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I–V characteristics and temperature

Key idea: Use V/I at an operating point to find resistance on an I–V curve. Heating raises a metal's resistance through scattering, but lowers an NTC thermistor's by increasing charge carriers.

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  • GCE A-Level H2 Physics 2027
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Enlarged diagram

A graph of current against p.d., the I–V characteristic, is a fingerprint of a component. A resistor gives a straight line, a lamp a curve that flattens, a diode a sharp one-way bend. Each shape comes from what happens to the charge carriers inside, and especially from how the component responds to being heated.

Reading an I–V characteristic

On an I–V graph, the current I is plotted against the p.d. V. For an ohmic component, such as a metal wire at constant temperature, the graph is a straight line through the origin with gradient 1/R.

For a curved graph, the resistance at any point is still R = V/I, using the coordinates of that point. That is the reciprocal of the gradient of the line from the origin to the point, not of the tangent there. The tangent describes how the current responds to a small change in p.d., which is a different quantity.

Operating resistance and local tangent

A curved lamp I–V trace passes through P at 4 V and 0.5 A. A dashed line joins the origin to P and a dotted local tangent through P has a smaller gradient.

Scroll across the figure to read all labels.

A curved lamp I–V trace passes through P at 4 V and 0.5 A. A dashed line joins the origin to P and a dotted local tangent through P has a smaller gradient.A curved lamp I–V trace passes through P at 4 V and 0.5 A. A dashed line joins the origin to P and a dotted local tangent through P has a smaller gradient.
At P, V = 4.00 V and I = 0.500 A, so V/I = 8.00 Ω. The origin-to-P line has gradient 1/8 A/V, while the tangent has gradient 1/16 A/V. This illustrative lamp model uses V = 4I + 16I³ with V in volts and I in amperes; it is not a universal filament law.
Open full-size graph
What to noticeCompare the tangent with the origin-to-P line. Operating resistance uses the coordinates V/I at P.
View figure data
Figure data: Operating resistance and local tangent
SeriesPotential difference, V (V)Current, I (A)
Illustrative lamp0.000000.0000
Illustrative lamp0.010000.0025
Illustrative lamp0.020000.0050
Illustrative lamp0.030010.0075
Illustrative lamp0.040020.0100
Illustrative lamp0.050030.0125
Illustrative lamp0.060050.0150
Illustrative lamp0.070090.0175
Illustrative lamp0.080130.0200
Illustrative lamp0.090180.0225
Illustrative lamp0.100250.0250
Illustrative lamp0.110330.0275
Illustrative lamp0.120430.0300
Illustrative lamp0.130550.0325
Illustrative lamp0.140690.0350
Illustrative lamp0.150840.0375
Illustrative lamp0.161020.0400
Illustrative lamp0.171230.0425
Illustrative lamp0.181460.0450
Illustrative lamp0.191710.0475
Illustrative lamp0.202000.0500
Illustrative lamp0.212320.0525
Illustrative lamp0.222660.0550
Illustrative lamp0.233040.0575
Illustrative lamp0.243460.0600
Illustrative lamp0.253910.0625
Illustrative lamp0.264390.0650
Illustrative lamp0.274920.0675
Illustrative lamp0.285490.0700
Illustrative lamp0.296100.0725
Illustrative lamp0.306750.0750
