notes for course ee1.1 circuit analysis 2004-05 topic 3 ...cas.ee.ic.ac.uk/people/dhaigh/ee1.1...

Notes for course EE1.1 Circuit Analysis 2004-05

TOPIC 3 – CIRCUIT ANALYSIS USING SUB-CIRCUITS

OBJECTIVES

1) To introduce the Source Transformation

2) To consider the concepts of Linearity and Superposition

3) To introduce Thevenin and Norton Equivalent Subcircuits

4) To consider Practical Sources and Matching

5) To introduce Voltage and Current Source Transportation

6) To introduce Voltage and Current Source Substitution

1 SOURCE TRANSFORMATION.

Consider a voltage source connected in series with a resistor:

We can test its v-i characteristic by attaching an independent source to its terminals, varying thatsource to all possible values, and recording the resulting value of the response variable; we havearbitrarily chosen a current source above but a voltage source would have worked just as well.

We can compute v very simply using KCL and Ohm's law:

v = vs + iRsLet's invert this relation and express i in terms of v:

i = −vsRs

+vRs

This can be interpreted as a KCL equation for the following subcircuit:

This subcircuit can be tested also with a current source as shown or a voltage source to obtain the i-v characteristic.

Since the equations describing the voltage source with series resistor and current source withparallel resistor are equivalent, it follows that these subcircuits are equivalent subcircuits.

≡

No test at the terminals can distinguish one from the other.

Topic 3 – Circuit Analysis using Sub-circuits

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The source equivalence is a powerful tool for analyzing circuits. Consider the following example:

Example 3.1

Find the voltage v and the current i in the following circuit using an equivalent subcircuit.

Solution

We see that there are two subcircuits consisting of a voltage source with a series resistor, asfollows:

We can replace the two subcircuits with equivalent subcircuits:

Notice that we have lost the branch that current i flowed in but the branch corresponding to voltagev is retained. We proceed to determine v:

We apply KCL equation at the top node (noting that this node voltage is v):

IoutResistors∑ = Iin

Sources∑

v6+v3+v6= 8 − 4

v 16+

13+

16

⎛⎝⎜

⎞⎠⎟= 4

v 23= 4

v = 6 VTo determine i, we redraw the original circuit with the voltage v = 6 V shown explicitly:

We can now write KCL at the top node to obtain:


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IoutResistors∑ = Iin

Sources∑

63+

6 − −24( )6

= −i

i = −2 − 5 = −7 AUsing the source transformation, we can analyze circuits that do not consist merely of a single loopor a single pair of nodes and that are neither series nor parallel.

Example 3.2

Find the current i in the following circuit:

Solution

We apply the source transformation to the two-terminal subcircuit on the left :

Analysis of this simple series circuit gives

i =veffectivereffective

=24 − −12( )

4 + 2=

366

= 6 A

2 LINEARITY AND SUPERPOSITION

Consider a simple circuit whose independent source values are variables, is and vs:

Using the source transformation on the parallel subcircuit consisting of the current source and itsnearest resistor, we derive the equivalent circuit shown:

The desired response is the current in one of those elements.

We easily analyze this single loop (series) circuit to obtain


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i =veffectivereffective

=2is + vs2 + 2 + 2

=13is +

16vs

Thus, i is a linear combination of the two source values, is and vs.

This is a property that holds in general for all voltage and current responses in circuits with onlyindependent sources and resistors.

It will be proved from first principles in the next topic.

Now we derive an important result; we can define current parameters i1 and i2 as follows:

i1 = i vs =0 =13is i2 = i is =0 =

16vs

We call i1 and i2 the partial responses.

We can now write the equation for i in the form:

i = i1 + i2In words, i1 is the response with the voltage source reduced to zero and i2 the response with thecurrent source reduced to zero.

When a source value, voltage or current, is set to zero, we say the the source is de-activated.

However, a deactivated voltage source is a short circuit and a deactivated current source is an opencircuit.

Hence the partial response expressions describe the circuits shown:

These two circuits are referred to as partial circuits.

We can now generalize.

Consider a circuit having only a number of resistors and n independent sources with values x1, x2, ..., xn.

Any response y, voltage or current, has the form:

y = a1x1 + a2x2 + ...+ anxnwhere ai are constants determined by the resistive portion of the circuit.

