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Resistance and Power Dissipation in Power Electronics: Ohm’s Law, Joule Heating, RMS, and Derating

Learn how resistors dissipate energy in power electronics and how to size them using Ohm’s law, Joule heating, RMS current, PWM, tolerance, TCR, derating, pulse energy, voltage limits, and thermal analysis.

Diego MunizDiego Muniz27 min read
Resistance and Power Dissipation in Power Electronics: Ohm’s Law, Joule Heating, RMS, and Derating

In power electronics, the resistor appears to be one of the simplest components in a circuit. However, once we begin working with high current, PWM, temperature, and energy pulses, it stops being merely the symbol R in an equation and becomes a thermal, electrical, and mechanical component that must be sized carefully.

A resistor marked 5 W, for example, is not saying that it can dissipate 5 W continuously under any condition. That rating depends on temperature, mounting, ventilation, construction technology, applied voltage, pulse duration, and the conditions defined by the manufacturer.

For this reason, the central question of this lesson is:

how do we turn voltage, current, RMS, tolerance, and temperature into a realistic power-resistor design?

Throughout this article, we will connect Ohm's law, Joule heating, RMS value, instantaneous power, pulse energy, tolerance, TCR, derating, maximum voltage, and thermal behavior. The goal is to develop a method that can be reused for resistive loads, precharge resistors, bleeders, shunts, snubbers, gate resistors, and power-dissipation banks.

1. A resistor is not just a number in ohms

In its simplest model, electrical resistance relates voltage and current:

R=viR=\frac{v}{i}

Its unit is the ohm:

1 Ω=1 VA1\,\Omega=1\,\frac{\mathrm{V}}{\mathrm{A}}

A real resistor, however, has many more parameters than its nominal resistance:

  • nominal resistance;
  • tolerance;
  • power rating;
  • temperature coefficient;
  • maximum working voltage;
  • overload and pulse capability;
  • temperature range;
  • parasitic inductance and capacitance;
  • mounting and ventilation constraints.

Therefore:

selecting a power resistor is not just choosing R and W; resistance, power, voltage, temperature, pulse capability, and construction must all be checked together.

2. What physically produces resistance?

When a potential difference is applied to a resistive material, an electric field is established. Charge carriers acquire an average drift velocity, but they continuously interact with the microscopic structure of the material.

These interactions transfer electrical energy to vibrations in the material structure. Macroscopically, we observe this transfer as heating.

For an approximately uniform conductor:

R=ρℓAR=\rho\frac{\ell}{A}

where:

  • ρ is the resistivity of the material;
  • ℓ is the effective length of the current path;
  • A is the cross-sectional area.

Thus:

  • increasing the length increases resistance;
  • increasing the cross-sectional area reduces resistance;
  • changing the material changes resistivity;
  • changing temperature can also change resistance.

3. Ohm's law: voltage, current, and resistance are not the same quantity

For a linear resistor whose value is approximately constant:

vR(t)=R iR(t)\boxed{v_R(t)=R\,i_R(t)}

Therefore:

iR(t)=vR(t)Ri_R(t)=\frac{v_R(t)}{R}

If we apply 12 V to a 100 Ω resistor:

I=12100=0.12 AI=\frac{12}{100}=0.12\,\mathrm{A}

or:

I=120 mAI=120\,\mathrm{mA}

Resistance does not “consume current.” Together with the applied voltage and the rest of the circuit, it determines the relationship between voltage and current.

4. Where do the resistor power equations come from?

Starting from the fundamental definition of power:

p(t)=v(t)i(t)p(t)=v(t)i(t)

For a resistor:

vR(t)=R iR(t)v_R(t)=R\,i_R(t)

Substituting:

pR(t)=(RiR(t))iR(t)p_R(t)=\left(Ri_R(t)\right)i_R(t)

Therefore:

pR(t)=R iR2(t)\boxed{p_R(t)=R\,i_R^2(t)}

We can also write:

iR(t)=vR(t)Ri_R(t)=\frac{v_R(t)}{R}

and obtain:

pR(t)=vR2(t)R\boxed{p_R(t)=\frac{v_R^2(t)}{R}}

Therefore, the three expressions:

pR=vi=i2R=v2Rp_R=vi=i^2R=\frac{v^2}{R}

do not represent different laws. They are equivalent ways of expressing the same rate of conversion of electrical energy into heat.

