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Why Does Power Electronics Exist? Energy, Power, Efficiency, and PWM

Power electronics efficiently converts and controls electrical energy using switching, PWM, MOSFETs, inductors, and capacitors while minimizing power losses.

Why Does Power Electronics Exist? Energy, Power, Efficiency, and PWM

Power electronics is present in virtually every modern system that needs to convert, control, or direct electrical energy.

It can be found inside a smartphone charger, a computer power supply, a solar inverter, an electric motor controller, an electric vehicle charger, an industrial power supply, a Class D amplifier, and thousands of other devices.

But there is a question more important than simply defining the field:

Why do we need power electronics?

The answer is directly related to four fundamental concepts:

  • energy;
  • power;
  • losses;
  • efficiency.

In this lesson, we will build these concepts from the ground up and see why, in power systems, controlling energy through switching is usually much more efficient than controlling energy through dissipation.

By the end, you should understand why an ideal electronic switch can control large amounts of energy while dissipating virtually none of it — and why real devices, such as MOSFETs, are designed to approach this behavior.


1. The question that will guide the entire course

Imagine that we have a power source and a load.

The source has a certain voltage and is capable of supplying a certain current. The load, however, may require completely different operating conditions.

For example:

  • the source provides 24 V, while the load requires 12 V;
  • the battery provides DC voltage, while the motor requires three-phase AC;
  • the mains supply is 60 Hz, but we want to control a motor's speed;
  • a solar panel produces a variable voltage, but we need to charge a battery;
  • a processor requires 1 V and tens of amperes, even though the board is supplied from 12 V.

The role of power electronics is to solve this problem.

We can represent the idea as follows:

flowchart LR; A["Energy source"] --> B["Electronic converter"]; B --> C["Adapted electrical energy"]; C --> D["Load"]; B -->|Losses| E["Heat, noise, and other losses"];
Power electronics adapts the available electrical energy to the requirements of the load.

The converter does not create energy.

It only changes characteristics such as:

  • voltage;
  • current;
  • polarity;
  • frequency;
  • waveform;
  • electrical isolation;
  • direction of power flow.

Therefore, a useful way to define the field is:

Power electronics is the engineering discipline concerned with the efficient processing of electrical energy through electronic devices.


2. Energy and power are not the same thing

Before studying MOSFETs, PWM, inductors, or converters, we need to separate two concepts that are often confused.

Energy

Energy represents the total amount of work that can be performed or transferred.

Its SI unit is the joule (J).

Power

Power indicates how quickly energy is being transferred.

Its unit is the watt:

1W=1Js1\,\mathrm{W}=1\,\frac{\mathrm{J}}{\mathrm{s}}

This means that a 100 W device transfers or converts energy at a rate of 100 joules per second.

A simple analogy is to think about water:

  • energy would be similar to the total amount of water moved;
  • power would be similar to flow rate, that is, the amount of water moved per second.

This analogy is limited, but it helps build intuition.


3. Where does the equation \(p(t)=v(t)i(t)\) come from?

This is one of the most important equations in the entire course.

We do not want to simply memorize it. We want to understand where it comes from.

Electrical voltage can be interpreted as energy per unit charge:

v(t)=dWdqv(t)=\frac{dW}{dq}

Electrical current represents the rate at which charge moves:

i(t)=dqdti(t)=\frac{dq}{dt}

Instantaneous power is the rate of energy transfer:

p(t)=dWdtp(t)=\frac{dW}{dt}

Applying the chain rule:

dWdt=dWdqdqdt\frac{dW}{dt}=\frac{dW}{dq}\frac{dq}{dt}

Substituting the definitions of voltage and current:

p(t)=v(t)i(t)\boxed{p(t)=v(t)i(t)}

This equation means that, at any instant, electrical power is related to the product of the voltage across a component and the current flowing through it.


4. The physical meaning of power

Consider a resistor with 12 V across it and 1 A flowing through it.

The power is:

P=VIP=VIP=121=12WP=12\cdot1=12\,\mathrm{W}

This means that the resistor is converting electrical energy into heat at a rate of 12 joules per second.