Thus, we can define n partial responses:

yi = aixi for i = 1, 2, ... , n

We can write:

y = y1 + y2 + ...+ ynwhere

yi = y xk =0 k = 1,2,...,n with k ≠ i


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Thus, we can compute each partial response yi from the circuit that results when all the independentsources other than the ith are deactivated.

This is known as the principle of superposition.

It is often a useful analysis tool.

By its means, we can split a complicated circuit analysis problem into a number of simpler ones.

Example 3.3

Find the indicated response current the following circuit using superposition:

Deactivation of the 12-A i-source and the 8-A i-source gives the following partial circuit:

Using parallel and series equivalents, Ohm's law, and voltage and current division, one finds thepartial response to be:

i1 = 2 A

Deactivation of the 12-A i-source and the v-source gives the second partial circuit:

We can determine that:

i2 = −5 A

We then allow the 12-A source to remain active and deactivate the 8-A i-source and the v-source:

This single source circuit can be analyzed to determine that:

i3 = 3 A

Finally, we add each of the three partial response currents to obtain the actual current in the originalcircuit:

i = i1 + i2 + i3 = 2 − 5 + 3 = 0 A

This surprising result can be verified by checking that KCL and KVL hold for all loops and nodes.


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3 THEVENIN AND NORTON EQUIVALENT SUBCIRCUITS

3.1 The Thevenin Equivalent

As an example, consider the following subcircuit:

We assume that this subcircuit is connected into a larger circuit; that is, there are elements externalto this subcircuit connected to it through the two terminals shown.

Therefore, both the terminal voltage v and the terminal current i have nonzero values in general.

If we apply a source transformation to the current source and its parallel 4 Ω resistor, we obtain theequivalent shown:

The two voltage sources and the two 4 Ω resistors are series connected and can be combined.

Elements in series, such as the two voltage sources may be moved around in the series loop becausethis does not change the KVL equation.

Now we apply one last source transformation to the elements in the shaded box:

We can now combine the two 8 Ω parallel resistors to obtain the final equivalent subcircuit:

Notice that all of our transformations have left i and v unaffected.


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We can now write one KCL equation at the top node to get:

v4= i + 1

2is +

18vs

v = 2is +12vs + 4i

We see that the voltage is a linear combination of the internal independent sources and the terminalcurrent i.

The first two terms on the right side of the equation for v are independent of i and that the third isdirectly proportional to the terminal current i.

Thus the equation may be represented as follows:

v = voc + iReq

where

voc = 2is +12vs

Req = 4

The simpler equation describes the series combination of a v-source and a resistor:

This subcircuit is, therefore, equivalent to the original because it has the same v-i characteristic. It iscalled the Thevenin equivalent (subcircuit) after the French telegraph engineer Charles Thevenin.

Since the superposition property applies to any two-terminal subcircuit composed of resistors andindependent sources, then any such subcircuit, however complex, has a valid Thevenin equivalentcircuit in this form.

We now seek to understand the significance of the two parameters voc and Req.

Consider the parameter voc:

Although only a part of the total terminal voltage v, it becomes the actual terminal voltage under thecondition that i = 0.

v = voc + iReq → v = voc

The condition i = 0 is imposed by simply removing the subcircuit from the elements to which it isconnected:

It is important to notice that this is precisely the same as the original subcircuit with only two smallnotational changes: we have explicitly specified that the terminal current is zero, and we havelabelled the terminal voltage with the symbol voc to denote that the voltage obtained is the voltage


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source in the Thevenin equivalent circuit. Note that voc depends on all the sources and also on theresistors.

Using superposition to analyse the sub-circuit for determining voc gives:

voc = 2is +12vs

which is correct.

In order to consider the significance of the term Req, consider again the equation for v in ourexample:

v = 2is +12vs + 4i

Note that if we deactivate the internal sources (set is and vs to zero in our example) the voc term willbe forced to zero, leaving only the term proportional to i:

v = 4iDeactivating the sources will make voc = 0 in the Thevenin equivalent circuit. The equationdescribing the Thevenin equivalent circuit then becomes:

v = voc + iReq → v = iReq

Thus Req is v/i for the sub-circuit with all its sources deactivated, which is as follows:

v/i is the equivalent resistance of the deactivated circuit. Analysis of the deactivated circuit yieldsReq = 4 Ω, which agrees with the previous figure.