5. Basic example: 12 V across 100 Ω

Applying 12 V DC to 100 Ω:

I=12100=0.12 AI=\frac{12}{100}=0.12\,\mathrm{A}

The power is:

P=VI=12⋅0.12=1.44 WP=VI=12\cdot0.12=1.44\,\mathrm{W}

Confirming with the alternative form:

P=122100=1.44 WP=\frac{12^2}{100}=1.44\,\mathrm{W}

The graph shows how power increases quadratically with voltage for a fixed 100 Ω resistor:

Power dissipated in a 100 ohm resistor as a function of applied voltage
Because P equals V squared divided by R, doubling the voltage quadruples the dissipated power.

Doubling the voltage across the same resistor quadruples the power.

6. RMS current determines resistive dissipation

When current varies with time, we cannot simply use average current to calculate heating.

The average power is:

PR=1T∫0TR iR2(t) dtP_R=\frac{1}{T}\int_0^TR\,i_R^2(t)\,dt

Using the definition of RMS current:

IRMS=1T∫0TiR2(t) dtI_{RMS}=\sqrt{\frac{1}{T}\int_0^Ti_R^2(t)\,dt}

we obtain:

PR=IRMS2R\boxed{P_R=I_{RMS}^2R}

or:

PR=VRMS2R\boxed{P_R=\frac{V_{RMS}^2}{R}}

This means that a pulsed current may have a relatively low average value while still producing significant heating.

7. Power in PWM: peak, average, and pulse energy

Consider a voltage pulse with amplitude Vpk, duty cycle D, and period T.

During the ON state:

Ppk=Vpk2RP_{pk}=\frac{V_{pk}^2}{R}

If the ON time is:

ton=DTt_{on}=DT

the energy absorbed in each pulse is:

Ep=Ppkton\boxed{E_p=P_{pk}t_{on}}

The average power is:

Pavg=EpT=DPpkP_{avg}=\frac{E_p}{T}=DP_{pk}

Therefore:

Pavg=DVpk2R\boxed{P_{avg}=D\frac{V_{pk}^2}{R}}

It is important to distinguish three quantities:

Quantity
Meaning
Ppk
instantaneous power during the pulse
Ep
energy absorbed in each pulse
Pavg
average power over repeated cycles

Low average power does not automatically guarantee that the resistor can survive the pulse. The component must also satisfy the manufacturer's pulse-overload curve.

8. Complete example: 47 Ω, 12 V PWM, and 40% duty cycle

Consider:

  • nominal resistor: 47 Ω;
  • tolerance: 5%;
  • PWM: 0–12 V;
  • frequency: 1 kHz;
  • duty cycle: 40%.

The following results are rounded for display; calculations use the unrounded values.

Period

T=11000=1 msT=\frac{1}{1000}=1\,\mathrm{ms}

ON time

ton=0.40⋅1 ms=0.40 mst_{on}=0.40\cdot1\,\mathrm{ms}=0.40\,\mathrm{ms}

Peak current

Ipk=1247=0.2553 AI_{pk}=\frac{12}{47}=0.2553\,\mathrm{A}

Average current

Iavg=0.40⋅0.2553=0.1021 AI_{avg}=0.40\cdot0.2553=0.1021\,\mathrm{A}

RMS current

IRMS=0.25530.40=0.1615 AI_{RMS}=0.2553\sqrt{0.40}=0.1615\,\mathrm{A}

Peak power

Ppk=12247=3.0638 WP_{pk}=\frac{12^2}{47}=3.0638\,\mathrm{W}

Average power

Pavg=0.40⋅3.0638=1.2255 WP_{avg}=0.40\cdot3.0638=1.2255\,\mathrm{W}

Confirming with RMS current:

