Power Electronics Waveforms: RMS, Average, Peak & PWM

Learn how to interpret power electronics waveforms and distinguish instantaneous, average, RMS, peak, and peak-to-peak values. The article explains PWM, power, ripple, and measurement techniques through calculations, graphs, circuits, and practical examples.

Diego MunizPublished: September 10, 2026Technical review: September 11, 202627 min read

Power Electronics Waveforms: RMS, Average, Peak & PWM

In power electronics, a single waveform can be described by several different numbers, and all of them may be correct at the same time.

A PWM signal, for example, may have an average value of 3 V, an RMS value of 6 V, a peak value of 12 V, and a peak-to-peak value of 12 V. Therefore, the right question is not simply “what is the voltage?”, but rather:

which waveform quantity answers the engineering problem I am trying to solve?

This distinction is fundamental when sizing resistors, MOSFETs, diodes, inductors, capacitors, PCB traces, cables, and measurement instruments. Using the average value when heating depends on RMS, or ignoring a voltage peak because the average is low, can turn an apparently correct design into a thermal or electrical failure.

In this lesson, we will build a systematic method for interpreting periodic signals, relating each metric to its physical meaning and practical use in power electronics.

1. A waveform can have several “voltages” at the same time

Consider a PWM waveform that switches between 0 V and 12 V, remains ON for 25% of the period, and OFF for the remaining 75%.

For this same signal, we can obtain:

average value: 3 V;
RMS value: 6 V;
peak value: 12 V;
peak-to-peak value: 12 V.

None of these values invalidates the others. Each one answers a different question:

Quantity	Engineering question
Instantaneous value	What is the signal value at this exact instant?
Average value	What is the DC component or algebraic balance of the signal?
RMS value	What DC value would produce the same resistive heating?
Peak value	What is the maximum instantaneous electrical stress?
Peak-to-peak value	What is the total excursion between the minimum and maximum values?

A good waveform-reading procedure follows a logical sequence:

flowchart TD; A["Define signal and reference"] --> B["Choose observation window"]; B --> C["Identify period and frequency"]; C --> D["Measure maximum, minimum, and peak-to-peak"]; D --> E["Calculate average and RMS"]; E --> F["Relate the metric to its physical effect"]; F --> G["Validate against power and component limits"];

Before using a waveform measurement, define the reference, the observation window, and the physical quantity of interest.

2. First read the time axis: period, frequency, and duty cycle

A periodic waveform repeats after a time interval T, called the period.

Frequency indicates how many repetitions occur per second:

f=\frac{1}{T}

If a PWM signal operates at 1 kHz:

f_s=1\,\mathrm{kHz}=1000\,\mathrm{Hz}

then its period is:

T_s=\frac{1}{1000}=1\,\mathrm{ms}

The duty cycle represents the fraction of the period during which the signal remains in its active state:

D=\frac{t_{on}}{T}

For a duty cycle of 25%:

D=0.25

and, for a 1 ms period:

t_{on}=DT=0.25\cdot1\,\mathrm{ms}=250\,\mu\mathrm{s}

t_{off}=(1-D)T=750\,\mu\mathrm{s}

The timing behavior can be visualized directly:

12 V PWM waveform with a 25 percent duty cycle — 1 kHz PWM: the output remains at 12 V for 250 us and at 0 V for 750 us during each period.

Frequency tells us how often the pattern repeats; duty cycle tells us how long the switch remains active within each repetition.

3. Instantaneous value: a snapshot of the signal

The instantaneous value is simply the voltage or current value at a specific instant.

For a voltage:

v(t_0)

is the voltage at instant t₀.

On an oscilloscope, each point on the trace represents a sample of the signal at that instant. This information is essential for observing:

rise time;
fall time;
overshoot;
undershoot;
ringing;
delays between voltage and current;
switching sequence.

However, a single instantaneous value does not describe the energy behavior of the signal over an entire cycle.

4. Maximum, minimum, peak, and peak-to-peak values

Within an observation window, we can define:

X_{max}=\max x(t)

X_{min}=\min x(t)

The absolute peak is the largest magnitude reached:

X_{pk}=\max |x(t)|

The peak-to-peak value is the distance between maximum and minimum:

\boxed{X_{pp}=X_{max}-X_{min}}

For a 0-to-12 V PWM waveform:

V_{max}=12\,\mathrm{V}

V_{min}=0\,\mathrm{V}

V_{pp}=12-0=12\,\mathrm{V}

For a symmetrical sine wave between +12 V and -12 V:

V_{pk}=12\,\mathrm{V}

V_{pp}=12-(-12)=24\,\mathrm{V}

This distinction is important: peak and peak-to-peak are not synonyms.

