2. DC Circuit Analysis

University of Aberdeen - School of Engineering

EG1008 – PRINCIPLES OF ELECTRONICS

3. DC CIRCUIT ANALYSIS

Series/parallel circuits. Kirchhoff's current and voltage laws. Current sources. Current/voltage source conversions. Circuits and theorems.

3.1 Series/parallel circuits

Many (but not all) circuits can be analysed, and designed, by considering them as combinations of series circuits and parallel circuits. The technique is simply:

identify parallel and series combinations of resistors

replace such combinations by their equivalent resistors

identify any parallel and series combinations of these equivalent resistors

continue until the circuit is reduced to a straightforward parallel or series circuit

finally use Ohm's law, or the voltage and current divider rules, to find the individual resistor voltages and currents if required.

EXAMPLES (i) Consider the circuit of Figure 4(a). What is the voltage across R3 and what current is drawn from the source?

The equivalent resistance of the parallel combination of R2 and R3 is R2||R3 = R2R3/(R2+R3) = 10´1/(10+1) = 0.909kW. Hence, by the voltage divider rule, the voltage across this combination (which is equal to V3) is

VS = 10 = 0.917V

and the equivalent resistance connected across the source is R1+(R2||R3) = 9.909kW, so that the current drawn is 10/9.909 = 1.009mA.

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Figure 4. (a) to (c) Series/parallel circuits. (d) A circuit which cannot be reduced to a series/parallel combination. Note the drawing convention in (c): T-junction means join, but X-“junction” actually means

no join. Alternatives are to show junctions with dots (as in the course text) or, less commonly, to show non-junctions as explicit “bridges”.

(ii) Derive an expression for the equivalent resistance of the circuit of Figure 4(b). If R2 = R4 = R5 = 1W and R1 = R3 = 2W, what is the value of this equivalent resistance?

R4 and R5 are in series, to give equivalent resistance R4 + R5. This is in parallel with R3 to give R3||(R4+R5) = R3(R4+R5)/(R3+R4+R5). This equivalent resistance is in series with R2 giving equivalent resistance R2 + R3 (R4+R5)/(R3+R4+R5) which, finally, is in parallel with R1 giving a final result for R:

= +

Substituting in the values given, 1/R = 0.5 + 1/[1 + 2´2/4] = 1S. Alternatively, and more directly, with these values, R4 + R5 = 2W, so R3||(R4+R5) = 2||2 = 1W, R2 + R3||(R4+R5) = 1 + 1 = 2W, and finally R1||[R2 + R3||(R4+R5)] = 2||2 = 1W.

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This procedure can work with quite large circuits, such as Figure 4(c):

R = [R1 + (R6||R7)]||[R2 + R3||(R4+R5)]

(which can be re-expressed if required, but in most practical cases would be better evaluated directly as in the alternative treatment for the Example given above). However, the procedure FAILS for Figure 4(d), even though at first glance the circuit may look less complicated than that of Figure 4(c): no sequence of parallel and series combinations can be found for Figure 4(d).

The methods so far described are important. They can be applied to give a complete analysis of a circuit in many cases (that is, they determine all voltages and currents). In such cases, it may be unneccesary to use any more advanced methods. However, there are important circumstances where they do not work - Figure 4(d) gives just one example. Other methods must therefore be explored.

3.2 Kirchhoff's current and voltage laws

These are just formal statements of rules which have already been used. They can, however, be used in turn to derive more advanced analysis methods. In parallel circuits, when adding currents flowing in parallel resistors and equating to the total current, we have already used:

Kirchoff's current law (KCL): at any node of a network, the sum of the currents flowing into the node is equal to the sum of the currents flowing out of the node.

Alternative statement of KCL: at any node of a network, the algebraic sum of the currents flowing is zero.

Note that a node is a point, or junction, where two or more circuit components are connected together. As in Figure 5, the point need not actually be drawn as a point, nor need it be a physical point in a circuit, as long as the currents are correctly identified. A network is simply an alternative word for a circuit (it tends to be used for more complicated patterns of interconnection). In the alternative statement of the law, currents flowing into the node and currents flowing out of the node are considered as having opposite signs: in Figure 5, currents flowing into a node are considered positive (but this is by no means a universal convention - either convention can be used as long is it is applied consistently).