In general, all independent v-sources are replaced by short circuits and all i-sources by opencircuits.

Example 3.4

Find the Thevenin equivalent subcircuit for the subcircuit shown:

Solution

We first determine voc and set i = 0:


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Since i = 0, the current in the resistor must equal the current in the current source, as shown. ThenKVL gives:

voc = −27 + 2 = −25 V

To compute Req we deactivate the subcircuit, thus replacing the 2-V source with an equivalent shortcircuit and the 3-A source with an equivalent open circuit:

Clearly, Req = 9 Ω. The Thevenin equivalent subcircuit is as follows:

Example 3.5

Check the Thevenin equivalent circuit we have derived by determining the v-i equation for the sub-circuit.

Solution

KCL at the left upper node tells us that the current in the 9 Ω resistor is i – 3 A (downwards). KVLaround the loop gives us:

v = 9 i − 3( ) + 2= −25 + 9i

This confirms the values of voc = –25 V and Req = 4 Ω obtained previously.

1.2 The Norton Equivalent

The Thevenin equivalent circuit is equivalent to any 2-terminal subcircuit containing resistors andindependent voltage and current sources. The Norton equivalent circuit is an alternative generalequivalent circuit which can be derived independently or it can be obtained from the Theveninequivalent circuit by applying the source transformation.

Application of the source transformation to the Thevenin subcircuit leads to the Norton equivalentcircuit as follows:

≡

In order for this subcircuit to be equivalent to the original circuit, we must have


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isc = vocReq

Notice that the terminal v-i characteristic for the Norton equivalent is:

i = vReq

− isc

Combining these two equations leads to the equation describing the Thevenin equivalent:

i = vReq

−vocReq

v = voc + iReq

Hence, the Thevenin and Norton equivalent circuits are equivalent to each other.

To understand the significance of the parameter isc, suppose the subcircuit is shorted - that is, a shortcircuit is placed across its terminals.

This is the same as shorting the Norton equivalent:

(Arrows on the short denote that it is merely a test connection.)

The terminal voltage and therefore the resistor voltage are both zero, so the resistor current is zeroalso. Thus, all of the source current flows out of the top terminal and down through the short circuit,as shown above.

Thus isc, or ishort-circuit, is the current that flows in the subcircuit when the terminals are shortedtogether.

The Thevenin and Norton equivalents involve three parameters: voc, Req and isc.

Req is common to both equivalents and in addition all three parameters are related via the sourcetransformation equation:

isc = vocReq

Hence, one need only compute two of the parameters from the circuit in order to determine al three.

We can see the connection graphically starting from the v-i characteristic for the Theveninequivalent:

v = voc + iReq

We can plot a graph of this equation as follows:

We have sketched this graph under the assumption that voc (and hence isc) is positive.


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It is clear that the vertical intercept is voc, the horizontal intercept is –isc and the gradient is Req.

Example 3.6

Find voc, isc, and Req by direct tests on the 2-terminal subcircuit shown:

Solution

When we determine Thevenin or Norton equivalent circuits, we are free to define the referencedirections as we wish.

The open circuit voltage can be computed by analyzing the following circuit where we haveimposed the condition i = 0:

We first deactivate the 36 V v-source and find the corresponding partial response from the partialsubcircuit:

We note that the two resistors are connected in parallel, giving an equivalent resistance of 2 Ω.Thus, we have voc1 = 6 × 2 = 12 V. Next, we deactivate the 6 A i-source, resulting in the otherpartial subcircuit:

The two resistors and the v-source form a series circuit and that the voltage across the 3 Ω resistor isthe partial terminal voltage. Thus, we have voc2 = 36 × 3/(3 + 6) = 12 V. Adding, we obtain the total(actual) open circuit voltage: voc = vocl + voc2 = 12 + 12 = 24 V.

In order to calculate the value of Req, we must first deactivate the entire subcircuit:

Hence, Req = 2 Ω.