Pavg=(12470.40)2⋅47≈1.2255 WP_{avg}=\left(\frac{12}{47}\sqrt{0.40}\right)^2\cdot47\approx1.2255\,\mathrm{W}

Energy per pulse

Ep=3.0638⋅0.40 ms=1.2255 mJE_p=3.0638\cdot0.40\,\mathrm{ms}=1.2255\,\mathrm{mJ}

Because there are 1000 pulses per second:

Pavg=Epfs=1.2255 mJ⋅1000=1.2255 WP_{avg}=E_pf_s=1.2255\,\mathrm{mJ}\cdot1000=1.2255\,\mathrm{W}

9. How does power vary with duty cycle?

For a 47 Ω resistor supplied by a 0–12 V PWM waveform:

Pavg=D12247P_{avg}=D\frac{12^2}{47}

Therefore, average power increases linearly with D:

Average power in a 47 ohm resistor supplied by 12 V PWM as a function of duty cycle
Average power increases linearly with duty cycle, while peak power during the ON state remains approximately 3.064 W.

This also shows why duty cycle is a direct power-control tool for resistive loads.

10. Why does using average current underestimate heating?

In the same example:

Iavg=0.1021 AI_{avg}=0.1021\,\mathrm{A}

If someone calculates:

Iavg2R=(0.1021)2⋅47=0.490 WI_{avg}^2R=(0.1021)^2\cdot47=0.490\,\mathrm{W}

the result is less than half the actual dissipation.

The correct answer is:

IRMS2R=(12470.40)2⋅47≈1.2255 WI_{RMS}^2R=\left(\frac{12}{47}\sqrt{0.40}\right)^2\cdot47\approx1.2255\,\mathrm{W}

The error comes from squaring the average current instead of averaging the square of the current.

11. Tolerance: the worst case depends on the type of source

A 47 Ω resistor with a 5% tolerance may have a value between:

Rmin=47(1−0.05)=44.65 ΩR_{min}=47(1-0.05)=44.65\,\OmegaRmax=47(1+0.05)=49.35 ΩR_{max}=47(1+0.05)=49.35\,\Omega

Voltage-driven excitation

With fixed voltage:

P=V2RP=\frac{V^2}{R}

The lower the resistance, the higher the power. Therefore, the worst case is Rmin.

For 12 V PWM with D = 40%:

Pmax,tol=0.4014444.65=1.290 WP_{max,tol}=0.40\frac{144}{44.65}=1.290\,\mathrm{W}

Current-driven excitation

With fixed current:

P=I2RP=I^2R

Now the larger resistance produces the larger power. Therefore, the worst case is Rmax.

This reasoning can be summarized as follows:

flowchart TD;
A["Define excitation type"] --> B{"Approximately fixed voltage?"};
B -->|Yes| C["Use Rmin for maximum power"];
B -->|No| D{"Approximately fixed current?"};
D -->|Yes| E["Use Rmax for maximum power"];
D -->|No| F["Model the complete source and load"];
With fixed voltage, the lowest resistance maximizes power; with fixed current, the highest resistance maximizes power.

12. Visualizing the effect of tolerance in the 47 Ω example

With Vpk = 12 V and D = 40%, the average power at the three resistance values is:

Effect of 5 percent tolerance on the average power of a nominal 47 ohm resistor
With fixed voltage, the resistor at the lower resistance limit dissipates the highest power.

A relatively small change in resistance directly affects the available thermal margin.