5. Average value: the DC component of the signal

The average value represents the algebraic balance of the waveform over the observation window.

It can be understood as the height of a rectangle having the same algebraic area as the waveform over one period.

For a periodic signal:

\boxed{X_{avg}=\frac{1}{T}\int_{t_0}^{t_0+T}x(t)\,dt}

For a sine wave perfectly symmetrical about zero, the positive area cancels the negative area. Therefore:

X_{avg}=0

This does not mean that there is no voltage, current, or heating. It only means that the algebraic balance over the period is zero.

Average value answers the question “what is the DC component?”, not “how much does it heat?”.

6. RMS: the value that connects the waveform to heating

RMS means root mean square.

Its physical meaning becomes especially clear when we analyze a resistor.

The instantaneous power dissipated by a resistance is:

p_R(t)=R\,i^2(t)

The average power is:

P_R=\frac{1}{T}\int_{t_0}^{t_0+T}R\,i^2(t)\,dt

Because R is constant:

P_R=R\left[\frac{1}{T}\int_{t_0}^{t_0+T}i^2(t)\,dt\right]

We then define an equivalent DC current that produces exactly the same heating:

P_R=R I_{RMS}^2

This leads to:

\boxed{I_{RMS}=\sqrt{\frac{1}{T}\int_{t_0}^{t_0+T}i^2(t)\,dt}}

For voltage:

\boxed{V_{RMS}=\sqrt{\frac{1}{T}\int_{t_0}^{t_0+T}v^2(t)\,dt}}

For a purely resistive load:

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

The key point is that squaring removes polarity. A current of -2 A produces the same resistive heating as +2 A.

7. A circuit that physically demonstrates the meaning of RMS

Consider two identical 12 Ω resistors.

In the first circuit, we apply a 0-to-12 V PWM waveform with a 25% duty cycle. In the second, we continuously apply 6 V DC.

Comparison between a resistive load supplied by PWM and a resistive load supplied by an equivalent DC voltage — A 0-to-12 V PWM waveform with a 25 percent duty cycle and a 6 V DC source both produce an average power of 3 W in 12 ohm resistors.

In the PWM circuit, when the switch is ON:

I_{pk}=\frac{12}{12}=1\,\mathrm{A}

and the instantaneous power in the resistor is:

p_{on}=12\cdot1=12\,\mathrm{W}

Because this occurs during only 25% of the period:

P_R=D\,p_{on}=0.25\cdot12=3\,\mathrm{W}

In the DC circuit:

P_R=\frac{6^2}{12}=3\,\mathrm{W}

Therefore, the PWM waveform produces the same heating as a 6 V DC signal. This is exactly why we say:

V_{RMS}=6\,\mathrm{V}

8. Deriving the average and RMS values of a PWM waveform

Consider a unipolar PWM waveform defined by:

v(t)=\begin{cases}V_g, & 0\le t<DT\\0, & DT\le t<T\end{cases}

The waveform repeats every period T.

Average value

V_{avg}=\frac{1}{T}\left(\int_0^{DT}V_g\,dt+\int_{DT}^{T}0\,dt\right)

V_{avg}=\frac{V_gDT}{T}

\boxed{V_{avg}=DV_g}

RMS value

V_{RMS}=\sqrt{\frac{1}{T}\left(\int_0^{DT}V_g^2\,dt+\int_{DT}^{T}0^2\,dt\right)}

V_{RMS}=\sqrt{\frac{V_g^2DT}{T}}

\boxed{V_{RMS}=V_g\sqrt D}

For 0 < D < 1:

D<\sqrt D

and therefore:

V_{avg}0,\ 0<D<1)

The graph makes this difference very clear:

Average and RMS values of a 0-to-12 V PWM waveform as a function of duty cycle — The average value increases linearly with D, while the RMS value increases with the square root of D. The peak remains at 12 V whenever a pulse is present.