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Figure 5. Example of Kirchhoff's current law. A node may be drawn either as a point or, often more conveniently, as an extended connection region: (b) is identical to (a), but redrawn. The law can also be applied to an arbitrary region of a network, as in (c). Note the alternative method of showing current, by

arrows on the conductors - this is more compact than separate arrows.

Similarly, in series circuits, when adding voltages across series resistors and equating to the driving e.m.f. we have already used:

Kirchhoff's voltage law (KVL): the algebraic sum of all the voltages across individual components around any loop of a circuit is zero.

As mentioned above, these laws have already been used in the treatment of series circuits (where it was assumed that the sum of resistor voltages is equal to source voltage) and parallel circuits (total current equal to sum of individual resistor currents). The laws can also be used directly to find currents and voltages in some circuits which are not series/parallel circuits, or where not all source voltage and resistance values of resistors are given.

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Figure 6. Example of Kirchhoff's voltage law.

Figure 7. Solving problems using KCL and KVL directly.

EXAMPLES (i) Find the unknown currents in the circuit of Figure 7(a).

Using KCL at node A: 2 - 5 - I1 = 0 or I1 = -3A

(this means that I1 actually flows opposite to the direction shown by the arrow)

Using KCL at node B: 1 + 3 + I1 - I2 = 0

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thus 4 + (-3) - I2 = 0 so I2 = 1A

Using KCL at node C: I2 + I4 - 4 = 0 so I4 = 3A

Check - using KCL for whole circuit: 2 - 5 + 1 + 3 - 4 + I4 = 0

or I2 = 3A, as required.

Note that the solution to the problem has been made simpler by the choice of order of nodes - starting with node B or C would give an equation with two unknowns.

(ii) Find the unknown voltages in the circuit of Figure 7(b).

KVL for loop JABKHJ: 12 + 9 - V2 = 0 so V2 = 21V

KVL for loop KBCLK: -9 - V1 + 6 = 0 so V1 = -3V

(this means that the sense of V1 is actually reversed - point L is more positive than point C)

KVL for loop KLGHK: -6 - V3 + V2 = 0

thus -6 - V3 + 21 = 0 so V3 = 15V

KVL for loop JABCDEFGHJ: 12 - 3 - V5 = 0 so V5 = 9V

KVL for loop LCDEL: V1 - 3 + V4 = 0 so V4 = 6V

(loop LCDEL includes the "open circuit" EL, across which V4 is measured: KVL is still valid when sections of a loop are open circuits)

Check - using KVL for loop HKBCLGH:

V2 - 9 - V1 - V3 = 0

or 21 - 9 - (-3) - 15 = 0, as required.

(this is just one of the checks that can be made). Note that labels in this example are not all nodes.

3.3 Current sources and current/voltage source conversions

So far, we have only considered circuits containing voltage sources (including batteries) and resistors. There is an alternative form of electrical source, useful both in practical circuits and for formal analysis, known as a current source. An ideal current source is defined as a source which supplies an unchanged current to a load, independent of the value of the load. This behaviour is, in a sense, opposite to that of the ideal voltage source, which supplies an unchanged voltage to a load, independent of the value of the load. Notice, in particular, that the voltage across the current source must change as the load changes.

Figure 8 shows the symbol for such a source, together with two possible implementations. Neither implementation is ideal: that of Figure 8(b) will, however, provide less than 1% current variation for loads up to about 10kW, and the circuit of Figure 8(c) might be somewhat better, as well as more practical. More advanced electronic circuits can give better performance.

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In a non-ideal current source, the current supplied by the source will generally fall as the load resistance rises. This behaviour can be modelled by placing a suitable equivalent resistance in parallel with an ideal source - the combination of ideal source and parallel resistor then constitute the model of the non-ideal source. If a change of terminal voltage DV leads to a change of current DI supplied by the non-ideal source, the value of parallel resistor is DV/DI.

Figure 8. The current source. (a) Basic symbol and circuit. (b) Simple implementation. (c) A more practical implementation, using a bipolar transistor (you will learn about transistors later).

Now return to the voltage source. Recall that a non-ideal voltage source, or non-ideal battery, can be modelled as an ideal voltage source in series with a resistor (Figure 1(c)). This model, or equivalent circuit, accounts for the way the voltage across the source varies as the load varies. However, as Figure 8(b) shows, a voltage source in series with a resistor can also implement a (non-ideal) current source. Hence, while an ideal voltage source is quite different in behaviour from an ideal current source, a non-ideal voltage source and a non-ideal current source behave in an essentially similar fashion. In fact, any non-ideal source can be modelled either as an ideal voltage source in series with a resistor or as an ideal current source in parallel with a resistor (Figure 9). These two circuits are called the Thévenin and Norton equivalent circuits of the source respectively, and each can give an equally good description of the source as far as the load is concerned.