12

The short circuit current isc can be calculated from the voc and Req, but let us first determine it fromthe circuit. We place a short circuit across the subcircuit terminals and identify the short circuitcurrent:

The test short circuit makes the voltage in the 3 Ω resistor zero; hence its current is also zero byOhm's law.

isc is then the sum of the current source current and the current upward through the 6 Ω resistor..

Thus, we have isc = 6 + (36/6) = 12 A.

A quick check shows that, indeed, voc/Req = 24 V/2 Ω = 12 A, as expected.

We may verify these parameters by direct analysis of the subcircuit. We attach a test voltage sourcev to the subcircuit:

Application of KCL to the upper node gives:

i + 6 = v − 366

+v3

i = v2−12

where we have used Ohm's law in order to determine the resistor currents.

Comparison with the Norton equivalent circuit expression:

i = vReq

− isc

gives isc = 12 A and Req = 2 Ω as expected.

Rearranging the equation in terms of voltage:

v = 2i + 24v = iR + voc

confirms voc = 24 V.

The Thevenin and Norton equivalent subcircuits are as follows:


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4 PRACTICAL SOURCES AND MATCHING

4.1 Resistance Matching a Subcircuit

Consider the following problem. A crystal microphone can be assumed to possess a Theveninequivalent circuit as follows:

Furthermore, a loudspeaker can be modelled to a good degree of approximation by the resistor RL

shown.

The current in the speaker is

iL =1 V

1 MΩ +8 Ω≈ 1 µA

This is quite a problem, because for a loudspeaker to be audible, current needs to be severalmilliamps. The auditory power equals the electrical power absorbed:

PL = vLiL = iL2RL = 8 ×10−12 = 8 pW

Consider the same circuit with more general element values:

We first assume that the Thevenin equivalent parameters voc and Req are specified and we are askedto determine the value of RL that results in the power absorbed by RL being the maximum possible.

Before solving this analytically, let's carry out a numerical study. Suppose that voc = 12 V and Req =4 Ω. The power absorbed by RL is:

PL = vLiL = iL2RL =

vocReq + RL

⎛

⎝⎜

⎞

⎠⎟

2

RL =voc2 RL

Req + RL( )2=

144RL4 + RL( )2

If we evaluate this expression at RL = 0 Ω we get PL = 0 W; at RL = ∞ Ω, again PL = 0 W.

If we evaluate the equation at several values of RL between zero and infinity, we obtain thefollowing plot:

We see that the power absorbed by the load exhibits a maximum of around 9 W for a finite value ofRL.


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Now let's solve the problem analytically. We can re-write the expression for PL:

PL =voc2 RL

Req + RL( )2=

voc2

ReqRL

+ RL⎛

⎝⎜⎞

⎠⎟

2 =voc2

Req1

ReqRL

+ RLReq

⎛

⎝⎜

⎞

⎠⎟

2

Maximising PL is obtained by minimising the denominator.

RL appears only within the bracket and the bracket term is of the form x + x–1. It can easily beshown that x + x–1 has a minimum value of 2 for x = 1. Hence the term in the bracket has aminimum value of 2 for √(Req/RL) = 1, ie:

RL = Req

for which case, we have:

PL max( ) =voc2

4Req

Therefore, for maximum power to be absorbed from a source with given voc and Req, the value ofthe load resistance must be numerically the same as Req.

Now consider the problem that the load resistance RL and the Thevenin equivalent voltage voc arefixed and the problem is to pick Req such that the power absorbed by the load is maximized.

Consider the previous expression:

PL =voc2 RL

Req + RL( )2We see that the condition for maximum power in the load is:

Req = 0

This means that for maximum power in a given load, the subcircuit represented by the Theveninequivalent must be an ideal voltage source.

The problem of a transferring power from a source to a load such that power in the load ismaximised is known as the matching problem.

1.2 Practical Sources

In theory, the Thevenin and Norton equivalent circuits for a subcircuit are equivalent to each otherand therefore the choice of which to use is arbitrary. However, for certain practical sources one orthe other of the equivalents may be preferred.

Consider a device called a photodiode which converts light into an electrical signal; it is animportant component in optical communication systems.