13. Temperature coefficient: resistance changes as the resistor heats up

Near a reference temperature T0, we can use a linear approximation:

R(T)≈R(T0)[1+α(T−T0)]\boxed{R(T)\approx R(T_0)\left[1+\alpha(T-T_0)\right]}

If the temperature coefficient is specified in ppm/°C:

α1/∘C=αppm/∘C⋅10−6\alpha_{1/^{\circ}\mathrm{C}}=\alpha_{\mathrm{ppm}/^{\circ}\mathrm{C}}\cdot10^{-6}

For example:

200 ppm/∘C=0.0002/∘C200\,\mathrm{ppm}/^{\circ}\mathrm{C}=0.0002/^{\circ}\mathrm{C}

For a 100 Ω resistor referenced to 25 °C, using this TCR only as a mathematical example of the model:

R(75∘C)=100[1+0.0002(75−25)]=101 ΩR(75^{\circ}\mathrm{C})=100\left[1+0.0002(75-25)\right]=101\,\Omega

The graph shows the linear trend predicted by this model:

Example of the variation of a 100 ohm resistor with a temperature coefficient of 200 ppm per degree Celsius
The graph illustrates the local linear TCR model; the real behavior must be confirmed in the datasheet for the selected resistor series.

This is a local model. The exact behavior, valid temperature range, and actual TCR must be obtained from the selected component's datasheet.

14. Rated power is not universally available power

When a resistor is specified as 2 W, 5 W, or 10 W, that rating was defined under specific test conditions.

If PN is the rated power and the derating curve provides a factor kD:

Padm=kDPN\boxed{P_{adm}=k_DP_N}

The design must satisfy:

Pcalc,max≤PadmP_{calc,max}\le P_{adm}

If we also want an additional design margin mP:

Pcalc,max≤mPPadmP_{calc,max}\le m_PP_{adm}

There is no universal margin value. The choice depends on reliability requirements, ventilation, ambient temperature, calculation uncertainty, and the consequences of failure.

15. Preliminary power-rating selection example

In the 47 Ω example, the worst case due to tolerance under fixed-voltage excitation was:

Pmax,tol=1.290 WP_{max,tol}=1.290\,\mathrm{W}

If, only for this exercise, we adopt a conservative criterion limiting continuous use to 50% of the rated power:

PN≥1.2900.50=2.58 WP_N\ge\frac{1.290}{0.50}=2.58\,\mathrm{W}

A 3 W class would satisfy only this abstract criterion. A 5 W component would provide additional margin.

But the design is still not complete. We must also check:

  • the derating curve;
  • maximum working voltage;
  • pulse capability;
  • TCR;
  • ambient temperature;
  • mounting method;
  • ventilation;
  • distance from other hot components.

16. Maximum voltage is a limit independent of power

Even when calculated power is below the resistor's rating, the component may exceed its maximum working voltage.

Therefore, we must separately verify:

∣VR∣pk≤Vwork,max|V_R|_{pk}\le V_{work,max}

This is especially relevant in:

  • high-voltage dividers;
  • DC-link bleeder resistors;
  • snubbers;
  • discharge resistors;
  • series resistor strings.

In a series string, we should not automatically assume perfect voltage sharing. Tolerances and parasitic effects can change the actual voltage distribution.

17. Pulses: low average power does not guarantee survival

A resistor can receive, for a few microseconds, instantaneous power much higher than its continuous rating.

What must be evaluated is the combination of:

  • peak power;
  • pulse duration;
  • pulse energy;
  • repetition frequency;
  • initial temperature;
  • physical resistor construction.

For a rectangular pulse:

Ep=PpktpE_p=P_{pk}t_p

The pulse curve in the datasheet must be checked for the exact component part number.

Never extrapolate an overload curve into regions not published by the manufacturer without specific justification.

18. Thermal model: electrical power becomes temperature

At steady state, a first-order thermal model is:

TR≈Ta+PRRθ\boxed{T_R\approx T_a+P_RR_{\theta}}

where:

  • TR is the temperature of the modeled point in the resistor;
  • Ta is ambient temperature;
  • PR is the average dissipated power;
  • Rθ is the thermal resistance for the actual mounting condition.

Using a first-order dynamic approximation:

TR(t)≈Ta+PRRθ(1−e−t/τθ)T_R(t)\approx T_a+P_RR_{\theta}\left(1-e^{-t/\tau_{\theta}}\right)

The resistor does not reach its final temperature instantly. It has thermal inertia.

flowchart LR;
A["Electrical power"] --> B["Resistive element"];
B --> C["Temperature rise"];
C --> D["Conduction"];
C --> E["Convection"];
C --> F["Radiation"];
D --> G["Environment"];
E --> G;
F --> G;
Electrical power is converted into heat and removed through conduction, convection, and radiation until thermal equilibrium is reached.