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Figure 9. Thévenin and Norton equivalent circuits for a source.

Suppose a (non-ideal) voltage source is represented by its Thévenin equivalent. This is an ideal voltage source VT in series with a resistor RT (Figure 9). The open circuit voltage VOC (i.e. the terminal potential difference of the source when no load is connected) is just VT. The short circuit current ISC (i.e. the current flowing in a load of resistance equal to zero) is VT/RT:

VOC = VT

ISC = VT/RT

To represent the same source by its Norton equivalent, it is necessary to ensure the same values for VOC

and ISC. From Figure 9, the Norton current source current value IN and the parallel Norton resistance value RN are related to VOC and ISC by:

ISC = IN

VOC = INRN

Hence, for the Norton or current source representation to be equivalent to the Thévenin or voltage source representation:

VT = VOC = INRN

andIN = ISC = VT/RT

hence, given the Norton circuit, the equivalent Thévenin circuit values are:

RN = RT and IN = VT/RT

or, given the Norton circuit, the equivalent Thévenin circuit values are:

RT = RN and VT = INRN

As a check on this, the current into a load RL is:

IL(T)

= (Thévenin equivalent)

IL(N)

= (Norton equivalent)

(the current divider rule has been used here). But, using RN = RT and IN = VT/RT, the Norton equivalent form gives

IL(N)

= = IL(T)

as required.

These relations thus allow a (non-ideal) current source to be converted to a voltage source, and vice-versa. Notice that the behaviour is equivalent only as far as the load is concerned: the behaviour inside the source is quite different. For instance, if there is no load connected, the Norton (current source) form apparently dissipates heat inside the source, while the Thévenin equivalent does not. This discrepancy does not, however, affect the behaviour as far as the load is concerned.

The value of RT (or RN - since for any given source RN = RT) is frequently referred to as the source resistance, or the internal resistance or the output resistance of the source, or sometimes the resistance

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"looking into the terminals" of the source. Note that the ideal current source can now be considered as a source with infinite internal resistance (RN or RT), while the ideal voltage source has zero internal resistance.

Source conversions are sometimes of particular value in dealing with circuits containing multiple sources, including both current and voltage types. An example is shown in the next section (1.3.4).

EXAMPLES (i) A voltage source has an e.m.f. of 12V and a source resistance of 1W. What is the current source which is equivalent to this voltage source?

From the description of the source, its Thévenin equivalent has VT = 12V and RT = 1W. The equivalent current source is just the Norton equivalent circuit shown inside the dashed line in Figure 9(a). Using the relationships given above, the Norton equivalent values are

RN = RT = 1W and IN = VT/RT = 12A

(ii) A source produces an open circuit voltage of 10V, and a load current of 1mA when connected to a 1kW load resistor. Derive the Thévenin and Norton equivalents of this source.

Using the Thévenin equivalent circuit (Figure 9(b)) VT = VOC = 10V. Also, when RL = 1kW and IL = 1mA,

VL = ILRL = 1V

Hence the voltage drop across the equivalent source resistance RT is VT - VL = 10 - 1 = 9V. Thus ILRT = 9V, and so RT = 9kW. The Norton equivalents are thus RN = 9kW and IN = 10/9 = 1.11mA.

Note that the calculation of this source resistance is essentially the same as in Section 1.1.3 for a non-ideal voltage source. It is equally possible, but in this case slightly less convenient, to use the Norton equivalent circuit.

3.4 Thévenin theorems

It turns out that the results of the last section for voltage and current sources can be extended to more complicated circuits in a way which is straightforward but very useful.

Thévenin's theorem. As far as any load is concerned, any network consisting of ideal voltage sources, ideal current sources and resistors can be replaced by a single ideal voltage source VT in series with a resistance RT. The value of VT is equal to the open circuit voltage of the network. The value of RT is the resistance which would be measured between the terminals of the network if all ideal voltage sources were replaced by short circuits, and all ideal current sources by open circuits (the load being removed).

Note that, although formulated in terms of ideal sources, the fact that non-ideal sources can be represented by equivalent circuits involving ideal sources means this applies to all circuits we have met, or will meet. Apart from its extension to circuits rather than sources, the only new idea here is the procedure for finding RT.