If it remains energized for 10 seconds:

E=PΔtE=P\Delta tE=1210=120JE=12\cdot10=120\,\mathrm{J}

Therefore:

power tells us how quickly energy is transferred; energy tells us the total amount transferred.


5. Why can reducing voltage with a resistor be a problem?

Let us study a simple example.

We have:

  • source: 12 V;
  • required load voltage: 6 V;
  • load current: 0.5 A.

The useful power required by the load is:

Pout=VoutIoutP_{out}=V_{out}I_{out}Pout=60.5=3WP_{out}=6\cdot0.5=3\,\mathrm{W}

An apparently simple solution would be to place a resistor in series to “absorb” the excess 6 V.

The circuit is:

12 V source supplying a 12 ohm load through a 12 ohm series resistor
Dissipative voltage reduction: half of the power supplied by the source is dissipated in the series resistor.

The equivalent resistance is:

Req=RS+RL=12+12=24ΩR_{eq}=R_S+R_L=12+12=24\,\Omega

The current is:

I=VR=1224=0.5AI=\frac{V}{R}=\frac{12}{24}=0.5\,\mathrm{A}

The load receives:

Vout=IRLV_{out}=I R_LVout=0.512=6VV_{out}=0.5\cdot12=6\,\mathrm{V}

So far, everything appears correct.

However, observe the power dissipated by the series resistor:

PRS=I2RSP_{R_S}=I^2R_SPRS=0.5212=3WP_{R_S}=0.5^2\cdot12=3\,\mathrm{W}

The load receives 3 W, but the resistor also converts 3 W into heat.

The source supplies:

Pin=120.5=6WP_{in}=12\cdot0.5=6\,\mathrm{W}

Therefore, the efficiency is:

η=PoutPin=36=0.5\eta=\frac{P_{out}}{P_{in}}=\frac{3}{6}=0.5η=50%\boxed{\eta=50\%}

In other words:

for every 1 joule usefully delivered to the load, another 1 joule is wasted in the resistor.


6. The problem grows quickly as power increases

In the previous example, only 3 W was lost.

Now imagine a 3 kW system.

If it also had only 50% efficiency, we would need to draw:

Pin=6kWP_{in}=6\,\mathrm{kW}

from the source to deliver:

Pout=3kWP_{out}=3\,\mathrm{kW}

The other 3 kW would be converted into heat:

Ploss=3kWP_{loss}=3\,\mathrm{kW}

This would require:

  • larger heatsinks;
  • ventilation;
  • materials capable of withstanding high temperatures;
  • higher input power consumption;
  • larger physical volume;
  • higher cost;
  • greater thermal stress on components.

This is exactly why efficiency is one of the central concerns in power electronics.


7. So what is the alternative?

The fundamental idea is to avoid maintaining, at the same time:

  • a large voltage across the control element;
  • a large current through it.

This is where the electronic switch becomes important.

An ideal switch has two states.

ON state

When the switch is closed:

  • current can flow;
  • the voltage across the ideal switch is zero.

Therefore:

vS=0v_S=0

and:

pS=vSiSp_S=v_S i_SpS=0iS=0p_S=0\cdot i_S=0

OFF state

When the switch is open:

  • there is voltage across it;
  • the current is zero.

Therefore:

iS=0i_S=0

and:

pS=vS0=0p_S=v_S\cdot0=0

Thus:

PS,ideal=0\boxed{P_{S,\mathrm{ideal}}=0}

in both static states.

This is the principle that makes switched-mode converters so efficient.


8. Visualizing the ideal-switch concept

Consider the circuit:

Ideal switch controlling a resistive load supplied by a 12 V source
The switch alternately connects and disconnects the source from the load. In the ideal model, the switch has no power loss in either the ON or OFF state.

When the switch closes:

I=1212=1AI=\frac{12}{12}=1\,\mathrm{A}

The load receives:

PR=121=12WP_R=12\cdot1=12\,\mathrm{W}

But the ideal switch dissipates:

PS=0WP_S=0\,\mathrm{W}

When the switch opens:

I=0I=0

and:

PS=120=0WP_S=12\cdot0=0\,\mathrm{W}

Therefore, the switch controls the transfer of energy without continuously dissipating the excess energy.


