5

A circuit analysis toolkit

Introduction

This chapter is intended as a quick guide to the basic techniques of network analysis. The mathematical requirements are minimal; you need to be able to transpose equations and solve simultaneous equations by substitution. We also get to consider some characteristics of real voltage and current sources, which ties in nicely with some of the analysis tools presented.

We will take DC circuits for all of our examples, as this is the easiest way of demonstrating the techniques. Everything said applies equally to AC circuits as well, but you generally need to use complex numbers (see Chapter 4 and Appendix 4) to take account of phase if there are capacitors and inductors present, or sources at different phases.

Basic topologies

There are two basic shapes that electric and electronic circuits come in; series and parallel. Each has associated with it one of two electrical laws which tell us how to go about calculating the voltages and currents in that circuit.

The series circuit and Kirchhoff’s voltage law

A series circuit is represented in Figure 5.1. The components are an input voltage, V,10 V DC, and three resistors: R1, 1k; R2, 2k2; and R3, 4k7. We have also labelled on the drawing, for our convenience, the voltages across each resistor, V1, V2 and V3. The voltages are labelled with arrows. With DC voltages we follow a convention of putting the head of the arrow to the more positive end of the component.

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Figure 5.1 Series circuit

The key to analysing a series circuit is to remember that the current is identical through each component in the loop. (We can also say that connecting the components in a different order will not affect that current.) Hence the current in the loop is labelled just once, this time with an arrow on the loop, with the legend I next to it to identify it. The head of the arrow shows the direction of current flow.

Example 5.1: Suppose that we wish to calculate V1, V2 and V3. How do we do it? We need to find the value of current, I. Then we will be able to calculate each voltage using Ohm’s Law.

We can find the total resistance as described in Chapter 2. In our case:

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Now we can say that:

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Now we can use Ohm’s Law to find V1, V2 and V3:

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Kirchhoff’s voltage law states that, in any loop of voltages on a circuit, their sum will be zero if we take the clockwise ones as being positive, the anticlockwise ones as negative (or vice versa). Put another way, the clockwise ones equal the anticlockwise ones. This enables us to check our calculations: 1.27 + 2.78 + 5.95 = 10V – we did it right. In any mathematical analysis of a circuit, its always good to have a way to check our result, and using voltage loops is one way.

The parallel circuit and Kirchhoff’s current law

Example 5.2: Figure 5.2 shows a basic parallel circuit made with the same components. With parallel circuits the rule is that the voltage across all the branches is identical. Hence there is only one voltage to be labelled, the source voltage. But there are four currents: I, the current leaving the voltage source, and I1, I2 and I3, the currents through the resistors. We would like to calculate them.

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Figure 5.2 Parallel circuit

We can see that the voltage across each resistor is the same; the 10 V source voltage. Therefore we can apply Ohm’s Law directly to each branch of the circuit to find I1, I2 and I3:

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Kirchhoff’s current law states that the total of currents entering and leaving any ‘node’ (a node is any place where two or more conductors join) is zero. This enables us to calculate the current I:

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We know that the resistance between any two points is the ratio of the voltage across them and the current passing between them. So in this case the resistance is:

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We could have calculated the parallel resistance directly, by using the formula from Chapter 2:

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So:

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Again we have cross-checked our result, and it’s fine.

Arrow directions for voltages and currents

It is worth taking a moment to think about the rules that we apply when adding voltages and currents in using Kirchhoff’s two theorems, as it can be surprisingly easy to confuse ourselves.

Here’s the rules. If we follow these, things work:

1. The arrows for voltage across and current through a voltage or current source must point in the same direction.

2. The arrows for voltage across and current through a resistance must point in opposite directions.

3. For a voltage loop, the sum of arrows going clockwise equals the sum of arrows going anti-clockwise.

4. For a node, the sum of currents entering a node equals the sum of currents leaving the node, as denoted by the arrows’ directions.

In the above examples, it was obvious which end of the resistors would be more positive and in which direction currents would be flowing. Hence drawing the arrows was easy.

In other instances it will not be obvious, when labelling currents and voltages at the start of the analysis, which end of the component is the more positive. Such a situation is shown in Figure 5.3. We wish to know the voltage V2, but this voltage is affected by both voltage and current sources which have a tendency to create voltages of opposite polarity across R2. The way to approach this sort of problem is not to worry about the polarity of the voltage; if we label it in the wrong direction, the voltage will come out as negative in solving the equations. We simply get on with labelling the voltages and currents (following rules 1 and 2), then we can write equations for voltage loops (using rule 3, the voltage law) or nodes (with rule 4, the current law).

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Figure 5.3 Example 5.3

Example 5.3: Solve Figure 5.3.

For node A, we can write:

I1 = I2 + 10 mA

I2 = I1 – 10 mA (call this Eqn (A))

For the loop on the left we can write:

VIN = V1 + V2

V2 = VINV1

(but V1 = I1.R1 = 10k.I1 and V2 = I2.R2 = 2k2.I2)

    so:

2k2.I2 = 5 – 10k.I1

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I2 = 2.27 mA − 4.55 I1

We can substitute this back into Eqn (A), giving:

2.27 mA − 4.55.I1 = I1 −10 mA

I1 (1 + 4.55) = 2.27 mA + 10 mA

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So, again from Equation A:

I2 = 2.21 mA − 10 mA = −7.79 mA

and therefore

V2 = I2.R2 = −17.1V

and

V1 = 2.21mA × 10k = 22.1V

Our solution is: V1 = 22.1 V, I1 = 2.21 mA, V2 = −17.1 V, I2 = −7.79 mA. We can now redraw the circuit, reversing the direction of the arrows where our solution gave a minus sign (Figure 5.4). A few seconds of calculator-punching demonstrates that nothing went wrong with the number crunching, and the solution obeys the voltage and current laws. There are quicker ways to get to this result, but as a ‘swiss army knife’ technique, this is a good one to have. Soon we get some better ones still.