19. Electrothermal feedback

If resistance increases with temperature, the type of source changes the electrothermal tendency.

Approximately constant-current source

P=I2R(T)P=I^2R(T)

If R increases, power increases. This creates a positive-feedback tendency.

Approximately constant-voltage source

P=V2R(T)P=\frac{V^2}{R(T)}

If R increases, power decreases. This creates a local negative-feedback tendency.

This analysis describes only a physical tendency. It does not replace a complete thermal analysis and does not guarantee stability.

20. Two 100 Ω resistors can operate at very different temperatures

Consider two resistors:

  • 100 Ω / 2 W;
  • 100 Ω / 5 W.

Applying 12 V to either one:

I=12100=0.12 AI=\frac{12}{100}=0.12\,\mathrm{A}

and:

P=122100=1.44 WP=\frac{12^2}{100}=1.44\,\mathrm{W}

Both absorb the same electrical power.

However, they may reach different body temperatures because of:

  • physical dimensions;
  • surface area;
  • resistive technology;
  • thermal mass;
  • coating;
  • mounting method;
  • distance from the PCB;
  • airflow.

Dissipated power and temperature are not the same quantity.

21. Teaching circuit for a 12 V test

The circuit below allows current, voltage, and temperature to be measured on a 100 Ω resistor.

Power-dissipation test circuit using a 100 ohm resistor supplied from 12 V with an ammeter and voltmeter
The supply should be current-limited to 150 mA. The ammeter is connected in series and the voltmeter in parallel with the resistor under test.

The power supply should be set to 12 V with a current limit of approximately 150 mA.

The ammeter must be connected in series. The voltmeter must be connected in parallel.

Never connect an ammeter directly across the power supply.

22. How does power increase in the 100 Ω experiment?

The 6 V, 9 V, and 12 V operating points give:

Voltage
Current
Power
Utilization of a 2 W resistor
Utilization of a 5 W resistor
6 V
60 mA
0.36 W
18%
7.2%
9 V
90 mA
0.81 W
40.5%
16.2%
12 V
120 mA
1.44 W
72%
28.8%

The same curve was shown earlier using P = V²/R. These three points are ideal for a teaching experiment on temperature rise.

23. How should temperature be measured?

Some options are:

  • thermocouple: measures locally, but its wires can slightly affect the thermal path;
  • infrared thermometer or thermal camera: depends on emissivity and target size;
  • resistance change: can estimate average temperature when TCR is known;
  • touch: is not a measurement method and can cause burns.

Always record ambient temperature before the test.

After power is removed, allow the component to cool before changing connections.

24. Real resistors have parasitic elements

At high frequency, a resistor is not just R.

It also exhibits:

  • series inductance;
  • parasitic capacitance;
  • terminal resistance;
  • coupling to the PCB and nearby components.

Wirewound resistors, for example, may have significant inductance.

This may be acceptable for a DC load but unsuitable for:

  • fast shunts;
  • snubbers;
  • RF circuits;
  • very fast gate-drive paths.

In these applications, low-inductance construction becomes an important specification.

25. Cable, PCB-trace, and switch resistance also dissipate power

The same physics applies to resistances that were not deliberately added to the circuit.

For a cable:

Pcab=IRMS2RcabP_{cab}=I_{RMS}^2R_{cab}

Therefore, if current doubles while resistance remains approximately constant:

Pnew=(2I)2R=4I2RP_{new}=(2I)^2R=4I^2R

In other words:

doubling the current quadruples resistive loss.

This principle will be fundamental when studying MOSFETs, inductor windings, transformers, PCB traces, and busbars.

