Topic 1.4 Components in Sensing Circuits

Topic 1.4 – Components in Sensing Circuits.

Learning Objectives:

At the end of this topic you will be able to;

1.4.1 Resistors

understand that resistance can be increased by connectingresistors in series;

understand that resistance can be decreased by connectingresistors in parallel;

select and use the equationto perform calculations involving the combined resistanceof two resistors in series;

select and use the equationto perform calculations involving the combinedresistance of two resistors in parallel;

describe how fixed and variable resistors can be used involtage dividers;

describe how potentiometers can be used as variableresistors and voltage dividers;

use the colour and printed code to work out the value andtolerance of a resistor;

select appropriate preferred values from the E24 series;

1.4.2 Light Dependent Resistors (LDRs)

state that the resistance of an LDR falls as light intensityincreases (non-linear);

1.4.3 NTC Thermistors

state that the resistance of ntc thermistors decreases astemperature increases (non-linear);

1.4.4 Switches

distinguish between the following types of mechanicalswitches:

push, toggle, reed, micro, tilt, rotary

1.4.1 – Resistors

Using resistors to control and limit current

If you used the Alpha Kit during Topic 1.2, you would have used 6V, 0.06A bulbs in the Alpha Kit. Such bulbs are designed to work on a 6V supply. When 6V is applied across a bulb, its filament offers sufficient resistance to keep the current down to 0.06A and the bulb lights up to its specified brightness. At working temperature, the filament provides a resistance of about 100Ω.

If we were to connect the same bulb to a 12V battery, this resistance would only be sufficient to keep the current down to about 0.12A. This high current would probably burn out the filament, and the bulb would be destroyed. Extra resistance is required in the circuit to limit the current to 0.06A. This extra resistance could be provided by connecting two such bulbs in series across the supply (See fig 1a).

The second bulb provides an extra resistance of about 100Ω. The same effect could be produced by using a fixed resistor of value 100Ω (Fig 1b). The wide range of resistor values offered by manufacturers enable us to limit the current through a component to almost any desired value.

If the current flowing through a component has to be very precisely set, a variable resistor is used as shown below.

Potentiometers

Potentiometers can be used for dividing up a voltage into any value between zero and the full supply voltage. A potentiometer consists of a circular conducting track, made of carbon or resistance wire, over which a sliding contact moves.

The voltage to be divided is connected across the end of track tags and the output voltage is taken between one of these tags and the wiper tag.

When the wiper is at position A the full supply voltage is available at the output. At position B the output voltage is zero.

Potentiometers can also be set up to act as variable resistors in circuits. In this case the wiper tag is connected to one of the end of track tags and the unit is used as shown opposite.

The following pictures show some of the different types of potentiometers available.

Presets

These are similar to potentiometers but are usually smaller and have to be adjusted using a screwdriver. They are designed to be inserted into a circuit then adjusted to the required value. Once accurately set they are usually sealed so that they do not change from this value. The following pictures illustrate the difference between presets and the continuously variable type.

Selecting a resistor

If you turn to the resistor section in any electronics supplies catalogue you will find a wide range of values and types on offer. After calculating the ideal value of the resistor required in a circuit you must consider the following points before making your selection.

(a) Preferred values

It is very unlikely that you will be able to find your ideal value within the range of values. Manufacturers only produce certain preferred values. You have to select the nearest value of resistor within the range.

In the E24 series, the 24 preferred values are:

10, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 43, 47, 51, 56, 62, 68, 75, 82, 91

together with multiples of 10 of these values, up to about 10MΩ. The increase between values in E24 is about 10%.

If we multiply each of the values above by 10 we get the next 24 available resistor values:

100, 110, 120, 130, 150, 160, 180, 200, 220, 240, 270, 300, 330, 360,

390, 430, 470, 510, 560, 620, 680, 750, 820, 910

Followed by

1k, 1.1k, 1.2k …………… and so on up to 10MΩ.

