The practice of designing a new circuit, or understanding an existing circuit, is known as “circuit analysis”.

While circuits can be copied and even modified without thoroughly understanding how they work, any serious hobbyist should try to understand as much as they can about circuits and circuit analysis.

Circuits can be grouped into five types: simple, series, parallel, compound, and complex. The aim of this article is to help you understand the first four, so as to provide you with the tools for delving into the world of most circuits.

To recap from last month and previous issues of DIYODE, every circuit must form at least one complete circle of conductive material in order for the current to flow. A circuit consists of the source (of energy), path (the circuit) and sink (the load where the energy is converted into another form.) There can be many sources, paths and sinks in a circuit connected in series, parallel, or compound form.

Complex circuits comply with all of the rules of electricity, but the method of solving them is what earns them their name.

NODES AND LOOPS

Both nodes and loops have been discussed before, but for circuit analysis we need to be sure we’re on the same page. A simple circuit has only one loop, which is also called the “path”, and in a simple circuit it is the circuit. The loop has two nodes: one is the supply rail, or positive connection, and the other is the ground, or negative connection.

Series circuits still only have one loop or path, and all of the components are on that loop. Series circuits have more than two nodes – with one between each pair of components.

Switches and instruments do not normally require extra nodes as they are usually considered to be a short circuit, or an open circuit.

Our circuit however, show nodes between them; mostly to demonstrate more nodes [1]. Circuits for analysis would only show those things that are required to perform the calculations. The nodes have been highlighted with coloured blocks, representing the fact that any conductor/connection that joins all those components with an effective low resistance, approximately zero Ohms, counts as a single node.

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In analysis, a conductor with a great enough resistance to affect the circuit will be shown as a resistor. There are five nodes shown, but nodes 1, 2, and 3 are the same node, as long as the switch and ammeter have very low resistances compared to the resistors in the circuit.

A few loops are also shown in colour. There are two more loops not shown, but they’re for you to find. The reason for identifying the loops is that in any loop, all the voltages add up to zero – (∑V=0) – or in case that confuses you, Kirchhoff’s Law states “The sum of the applied EMFs must equal the sum of the voltage drops”: (∑E=∑V).

Loop analysis recognises the first rule, including within loops that have no applied EMF. The middle loop, for example, which would normally be numbered, contains R1, R2 and R3 all in series, but no power supply. The voltage across R1 is equal to the voltage across R2 added to R3, and that satisfies the first rule, as within the loop the voltages add up to zero.

EQUIVALENT CIRCUIT

The first step in analysing a circuit is to remove distractions not involved in the maths. For example, we can show the same circuit without the switch or instruments [2], as they are not expected to alter the circuit. Just remember that a perfect voltmeter would have an impedance (resistance) of infinity, and a perfect ammeter would be expected to have zero resistance.

The switch is either a short circuit, zero resistance, or an open circuit – but we’re not interested in the circuit values without power!

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Two arrow symbols are applied to each component in the circuit, at least while learning about circuit analysis, which are very important to complex circuits.

An arrowhead is placed on a wire to show the direction of current flow. This is worth confirming in your own mind, and a good practice to adopt. Some would add one direction arrow for each current loop, while others prefer one for each component. What you prefer is individual, but as current can only flow one way at any time in a series of components, only one method is essential.

The other arrow is placed beside the component to represent the voltage across that component. Remember that the positive (+ve) represents the most positive end of the arrow and, therefore, the arrowhead.

An applied voltage or EMF (E) arrow points the same direction as the current arrow, but a voltage drop (V or PD) arrow points against the current flow, which means that current flowing through that component generates the opposing voltage, or voltage drop.

The only other nomination on this diagram are the (+) and (–) signs at the point expected to be the positive reference, and the negative or ground reference. This is not essential, but is also good practice.

Complex circuits with multiple power supplies will often have the voltage nominated; for example (+12V) or (+9V).

In complex circuits, the current direction in some components may not be determined before the maths is performed; therefore, the arrows may both point in the wrong direction. You may only realise it when you calculate a voltage or current with a negative value.

This is part of working with complex circuits; just don’t start changing arrow directions yet as that is often the cause of many confusing errors.

In equivalent circuits, some other changes are made for simplicity. A simplified circuit may use three cells made into a battery [3]. The equivalent circuit should show a single battery symbol.

The lamp in the circuit can be made into a SPICE model, but we can simply treat it as a resistor for this circuit; hence, we draw a resistor instead, and treat the lamp as a simple resistor.

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SERIES CIRCUITS

If you read this series you may remember we explained how to use VIRP tables to calculate the values for voltage, current, resistance and power for a circuit [4]. Rather than repeat this information, we’ll soon do it to the parallel circuit that follows, so you can see how to use VIRP for series, even if you missed the last issue.

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Please do convert this circuit to an equivalent circuit, and add arrows and any other notations that will help you calculate the missing values.

Replace the cells with a single battery of the appropriate voltage. Remove the ammeter and voltmeter but note the current and voltages against the correct components.