9. How do we control power using ON and OFF states?

The solution is to alternate the switch rapidly between the two states.

This technique is the basis of PWM — Pulse Width Modulation.

Consider:

  • source voltage: 12 V;
  • switching frequency: 1 kHz;
  • duty cycle: 25%.

The frequency is:

fs=1kHzf_s=1\,\mathrm{kHz}

The period is:

Ts=1fsT_s=\frac{1}{f_s}Ts=11000=1msT_s=\frac{1}{1000}=1\,\mathrm{ms}

With a 25% duty cycle:

D=0.25D=0.25

The ON time is:

ton=DTst_{on}=DT_ston=0.251ms=250μst_{on}=0.25\cdot1\,\mathrm{ms}=250\,\mu\mathrm{s}

The OFF time is:

toff=750μst_{off}=750\,\mu\mathrm{s}

We can visualize this sequence:

flowchart LR; A["Start of period"] --> B["ON for 250 us"]; B --> C["OFF for 750 us"]; C --> D["Next period"]; D --> B;
During each period, the switch remains ON for 25% of the time and OFF for 75%.

The graph below shows the 12 V waveform at 1 kHz with a 25% duty cycle:

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.

10. Average value is not the same as RMS value

This point is essential.

If we apply pulses from 0 to 12 V with a 25% duty cycle, the average value is:

Vavg=DVgV_{avg}=DV_gVavg=0.2512=3VV_{avg}=0.25\cdot12=3\,\mathrm{V}

However, the RMS value is:

VRMS=VgDV_{RMS}=V_g\sqrt{D}VRMS=120.25=6VV_{RMS}=12\sqrt{0.25}=6\,\mathrm{V}

Therefore:

  • average value = 3 V;
  • RMS value = 6 V;
  • peak value = 12 V.

These values describe different properties of the same waveform.

For a resistive load, heating is related to the RMS value.

The average power is:

PR=VRMS2RP_R=\frac{V_{RMS}^2}{R}PR=6212=3WP_R=\frac{6^2}{12}=3\,\mathrm{W}

Notice something important:

a 0-to-12 V PWM waveform with a 25% duty cycle has an average value of 3 V, but it produces the same heating in a 12 Ω resistor as a 6 V DC source.

We can also visualize how the average and RMS values evolve as the duty cycle increases:

Comparison between average voltage and RMS voltage as a function of duty cycle
For a 0-to-12 V PWM waveform, the average value increases linearly with D, while the RMS value follows the square root of D.

The difference between average, RMS, and peak values will be studied in greater depth in Lesson 002.


11. The switch does not create an intermediate DC voltage by itself

A common mistake is to imagine that a PWM switch operating at 25% “converts 12 V into 3 V.”

That is not strictly correct.

At the switch output, we find a waveform that alternates between:

0V0\,\mathrm{V}

and:

12V12\,\mathrm{V}

The average value may be 3 V, but the output is not a 3 V DC source.

To convert this pulsating energy into a relatively continuous DC voltage, we use components that store energy:

  • inductors;
  • capacitors;
  • transformers, when required.

A typical conceptual structure is:

flowchart LR; A["DC source"] --> B["Electronic switch"]; B --> C["Pulsed energy"]; C --> D["Inductor and capacitor"]; D --> E["Filtered output"]; E --> F["Load"];
The switch controls when energy is transferred, while the filter organizes that energy to produce a useful output.

This is the idea that leads directly to buck, boost, buck-boost, flyback, forward, and many other converter topologies.


12. Why do we use inductors and capacitors?

Resistors dissipate energy.

In the ideal model, inductors and capacitors temporarily store energy and can return it to the circuit.

The energy stored in a capacitor is:

EC=12CV2E_C=\frac{1}{2}CV^2

The energy stored in an inductor is:

EL=12LI2E_L=\frac{1}{2}LI^2

This ability to store and return energy allows the converter to reorganize the pulsating energy generated by the switch.