26. Where are resistors used in power electronics?

Application
Main function
Critical parameters
Precharge
Limit DC-link charging current
pulse energy, power, voltage
Bleeder
Discharge capacitors
continuous power, voltage, discharge time
Shunt
Convert current into voltage
tolerance, TCR, power, Kelvin connection
Snubber
Damp oscillations and dissipate energy
pulse, frequency, parasitics
Gate resistor
Control gate current and switching speed
pulse capability, inductance, resistance value
Test load
Dissipate a known amount of energy
power, temperature, ventilation
Braking
Absorb regenerative energy
energy, continuous power, pulse power

27. Practical sizing method

A minimum sequence for selecting a power resistor is:

flowchart TD;
A["Define voltage, current, and waveform"] --> B["Calculate peak, RMS, and average power"];
B --> C["Calculate pulse energy when applicable"];
C --> D["Apply worst-case tolerance"];
D --> E["Apply derating and margin"];
E --> F["Check maximum voltage"];
F --> G["Check pulse curve"];
G --> H["Evaluate mounting and temperature"];
H --> I["Validate in the laboratory"];
Correct sizing combines electrical stress, tolerance, derating, maximum voltage, pulse capability, and thermal validation.

As a checklist:

  1. define the actual voltage or current;
  2. determine the waveform;
  3. calculate RMS;
  4. calculate average power;
  5. calculate peak power;
  6. calculate pulse energy;
  7. apply tolerance;
  8. check TCR;
  9. apply derating;
  10. check maximum voltage;
  11. check pulse capability;
  12. evaluate mounting and ventilation;
  13. measure temperature and current in a controlled test.

28. Series and parallel resistor combinations

For n identical resistors in series:

Req=nRR_{eq}=nR

They all carry the same current.

For n identical resistors in parallel:

Req=RnR_{eq}=\frac{R}{n}

In the ideal model, current divides equally. In practice, tolerance and temperature can alter current sharing.

In power resistor banks, the goal is not only to obtain the correct equivalent resistance, but also to distribute voltage, current, power, and heat safely.

29. Load-bank example: approximately 8 W at 24 V

To dissipate approximately 8 W at 24 V, the desired equivalent resistance is:

Req=2428=72 ΩR_{eq}=\frac{24^2}{8}=72\,\Omega

Two 150 Ω resistors in parallel would result in:

Req=75 ΩR_{eq}=75\,\Omega

and:

Ptotal=24275=7.68 WP_{total}=\frac{24^2}{75}=7.68\,\mathrm{W}

But each resistor would dissipate:

Peach=242150=3.84 WP_{each}=\frac{24^2}{150}=3.84\,\mathrm{W}

This exceeds the rating of a 3 W resistor.

One solution presented in the exercise is to use four branches, each branch containing two 150 Ω resistors in series. Each branch is 300 Ω, and four branches in parallel produce 75 Ω.

75 ohm load bank built from eight 150 ohm resistors
Four parallel branches, each containing two 150 ohm resistors in series, distribute the power among eight components.

The power per resistor in the nominal case is approximately:

Peach=7.688=0.96 WP_{each}=\frac{7.68}{8}=0.96\,\mathrm{W}

In the worst case from the exercise, with all resistors at 142.5 Ω:

Req,min=71.25 ΩR_{eq,min}=71.25\,\OmegaPtotal,max=24271.25=8.084 WP_{total,max}=\frac{24^2}{71.25}=8.084\,\mathrm{W}

Even after this calculation, we still need to check derating, maximum voltage, mounting, spacing, and mutual heating.

30. Suggested SPICE simulation

The 47 Ω PWM case can be simulated using generic elements:

* Lesson 003 - resistor under PWM with tolerance sweep
.param RVAL=47
V1 in 0 PULSE(0 12 0 1u 1u 400u 1m)
R1 in 0 {RVAL}
.step param RVAL list 44.65 47 49.35
.tran 0 5m 0 1u
.meas tran V_RMS RMS V(in) FROM 1m TO 5m
.meas tran I_RMS RMS I(R1) FROM 1m TO 5m
.meas tran P_AVG AVG V(in)*I(R1) FROM 1m TO 5m
.meas tran P_PEAK MAX V(in)*I(R1) FROM 1m TO 5m
.end

Expected nominal results:

  • V RMS ≈ 7.59 V;
  • I RMS ≈ 161.5 mA;
  • average power ≈ 1.226 W;
  • peak power ≈ 3.064 W.