9. Complete example: 24 V PWM across an 8 Ω load

Now consider an 8 Ω resistive load supplied by an ideal PWM waveform between 0 and 24 V, with:

D=0.36

f_s=20\,\mathrm{kHz}

Period

T=\frac{1}{20\times10^3}=50\,\mu\mathrm{s}

ON time

t_{on}=DT=0.36\cdot50\,\mu\mathrm{s}=18\,\mu\mathrm{s}

Average voltage

V_{avg}=0.36\cdot24=8.64\,\mathrm{V}

RMS voltage

V_{RMS}=24\sqrt{0.36}=14.4\,\mathrm{V}

Peak and peak-to-peak voltage

V_{pk}=V_{pp}=24\,\mathrm{V}

Peak current

I_{pk}=\frac{24}{8}=3\,\mathrm{A}

Average current

I_{avg}=DI_{pk}=0.36\cdot3=1.08\,\mathrm{A}

RMS current

I_{RMS}=I_{pk}\sqrt D=3\sqrt{0.36}=1.8\,\mathrm{A}

Average power

P_R=I_{RMS}^2R

P_R=(1.8)^2\cdot8=25.92\,\mathrm{W}

The same answer is obtained using RMS voltage:

P_R=\frac{(14.4)^2}{8}=25.92\,\mathrm{W}

or directly from the switching states:

P_R=D\frac{V_g^2}{R}=0.36\frac{24^2}{8}=25.92\,\mathrm{W}

Three different methods lead to the same result. This agreement is an excellent validation tool.

10. Why can multiplying average voltage by average current give the wrong answer?

In the previous example:

V_{avg}=8.64\,\mathrm{V}

I_{avg}=1.08\,\mathrm{A}

If we multiply the averages:

V_{avg}I_{avg}=8.64\cdot1.08=9.3312\,\mathrm{W}

But we have just shown that the actual average power is 25.92 W.

The error occurs because power must be formed before averaging:

\boxed{P=\frac{1}{T}\int_{t_0}^{t_0+T}v(t)i(t)\,dt}

In general:

\boxed{P\ne V_{avg}I_{avg}}

In a resistive PWM load, voltage and current rise and fall together. Multiplying their separate averages loses this time correlation.

flowchart LR; A["Measure v(t)"] --> C["Multiply point by point"]; B["Measure i(t)"] --> C; C --> D["p(t) = v(t)i(t)"]; D --> E["Calculate the average of p(t)"]; E --> F["Average power P"];

Instantaneous power must be calculated by multiplying voltage and current at the same instant; only then should the time average be taken.

11. Energy per cycle: another way to verify the result

For the 24 V, 8 Ω, 36% duty-cycle, 20 kHz example, the power during the ON interval is:

p_{on}=V_gI_{pk}=24\cdot3=72\,\mathrm{W}

The ON time is 18 μs. Therefore, the energy dissipated during one cycle is:

E_T=p_{on}t_{on}

E_T=72\cdot18\,\mu\mathrm{s}=1.296\,\mathrm{mJ}

Because the period is 50 μs:

P=\frac{E_T}{T}=\frac{1.296\,\mathrm{mJ}}{50\,\mu\mathrm{s}}=25.92\,\mathrm{W}

Again, the result matches the RMS-based calculation.

12. Common waveforms and their characteristic values

For a sine wave:

x(t)=X_{pk}\sin(\omega t)

with:

\omega=2\pi f

over one complete period:

X_{avg}=0

X_{RMS}=\frac{X_{pk}}{\sqrt2}

X_{pp}=2X_{pk}

Waveform	Average value	RMS value	Peak-to-peak
Constant DC X₀	X₀	\|X₀\|	0
Unipolar PWM 0/X_pk	DX_pk	X_pk√D	X_pk
Bipolar square wave ±X_pk	0	X_pk	2X_pk
Sine wave X_pksin(ωt)	0	X_pk/√2	2X_pk
Full-wave rectified sine wave	2X_pk/π	X_pk/√2	X_pk

13. DC component and ripple

In switched-mode power supplies, the output is rarely a perfectly flat line. It usually contains a desired DC component and a smaller varying component called ripple.

We can write:

x(t)=X_{avg}+x_{ripple}(t)

By definition, the ripple has zero average over the analyzed period:

\frac{1}{T}\int_{t_0}^{t_0+T}x_{ripple}(t)\,dt=0

The relationship between total RMS, average component, and ripple RMS is:

\boxed{X_{RMS}^2=X_{avg}^2+X_{ripple,RMS}^2}

Therefore:

\boxed{X_{ripple,RMS}=\sqrt{X_{RMS}^2-X_{avg}^2}}

This equation is especially important for capacitors, because RMS ripple current contributes directly to their internal losses.