In a buck converter, for example:

  1. the switch connects the source to the inductor;
  2. the inductor current increases;
  3. energy is stored in the magnetic field;
  4. the switch opens;
  5. the inductor continues to supply current;
  6. the capacitor helps maintain the load voltage;
  7. the cycle repeats thousands or millions of times per second.

This behavior will be studied in depth later in the course.


13. A real converter still has losses

So far, we have studied an ideal switch.

A real MOSFET, however, is not perfect.

During conduction

When it is ON, the MOSFET presents an effective resistance:

RDS(on)R_{DS(on)}

A first approximation for conduction loss is:

PcondIRMS2RDS(on)P_{cond}\approx I_{RMS}^2R_{DS(on)}

During switching

The transition between ON and OFF is not instantaneous.

For a few nanoseconds or microseconds, we may simultaneously have:

  • significant voltage;
  • significant current.

During this interval:

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

is not zero.

The energy lost during a transition can be represented by:

Esw=ΔtswvS(t)iS(t)dtE_{sw}=\int_{\Delta t_{sw}}v_S(t)i_S(t)\,dt

If the switch commutates \(f_s\) times per second:

Psw=fsEswP_{sw}=f_sE_{sw}

This creates an important design tradeoff:

  • higher switching frequency allows smaller magnetic components;
  • however, it generally increases switching losses and EMI.

14. Other sources of loss

A real converter can lose energy in:

  • MOSFETs;
  • diodes;
  • PCB trace resistance;
  • cable resistance;
  • capacitor ESR;
  • winding resistance in inductors and transformers;
  • magnetic core losses;
  • gate drive;
  • snubbers;
  • auxiliary circuits;
  • control and sensing circuits.

In steady state:

Pin=Pout+PlossP_{in}=P_{out}+P_{loss}

Therefore:

Ploss=PinPoutP_{loss}=P_{in}-P_{out}

Efficiency is:

η=PoutPin\eta=\frac{P_{out}}{P_{in}}

or, as a percentage:

η%=100PoutPin\eta_{\%}=100\frac{P_{out}}{P_{in}}

15. A very important physical test: efficiency can never exceed 100%

Consider a converter measured with:

  • input: 12 V and 0.25 A;
  • output: 6 V and 0.55 A.

The input power would be:

Pin=120.25=3.0WP_{in}=12\cdot0.25=3.0\,\mathrm{W}

The output power would be:

Pout=60.55=3.3WP_{out}=6\cdot0.55=3.3\,\mathrm{W}

The calculated efficiency would be:

η=3.33.0=1.10\eta=\frac{3.3}{3.0}=1.10η=110%\eta=110\%

This violates the energy balance for a converter with no auxiliary energy source.

Therefore, the calculation does not demonstrate a “super-efficient” converter. Instead, it indicates a problem with:

  • measurement;
  • timing or synchronization;
  • measurement range;
  • interpretation;
  • instrumentation;
  • the definition of the system energy boundary.

This type of consistency check is extremely important in the laboratory.


16. Complete comparison: resistor versus switched-mode converter

Consider again:

  • input: 12 V;
  • output: 6 V;
  • load: 12 Ω;
  • output current: 0.5 A.

Load

Iout=612=0.5AI_{out}=\frac{6}{12}=0.5\,\mathrm{A}Pout=60.5=3WP_{out}=6\cdot0.5=3\,\mathrm{W}

Dissipative solution

Pin=120.5=6WP_{in}=12\cdot0.5=6\,\mathrm{W}Ploss=63=3WP_{loss}=6-3=3\,\mathrm{W}η=50%\eta=50\%

Ideal converter

Ploss=0P_{loss}=0Pin=Pout=3WP_{in}=P_{out}=3\,\mathrm{W}

The average input current would be:

Iin=312=0.25AI_{in}=\frac{3}{12}=0.25\,\mathrm{A}

Notice the transformation:

12V×0.25A=6V×0.5A12\,\mathrm{V}\times0.25\,\mathrm{A}=6\,\mathrm{V}\times0.5\,\mathrm{A}

On both sides, we have:

3W3\,\mathrm{W}

The converter does not create energy. It changes the voltage-current relationship.