The simulation does not automatically determine temperature or prove that a real resistor can withstand the pulses. Those checks belong to the datasheet and the physical test.

31. Laboratory procedure

The proposed experiment compares 100 Ω / 5 W and 100 Ω / 2 W resistors at low voltage.

Materials

  • isolated DC power supply with current limiting;
  • 100 Ω / 5 W resistor;
  • 100 Ω / 2 W resistor;
  • multimeter;
  • thermocouple or suitable thermometer;
  • stopwatch;
  • nonflammable support.

Procedure

  1. identify the manufacturer, series, and datasheet for each resistor;
  2. measure the cold resistance with the circuit de-energized;
  3. mount the 5 W resistor with adequate ventilation space;
  4. set the supply to 6 V with a 150 mA current limit;
  5. record voltage, current, and temperature;
  6. repeat at 9 V and 12 V;
  7. allow the resistor to cool completely;
  8. repeat with the 2 W resistor only if the datasheet and safety conditions allow it.

Using the teaching criterion described in the lesson, stop the test at a measured surface temperature of 70 °C or earlier if the datasheet or laboratory rules specify a lower limit. Stop immediately if there is odor, smoke, discoloration, unexpected current, or instability.

32. Measuring correctly

Resistance

  • power supply off;
  • resistor cool;
  • preferably removed from the circuit or with one terminal isolated;
  • never use an ohmmeter on an energized circuit.

Voltage

The voltmeter is connected in parallel with the resistor.

Current

The ammeter is connected in series.

Temperature

Whenever possible, position the temperature sensor before energizing the circuit. Also record ambient temperature.

Power

In DC:

PR=VRIRP_R=V_RI_R

For a resistive PWM waveform:

PR=VR,RMS2RP_R=\frac{V_{R,RMS}^2}{R}

33. What can go wrong?

Symptom
Likely cause
Diagnosis / correction
Current is much higher than expected
Wrong resistor, short circuit, or wiring error
Power off and measure resistance outside the circuit
Power supply enters CC mode
Current limit reached
Use the actual measured voltage, not the programmed voltage
Resistor heats more than expected
Incorrect RMS calculation or poor thermal mounting
Recalculate power and review ventilation
Two resistors with the same rating heat differently
Different technologies and thermal paths
Compare datasheets and mounting conditions
Resistance drifts after the test
Overtemperature or overload
Compare cold resistance before and after the test
Infrared temperature reading appears too low
Incorrect emissivity setting
Review the thermal measurement method
PCB darkens
Excessive heat transferred to the board
Review spacing, mounting, and power distribution

34. Exercises

Exercise 1

A 220 Ω resistor is supplied with 10 V DC. Calculate current, power, and energy dissipated in 3 minutes.

Exercise 2

A 20 Ω resistor receives a 0–10 V PWM waveform with a 25% duty cycle and a frequency of 2 kHz. Calculate peak current, RMS current, peak power, average power, ON time, and pulse energy.

Exercise 3

A nominal 1 kΩ resistor with 1% tolerance receives 30 V DC. Determine the resistance range, the maximum possible power due to tolerance, and evaluate a nominal 1 W component using a maximum-utilization criterion of 60% before thermal derating.

Exercise 4

Design an approximately 8 W load for 24 V using only 150 Ω / 3 W resistors with 5% tolerance.