(b) Tolerance

This provides an indication of how much above, or below, the stated value the resistor might be. A 1.5kΩ resistor with a tolerance of ±5% could be as low as 1425Ω or as high as 1575Ω, since 5% of 1500 is 75Ω.

Compare the tolerance of carbon film and metal film resistors in your catalogue.

Carbon Film Tolerance = ......

Metal Film Tolerance = ......

Which type of resistor offers the closest tolerance? ......

You will find that the same type, and value, of resistor is offered at different wattage. The resistor with a power rating just above your required power rating should be selected. A power rating of 0.25W is sufficient for most of your practical work.

Compare the size of similar resistors but with different power rating.

(d) Stability

This gives an indication of how well the resistor behaves when physical conditions change.

Types of Resistor

The following types of resistors are commonly found in electronic systems.

(a) Wire wound

These are made by winding a piece of resistance wire e.g. constantan or nichrome, on to a ceramic former and coating it with an insulating material such as varnish.

Advantages / Disadvantages
1.Can be made very accurate. Tolerance of 0.1% available. / 1.Tend to be bulky and heavy.
2.Value does not change much when resistor heats up. / 2.Rather expensive.
3.Capable of dissipating high power.

This type of resistor is used where close tolerance or high power dissipation is required. Values are available up to about 22kΩ with a power dissipation capability of up to 50W.

(b) Carbon film

In this type of resistor a film of graphite is deposited on a ceramic former. A helical groove is cut into the carbon. The value of the resistor is determined by the size of the groove, the thickness of the film and the size of the former.

Advantages / Disadvantages
1.Easy to manufacture, and cheap. / 1.Rather poor temp stability
2.Good tolerance e.g. 5%

These are manufactured in a similar way to the carbon film resistors but the conducting film is made from metal (e.g. nichrome) or metal oxide.

They have all the advantages of the carbon film type and have much better tolerance and temperature stability than carbon film type.

Resistor Colour Code

The value of the resistor and its tolerance can be worked out from four colour bands on its body.

N.B. Some resistors especially metal film resistors use a five band colour code. Details can be found in suppliers catalogues. Only the four band code will be tested in the examinations.

The tolerance band is a single band near one end of the resistor and is normally gold or silver. A gold band indicates a tolerance of ±5%, while a silver band indicates ±10%. If there is no fourth band then the tolerance will be ±20%.

The value of the resistor (in ohms) can be worked out by looking at the three other coloured bands and using the colour code table.

Colour / Value / Colour / Value
Black / 0 / Green / 5
Brown / 1 / Blue / 6
Red / 2 / Violet / 7
Orange / 3 / Grey / 8
Yellow / 4 / White / 9

Examples:

Value = 2 2 (three 0’s)Ω

= 2 2 000 Ω = 22kΩ

Tolerance = ± 5%

Value = 4 7 00000 Ω = 4700kΩ = 4.7MΩ

Tolerance = ± 20%

Value = 1 2 (no 0’s) Ω = 12Ω

Tolerance = ± 10%

4.Complete the following diagrams by showing the colour code required for the following resistors:

a)75kΩ, ±5% resistor.

75kΩ = 75000Ω = 7 5 000 ±5%

b) 18Ω ±10% resistor.(1 8 _ ±10%)

c)360Ω (±5%) resistor.(3 6 0 ±5%)

d)2.4MΩ (±5%) resistor.(2 4 00000 ±5%)

Printed value

Equipment manufacturers’ circuit diagrams often use the following code for indicating resistor values. The letters give multiples and the position of the decimal point.

Examples:

Marking / Resistor Value
R33 / 0.33Ω
3R3 / 3.3Ω
33R / 33Ω
330R / 330Ω
3k3 / 3.3kΩ
33k / 33kΩ
3M3 / 3.3MΩ

Self Assessment Test.

1.Use the colour code to find the value of the resistors shown below.

......

c.What is the highest value that the resistor (b) is likely to have?

......

d.Using the resistor printed code – what are the values of the following resistors.

i)470R......

ii)2k2......

iii)5M6......