Illustrative lamp0.317450.0775
Illustrative lamp0.328190.0800
Illustrative lamp0.338980.0825
Illustrative lamp0.349830.0850
Illustrative lamp0.360720.0875
Illustrative lamp0.371660.0900
Illustrative lamp0.382660.0925
Illustrative lamp0.393720.0950
Illustrative lamp0.404830.0975
Illustrative lamp0.416000.1000
Illustrative lamp0.427230.1025
Illustrative lamp0.438520.1050
Illustrative lamp0.449880.1075
Illustrative lamp0.461300.1100
Illustrative lamp0.472780.1125
Illustrative lamp0.484330.1150
Illustrative lamp0.495960.1175
Illustrative lamp0.507650.1200
Illustrative lamp0.519410.1225
Illustrative lamp0.531250.1250
Illustrative lamp0.543160.1275
Illustrative lamp0.555150.1300
Illustrative lamp0.567220.1325
Illustrative lamp0.579370.1350
Illustrative lamp0.591590.1375
Illustrative lamp0.603900.1400
Illustrative lamp0.616300.1425
Illustrative lamp0.628780.1450
Illustrative lamp0.641340.1475
Illustrative lamp0.654000.1500
Illustrative lamp0.666750.1525
Illustrative lamp0.679580.1550
Illustrative lamp0.692510.1575
Illustrative lamp0.705540.1600
Illustrative lamp0.718660.1625
Illustrative lamp0.731870.1650
Illustrative lamp0.745190.1675
Illustrative lamp0.758610.1700
Illustrative lamp0.772130.1725
Illustrative lamp0.785750.1750
Illustrative lamp0.799480.1775
Illustrative lamp0.813310.1800
Illustrative lamp0.827250.1825
Illustrative lamp0.841310.1850
Illustrative lamp0.855470.1875
Illustrative lamp0.869740.1900
Illustrative lamp0.884130.1925
Illustrative lamp0.898640.1950
Illustrative lamp0.913260.1975
Illustrative lamp0.928000.2000
Illustrative lamp0.942860.2025
Illustrative lamp0.957840.2050
Illustrative lamp0.972950.2075
Illustrative lamp0.988180.2100
Illustrative lamp1.003530.2125
Illustrative lamp1.019010.2150
Illustrative lamp1.034630.2175
Illustrative lamp1.050370.2200
Illustrative lamp1.066240.2225
Illustrative lamp1.082250.2250
Illustrative lamp1.098390.2275
Illustrative lamp1.114670.2300
Illustrative lamp1.131090.2325
Illustrative lamp1.147650.2350
Illustrative lamp1.164340.2375
Illustrative lamp1.181180.2400
Illustrative lamp1.198170.2425
Illustrative lamp1.215300.2450
Illustrative lamp1.232570.2475
Illustrative lamp1.250000.2500
Illustrative lamp1.267580.2525
Illustrative lamp1.285300.2550
Illustrative lamp1.303180.2575
Illustrative lamp1.321220.2600
Illustrative lamp1.339410.2625
Illustrative lamp1.357750.2650
Illustrative lamp1.376260.2675
Illustrative lamp1.394930.2700
Illustrative lamp1.413760.2725
Illustrative lamp1.432750.2750
Illustrative lamp1.451910.2775
Illustrative lamp1.471230.2800
Illustrative lamp1.490720.2825
Illustrative lamp1.510390.2850
Illustrative lamp1.530220.2875
Illustrative lamp1.550220.2900
Illustrative lamp1.570400.2925
Illustrative lamp1.590760.2950
Illustrative lamp1.611290.2975
Illustrative lamp1.632000.3000
Illustrative lamp1.652890.3025
Illustrative lamp1.673960.3050
Illustrative lamp1.695220.3075
Illustrative lamp1.716660.3100
Illustrative lamp1.738280.3125
Illustrative lamp1.760090.3150
Illustrative lamp1.782100.3175
Illustrative lamp1.804290.3200
Illustrative lamp1.826670.3225
Illustrative lamp1.849250.3250
Illustrative lamp1.872020.3275
Illustrative lamp1.894990.3300
Illustrative lamp1.918160.3325
Illustrative lamp1.941530.3350
Illustrative lamp1.965090.3375
Illustrative lamp1.988860.3400
Illustrative lamp2.012840.3425