The photodiode is usually given the Norton equivalent circuit shown, where the current of 1 mA isthe value for a given light level:


15

The current source is the primary element in the equivalent circuit as its current represents the lightsignal applied to the photodiode and is amplified by the following circuitry. The followingamplifier is designed so that as much of the 1 mA current goes into the amplifier and is amplifiedand as little as possible goes into the 100 MΩ resistor.

The 100 MΩ resistor is not very significant and is a parasitic or unwanted element. If the 100 MΩresistor is removed, the model will be less accurate but voltages and currents in the whole systemwill not change that much.

We know that any Norton equivalent source can be turned into a Thevenin equivalent source usingthe source transformation. This leads to the following equivalent circuit for the photodiode:

(Note that voc = iscReq, = 1 mA × 100 MΩ = 100 kV)

The voltage of 100 kV does not exist in the physical device and is a mathematical artefact of theequivalent circuit. Also the 100 MΩ resistor is no longer a parasitic element and can not beremoved without destroying the operation of the circuit completely.

Thus, although this equivalence is valid, we can say that for the photodiode, the Norton equivalentis more natural than the Thevenin equivalent and closer to the physical mechanism of the device.

Consider now the car battery. Assuming that the terminal voltage has an open circuit values of 12V and falls to 6 V at a current of 600 A, we can determine the parameters of the Theveninequivalent circuit:

12 V

0.01 Ω

The 0.001 Ω resistor is a parasitic or unwanted element. If its value is set to zero, the model is lessaccurate but still represents the real situation approximately. Let us derive the Norton equivalentcircuit using the source transformation:

0.01 Ω1200 A

(Note that isc = voc/Req, = 12 V / 0.01 Ω = 1200 A)

Although this equivalent circuit is valid, there is nothing in the physical battery that corresponds tothe huge current of 1200 A which flows into the resistor even when the terminal current is zero. The0.01 Ω resistor is not a parasitic element and cannot be removed from the circuit without completelydestroying the equivalence.

Thus, in the case of the car battery, the Thevenin equivalent circuit is closer to the physicalconstruction and operation of the real device.


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In general, we see that although the Thevenin and Norton equivalent circuits are strictly equivalentto each other, when we represent a real device, such as a battery or a photodiode, one or other of theequivalents may be more appropriate than the other.

5 SOURCE TRANSPORTATION

Consider a 3-terminal subcircuit:

We have used the double subscript notation for the voltages between terminals 1 and 3 and between2 and 3.

By KVL, we see that these two voltages suffice to determine any other voltage relative to the threeterminals.

Similarly, KCL shows that the terminal currents i1 and i2 suffice to determine the current at terminal3.

Thus relationships between v13, v23, i1 and i2 suffice to completely describe the subcircuit.

5.1 Voltage Source Transportation

Consider now the following 3-terminal subcircuit:

For this circuit:

v13 = v23 = vsThese equations apply independently of the currents i1 and i2.

Consider now the following subcircuit:

Analysis leads to the same expression as the single source circuit. Thus, the two are equivalent.

The equivalence we have just derived is called voltage source transportation because we are"transporting" a v-source through a node.

Note that each circuit has the same 3 nodes. Connections of other elements to these 3 nodes are notchanged by this equivalence.

Note that for Thevenin and Norton equivalents, nodes inside the transformed subcircuit are changed(in most cases, they disappear).

Example 3.5-1


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Find the current i in the circuit shown using v-source transportation.

Solution

This circuit is neither a parallel circuit nor a series circuit.

We first apply v-source transportation to the v-source:

This circuit can be redrawn in a more familiar form:

We now apply the Thevenin equivalent transformation to the v-sources and their closely associatedresistors to obtain an equivalent form:

This is a series circuit that we can quickly analyze to determine:

i =veffectiveReffective

=−66

= −1 A

5.2 Current Source Transportation

Consider the following 3-terminal subcircuits containing current sources attached to two nodes:

Both circuits are described by:

i1 = −is i2 = is i3 = 0


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These equations apply independently of the applied voltages, v13 and v23.

Because the v-i characteristics are the same, the two 3-terminal current source subcircuits areequivalent as far as any external elements connected to the terminals are concerned.

This equivalence is called current source transportation.

In its usual application, a single current source between two nodes is replaced by two equal currentsources of correct polarity between that node and any other node.