2.Complete the following diagrams by showing the colour code required for the following resistors:

i)270 Ω, ±5% resistor.

ii) 10kΩ ±10% resistor.

iii)3.9kΩ±10% resistor.

iv)8.2MΩ ±5% resistor.

Self Assessment Test

1.a.27000Ω = 27kΩ ±5%b.56Ω ±10%

c.d.i)470Ω; ii) 2.2kΩ iii)5.6MΩ

2.i)Red Violet Brown Gold.

ii) Brown Black Orange Silver.

iii)Orange White Red Silver.

iv)Grey Red Green Gold

Calculatingthevalue of a current limiting resistor

Suppose we want to operate a 2.5V, 0.25A bulb on a 6V supply.

For the bulb to operate at its specified brightness, it must have 2.5V dropped across it. The difference between this voltage and the supply voltage can be dropped across a series resistor. The resistor value selected should allow 0.25A to flow through it when there is a voltage of (6-2.5)V across it.

Applying Ohm’s Law to the resistor:

There are no 14Ω resistors available in the E24 series. This provides us with a dilemma since we have to choose between a 13Ω, or a 15Ω resistor. Let us look at the effect of each.

If we choose a 13Ω resistor, this will mean that the circuit resistance will be less than we needed. A larger current than expected will flow and will therefore put a greater strain on the bulb, and this will reduce its operating life time.

If we choose a 15Ω resistor, this will mean that the circuit resistance is slightly higher than that required, which will reduce the current flowing below 0.25A. The result will be a bulb operating at slightly less than full brightness, but within its maximum value.

The most suitable preferred value resistor is this case would be 15Ω.

Resistors in series

The following 2 circuits have been set up on a circuit simulator.

Look at Circuit 1 and you will see that the ammeter reading is 6.00mA. We can apply Ohm’s Law to the circuit to find the total resistance of the circuit.

If we add up the values of R1 and R2 from Circuit 1 we get

R1 + R2 = 1kΩ+ 1kΩ= 2kΩ

Which is exactly the same answer as we got using Ohm’s Law?

Look at Circuit 2 and you will see that the ammeter reading is 99.88mA. We can apply Ohm’s Law to this circuit to find

If we add up the values of R3, R4 and R5 from Circuit 2 we get

R3 + R4 + R5 = 47Ω+ 33Ω+ 20Ω = 100.

Which is nearly but not exactly the same answer as we got using Ohm’s Law?

The difference in this case of 0.12 is due to very small rounding errors that occur when the simulator is displaying current flow.

So we can see that the total or effective resistance Rs of resistors in series is given by the general equation:

Rs = R1 + R2 + R3+ ......

Therefore:

if R1 = 10Ω and R2 = 40Ω, then Rs = 10 + 40 = 50Ω.

if R1 = 15kΩ, R2 = 25kΩ, and R3 = 75kΩ

then Rs = 15k + 25k+ 75k = 115kΩ.

Resistors in Parallel

Look at Circuit 1 and you will see that the ammeter reading is 23.99mA. We can apply Ohm’s Law to the circuit to find the total resistance of the circuit.

Look at Circuit 2 and you will see that the ammeter reading is 513.12mA. We can apply Ohm’s Law to this circuit to find

There does not seem to be an obvious relationship between the effective resistances and the resistor values used, other than the effective resistance in each case is smaller than the individual parallel resistor values.

It can be shown that the total or effective resistance Rp of resistors in parallel is given by the general equation:

For only 2 resistors in parallel this can be simplified to

For example,

If R1 = R2 = 1kΩ, (as given in circuit 1) then

If R1 = 33Ω and R2 = 47Ω(as given in circuit 2) then

Note

1.You should always check you answer when using the formula to make sure that the effective resistance of 2 resistors in parallel is smaller than the individual resistor values.

2.When 2 resistors of the same value are connected in parallel the effective resistance is 1/2 (one half) of their individual values.