Illustrative lamp2.037020.3450
Illustrative lamp2.061400.3475
Illustrative lamp2.086000.3500
Illustrative lamp2.110810.3525
Illustrative lamp2.135820.3550
Illustrative lamp2.161050.3575
Illustrative lamp2.186500.3600
Illustrative lamp2.212160.3625
Illustrative lamp2.238030.3650
Illustrative lamp2.264130.3675
Illustrative lamp2.290450.3700
Illustrative lamp2.316990.3725
Illustrative lamp2.343750.3750
Illustrative lamp2.370740.3775
Illustrative lamp2.397950.3800
Illustrative lamp2.425390.3825
Illustrative lamp2.453070.3850
Illustrative lamp2.480970.3875
Illustrative lamp2.509100.3900
Illustrative lamp2.537470.3925
Illustrative lamp2.566080.3950
Illustrative lamp2.594920.3975
Illustrative lamp2.624000.4000
Illustrative lamp2.653320.4025
Illustrative lamp2.682880.4050
Illustrative lamp2.712690.4075
Illustrative lamp2.742740.4100
Illustrative lamp2.773030.4125
Illustrative lamp2.803570.4150
Illustrative lamp2.834370.4175
Illustrative lamp2.865410.4200
Illustrative lamp2.896700.4225
Illustrative lamp2.928250.4250
Illustrative lamp2.960050.4275
Illustrative lamp2.992110.4300
Illustrative lamp3.024430.4325
Illustrative lamp3.057010.4350
Illustrative lamp3.089840.4375
Illustrative lamp3.122940.4400
Illustrative lamp3.156310.4425
Illustrative lamp3.189940.4450
Illustrative lamp3.223830.4475
Illustrative lamp3.258000.4500
Illustrative lamp3.292440.4525
Illustrative lamp3.327140.4550
Illustrative lamp3.362120.4575
Illustrative lamp3.397380.4600
Illustrative lamp3.432910.4625
Illustrative lamp3.468710.4650
Illustrative lamp3.504800.4675
Illustrative lamp3.541170.4700
Illustrative lamp3.577820.4725
Illustrative lamp3.614750.4750
Illustrative lamp3.651970.4775
Illustrative lamp3.689470.4800
Illustrative lamp3.727260.4825
Illustrative lamp3.765350.4850
Illustrative lamp3.803720.4875
Illustrative lamp3.842380.4900
Illustrative lamp3.881340.4925
Illustrative lamp3.920600.4950
Illustrative lamp3.960150.4975
Illustrative lamp4.000000.5000
Illustrative lamp4.040150.5025
Illustrative lamp4.080600.5050
Illustrative lamp4.121360.5075
Illustrative lamp4.162420.5100
Illustrative lamp4.203780.5125
Illustrative lamp4.245450.5150
Illustrative lamp4.287440.5175
Illustrative lamp4.329730.5200
Illustrative lamp4.372330.5225
Illustrative lamp4.415250.5250
Illustrative lamp4.458480.5275
Illustrative lamp4.502030.5300
Illustrative lamp4.545900.5325
Illustrative lamp4.590090.5350
Illustrative lamp4.634590.5375
Illustrative lamp4.679420.5400
Illustrative lamp4.724580.5425
Illustrative lamp4.770060.5450
Illustrative lamp4.815860.5475
Illustrative lamp4.862000.5500
Illustrative lamp4.908470.5525
Illustrative lamp4.955260.5550
Illustrative lamp5.002390.5575
Illustrative lamp5.049860.5600
Illustrative lamp5.097660.5625
Illustrative lamp5.145790.5650
Illustrative lamp5.194270.5675
Illustrative lamp5.243090.5700
Illustrative lamp5.292250.5725
Illustrative lamp5.341750.5750
Illustrative lamp5.391600.5775
Illustrative lamp5.441790.5800
Illustrative lamp5.492330.5825
Illustrative lamp5.543230.5850
Illustrative lamp5.594470.5875
Illustrative lamp5.646060.5900
Illustrative lamp5.698010.5925
Illustrative lamp5.750320.5950
Illustrative lamp5.802980.5975
Illustrative lamp5.856000.6000
Origin-to-P line0.000000.0000
Origin-to-P line6.000000.7500
Tangent at P2.000000.3750
Tangent at P6.000000.6250
P4.000000.5000