Real converter with 90% efficiency

Pin=Poutη=30.90=3.333WP_{in}=\frac{P_{out}}{\eta}=\frac{3}{0.90}=3.333\,\mathrm{W}Ploss=3.3333=0.333WP_{loss}=3.333-3=0.333\,\mathrm{W}

The average input current is:

Iin=3.33312=0.278AI_{in}=\frac{3.333}{12}=0.278\,\mathrm{A}

Comparison:

Method
Useful power
Losses
Input power
Efficiency
Series resistor
3 W
3 W
6 W
50%
Ideal converter
3 W
0 W
3 W
100%
Real converter
3 W
0.333 W
3.333 W
90%

The power balance of the three solutions can be viewed graphically:

Comparison of useful power and losses for a series resistor and switched-mode converters
The load receives 3 W in every case; the main difference is the amount of power dissipated as loss.

The loss reduction relative to the series resistor is:

30.333310088.9%\frac{3-0.333}{3}\cdot100\approx88.9\%

17. Efficiency is also a thermal problem

Lost electrical energy usually ends up as heat.

A simplified thermal model is:

T=Ta+PlossRθT=T_a+P_{loss}R_{\theta}

where:

  • \(T\) is the estimated component temperature;
  • \(T_a\) is ambient temperature;
  • \(P_{loss}\) is dissipated power;
  • \(R_{\theta}\) is thermal resistance.

Suppose:

Ploss=0.333WP_{loss}=0.333\,\mathrm{W}

and:

Rθ=30C/WR_{\theta}=30\,^\circ\mathrm{C/W}

Then:

ΔT=PlossRθ\Delta T=P_{loss}R_{\theta}ΔT=0.3333010C\Delta T=0.333\cdot30\approx10\,^\circ\mathrm{C}

If ambient temperature is 25 °C:

T35CT\approx35\,^\circ\mathrm{C}

This calculation is only a first-order model, but it demonstrates a direct consequence:

lower electrical loss generally means lower thermal stress.

With the simplified thermal model used in this example, the relationship between power dissipation and temperature rise is linear:

Temperature rise as a function of dissipated power
Illustrative model using a thermal resistance of 30 degrees Celsius per watt.

18. What happens when we switch very quickly?

When current changes rapidly, parasitic inductances begin to matter.

The fundamental inductor relationship is:

v=Ldidtv=L\frac{di}{dt}

If a trace or wire has a small parasitic inductance \(L_{par}\), then a rapid current change produces:

vpar=Lpardidtv_{par}=L_{par}\frac{di}{dt}

Even a very small inductance can generate a significant voltage when \(di/dt\) is high.

This is why, in power electronics:

  • trace length matters;
  • loop area matters;
  • capacitor placement matters;
  • grounding matters;
  • driver placement matters;
  • gate interconnection matters.

PCB layout is part of the circuit.


19. The high-current loop

In the switch-and-resistor example, the current path is:

flowchart LR; A["Positive source terminal"] --> B["Switch"]; B --> C["Load"]; C --> D["Return"]; D --> A;
The source-switch-load-return path should be kept physically compact in fast-switching circuits.

The larger the area enclosed by this loop, the greater its parasitic inductance tends to be.

An extremely important design rule is:

high-frequency current should travel through the smallest possible loop.

Later, we will see that this principle is essential in buck converters, half bridges, H bridges, and gate-driver circuits.


20. Where is power electronics used?

Switched-mode power supplies

A computer does not use the 127 V or 220 V mains voltage directly.

Its power supply performs several conversion stages to generate voltages such as:

  • 12 V;
  • 5 V;
  • 3.3 V;
  • even lower voltages in point-of-load converters on the motherboard.

Vehicles

Modern automotive systems may use:

  • 12 V;
  • 24 V;
  • 48 V;
  • hundreds of volts in electric vehicles.

Power converters are required to integrate these different voltage buses.

Solar energy

A solar panel produces voltage and current that depend on:

  • irradiance;
  • temperature;
  • operating point.

The power system must find the maximum power point and then adapt the energy for a battery or the electrical grid.

Motors

Inverters can control:

  • frequency;
  • amplitude;
  • phase;
  • voltage sequence.