3.If 3 resistors of the same value are connected in parallel then the effective resistance is 1/3 (one third) of their individual values. E.g. If three 10k resistors are connected in parallel their effective resistance = 10k/3 = 3.333kΩ.

4.In general if ‘n’ resistors of the same value are connected in parallel then the effective resistance is 1/n (one ‘n’th) of their individual values. E.g. If ‘n’ 10k resistors are connected in parallel their effective resistance = 10k/’n’ Ω.

Worked examples

Find

(i)the current in the 2Ω resistor,

(ii)the current in the 4Ω resistor,

(iii)the voltage across the 2Ω resistor,

(iv)the voltage across the 4Ω resistor, and

(v)the supply voltage.

Solution:

(i)2A (Since this is a series circuit so current is same everywhere)

(ii)2A (Same reason as (i))

(iii)Apply V = I x R to 2Ω resistor.

V2Ω = 2 x 2 = 4V.

(iv)Apply V = I x R to 4Ω resistor.

V4Ω = 2 x 4 = 8V

(v)Supply voltage V = V2Ω + V4Ω = 4 + 8 = 12V.

Find(a)the current in the 2Ω resistor,

(b)the current in the 2Ω resistor,

(c)the voltage across the 2Ω resistor,

(d)the voltage across the 4Ω resistor, and

(e)the effective resistance of the parallel circuit.

Solution:

(a)2A (because the current in the 2Ω resistor will be twice that in the 4Ω resistor.)

(b)1A (because the current in the 4Ω resistor will be half that in the 2Ω resistor.) Or (Since 3A enters the network, and 2A goes through the other resistor only 1A is left).

(c)Apply V = I x R to 2Ω resistor.

V2Ω = 2 x 2 =4V

(d) Apply V = I x R to 4Ω resistor.

V4Ω= 1 x 4 = 4VOr by inspection V4Ω = 4V since resistors are in parallel and therefore the voltage must be the same as V2Ω.

(e)

3.For the network shown below, calculate:

(a)the combined resistance Rp of R1 and R2 in parallel.

(b)the total resistance RT of the network.

(c)I.

(d)V1 and V2.

(e)I1 and I2.

(f)What is the nearest preferred value to RTin the E24 series.

Solution:

(a)Rp = 20 /2 = 10 equal resistors in parallel)

(b)RT = R3 + R1 = 10 +30 = 40

(c)The voltage V across the whole network is 6V and its total resistance RT is 40Ω therefore

(d) V1 = I x R3 = 0.15A x 30Ω = 4.5V

But V = V1 + V2 therefore V2 = V–V1= 6 – 4.5 = 1.5V

(e)I1 = I2 = ½I (since R1 = R2) therefore I1 = ½ x 0.15A = 0.075A.

(f)The 2 nearest preferred values to 40 are 39 and 43 so in this case choose 39

Summary 

Resistors usually exist in combinations of series and parallel components.
The effective resistance Rs of series resistors is given by the following formula.

The effective resistance Rp of two resistorsin parallel is given by the formula.

Homework Questions 1

Draw a diagram to show how you would connect two 10Ω resistors to give a total resistance of (a) 20Ω, (b) 5Ω.

(a)(b)

[2]

2.In the circuit below what is

(a)the current in the 3Ω resistor. ……………………… [1]

(b)the current in the 6Ω resistor. ……………………… [1]

(c)voltage across the 3Ω resistor.

…………………………………………………………………………………………………………..……………………………… [2]

(d)voltage across the 6Ω resistor.

…………………………………………………………………………………………………………..……………………………… [2]

(e)the supply voltage.

………………………………………………………………………………………………..………………………………………… [1]

3.In the circuit below, calculate

(a)the current in the 3Ω resistor.

………………………………………………………………………………………………..…………………………………… [1]

(b)the current in the 6Ω resistor.

………………………………………………………………………………………………..…………………………………… [1]

(c)the voltage across the 3Ω resistor.

………………………………………………………………………………………………..…………………………………… [1]

(d)the voltage across the 6Ω resistor.