Worked example 1

Point or tangent?

Problem

On the lamp graph above, point P is at V = 4.00 V, I = 0.500 A. The tangent at P has a gradient of 0.0625 A V⁻¹. Find the resistance of the lamp at P.

Show full solution

  1. Use the coordinates

    Method

    Divide the p.d. at P by the current at P.

    Reason

    Resistance is defined as V/I at the operating point.

    Working

    Answer

    R = 4.00/0.500 = 8.00 Ω

  2. The tangent

    Working

    Answer

    1/0.0625 = 16.0 Ω is not the resistance at P; it describes the response to a small change near P.

Guided practice 2

Another curved graph

About 2 min

Problem

At one point on a component’s I–V graph, V = 6.0 V and I = 0.20 A. The tangent there has a gradient of 0.020 A V⁻¹. Find the resistance of the component at that point.

Find R

Show the steps Optional working
Unit: A/V
Unit: Ω

Hints

Hint 1

definition

R = V/I at the point. The tangent gradient is not needed.

Show full solution

  1. Resistance

    Working

    Answer

    R = 6.0/0.20 = 30 Ω. The tangent would give 50 Ω, which is not the resistance.

Four components

A-Level current–voltage characteristicsFour qualitative current against potential difference graphs. An ohmic conductor is a straight line through the origin. A filament lamp becomes less steep as voltage magnitude rises. A diode carries almost no reverse current and rises steeply in forward bias. An NTC thermistor becomes steeper as self-heating reduces resistance.Ohmic conductorVIFilament lampVISemiconductor diodeVINTC thermistorVIforward risewarmer, lower R
A-Level current–voltage characteristicsFour qualitative current against potential difference graphs. An ohmic conductor is a straight line through the origin. A filament lamp becomes less steep as voltage magnitude rises. A diode carries almost no reverse current and rises steeply in forward bias. An NTC thermistor becomes steeper as self-heating reduces resistance.Ohmic conductorVIFilament lampVISemiconductor diodeVIforward riseNTC thermistorVIwarmer, lower R
Qualitative I–V curves: use R = V/I at each point. The lamp and thermistor curves assume self-heating.

Ohmic resistor. At constant temperature, I is proportional to V: a straight line through the origin.

Filament lamp. As the current increases, the filament gets hotter and its resistance rises. Each extra volt adds less current than the one before, so the graph curves and flattens. It is the same for either direction of current.

Semiconductor diode. In one direction (forward bias) the current is tiny until the p.d. reaches a few tenths of a volt, then rises very steeply. In the other direction (reverse bias) the current is almost zero. In its normal operating range it conducts readily in forward bias and has only a tiny reverse current; large reverse p.d.s can cause breakdown.

Semiconductor diode I–V characteristic

An illustrative diode characteristic with negligible reverse current and a rapidly increasing forward current after the knee region.

Scroll across the figure to read all labels.

An illustrative diode characteristic with negligible reverse current and a rapidly increasing forward current after the knee region.An illustrative diode characteristic with negligible reverse current and a rapidly increasing forward current after the knee region.
Illustrative values only: the required feature is strongly asymmetric conduction, not a universal fixed turn-on voltage.
Open full-size graph
View figure data
Figure data: Semiconductor diode I–V characteristic
Potential difference, V (V)Diode
-2.0-0.10
-1.0-0.05
0.00.00
0.30.10
0.50.70
0.63.00
0.79.00
0.818.00

NTC thermistor. Its resistance falls as its temperature rises (negative temperature coefficient). If the current heats it noticeably, its resistance falls as the p.d. increases, so the graph curves upwards, getting steeper. At a fixed temperature, with currents too small to heat it, its I–V graph is close to a straight line, and the line is steeper at a higher temperature.

