036.TSC


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{Using superposition theorem}

V1:=Vs1*R2*R4/(R2+R4)/(R1+R2*R4/(R2+R4));

V1=[50]

V2:=Is1*R2*R4*R1/(R2+R4)/(R1+R2*R4/(R2+R4));

V2=[10]

V3:=Vs2*R1*R2/(R1+R2)/(R4+R1*R2/(R1+R2));

V3=[60]

V:=V1+V2+V3;

V=[120]

You can see that using the superposition theorem for circuits containing morethen two sources is pretty complicated. The more sources there are in the circuit,the more steps are required. This is not necessarily the case with the other, moreadvanced methods described in later chapters. If superposition requires you to


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analyze a circuit three or more times, it is all too easy to mix up a sign or makesome other mistake. So if the circuit has more than two sources--unless it is verysimple--it is better to use Kirchhoffs equations and its simplified versions, the

methods of nodal voltages or mesh currents described later.

While the superposition theorem can be useful for solving simple practical problems, its main use is in the theory of circuitanalysis, where it is employed in proving other theorems.

MAXIMUM POWER TRANSFER THEOREM KIRCHHOFF'S LAWS

Copyright 2011 DesignSoft, Inc. All rights reserved.

Circuit Diagram For Half-wave Rectifier:
http://www.tina.com/English/tina/course/11maxim/maximhttp://www.tina.com/English/tina/course/13kirch/kirchhttp://2.bp.blogspot.com/_Jt8jI5P6sEU/SZsk-yXxSII/AAAAAAAAC6c/pUxEMVTd8hU/s1600-h/Half-wave+Rectifier.JPGhttp://www.tina.com/news.rsshttp://www.tina.com/English/tina/course/coursehttp://www.tina.com/English/tina/course/11maxim/maximhttp://www.tina.com/English/tina/course/13kirch/kirch


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The bridge rectifier circuit and their input and output voltage as a function of time isshown below. Peak voltage across each diode when it is reverse-biased

PIV = Vm VDODC Voltage, VDC = 2Vm/ 2 VDORipple Factor: -

A rectifier converting alternating currents into a unidirectional current, periodically

fluctuating components still remaining in the output wave. A measure of the fluctuatingcomponent is given by the ripple factor r, which is defined as

R = rms value of alternating components of wave/Average value of wave

= I rms/Idc = V rms/VdcWhere, I rms and V rms denote the rms value of the ac components of the current and

voltage, respectively.

For a half-wave rectifier, r = 1.21 and for a full wave rectifier, r = 0.482Calculating Ripple Factor for Half-wave Rectifier: -

For C = 1F,

The DC value is 0.5VThe rms value is 0.5V

So the Ripple Factor is 0.5/0.5 = 1

For C = 47F,

The DC value is 5VThe rms value is 0.9V

So the Ripple Factor is 0.9/5 = 0.18Calculating Ripple Factor for Full-wave Rectifier: -

For C = 1F,

The DC value is 0.9V

The rms value is 0.22VSo the Ripple Factor is 0.22/0.9 = 0.24

For C = 47F,

The DC value is 8.78V

The rms value is 6.18V


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So the Ripple Factor is 6.18 /8.78 = 0.7Procedure: -

1. Construct circuit of Figure-1 without thecapacitor. Observe Vi and Vo simultaneously on the oscilloscope. Sketch input

and output waveforms. Measures Vo with multimeter in dc and ac mode.

2.

Connect 1F capacitor across the loadresistor. BE CAREFUL about the polarity of the capacitor. Sketch input and

output waveforms. Measure Vo with multimeter.

3.Replace 1F capacitor with 47F and repeatstep-2.

4.Construct the circuit of Figure-2 without thecapacitor. Observe and sketch Vi, Vo. DO NOT TRY to observe Vi, Vo

simultaneously. Measure AC and DC components of Vo with multimeter.5.Connect 1F capacitor as shown in Figure-2

and repeat step-4.

6.Replace 1F capacitor by 47F for Figure-2and repeat step-4.

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