This allows us to control speed and torque.

Processors

A processor may operate at approximately 1 V while drawing tens or even hundreds of amperes.

This requires extremely efficient converters located very close to the load.


21. Typical architecture of a power system

A real system often contains:

flowchart LR; A["Source"] --> B["Power stage"]; B --> C["Filter"]; C --> D["Load"]; D --> E["Sensor"]; E --> F["Controller"]; H["Reference"] --> F; F --> G["Driver"]; G --> B;
The controller measures the output, compares it with the reference, and adjusts the power stage.

Notice the separation:

Power path

This path carries energy:

  • source;
  • MOSFET;
  • diode;
  • inductor;
  • transformer;
  • capacitor;
  • load.

Control path

This path decides when and how the semiconductors operate:

  • controller;
  • PWM;
  • logic buffer;
  • gate driver;
  • sensors;
  • feedback.

This separation will be very important throughout the course.


22. Where do the 74HC14, TC4424, and IRF3205 fit?

Later in the course, these components will form a very useful teaching chain.

We can visualize it as follows:

flowchart LR; A["Microcontroller PWM"] --> B["74HC14"]; B --> C["TC4424"]; C --> D["IRF3205 gate"]; D --> E["Power load"];
Example architecture that will be studied in later lessons of the course.

The conceptual function of each block is:

74HC14

It can be used as a:

  • logic buffer;
  • edge restorer;
  • Schmitt trigger;
  • simple logic-stage interface between circuit blocks.

TC4424

This is a gate-driver IC.

Its purpose is to source and sink gate charge currents that are much higher than a typical GPIO can provide.

IRF3205

This is a power MOSFET.

Its function is to control load current through switching.

However, using these components correctly requires first understanding:

  • gate voltage;
  • gate charge;
  • gate resistance;
  • driver current;
  • rise time;
  • fall time;
  • switching losses;
  • \(R_{DS(on)}\);
  • \(V_{GS}\);
  • \(V_{DS}\);
  • Miller effect;
  • parasitic inductances.

For this reason, this chain will be built progressively throughout the course.


23. Safe teaching experiment

In this first lesson, the objective is not yet to build a complete MOSFET converter.

The experiment will compare:

  1. voltage reduction using a resistor;
  2. voltage reduction using a commercial buck converter module.

Materials

  • 12 V bench power supply;
  • 12 Ω / 10 W resistor;
  • 12 Ω / 10 W load resistor;
  • adjustable buck converter module;
  • multimeters;
  • appropriate test leads.

Part A — series resistor

Build the following circuit:

Experiment using a series resistor to reduce 12 V to approximately 6 V across a 12 ohm load
Experimental setup for the dissipative solution.

Measure:

  • input voltage;
  • source current;
  • load voltage.

Calculate:

Pin=VinIinP_{in}=V_{in}I_{in}Pout=Vout2RLP_{out}=\frac{V_{out}^2}{R_L}η=100PoutPin\eta=100\frac{P_{out}}{P_{in}}

You should obtain a value close to 50%.

Part B — buck converter

Adjust the module to approximately 6 V.

Measure:

  • \(V_{in}\);
  • \(I_{in}\);
  • \(V_{out}\);
  • \(I_{out}\).

Then calculate:

Pin=VinIinP_{in}=V_{in}I_{in}Pout=VoutIoutP_{out}=V_{out}I_{out}Ploss=PinPoutP_{loss}=P_{in}-P_{out}η=100PoutPin\eta=100\frac{P_{out}}{P_{in}}

Compare the result with the resistive method.


24. Laboratory precautions

Even when working with only 12 V, there are still risks of:

  • short circuit;
  • overheating;
  • instrument damage;
  • converter-module damage.

Some important rules:

  1. make or change connections with the power supply turned off;
  2. use the current-limit function;
  3. confirm the module polarity;
  4. check the multimeter lead positions before energizing the circuit;
  5. never place a multimeter in current mode directly across a voltage source;
  6. keep power resistors away from flammable materials;
  7. allow resistors to cool before touching them.

A low-voltage power source may appear harmless, but high currents can still heat wires and components very quickly.