Kirchhoff’s Voltage Law (KVL) is the most important law for series circuits or components in series, or in a loop. However, as the current in a series circuit is the same in every element in the series circuit, Ohm’s Law makes it clear that the resistances in a series circuit add up to the total resistance, which equates to the equivalent resistance of the whole circuit.

In addition, since power is calculated from voltage and current, the power in the series circuit will also add up to the total power used by the circuit.

All of these relationships can be summarised into one simple diagram [5], which my electrical students were always instructed to copy into their notes.

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You may be wondering why we included a component named “Rn”. The notation “n” represents any number. That means that 100 resistors could be added together if they are in series, but even better if they are the same value, as you could simply multiply that value by the number of resistors.

High voltage probes often have a number of resistors in series, partly for higher power rating, but mostly for a higher voltage rating than a single resistor. A string of 1MΩ or even 10MΩ resistors might be used for measuring the voltage from an electric fence, if you are keen to do such a thing!

PARALLEL CIRCUITS

If you have parallel circuit, as you might find in a book [6], you can begin by converting it to an equivalent circuit, using the process we described for a series circuit.

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Remove the switch and instruments and use a simple symbol for the battery (i.e., draw it as a cell). The parallel circuit has the same voltage across each component. Therefore, students of electricity can simply treat the parallel circuit as, in this case, three separate circuits, all powered by a single supply. However, what if the third resistor did not provide any details – at least not on the component itself? What if, instead of current for this resistor, a total circuit current was given?

Kirchhoff’s Current Law comes into use here by telling us that “The sum of the currents entering a node is equal to the sum of the currents leaving a node”; or more simply, “The sum of all current at any node is zero” – i.e., (∑I=0).

It makes sense if you remember that current is moving, so it cannot loiter in a node; it must either enter or leave.

By calculating the current in the other branches of the parallel circuit, and assuming you were given the voltage, and either the total current or total power, Kirchhoff would provide a means of calculating the missing current as:

(I_total = I₁ + I₂ + I₃)

Here's another formulae summary straight off a TAFE blackboard in an Electrical Apprentices’ class [7]. They didn’t need to learn about nodes, but you can add to the bottom of the diagram, for your own memory’s sake: “Parallel circuits have two nodes and two or more paths”.

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The formula for total resistance can be memorised, but is a good exercise in maths to transpose it from other formulae:

If I_T = I₁ + I₂ + I₃ + … I_n and I = V/R

then V_T/R_T = V₁/R₁ + V₂/R₂ + V₃/R₃ + … V_n/R_n

but V_T = V₁ = V₂ = V₃ = … V_n

therefore V_T/R_T = V_T(1/R₁ + 1/R₂ + 1/R₃ + … 1/R_n)

and 1/R_T = 1/R₁ + 1/R₂ + 1/R₃ + … 1/R_n (V_T cancels)

or R_T = 1/( 1/R₁ + 1/R₂ + 1/R₃ + … 1/R_n)

If there are only two resistors, the total can be easily found without a calculator:

R_T = 1/(1/R₁ + 1/R₂) = R₁ x R₂/(R₁+R₂).

Note also that the equivalent resistance of any number of same value parallel resistors, can be found by dividing that value by the number of resistors:

R_T = Rn/n.

VIRP TABLE FOR A PARALLEL CIRCUIT

The VIRP table provided uses five columns. Some prefer to place the power supply first, but as it is usually the total voltage, current, resistance and power, others often prefer it to appear on the last line. Either way works as long as you know what each column and row represents.

Begin by making a blank table and adding the known values from the circuit.

	V	I	R	P
E	6
R1			10
R2				6
R3		0.16

We know the voltage is the same for every component in a parallel circuit so all components will have 6V across them. Fill the V column with 6V.

Then every row, except the first will have two values in them. Ohm’s Law and the Power Rule can be used to find the missing values for each row.

	V	I	R	P
E	6	1.76	3.4	10.56
R1	6	0.6	10	3.6
R2	6	1	6	6
R3	6	0.16	37.5	0.96

The total values are found by adding the values in the “I” and “P” columns, and by then using the parallel resistor formula to find total resistance. Since we all have calculators, a simple procedure for parallel resistors can be performed on any calculator using the inverse key:

[1/x] button.

Press the keys according to your calculator:

“R1”, [1/x], +, “R2”, [1/x], +, “R3”, [1/x], [=], [1/x], [=] (if required).

It is very important you remember the last [1/x]. Also, some calculators won’t do it all unless you press [enter] or [=] to conclude.

VIRP FOR A COMPOUND CIRCUIT

Compound circuits are simply a mix of series and parallel elements. They can be worked out by VIRP tables, but once the basics have been learned, it’s ideal to move on from the table method. VIRP tables are a way to make the work obvious, by having obvious blank spaces to encourage you to find the value.

Even though we are moving forwards quite fast, I won’t be using them here. Consider a simple compound circuit with three values of resistor, and a battery [8].

