Transcript Slide 1

2.2 Resistance

G482 Electricity, Waves & Photons

2.2.2 EMF

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2.1.1 Prior Learning

I + wire + V I + lamp + V

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2.2.3 Resistance

Students can carry out practical work to investigate the I–V characteristics of a resistor, lamp and different coloured LEDs. Students can discuss low-energy lighting, eg LED torches.

(a) define resistance; (b) select and use the equation for resistance V = IR (c) define the ohm; (d) state and use Ohm’s law; (e) describe the I–V characteristics of a resistor at constant temperature, filament lamp and light-emitting diode (LED); (f) describe an experiment to obtain the I–V characteristics of a resistor at constant temperature, filament lamp and light-emitting diode (LED); (g) describe the uses and benefits of using light emitting diodes (LEDs). (Homework Literacy Task – research and present on 1 A4 page)

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(a) define resistance; (b) select and use the equation for resistance V = IR (c) define the ohm; (d) state and use Ohm’s law;

The

electrical resistance

of an electrical element is the opposition to the passage of an electric current through that element. Electrical resistance shares some conceptual parallels with the mechanical notion of The SI unit of electrical resistance is the ohm ( Ω ).

friction . All materials show some resistance, except for superconductors , which have a resistance of zero.

( The resistance (R) of an object is defined as the ratio of voltage

V

) to current through it (

I

) across it For a wide variety of materials and conditions, proportional to each other, and therefore

R

is

V

and

I

constant are directly .

This proportionality is called called "Ohmic" materials.

Ohm's law , and materials that satisfy it are http://en.wikipedia.org/wiki/Electrical_resistance_and_conductance

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d) Resistor

As the voltage increases the current also increases at the same rate.

This is what is called “ohms law”

True for resistor at a constant temperature

It is directly proportional

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e) Potential Divider or Variable Resistor

Pot Div Variable Resistor Voltage: Pot div circuit can provide the full range of voltage from V -> 0V, while a variable resistor circuit will not reach 0V.

Current: In a pot div circuit the load resistance (bulb) is in parallel with the variable resistor which means that the overall resistance is less and more current flows in the circuit. Energy: In a pot divider circuit the current flow is more as there are two pathways for current to flow. Hence energy flow is more!

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e)

I + TASK: Draw out the axes and on an A4 sheet leaving space copy traces and add any explanations… I + wire + V I + High temp low temp Amps I + lamp + V + thermistor V + V Milli-amps 0.6 V diode -

e) Components and their characteristics

For metals:

   

resistance increases with temperature positive ions vibrate more conduction electrons movement is impeded positive temperature coefficient

For intrinsic semiconductors:

     resistance decreases with temperature more charge carriers become available negative temperature coefficient much larger change in resistance than with metals thermistors used as temperature sensitive component in a sensor I + low temp High temp + V -

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Filament Lamp / Bulb

 The resistance of a filament lamp increases as the temperature of the filament increases.

 i.e. the resistance changes as the temperature of the wire changes.

 The gradient of the graph represents the resistance

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Mini Resistance Question….

The filament lamp is one example of a non-ohmic device.

(i) State what is meant by the term

non-ohmic.

(ii) Name one other example of a non- ohmic device.

2 a) The characteristic shown below is that of a filament lamp. Explain why, as the voltage is increased either positively or negatively from zero, the characteristic has the form shown in the figure. b) At a certain point on the characteristic, the power developed in the lamp is 20 W and the current is 90 mA. Calculate the resistance of the filament at this point on the characteristic.

I

(i) V is not directly proportional to I [or resistance is constant] (1) (ii) e.g. semiconductor diode

(1)

0

V

Basic iSlice

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Semi Conducting Diode

The current through a diode flows in one direction only. The diode has a very high resistance in the reverse direction.

Often used in transformers to change A.C to D.C. currents or computers!

Why does this happen?

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Diode – example...