25. Measuring correctly with an oscilloscope

On many bench oscilloscopes, the probe ground clip is electrically connected to the protective earth conductor.

Therefore:

never connect the ground clip indiscriminately to a floating node or a high-side switching node.

In an isolated low-voltage DC circuit, the ground clip can normally be connected to circuit GND.

Later, in half-bridge and H-bridge circuits, we will use:

  • differential probes;
  • current probes;
  • measurement techniques appropriate for floating nodes.

Measurement is part of the design process.


26. Suggested SPICE simulation

The example below can be adapted for LTspice or ngspice.

* Lesson 001 — resistive reduction and approximate ideal switching
V1 in 0 12
* Dissipative reduction
Rseries in out_linear 12
Rload_linear out_linear 0 12
* Switching
Vpwm ctrl 0 PULSE(0 5 0 10n 10n 250u 1m)
S1 in out_pwm ctrl 0 SW_IDEAL
Rload_pwm out_pwm 0 12
.model SW_IDEAL SW(Ron=10m Roff=1Meg Vt=2.5 Vh=0.1)
.tran 0 5m 0 1u
.meas tran VLINEAR AVG V(out_linear) FROM=1m TO=5m
.meas tran VPWM_AVG AVG V(out_pwm) FROM=1m TO=5m
.meas tran VPWM_RMS RMS V(out_pwm) FROM=1m TO=5m
.end

What to observe:

  • V(out_linear) close to 6 V;
  • V(out_pwm) alternating between 0 and approximately 12 V;
  • average value close to 3 V;
  • RMS value close to 6 V.

This simulation helps reinforce one of the most important ideas in this lesson:

a waveform can have completely different average, RMS, and peak values.


27. What can go wrong?

Symptom
Likely cause
Interpretation
Resistive output is not 6 V
resistor tolerance
the real divider differs from the ideal one
Resistor becomes very hot
3 W dissipation
electrical energy is being converted into heat
Power supply enters current limit
short circuit or excessive load
requested current exceeds the configured limit
Buck converter does not maintain 6 V
thermal or current limitation
converter is outside its valid operating region
Calculated efficiency exceeds 100%
measurement error
energy balance is inconsistent
Multimeter fuse blows
current measurement performed in parallel
the ammeter effectively created a short circuit
PWM waveform appears to be DC
instrument displays only an aggregate quantity
the waveform itself must be observed

28. Exercises

1. Energy

A load operates at:

  • 24 V;
  • 2 A;
  • for 5 minutes.

Calculate the power.

P=VIP=VI

Then determine the energy in joules and Wh.


2. Efficiency

A converter absorbs:

Pin=120WP_{in}=120\,\mathrm{W}

and delivers:

Pout=108WP_{out}=108\,\mathrm{W}

Determine:

  • power loss;
  • efficiency.

3. PWM

A switch applies pulses from 0 to 12 V across a 12 Ω resistor.

Duty cycle:

D=0.36D=0.36

Determine:

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

4. Design comparison

A load requires:

  • 12 V;
  • 2 A;

and the available source is:

  • 24 V.

Compare:

  1. an idealized linear solution;
  2. a switched-mode converter with 92% efficiency.

For each solution, determine:

  • output power;
  • input power;
  • losses;
  • average input current.

29. Exercise answers

Exercise 1

P=242=48WP=24\cdot2=48\,\mathrm{W}

Five minutes corresponds to:

t=300st=300\,\mathrm{s}

Therefore:

E=PtE=PtE=48300=14400JE=48\cdot300=14\,400\,\mathrm{J}

In Wh:

E=48560=4WhE=48\cdot\frac{5}{60}=4\,\mathrm{Wh}

Exercise 2

Ploss=120108=12WP_{loss}=120-108=12\,\mathrm{W}η=108120=0.90\eta=\frac{108}{120}=0.90η=90%\boxed{\eta=90\%}

Exercise 3

Peak current:

Ipk=1212=1AI_{pk}=\frac{12}{12}=1\,\mathrm{A}

Average current:

Iavg=DIpkI_{avg}=DI_{pk}Iavg=0.36AI_{avg}=0.36\,\mathrm{A}

RMS current:

IRMS=IpkDI_{RMS}=I_{pk}\sqrt DIRMS=10.36=0.6AI_{RMS}=1\sqrt{0.36}=0.6\,\mathrm{A}

Average voltage:

Vavg=DVgV_{avg}=DV_gVavg=0.3612=4.32VV_{avg}=0.36\cdot12=4.32\,\mathrm{V}

RMS voltage:

VRMS=VgDV_{RMS}=V_g\sqrt DVRMS=120.6=7.2VV_{RMS}=12\cdot0.6=7.2\,\mathrm{V}

Power:

P=IRMS2RP=I_{RMS}^2RP=0.6212=4.32WP=0.6^2\cdot12=4.32\,\mathrm{W}

Exercise 4

Useful output power:

Pout=122=24WP_{out}=12\cdot2=24\,\mathrm{W}

Linear solution

Iin=2AI_{in}=2\,\mathrm{A}Pin=242=48WP_{in}=24\cdot2=48\,\mathrm{W}Ploss=4824=24WP_{loss}=48-24=24\,\mathrm{W}η=50%\eta=50\%

Switched-mode converter at 92%

Pin=240.92=26.087WP_{in}=\frac{24}{0.92}=26.087\,\mathrm{W}Ploss=26.08724=2.087WP_{loss}=26.087-24=2.087\,\mathrm{W}Iin=26.08724=1.087AI_{in}=\frac{26.087}{24}=1.087\,\mathrm{A}

The reduction in losses is:

242.0872410091.3%\frac{24-2.087}{24}\cdot100\approx91.3\%

30. What you should have learned

After completing this lesson, you should be able to answer the following questions.

Why does power electronics exist?

Because the available electrical energy almost never has exactly the characteristics required by the load.

Why does efficiency matter?

Because any input power that is not delivered to the load must appear in another form, usually heat.

Why is a switch more efficient than a linear element?

Because an ideal switch avoids sustaining significant voltage and significant current simultaneously.

Why do we use PWM?

Because we can control the average amount of energy transferred by varying how long the switch remains ON.

Why are inductors and capacitors important?

Because they allow energy to be stored temporarily and help transform a switched waveform into voltages and currents suitable for the load.

Why does a real circuit still lose energy?

Because MOSFETs, diodes, capacitors, inductors, transformers, PCB traces, and drivers all have non-ideal behavior.


31. One idea to remember

If you remember only one idea from this lesson, remember this:

Power electronics is not simply about changing voltage. It is about controlling the flow of energy efficiently.

The designer's objective is not merely to make the load operate.

The system must also achieve:

  • low loss;
  • acceptable temperature;
  • good reliability;
  • predictable behavior;
  • protection;
  • good electromagnetic compatibility;
  • appropriate cost.

This is why MOSFETs, diodes, inductors, capacitors, transformers, drivers, and controllers must be studied as parts of one integrated system.


32. Preparing for the next lesson

In the next lesson, we will study how to correctly describe an electrical waveform.

You will learn to distinguish:

  • instantaneous value;
  • average value;
  • RMS value;
  • peak value;
  • peak-to-peak value;
  • frequency;
  • period;
  • duty cycle.

This distinction will be essential for correctly understanding:

  • heating;
  • power;
  • MOSFET current;
  • inductor current;
  • ripple;
  • oscilloscope measurements;
  • multimeter measurements.

The question for the next lesson will be:

If a voltage continuously changes over time, which number actually represents that voltage?


References

  1. ERICKSON, Robert W.; MAKSIMOVIĆ, Dragan. Fundamentals of Power Electronics. Introductory concepts of power processing, efficiency, and the ideal switch.
  2. MOHAN, Ned; UNDELAND, Tore M.; ROBBINS, William P. Power Electronics: Converters, Applications, and Design. 3rd ed. Wiley, 2003.
  3. HURLEY, W. G.; WÖLFLE, W. H. Transformers and Inductors for Power Electronics: Theory, Design and Applications. Wiley, 2013.

Next lesson

Lesson 002 — Instantaneous, average, RMS, peak, and peak-to-peak values in electrical signals.

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