Check your understanding 1 Separate heating from the temperature setting

An NTC thermistor is held at a constant temperature while small currents are used. Predict the shape of its I–V graph. Then predict how that graph changes when the thermistor is held at a higher temperature.

At one fixed temperature the graph is approximately
At the higher fixed temperature it becomes
Show answer

At each fixed temperature the low-current graph is approximately straight through the origin. The warmer graph is steeper because R is smaller. An upward-curving sweep can occur when the measuring current itself changes the temperature.

Why temperature changes resistance

The current in a conductor is I = nAvq, where n is the number of charge carriers per unit volume and v is their drift velocity. Compare the current produced by the same p.d. at two temperatures.

Metals (filament lamp). A metal already has a huge number of free electrons, and heating hardly changes it. But the atoms of a hotter metal vibrate more, so the electrons collide with them more often and their drift velocity is lower. The current is smaller for the same p.d., so the resistivity rises.

Semiconductors (NTC thermistor). A semiconductor has far fewer free charge carriers. Heating gives more electrons enough energy to break free, so the number density of carriers increases sharply. This outweighs the extra collisions, so the current is larger for the same p.d. and the resistivity falls.

Check your understanding 2 Reverse the graph axes

A graph plots V vertically and I horizontally. At point P the coordinates are I = 0.40 A and V = 8.0 V, while the tangent gradient is 35 V A⁻¹. Find the operating resistance. On paper, draw the line from the origin to P and explain how its gradient gives the resistance for these axes.

Unit: Ω
Show answer

R = 8.0/0.40 = 20 Ω. The origin-to-P line has gradient 20 V A⁻¹; for these reversed axes it equals R, rather than 1/R. The tangent gradient is not the operating resistance.

Check your understanding 3 Explain what the axes change

Compare how you find operating resistance from an I-against-V graph and a V-against-I graph. Compare the line from the origin with the tangent. Keep your sketch on paper.

Show answer

Check the axes first. With I vertical the origin-to-point gradient is I/V = 1/R. With V vertical it is V/I = R. On a curved graph the tangent describes small changes and cannot replace the coordinate ratio.

Exam-style question 1

Warming a metal and a thermistor

5 marks

Examination question

A metal wire and an NTC thermistor are both warmed.

(a) Explain, in terms of charge carriers, why the resistivity of the metal increases. [2 marks]

(b) Explain, in terms of charge carriers, why the resistivity of the thermistor decreases. [2 marks]

(c) At a fixed temperature the thermistor is ohmic. State how its I–V graph changes when it is warmed. [1 mark]

Write your answer before viewing the mark scheme

Show solution step by step

  1. (a) Metal

    2 marks

    Method

    Use I = nAvq and decide which factor changes.

    Reason

    In a metal the number of free electrons hardly changes, so the effect must come from the drift velocity.

    Working

    Answer

    The atoms vibrate more, so electrons collide with them more often and their drift velocity is lower for the same p.d. The current is smaller, so the resistivity is larger.

  2. (b) Thermistor

    2 marks

    Reason

    In a semiconductor the number of carriers is very sensitive to temperature.

    Working

    Answer

    More charge carriers are freed, so the number density n increases sharply, outweighing the extra collisions. The current is larger for the same p.d., so the resistivity is smaller.

  3. (c) Graph

    1 mark

    Working

    Answer

    It is still a straight line through the origin, but steeper, because its gradient is 1/R and R is smaller.

Common mistakes

  • Finding the resistance at a point on a curve from the gradient of the tangent instead of V/I.
  • Explaining a lamp’s rising resistance by “fewer electrons”; the number stays about the same, but their drift velocity falls.
  • Explaining a thermistor’s falling resistance by “electrons move faster”; it is the number of charge carriers that rises.
  • Drawing a diode that conducts equally well in both directions.

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Syllabus and review details

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