Lesson 4 of 4
Temperature and I-V characteristics
I can describe how increasing temperature affects metallic resistance.Syllabus
Syllabus K323, 14(j). Describe how increasing temperature affects metallic resistance.
I can sketch and interpret I-V characteristics for an ohmic conductor, filament lamp and semiconductor diode.Syllabus
Syllabus K323, 14(k). Sketch and interpret I-V characteristics for an ohmic conductor, filament lamp and semiconductor diode.
Make a guess. You are not marked.
A filament lamp glows more brightly as the p.d. across it is increased. What happens to its resistance?
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It increases, because the hotter filament obstructs the current more.
A hotter metal filament has ions vibrating more, which obstruct the moving electrons more. So the I-V graph curves, and V/I rises as the lamp gets brighter.
An I-V graph shows how a component's current responds to different potential differences. Its shape tells you whether the resistance stays constant.
At any nonzero-current operating point, R = V/I. Read both values from that same point. The graphs here put current I in amperes on the vertical axis and p.d. V in volts on the horizontal axis.
Why a hotter metal has greater resistance
When a metallic conductor's temperature rises, the ions in its structure vibrate more vigorously. Their increased vibration hinders the drift of mobile electrons, so the resistance increases in the normal operating range considered here.
At the same p.d., the hotter conductor therefore carries a smaller current. This explanation concerns metals; it is not a rule that every material or component has increasing resistance with temperature.
Read and sketch three characteristic shapes
Start with labelled axes and an origin. Positive and negative values represent opposite voltage polarities and current directions under a fixed reference convention.
Read current vertically and p.d. horizontally
The conductor and lamp use the same axes and scales. Dots mark the supplied positive readings; dashed negative branches show the corresponding idealised symmetric behaviour.
Metallic conductor at constant temperature
Marked readings (V, I):
- (2 V, 0.10 A)
- (4 V, 0.20 A)
- (6 V, 0.30 A)
The straight line through the origin has constant I/V. At each nonzero point, V/I = 20 Ω.
Filament lamp: the curve becomes shallower
Marked readings (V, I):
- (2 V, 0.20 A)
- (4 V, 0.30 A)
- (6 V, 0.36 A)
As the filament heats, the graph becomes shallower and V/I increases. Use the point's V/I for resistance; the local tangent gradient is a different quantity.
Semiconductor diode: forward and reverse behaviour differ
Positive V is the forward direction chosen for this graph. Reverse current is negligible over the range shown; the curve does not imply that it stays negligible at every possible reverse voltage.
- Metallic conductor at constant temperature
- Sketch a straight line through the origin, extending into both positive and negative quadrants. Current is proportional to p.d., so V/I is constant. This is an ohmic conductor under those conditions. A wire that heats significantly is no longer being compared at constant temperature.
- Filament lamp
- Sketch a curve through the origin that becomes less steep as the voltage magnitude increases, with a corresponding branch for reversed polarity. Greater current heats the metal filament, increasing its resistance. Doubling the p.d. therefore need not double the current.
- Semiconductor diode
- Identify the forward and reverse polarities. In the forward direction, current rises strongly over the illustrated higher-voltage part of the curve. In reverse, it is negligible over the range shown. Sketch this asymmetric behaviour; there is no single exact turn-on voltage for every diode.
The diode's forward and reverse directions describe its orientation in the circuit. They are not a claim that all positive voltages in every drawing are forward: check the stated reference and component orientation. The displayed small reverse-current range does not establish zero current at every possible reverse voltage.
Calculate resistance at a point
Compare V/I
A constant-temperature wire and a lamp
The wire's supplied readings are 2.0 V with 0.10 A, 4.0 V with 0.20 A, and 6.0 V with 0.30 A. Each gives R = 20 Ω.
For the illustrative lamp readings:
- At 2.0 V and 0.20 A: R = 2.0/0.20 = 10 Ω.
- At 4.0 V and 0.30 A: R = 4.0/0.30 = 13.3 Ω.
- At 6.0 V and 0.36 A: R = 6.0/0.36 = 16.7 Ω.
The lamp's resistance rises as the filament heats. At the corresponding reversed point, -6.0 V and -0.36 A still give a positive resistance: (-6.0)/(-0.36) = 16.7 Ω.
For the straight I-V line through the origin, the gradient is I/V = 1/R. Here 0.30/6.0 = 0.050 A/V, whose reciprocal is 20 Ω. A steeper straight line on the same scales means smaller resistance.
For a curved graph, use the point's V/I. A tangent gradient describes how current changes close to that point; its reciprocal is not the resistance V/I there. Likewise, a ratio of changes between two separated points does not generally give either point's resistance. If the axes are exchanged to V vertically and I horizontally, the constant-resistance line instead has gradient R.
Collect readings that can distinguish the behaviours
Measure current through and p.d. across the component while varying a suitable low-voltage supply. Record paired readings, then plot I against V. For the constant-temperature wire comparison, limit heating and allow the wire to return to the same conditions between readings where needed.
For a lamp characteristic, let each reading settle at its operating condition; the changing filament temperature is part of the behaviour being observed. Use suitable component ratings and current limits. To investigate the opposite polarity, switch off before reversing the source connections and record the resulting signs consistently.
State the graph's axes and conditions. A straight line through the origin supports constant resistance over the range shown. A changing gradient on a lamp curve must be interpreted with the filament's changing temperature.
Try it step by step
Work it out in steps: Resistance at the lower p.d., V / I, in ohm, then Resistance at the higher p.d., V / I, in ohm.
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Problem 1.
Resistance at the lower p.d., V / I, in ohm: R = 2.0 / 0.20 = 10 ohm
Resistance at the higher p.d., V / I, in ohm: R = 6.0 / 0.30 = 20 ohm. The hotter filament has the greater resistance.
Problem 2.
Resistance at the lower p.d., V / I, in ohm: R = 1.5 / 0.25 = 6.0 ohm
Resistance at the higher p.d., V / I, in ohm: R = 4.5 / 0.45 = 10 ohm
Problem 3.
Resistance at the lower p.d., V / I, in ohm: R = 3.0 / 0.40 = 7.5 ohm
Resistance at the higher p.d., V / I, in ohm: R = 9.0 / 0.60 = 15 ohm
Check your understanding
A filament lamp carries 0.20 A at 2.0 V and 0.36 A at 6.0 V. What is its resistance at 6.0 V, and what do these readings show?
Show the answer
About 16.7 ohm; the hotter filament has a greater resistance.
R = 6.0 / 0.36 = 16.7 ohm, compared with 10 ohm at the lower-voltage point. Heating increases the metallic filament's resistance, so current does not rise in direct proportion to voltage.
Practise an exam question
Write your answer first, then tick the mark points it makes.
Exam-style question
Sketch the current-voltage (I-V) graph for a filament lamp, with I on the vertical axis. Explain the shape of the graph. [3]
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