35. Exercise answers

Exercise 1

I=10220=45.45 mAI=\frac{10}{220}=45.45\,\mathrm{mA}P=102220=0.4545 WP=\frac{10^2}{220}=0.4545\,\mathrm{W}

For 3 minutes:

t=180 st=180\,\mathrm{s}E=Pt=0.4545⋅180=81.8 JE=Pt=0.4545\cdot180=81.8\,\mathrm{J}

Exercise 2

Ipk=1020=0.5 AI_{pk}=\frac{10}{20}=0.5\,\mathrm{A}IRMS=0.50.25=0.25 AI_{RMS}=0.5\sqrt{0.25}=0.25\,\mathrm{A}Ppk=10220=5 WP_{pk}=\frac{10^2}{20}=5\,\mathrm{W}Pavg=0.25⋅5=1.25 WP_{avg}=0.25\cdot5=1.25\,\mathrm{W}T=12000=0.5 msT=\frac{1}{2000}=0.5\,\mathrm{ms}ton=0.25⋅0.5 ms=0.125 mst_{on}=0.25\cdot0.5\,\mathrm{ms}=0.125\,\mathrm{ms}Ep=5⋅0.125 ms=0.625 mJE_p=5\cdot0.125\,\mathrm{ms}=0.625\,\mathrm{mJ}

Exercise 3

Rmin=990 ΩR_{min}=990\,\OmegaRmax=1010 ΩR_{max}=1010\,\Omega

With a fixed 30 V source:

Pmax=302990=0.909 WP_{max}=\frac{30^2}{990}=0.909\,\mathrm{W}

A 60% utilization criterion on a 1 W component would allow only:

Plim=0.60 WP_{lim}=0.60\,\mathrm{W}

Therefore, 1 W does not satisfy this criterion.

The minimum nominal power before derating would be:

PN≥0.9090.60=1.515 WP_N\ge\frac{0.909}{0.60}=1.515\,\mathrm{W}

Exercise 4

The desired equivalent resistance is:

Req,desired=2428=72 ΩR_{eq,desired}=\frac{24^2}{8}=72\,\Omega

Two 150 Ω resistors in parallel provide 75 Ω, but each component would dissipate 3.84 W and exceed a 3 W rating.

The eight-resistor arrangement shown earlier distributes the power while maintaining approximately 75 Ω.

36. What you should remember from this lesson

  • Ohm's law relates voltage, current, and resistance.
  • Instantaneous resistor power can be written as vi, i²R, or v²/R.
  • For time-varying waveforms, resistive loss depends on RMS.
  • In PWM, peak power, pulse energy, and average power are different quantities.
  • With fixed voltage, Rmin maximizes power.
  • With fixed current, Rmax maximizes power.
  • Rated power depends on specified thermal conditions.
  • Derating, maximum voltage, and pulse capability are independent checks.
  • TCR causes resistance to vary with temperature.
  • Resistors with the same resistance can operate at very different temperatures.
  • Parasitic resistance in cables, PCB traces, and semiconductors also produces I²R losses.

The resistor is simple in the equation, but real-world sizing is an electrothermal problem.

37. Next lesson

In Lesson 004, the focus will be the capacitor in the time domain.

The resistance studied here will reappear in three important places:

  1. the resistor that limits charging current;
  2. the capacitor ESR;
  3. the discharge resistor.

The central physical distinction will be:

the resistor dissipates energy; the ideal capacitor stores energy in the electric field.

References

  1. ERICKSON, Robert W.; MAKSIMOVIĆ, Dragan. Fundamentals of Power Electronics. 2nd ed. Kluwer Academic Publishers, 2001.
  2. MOHAN, Ned; UNDELAND, Tore M.; ROBBINS, William P. Power Electronics: Converters, Applications, and Design. 3rd ed. John Wiley & Sons, 2003.
  3. HURLEY, W. G.; WÖLFLE, W. H. Transformers and Inductors for Power Electronics: Theory, Design and Applications. Wiley, 2013.
  4. For the component used in the laboratory, always consult the official datasheet for the exact series and part number: rated power, mounting conditions, derating, maximum voltage, tolerance, TCR, pulse capability, and temperature range.
  5. Also consult the manuals for the power supply, multimeter, and thermal instrument used in the experiment.

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Power Resistors: RMS, Dissipation, TCR, and Derating | Diego Muniz