Both the method of application and the value of Thévenin's theorem can be illustrated by examples.

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Figure 10. (a) Example for use of Thévenin's theorem.

EXAMPLES (i) Consider the circuit of Figure 10(a). Find the Thévenin equivalent for the network connected to load resistor R.

First, mark the terminals to distinguish the network for which the equivalent is to be calculated.

Second, find RT. Remove the load, replace the (ideal) voltage source by a short circuit and the (ideal) current source by an open circuit (Figure 10(a.ii)). In this case RT is a straightforward series/parallel combination, as given in the diagram.

Third, find VT. Restore both the sources. Leave the load disconnected, since VT is found as the open circuit output voltage VOC. In this case VOC is most easily found by converting the voltage source (15V in series with 5kW) into the equivalent current source (15/5 = 3mA in parallel with 5kW). The two current sources can then be combined as a single current source with strength equal to the algebraic sum of the individual strengths, in parallel with the parallel combination of the individual resistors, as in Figure 10(a.iii). Note that there is no current through, and hence no voltage drop across, the 2kW resistor when the network is open circuited.

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Figure 10. (b) Thévenin equivalent of potential divider.

(ii) In the potential divider circuit of Figure 10(b), the output voltage is reduced from its intended open circuit value if a load is connected. What value of load resistance RL will reduce the output voltage by 50%?

Replace the potential divider circuit, to the right of the marked terminals in Figure 10(b.i), by its Thévenin equivalent (strictly, only RT is needed for this question). It is now obvious that a value of RL equal to RT = 0.8kW will reduce the output voltage to 50% of its intended value (which is the open circuit voltage 0.2Vin).

(iii) Figure 10(c) shows a transistor amplifier. If the base current IB is 10mA, what is the base voltage VB?

Separate the base network from the “load”, which consists of the transistor, and resistors RE and RC. The base network is shown in Figure 10(c.ii). The base network can now be replaced by its Thévenin equivalent. (The “load” is non-linear, but this does not affect the validity of the theorem, since the base network is linear.) The calculation of VT and RT is carried out as usual. In this case, it is

convenient to replace both voltage sources by current source equivalents (20mA in parallel with 50kW, and 30mA in parallel with 500kW, respectively) so that RT = 50||100||500 = 31.25kW and VT = (0.03+0.02)´31.25 = 1.5625V. The base current now flows through RT so that the base voltage

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VB = VT - IBRT = 1.5625 - 0.01´31.25 = 1.25V

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4. SYSTEMATIC SOLUTION OF DC NETWORKS – NODAL ANALYSIS

4.1 Introduction

So far, all circuits encountered with the exception of Figure 4(d) have been possible to analyse using Ohm's law and the rules for combining resistors in series and parallel. Other rules, such as the voltage divider rule and Thévenin's theorem, provide considerable simplification, but the basis of solution is the same. Solutions have, in fact, consisted of identifying suitable rules to use, and choosing suitable series/parallel combinations of resistors.

The aim of this section is to introduce systematic methods of solution which will allow the analysis of any arbitrary arrangement of sources and resistors, and will provide a general way of doing so which does not rely on identifying special rules or choosing particular approaches. This means that, given a circuit with specified source and resistor values, the method should allow all currents and voltages to be found with as straightforward a procedure as possible.

Two common methods are nodal analysis and mesh analysis. Both rely on systematic application of Kirchhoff's laws. Both are well suited to manual analysis and, because of their systematic nature, also to computer analysis. Nodal analysis provides the basis of all electronic CAD systems such as SPICE.

4.2 Nodal analysis

Recall that a node is a junction, that is a point or extended connection region where two or more components are connected together - see Figure 5. A branch is simply a section of the network between two nodes - it will generally comprise a source or a resistor. Having identified all nodes of the circuit, the procedure for nodal analysis is as follows:

choose one node as a reference node, which will be considered to have zero voltage. The choice should not affect the final results, but there is often a fairly obvious convenient choice e.g. a junction of several resistors, or of two or more voltage sources.

assign node voltages V1, V2, V3, etc. to all the nodes other than the reference node (the labelling is arbitrary).

at any known node whose voltage is fixed by an ideal voltage source, the corresponding node voltage may be written down at once.

at all nodes other than the reference node or known nodes, write down Kirchhoff's current law in terms of the branch currents flowing into and out of the node. These currents are found from Ohm's law in the form (Vi - Vj)/Rij where i, j are the numbers labelling the nodes and Rij the resistance of the branch between them - unless a current is fixed by a current source.

solve the resulting set of simultaneous equations (the nodal equations) for the unknown node voltages in terms of the known voltage and current sources (and the known resistance values).

complete the analysis (if required) by working out the branch voltages and branch currents.