Note that node 3 must be connected to elements through which current can flow; otherwise we havean invalid connection of two current sources in series.

Example 3.8

Find the current i in the following circuit:

Solution

We apply current source transportation to the 18-A source using the ground node as the 3rd node:

Combining the parallel i-sources, we have:

We now can apply the source transformation twice:

We easily find:


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i =veffectiveReffective

=6 − 28

11=−2211

= −2 A

6 SOURCE SUBSTITUTION

6.1 Voltage Source Substitution

Consider a portion of a circuit as shown:

We have a 2-terminal element e connected between two nodes, i and j, to which other elements (notshown) are connected, as indicated by conductors at these nodes.

We assume that the voltage v across e is a known quantity.

We now connect one lead of a voltage source to node j, leaving its other terminal free, and adjust itsvalue to be exactly v; this creates a floating node, i':

Because of the floating node the current through the added source is clearly zero - hence its additiondoes not affect any of the voltages or currents in the circuit.

Now let us attach a resistor having a very large value, say R = 1015 Ω, between nodes i and i':

What is the current through this resistor? Of course, one would expect it to be small because of thelarge value of resistance. But in fact the current is identically zero because the voltage across theresistor is identically zero!

This means that we can decrease the value of this resistance to zero and the resistor current willcontinue to be zero. We have neither added nor subtracted any current at nodes i and j, so we seethat our experimentation thus far has not affected any of the voltages or currents in the circuit in theslightest manner.

When the resistance has reduced to zero, thus producing a short circuit, we have the folllowing:

But a two-terminal subcircuit consisting of a voltage source and element in parallel can be replacedby the voltage source alone:


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Hence we have the following:

If the voltage across any two-terminal element or subcircuit is known, it can be removed andreplaced by a voltage source having that known value, without affecting any of the other voltagesor currents in the circuit.

This statement is often called the voltage form of the substitution theorem

The theorem applies to the case when the original element is a current source with a known voltageacross it.

1.2 Current Source Substitution

Consider the case of an element with known current i flowing in it:

Suppose that we add a shorted independent i-source and adjust its value to be i:

This has no effect on the circuit, but note that the current in the segment of conductor between the i-source connections is zero!

Since this current is identically zero, the conductor segment can be clipped out and replaced by anopen circuit without affecting any of the other voltages or currents in the rest of the circuit:

However, any subcircuit consisting of an i-source connected in series with any other element isequivalent to the current source alone:

Hence, we have the current source form of the substitution theorem:

If the current in any 2-terminal element or subcircuit is known, that element or subcircuit can bereplaced by a current source having the known value, as far as the external voltages and currentsare concerned.

The theorem applies to the case when the original element is a voltage source with a known currentacross it.

Example 3.9

Find the value of vs required to adjust the current i to zero in the following circuit:


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Solution

We first use current source substitution.

Suppose that vs has been adjusted to the value required to make i = 0. Then, because i = 0, the 2 Ωresistor can be replaced by an i-source whose value is zero, which is an open circuit:

We also know that the voltages va and vb should be identical.

The voltage divider rule gives va = 10 V and vb = (2/3)vs. In order that these voltages are equal, wehave vs = 15 V.

Now we use voltage source substitution.

Since we require i = 0 in the original circuit and i flows in a resistor, it follows that the voltageacross the resistor is zero. Hence, by the theorem, the resistor may be replaced by a voltage sourceof zero value, which is a short circuit:

But we also know that the current flowing between nodes a and b must be zero.

It follows from the voltage divider rule that va = 10 V. Also, vb = 10 V and by the voltage dividerrule, vs = 15 V.

2 CONCLUSIONS

In this topic we have introduced a number of equivalences, transformations and theorems:

• Source Transformation

• Thevenin and Norton Equivalent Subcircuits

• Voltage and Current Source Transportation

• Voltage and Current Source Substitution

These theorems extend the complexity of the circuits which we can analyse.

However, there is a limit on the extent to which we can extend such a piecemeal approach. We nowhave the background to be able, in the next topic, to develop a method for systematic analysis of acircuit of arbitrary complexity and any topology.

notes for course ee1.1 circuit analysis 2004-05 topic 3 ...cas.ee.ic.ac.uk/people/dhaigh/ee1.1...

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