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As previously, we begin by drawing our own equivalent circuit. Ignore the switch, ammeter, and voltmeter for now, and draw the circuit just for the maths. This is often the most important step, as drawing the circuit causes you to look at every component, and every connection. Be very careful to include all components and connections.

Add values to the components to which they refer: battery voltage, resistance of R1, power given for R2, and current in R3.

In this circuit [9] the battery has the +ve at the top, so draw the potential arrow vertical pointing up. The current will flow in the same direction as this arrow. In some circuits, the battery may be getting charged, but draw the voltage arrow and current arrowhead upwards, and allow the calculations to show the battery being charged by giving a negative value.

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The current will flow clockwise back to the battery, being split into R1 and R2 circuits. As loads, the resistor voltages will be voltage drops, so the voltage arrows will be opposite to the current flow arrows.

Now, just like in the VIRP table, add the values that you know due to series and parallel rules, and KVL and KCL. That is, voltages the same for parallel components, and current for series components. [10] shows R1 marked at 6V, due to being in parallel with the battery, but consider also that the loop has a battery of 6V, so the only other component in that loop must have a voltage drop of 6V.

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R2 is in series with R3 and therefore, they must have the same current flow, which means R2 current is also 100mA.

Next comes a little application of Ohm’s Law and the Power Rule. R1 resistance and voltage means that current can be calculated as 0.6A or 600mA [11]. R2 has both power and current known, so the voltage can be calculated as 2V, and the resistance can be calculated as 10Ω.

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The only unknown remaining is R3, but voltage can be calculated from KVL as 6 – 2 = 4V. Therefore R3 = 4V/0.1A = 40Ω.

You can tidy up and prove your answers by calculating our, as yet, uncalculated values, and then using them to cross-check any value, including the values originally provided.

TOPOLOGIES

Teachers tend to continually seek new ways to connect circuits that students are asked to solve. Although circuits can be drawn in convoluted ways, perhaps by swapping a resistor and the battery, these eight circuits [12] are the eight topologies that can be drawn from four components and a battery, which is the suggested limit for TAFE electrical apprentices.

For four components, there can only be three relationships between the pairs, and [12] is a binary list of the eight circuits. Programmers may be able to see how it’s made. What the list introduces is a means of writing circuits as a formula.

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Resistors are added in series circuits, so the + sign represents a series relationship between two components.

Parallel is a little more complex, but knowing how to add two resistors in parallel allows us to use a common abbreviation for parallel. The “//” symbol has long been used to represent two parallel lines in geometry.

In the electrical sense, “//” simply means that the two components must be calculated as parallel values. For example, RT = R1 x R2/(R1+R2) or RT = 1/(1/R2 + 1/R2).

Looking at the circuits and at the formulae should help clarify this concept. Avid circuit analysers, and aspiring electricians and engineers might like to try themselves out with these circuits and values.

An Exercise In Resistance

This circuit was drawn up to challenge those who think they have mastered non-complex circuits. If you have read this article and the previous issue, you should be able to solve it. The aim is to calculate the total (equivalent) resistance of the complete circuit. It is meant to seem impossible, but very few of the students I have taught since 1985, have failed to find an answer. It is a great exercise for seeing how convoluted circuits can be, and for exploring tactics to find the total resistance.

VOLTAGE DIVIDER CIRCUITS

One common circuit in electronics is the “voltage divider” circuit. Voltage dividers are used for establishing a voltage of a fixed proportion to the supply voltage; perhaps for a transistor bias, or power supply reference voltage, or similar. Typically voltage dividers need to be a stable voltage, as stable as the supply it is connected to. The current flow in the divider circuit (called the “bleed current”) is typically 10 to 20 times the current taken from the junction.

Here we have a this deceptively simple circuit [13], and a reference formula for calculating the voltage based on the resistor values. However, you would most likely have the voltages you need, and have to calculate the resistances to achieve that voltage for a given current.

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If the circuit is to provide a 5V centre voltage, measured on the voltmeter, from a regulated 12V supply, and provide 5mA to the load, the bleed current must be 20 times the load current. So what resistors do you need?

THE OVERALL TAKEAWAYS

So all of the first four circuit types are solved using Ohm's Law, the Power Rule and Kirchhoff's Laws, KVL and KCL. The VIRP table helps learn how to apply these to the circuits, particularly series and parallel circuits, and can be used on compound circuits, but you should learn to work without the tables as your experience grows.

This is all that most of you will need, or even use, and all that is required for the old post-trade electronics courses. However, recognising that you are learning this voluntarily, as did I forty odd years ago, I know that there is often a hunger to learn more.

First, please do practice using what you have learned, and you might find circuit analysis as satisfying as completing a crossword puzzle for those without maths skills.

There are other skills that can help you with simple through to compound circuits, and we shall cover those skills next month, and in doing so will provide a path to more complex analysis.

Remember that DC is not the only supply option, nor are resistors the only passive components, and then there are active components that change their behaviour in response to changing physical values, voltage, current, temperature etc.