EXAMPLES (i) Carry out a nodal analysis of the potential divider circuit (of Figure 10(b) and 11) and hence obtain the output voltage Vout.

The circuit has just three nodes, as shown shaded in Figure 12, which also shows a convenient choice of reference node.

For the purposes of the nodal analysis, all resistors including RL are

treated as known, as is the one source Vin.

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Clearly, V1 = Vin (i.e. node 1 is a “known” node).

Figure 12. Nodal analysis for potential divider circuit. Note the symbol for the reference node.

We therefore need to write KCL for node 2. The relevant branch currents are labelled in Figure 12 (note the subscripts for the branch currents have no direct relation to the node subscripts). In terms of the node voltages:

I1 = (Vin - V2)/R1

I2 = (V2 - 0)/R2

I3 = (V2 - 0)/RL

and so KCL for node 2:

I1 - I2 - I3 = 0or

- - = 0or

V2 = Vin

This is the required result, since Vout = V2. Individual branch currents I1, I2, I3 can be written down explicitly in terms of Vin if required.

Note that it would be almost as convenient to choose node 1 as reference, but then the calculation of V out

requires two node voltages. Choice of node 2 as reference would be much less convenient, since it would then be impossible to write down any known node voltages directly. In this case, nodal analysis, although giving a complete result, gives it in a less helpful form than the analysis using the Thévenin equivalent.

Nodal analysis works (always) because the assignment of node voltages automatically satisfies KVL. Writing KCL at each of N (unknown) nodes then gives N equations in the N unknown node voltages.

(ii) Find the current drawn from the source in the circuit of Figure 13.

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This looks at first sight a familiar problem, similar to those already solved using series/parallel resistor combinations. However, the circuit is the network first encountered in Figure 4(d), which cannot be considered as a series/parallel combination. Therefore, even though only one current is to be evaluated, a full nodal analysis is required. This is not particularly hard, but there is no obvious short cut.

Figure 13. (a) Example for nodal analysis. (b) Circuit labelled to show reference node, nodal voltages V1 = VS, V2 and V3, and branch currents IS, I1, I2, I3, I4, I5.

Figure 13 shows the (four) nodes, a convenient choice of reference node, the nodal voltages, and the branch currents. Node 1 is a known node V1 = Vs = 10V, so we need to write down KCL for nodes 2 and 3:

KCL for node 2:

I2 - I3 - I4 = 0or

- - = 0or

V2 - V3 = VS (i)

KCL for node 3:

I1 + I4 - I5 = 0or

+ - = 0or

- V2 + V3 = VS (ii)

Substituting values for VS and R1 to R5 as given, we have

V2 - V3 = 10 from (i)

- V2 + V3 = 10 from (ii)or

V2 - 0.25 V3 = 10 ´ 0.5and

- 0.25 V2 + V3 = 10

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or0.85 V2 - 0.25 V3 = 5 (i’)

and- 0.25 V2 + 1.45 V3 = 10 (ii’)

This pair of simultaneous equations can be solved by elimination (or any other convenient means). Multiplying (i’) by (-0.25/0.85) and subtracting the result from (ii’):

[1.45 - (-0.25)´(-0.25/0.85)] V3 = 10 - 5´(-0.25/0.85)

or (working to 5 significant figures)

1.3765 V3 = 11.471 or V3 = 8.3332V

and, back-substituting into (i’):

0.85 V2 - 0.25´8.3332 = 5 or V2 = 8.3333V

This completes the nodal analysis. Finally, we need

IS = I1 + I2 (KCL at node 1)

= +

= + = 2.5002mA

In this case, because of the particular resistor values, it turns out there was a short cut: R 1 and R5 form a potential divider having the same division ratio as R2 and R3. Hence no current flows in R4, and V2 = V3

= (5/6)´10 = 8.3333V. The circuit is, in fact, a "balanced bridge". In general, however, non-zero current will flow in R4, and this trick will not work.

To summarise: nodal analysis is “slow but sure”.

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