PN junction is destroyed and current flows mA Scale High Resistance Low Resistance About 30V  A Scale

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Semiconductors (Extension)

Carbon, silicon and germanium (germanium, like silicon, is also a semiconductor) have a unique property in their electron structure - each has four electrons in its outer orbital. This allows them to form nice crystals. The four electrons form perfect covalent bonds with four neighbouring atoms, creating a lattice . In carbon, we know the crystalline form as diamond . In silicon, the crystalline form is a silvery, metallic-looking substance. Which is a near perfect insulator .

Well on its own that is not amazing but a semi-conductor is made from a silicon lattice with an

impurity

which will enable it to conduct. We often refer to these types of materials as intrinsic semiconductors .

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Doping p or n?

(Extension)

In N-type doping, phosphorus or arsenic is added to the silicon in small quantities. Phosphorus and arsenic each have five outer electrons, so they're out of place when they get into the silicon lattice. The fifth electron has nothing to bond to, so it's free to move around. It takes only a very small quantity of the impurity to create enough free electrons to allow an electric current to flow through the silicon. N-type silicon is a good conductor. Electrons have a negative charge , hence the name N-type. In P-type doping, boron or gallium is the doping agent. Boron and gallium each have only three outer electrons. When mixed into the silicon lattice, they form

"holes"

in the lattice where a silicon electron has nothing to bond to. The absence of an electron creates the effect of a positive charge, hence the name P-type. Holes can conduct current. A hole happily accepts an electron from a neighbour, moving the hole over a space. P-type silicon is a good conductor. p-type n-type

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PN junctions?

N-type and P-type silicon are not that amazing by themselves; but when you put them together, you get some very interesting behaviour at the junction. That's what happens in a diode. The place where they meet in the diagram forms what is called a depletion zone. or Band Gap Where free electrons have filled positive “holes” to form an area where there are no free charge carriers.

P N P N P N

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PN junctions - behaviour to applied pd

This typically takes 0.7V to overcome and bridge the gap and make it conduct!

V< 0.7V

V> 0.7V

P N P N P If we reverse the flow then the gap gets larger and does not conduct (up to 30V)

V< 0V V<< 0V

N P N P N P

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N

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(Extension)

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Mini Diode Question….

Explain what happens to the diode in terms of the band gap to explain this shape.. (4 marks) When voltage is in

forward bias

the resistance is large below 0.7V as

band gap is large

. When voltage is

above 0.7

the band gap is closed and the

diode conducts freely

and resistance is low.

Basic

If a diode is placed in a circuit in the forward bias direction the resistance of the diode changes. Use the graph to explain the difference for each part of the graph and link this to the band gap in each case (4). There are four areas to explain… When voltage is in

forward bias

the resistance is large below 0.7V as

band gap is large

. Current is in mA.

When voltage is

above 0.7

the band gap is closed and the

diode conducts freely

resistance is low.

and When in reverse bias direction the current flow is

very small

A

as band gap increases.

Above the breakdown voltage V b the band gap is very large the PN junction breaks down permanently.

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Connection

• • • Connect your learning to the content of the lesson Share the process by which the learning will actually take place Explore the outcomes of the learning, emphasising why this will be beneficial for the learner

Activation

• • • • Construct problem-solving challenges for the students Use a multi-sensory approach – VAK Promote a language of learning to enable the students to talk about their progress or obstacles to it Learning as an active process, so the students aren’t passive receptors

Demonstration

• • • • Use formative feedback – Assessment for Learning Vary the groupings within the classroom for the purpose of learning – individual; pair; group/team; friendship; teacher selected; single sex; mixed sex Offer different ways for the students to demonstrate their understanding Allow the students to “show off” their learning

Consolidation

• • • • • Structure active reflection on the lesson content and the process of learning Seek transfer between “subjects” Review the learning from this lesson and preview the learning for the next Promote ways in which the students will remember A “news broadcast” approach to learning

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