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Figure 5.4 Solution to Example 5.3

More about voltage and current sources

The voltage and current sources that we considered above were ‘ideal’ – that is, we consider that they will supply their correct voltage or current, regardless of what load is placed on them. (As such, they have no real existence.) For the ideal voltage source, we could say that the change in voltage across it is zero for any change in output current. This means that a notional quantity which we call its ‘internal resistance’ is also zero, by Ohm’s Law. For the current source we could say that the change in output current is zero for any change in output voltage. Hence its internal resistance is infinite. Real voltage and current sources are not ideal, and we have a way to represent this in our circuits (see Figure 5.5).

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Figure 5.5 Representation of non-ideal sources

We give our ideal voltage source a series resistor, Rs. The closer this resistor is to a short circuit the closer the source is to ideal. If we connect a real voltage source to too low a load resistance the voltage across the load will drop (and the source could be damaged).

The current source is given a parallel resistor, RP; the closer this is to an open circuit, the better the source. If we connect a real current source to too high a load resistance the current through the load will drop.

These circuit representations of real voltage and current sources are invaluable for circuit analysis, as we will now see. It is possible for any network of voltage sources, current sources and resistances to be replaced by either a single ideal voltage source with a series resistor, or a single ideal current source with a parallel resistor.

Thévenin’s and Norton’s equivalent circuits

The techniques that we used in the last example of circuit analysis are fine for instances where we can get a solution for the circuit by writing two or three simultaneous equations and solving them. If the network gets much more complicated than that, however, the amount of maths involved can get pretty tedious, and it can get time consuming to arrive at the correct result without making a numerical error. Also, we have only solved the network for one value of load resistance; if we want to see what happens with a different value, we must do the calculations again. This is where replacing whole lumps of circuit with an equivalent non-ideal current or voltage source is great.

Here’s how we go about finding the right values for our equivalent circuit:

Replace with a voltage source (Thévenin’s equivalent circuit):

1. Cut the network at two places to isolate the section to be replaced.

2. Calculate the voltage at the load terminals with no load connected (called the ‘open circuit voltage’ or VTH).

3. Calculate the equivalent resistance looking into the load terminals when any voltage sources are replaced with a short circuit and any current sources with an open circuit (called the ‘output resistance’ or RTH).

Example 5.4: What is the voltage across RL in Figure 5.6?

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Figure 5.6 Example 5.4

First we separate out the 12 Vand the 3 mA source, and combine then into a single voltage source(– Figure 5.7). The points where we cut are labelled A and B. Replacing the 12 V source with a short and the 3 mA one with an open, we get RTH = 1k + 1k = 2k (Figure 5.8). When we calculate the open circuit voltage, we get 3 mA × 1k = 3 V across R2, and nothing across R3, which has no current flowing in it. Hence VTH is 12 − 3 = 9 V.

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Figure 5.7 Figure 5.6 redrawn before simplifying

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Figure 5.8 Finding Rth and Vth

The equivalent circuit is replaced into the original in Figure 5.9. Things are getting easier! Now we can cut and replace at the load terminals themselves, L1 and L2 This time we’ll replace with a current source.

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Figure 5.9 Vth and Rth replaced

Looking into L1 and L2 with both sources replaced by a short, we get 2k in parallel with 2k, so RN is 1k. (Figure 5.10).

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Figure 5.10 Finding IN and RN

To calculate short circuit current, we replace RL with a short, and then calculate the current through the two resistors; 2.5 mA and 4.5 mA as shown on the drawing, taking care to label directions correctly. IN is then clearly the difference between the two (by Kirchhoff’s current law), 2 mA. So Figure 5.11 shows the equivalent circuit with R1 now connected. If R1 is 1 kΩ, parallel resistance with RN is 500 Ω, and voltage across R1 is 1 V, terminal L1 being most positive.

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Figure 5.11 Equivalent circuit to Figure 5.6

Naturally we want to check, and the easiest way to do this is to redraw the original circuit, labelling in voltages and currents, using Ohm’s Law and voltage and current laws, and working backwards from R1. If there is a problem it will become apparent. We find we are OK (see Figure 5.12). We now could find the current through any other load using our existing solution – no need to do all that number-crunching again.

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Figure 5.12 Solution of Figure 5.6

The network can now be replaced with an ideal voltage source equivalent to VTH, in series with a resistor equivalent to RTH.

Replace with a current source (Norton’s equivalent circuit):

1. Cut the network at two places to isolate the section to be replaced.

2. Calculate the current at the load terminals when they are short circuited (called the ‘short circuit current’ or IN).

3. Calculate the equivalent resistance looking into the load terminals when any voltage sources are replaced with a short circuit and any current sources with an open circuit (called the ‘output resistance’ or RN).

The network can now be replaced with an ideal current equivalent to IN, in parallel with a resistor equivalent to RN.

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