Peak-to-peak ripple and RMS ripple are different quantities. A fixed conversion factor exists only when the ripple waveform shape is known.

14. Crest factor: why narrow pulses are difficult to measure

Crest factor compares the largest peak with the RMS value:

CF=\frac{X_{pk}}{X_{RMS}}

For a unipolar PWM waveform:

CF=\frac{1}{\sqrt D}

As duty cycle decreases, the waveform can have a high peak but a relatively low RMS value. For example, at 1%:

D=0.01

CF=\frac{1}{\sqrt{0.01}}=10

The graph shows how crest factor increases rapidly for narrow pulses:

PWM crest factor as a function of duty cycle — Narrow pulses have a high crest factor, which can exceed the RMS measurement limits of some instruments.

This is one reason why a true RMS multimeter can still produce errors with highly pulsed signals: in addition to voltage range, its crest-factor limit and bandwidth must also be respected.

15. The measurement window is part of the answer

Average and RMS values always belong to a time window.

For periodic signals in steady state, the ideal approach is to calculate over:

one complete period; or
an integer number of periods.

If the window starts in the middle of a pulse and ends before an integer number of cycles is completed, the calculated average may change.

The correct measurement workflow can be summarized as follows:

flowchart TD; A["Define a safe GND reference"] --> B["Configure scale and probe attenuation"]; B --> C["Stabilize the trigger"]; C --> D["Capture complete cycles"]; D --> E["Measure Vmax, Vmin, and Vpp"]; E --> F["Measure average and RMS"]; F --> G["Compare with the theoretical calculation"];

A consistent measurement requires a correct reference, stable acquisition, and a window containing complete cycles.

16. Bandwidth: looking only at the PWM frequency is not enough

A 1 kHz PWM signal may appear to be “slow,” but its edges may occur within tens of nanoseconds.

A commonly used first-order estimate is:

BW\approx\frac{0.35}{t_r}

If the rise time is 20 ns:

BW\approx\frac{0.35}{20\,\mathrm{ns}}=17.5\,\mathrm{MHz}

This shows why choosing an oscilloscope based only on the PWM repetition frequency is inadequate. The spectral content of the edges may extend many orders of magnitude above the fundamental frequency.

17. Parasitics can change exactly the values that matter most: the peaks

In power electronics, wires, PCB traces, and component terminals have parasitic inductance.

When current changes rapidly:

v_p=L_p\frac{di}{dt}

Parasitic capacitances also conduct current when voltage changes rapidly:

i_p=C_p\frac{dv}{dt}

These effects can produce:

overshoot;
undershoot;
ringing;
ground bounce;
EMI.

This problem is particularly important because average and RMS values may change only slightly while the peak can exceed the component's absolute maximum rating.

The power loop and the measurement loop must be treated separately:

flowchart LR; A["Source"] --> B["Switch"]; B --> C["Load"]; C --> D["Power return"]; D --> A; C --> E["Oscilloscope probe"]; E --> F["Ground clip"]; F --> D;

The power loop carries the switched current; the oscilloscope probe loop may pick up or introduce parasitic signals if its area is too large.

18. Safe experiment: 0–5 V, 1 kHz, and a 1 kΩ resistor

A simple experiment allows us to observe all of these quantities without working at high power.

Materials

bench function generator;
1 kΩ / 0.25 W resistor;
oscilloscope;
10× passive probe;
digital multimeter;
breadboard and short wires.

Configuration

waveform: 0–5 V pulse;
frequency: 1 kHz;
initial duty cycle: 25%;
function-generator output: High-Z mode.

The expected average value is:

V_{avg}=0.25\cdot5=1.25\,\mathrm{V}

The expected total RMS value is:

V_{RMS}=5\sqrt{0.25}=2.5\,\mathrm{V}

The peak current is:

I_{pk}=\frac{5}{1000}=5\,\mathrm{mA}

And the average power is:

P_R=\frac{(2.5)^2}{1000}=6.25\,\mathrm{mW}

19. What should the multimeter and oscilloscope display?

On the oscilloscope, with DC coupling, you should observe approximately:

V_max = 5 V;
V_min = 0 V;
V_pp = 5 V;
period = 1 ms;
high time = 250 μs;
average = 1.25 V;
total RMS = 2.5 V.

In DC voltage mode, the multimeter will tend to indicate approximately the average component: 1.25 V.