………………………………………………………………………………………………..…………………………………… [1]

(e)the supply voltage.

………………………………………………………………………………………………..…………………………………… [1]

4. For the network shown below calculate the total resistance between

(a)X and Y,

……………………………………………………………………………………………………………

………………………………………………………………………………………………..…………………………………… [1]

(b) Y and Z,

……………………………………………………………………………………………………………

………………………………………………………………………………………………..…………………………………… [1]

……………………………………………………………………………………………………………

………………………………………………………………………………………………..…………………………………… [1]

(d)Use the list of E24 preferred values to select a single resistor to replace the network of 4 resistors.

………………………………………………………………………………………………..…………………………………… [1]

Voltage dividing chains

You should remember theVoltage Divider Rulefrom topic 1.3. It states:

If we connect a load across the output it could change the voltage level set by the resistor chain. If the current taken by the load is greater than 0.1I there will be a significant change in the output voltage.

We will investigate this in the following assignments.

Assignment 1.4A

Investigating resistors

Activity 1:

In this activity you will be investigating the use of resistors to control the current flowing in a circuit.

1a.Set up the following circuit using your circuit simulator.

1b.Close the switch. Note the brightness of the bulb, and record the reading from the ammeter.

Brightness : ......

Reading on ammeter = ...... (A, mA or µA)

Save your circuit as “E1-Circuits2-Act1”

1c.Open the switch then modify the circuit so that it has a 100Ω resistor in series with the bulb, as shown below:

1d.Close the switch. Note the brightness of the bulb, and record the reading from the ammeter.

Brightness : ......

Reading on ammeter = ...... (A, mA or µA)

What effect does the resistor have upon the brightness of the bulb and the current flowing through the filament ?

......

Save your circuit as “E1-Circuits2-Act2”

1e.Remove the 100Ω resistor and replace it with a 1 kΩ resistor. Close the switch. Note the brightness of the bulb, and record the reading from the ammeter.

Brightness : ......

Reading on ammeter = ...... (A, mA or µA)

Report on any change from 1c.

......

Save your circuit as “E1-Circuits2-Act3”

Complete the following:

When a resistor is connected in series with a bulb, it ______the current flowing through the filament.

The larger the value of the resistor, the ______the current flowing through the bulb.

Activity 2:

Let us now investigate the use of a potentiometer, set up as a variable resistor to control current flow in a circuit.

2a.Set up the following circuit in your simulator.

2b.Close the switch then move the slider on the potentiometer by sliding the grey box from left to right. Describe what you observe.

......

Explain this effect : ......

......

2c.Record the maximum and minimum values of current possible in this circuit.

Minimum Current = ...... (A, mA, µA)

Maximum Current = ...... (A, mA, µA)

2d.Connect a voltmeter across the bulb. Move the slider control so that you can measure the following:

Minimum voltage across the bulb=...... (V, mV, µV)

Maximum voltage across bulb=...... (V, mV, µV)

Complete the following:

The bulb is at its dimmest when the variable resistor is set to its

...... value.

Save your circuit as “E1-Circuits2-Act4”

Activity 3:

We shall now investigate the use of two resistors in voltage dividing chains.

3a.Set up the following circuit in your simulator. Note that the supply voltage is at 12V.

3b.Close the switch and record the voltage provided at the output.

Output voltage =...... volts

Save your circuit as “E1-Circuits2-Act5”

3c.Open the switch and connect a 6V, 0.06A bulb across the output, as shown below.

3d.Close the switch, you will probably find that the bulb does notlight up. Careful examination of the voltmeter might give you a clue as to why the bulb does not light up.

Voltage across bulb=...... (V, mV, µV)

Try to explain why the bulb does not light up. This is quite difficult. If you cannot answer ask your tutor for help.

......

Save your circuit as “E1-Circuits2-Act6”

3e.Switch off the circuit and replace the 10kΩ resistors with 18Ω resistors as shown below.

3f.Switch on the circuit and comment on your observations.

......