In AC voltage mode, the behavior depends on the instrument design. Many multimeters remove the DC component and measure only the alternating component.

If we know the total RMS value and the average value, we can calculate the RMS ripple:

V_{ripple,RMS}=\sqrt{2.5^2-1.25^2}=2.165\,\mathrm{V}

Therefore, directly comparing an AC multimeter reading with the oscilloscope's total RMS value may mean comparing two different quantities.

20. Variation with duty cycle: 25%, 50%, and 75%

Keeping the PWM amplitude at 0–5 V:

Duty cycle	Average voltage	RMS voltage	Peak voltage	Power in 1 kΩ
25%	1.25 V	2.50 V	5 V	6.25 mW
50%	2.50 V	3.54 V	5 V	12.50 mW
75%	3.75 V	4.33 V	5 V	18.75 mW

Graphically:

Comparison of average voltage, RMS voltage, and power for duty cycles of 25, 50, and 75 percent — For a fixed 5 V amplitude, average, RMS, and power evolve differently as duty cycle increases.

Notice that, for a resistive load with fixed amplitude, average power is proportional to duty cycle, while RMS voltage increases with √D.

21. How to configure the oscilloscope

For the 0–5 V experiment:

probe tip: upper terminal of the resistor;
ground clip: GND of the isolated circuit;
coupling: DC;
probe: 10×;
channel: also configured for 10× attenuation;
initial vertical scale: approximately 1 V/div;
time base: approximately 200 μs/div;
trigger: rising edge, near 2.5 V.

Measure:

V_max;
V_min;
V_pp;
average;
RMS;
period;
frequency;
duty cycle.

Then, if you want to observe only the varying component, temporarily switch to AC coupling. To record the complete waveform, return to DC coupling.

22. Pay attention to High-Z and 50 Ω function-generator settings

Many function generators calculate the displayed amplitude assuming that the output is terminated in 50 Ω.

If the generator is configured for 50 Ω but connected to a high-impedance load, the measured voltage may be approximately twice the expected value.

Therefore, before performing any calculation:

confirm the load setting on the function generator;
measure V_max and V_min on the oscilloscope;
use the actual measured values in your calculations.

23. Suggested SPICE simulation

The same experiment can be reproduced in LTspice, ngspice, PSpice, or QSPICE:

* Lesson 002 - 0 to 5 V PWM across a resistive load
V1 out 0 PULSE(0 5 0 100n 100n 250u 1m)
R1 out 0 1k
.tran 0 5m 0 1u
.meas tran VAVG AVG V(out) FROM 1m TO 5m
.meas tran VRMS RMS V(out) FROM 1m TO 5m
.meas tran VMAX MAX V(out) FROM 1m TO 5m
.meas tran VMIN MIN V(out) FROM 1m TO 5m
.meas tran PAVG AVG V(out)*V(out)/1000 FROM 1m TO 5m
.end

Expected results:

VAVG ≈ 1.25 V;
VRMS ≈ 2.50 V;
VMAX ≈ 5 V;
VMIN ≈ 0 V;
PAVG ≈ 6.25 mW.

24. What can go wrong during measurement?

Symptom	Likely cause	Correction
Amplitude appears doubled	Generator configured for 50 Ω while driving a High-Z load	Select High-Z or use the proper termination
Oscilloscope shows 0.5 V instead of 5 V	10× probe with channel configured as 1×	Match the attenuation setting on the channel and probe
Average changes when the acquisition window is shifted	Window does not contain complete cycles	Measure over an integer number of periods
Multimeter RMS differs from oscilloscope RMS	AC mode removes DC or the instrument has limited bandwidth	Check the manual and the exact measurement definition
Ringing changes when the ground clip is moved	Measurement loop is excessively large	Use a short ground connection
Waveform looks rounded	Insufficient bandwidth or sampling rate	Review bandwidth, sample rate, and active filters
Resistor heats more than expected	Power was calculated using average current	Recalculate the loss using RMS current

25. How do you choose which metric to use in a design?

Engineering problem	Primary metric
Resistor heating	RMS current or RMS voltage
Ohmic loss in a winding	RMS current
DC output component	Average value
Maximum MOSFET voltage	Peak value including overshoot
Maximum semiconductor current	Peak and RMS, depending on the limit being analyzed
Output ripple	Peak-to-peak and/or RMS
Capacitor ESR loss	RMS ripple current
Arbitrary average power	Average of v(t)i(t)

26. Exercises

Exercise 1 — DC source

A constant 9 V source supplies an 18 Ω resistor. Determine:

average voltage;
RMS voltage;
peak-to-peak voltage;
current;
power.

Exercise 2 — unipolar PWM

A PWM waveform switches from 0 to 20 V with a 25% duty cycle across a 10 Ω load. Determine:

average voltage;
RMS voltage;
peak current;
average current;
RMS current;
average power.

Exercise 3 — bipolar square wave

A waveform alternates between +15 V and -15 V, remaining at each level for half the period, across a 30 Ω resistor. Determine average voltage, RMS voltage, peak-to-peak voltage, and average power.

Exercise 4 — PWM load design

Approximately 30 W of average power is required in a resistive load driven by a 0–24 V PWM waveform with D = 60%. Determine the ideal resistance and compare the standard values of 12 Ω and 15 Ω.

27. Exercise answers

Exercise 1

Because the signal is constant DC:

V_{avg}=V_{RMS}=9\,\mathrm{V}

V_{pp}=0

I=\frac{9}{18}=0.5\,\mathrm{A}

P=9\cdot0.5=4.5\,\mathrm{W}

Exercise 2

V_{avg}=0.25\cdot20=5\,\mathrm{V}

V_{RMS}=20\sqrt{0.25}=10\,\mathrm{V}

I_{pk}=\frac{20}{10}=2\,\mathrm{A}

I_{avg}=0.25\cdot2=0.5\,\mathrm{A}

I_{RMS}=2\sqrt{0.25}=1\,\mathrm{A}

P_R=1^2\cdot10=10\,\mathrm{W}

Exercise 3

V_{avg}=0

V_{RMS}=15\,\mathrm{V}

V_{pp}=30\,\mathrm{V}

P_R=\frac{15^2}{30}=7.5\,\mathrm{W}

Exercise 4

From:

P=D\frac{V_g^2}{R}

we obtain:

R=\frac{DV_g^2}{P}=\frac{0.60\cdot24^2}{30}=11.52\,\Omega

The 12 Ω standard value is the closest option:

P=0.60\frac{24^2}{12}=28.8\,\mathrm{W}

With 15 Ω:

P=0.60\frac{24^2}{15}=23.04\,\mathrm{W}

For 12 Ω:

I_{pk}=\frac{24}{12}=2\,\mathrm{A}

I_{RMS}=2\sqrt{0.60}\approx1.55\,\mathrm{A}

The resistor power rating must be selected with margin and then validated against the manufacturer's derating curve, ambient temperature, and mounting method.

28. What you should remember from this lesson

Instantaneous value describes the signal at one instant.
Average value represents the DC component or algebraic balance.
RMS value connects the waveform to equivalent resistive heating.
Peak value describes the largest instantaneous magnitude.
Peak-to-peak value is the excursion between maximum and minimum.
For a 0–V_pk PWM waveform, V average = DV_pk.
For a 0–V_pk PWM waveform, V RMS = V_pk√D.
Average power is the average of the instantaneous product v(t)i(t).
In general, average voltage × average current is not average power.
For a fixed resistance, heating depends on I RMS²R.
The observation window, bandwidth, probe attenuation, and instrument mode are part of the measurement.

A waveform does not have one single “true” number. It has several quantities, and each one describes a different physical property.

29. Next lesson

In Lesson 003, we will apply these concepts directly to the study of resistance and power dissipation.

The central relationship will be:

P_R=I_{RMS}^2R=\frac{V_{RMS}^2}{R}

From this relationship, we will study:

Joule heating;
power rating;
tolerance;
temperature coefficient;
power pulses;
derating;
thermal failures;
practical resistor sizing in power circuits.

References

ERICKSON, Robert W.; MAKSIMOVIĆ, Dragan. Fundamentals of Power Electronics. 2nd ed. Kluwer Academic Publishers, 2001.
MOHAN, Ned; UNDELAND, Tore M.; ROBBINS, William P. Power Electronics: Converters, Applications, and Design. 3rd ed. John Wiley & Sons, 2003.
HURLEY, W. G.; WÖLFLE, W. H. Transformers and Inductors for Power Electronics: Theory, Design and Applications. Wiley, 2013.
Also consult the specific manuals for the oscilloscope, probe, and multimeter used in the laboratory for voltage limits, bandwidth, sampling, RMS measurement, and crest-factor specifications.