GATE EE Power Electronics Formula Sheet: Key Formulas and Traps

The key GATE EE power electronics formulas are the converter gains $D$, $1/(1 - D)$ and $-D/(1 - D)$, the rectifier averages in $\cos\alpha$ or $(1 + \cos\alpha)$, the commutation times $\pi\sqrt{LC}$ and $CV/I$, power factor as distortion factor times displacement factor, and the inverter fundamentals. Each holds only under a stated condition, usually continuous conduction. This sheet teaches them in learning order, each with a worked example and its trap.

Key takeaways

• Every DC-DC converter result comes from volt-second balance, charge balance and power balance.
• The converter gains hold only in continuous conduction, so check the inductance against its critical value first.
• Rectifier averages use the peak voltage $V_m$, while supplies are quoted as RMS, so convert once at the start.
• A fully controlled converter inverts above 90 degrees; a semi-converter never can.
• With a sinusoidal supply only the fundamental current carries power.
• Inverter questions separate waveform RMS from fundamental RMS, and phase values from line values.

The terms this sheet uses

Most lost marks come from misreading a symbol, so fix the notation first. The duty ratio $D = t_{on}/T$ is the fraction of each switching period for which the switch is on; $f = 1/T$ is the switching frequency.

The firing angle $\alpha$ is the delay from the instant a thyristor becomes forward-biased to the instant its gate is pulsed. A diode acts like a thyristor fired at $\alpha = 0$.

$V_m$ is always a peak: a supply of $V$ RMS has $V_m = \sqrt{2}\,V$. For a three-phase bridge, $V_{mL}$ is the line peak and $V_L$ the line RMS.

Continuous conduction means the inductor or load current never reaches zero in a cycle. $I_d$ is a ripple-free DC load current, held level by a large inductor. The overlap angle $\mu$ is the extra commutation angle caused by source inductance.

How each switching device behaves and where its losses come from

A thyristor turns on with a gate pulse once forward-biased, and turns off only when its current falls below the holding current and reverse voltage is held for longer than its turn-off time $t_q$. In speed, the MOSFET leads, then the IGBT, the GTO and the SCR.

Losses come from a straight-line on-state model: a fixed drop $V_0$ in series with a resistance $r$. The drop multiplies the average current; the resistance multiplies the squared RMS current.

Worked example. A diode with $V_0 = 0.8$ V and $r = 10\ \text{m}\Omega$ carries 30 A pulses at one-third duty: $I_{avg} = 10\ \text{A},\ I_{rms}^2 = 900/3 = 300,\ P = 0.8 \times 10 + 0.01 \times 300 = 11\ \text{W}$.

The trap is using 10 A in both terms, which gives 9 W. Pulsed current heats the resistance more than its average suggests.

DC-DC converters: gains, ripple and the edge of continuous conduction

In steady state the inductor voltage averages to zero over a period, which is volt-second balance, and the capacitor current averages to zero, which is charge balance. Add the lossless power balance $V_{in} I_{in} = V_o I_o$ and the whole table follows.

Ripples are peak to peak. The critical inductance is the value at which the inductor current just touches zero each period; below it, conduction is discontinuous.

Worked example (buck). With $V_{in} = 24$ V, $D = 0.5$, $f = 100$ kHz, $L = 60\ \mu$H, $C = 10\ \mu$F and $R = 6\ \Omega$: $\displaystyle V_o = 12\ \text{V},\ I_o = 2\ \text{A},\ \Delta I_L = \frac{12 \times 0.5}{60 \times 10^{-6} \times 10^{5}} = 1\ \text{A},\ I_{L,\min} = 1.5\ \text{A} > 0,\ \Delta V_o = \frac{1}{8 \times 10^{-5} \times 10^{5}} = 0.125\ \text{V}$.

Worked example (boost). With $V_{in} = 12$ V, $D = 0.6$, $R = 30\ \Omega$ and $f = 50$ kHz: $\displaystyle V_o = \frac{12}{0.4} = 30\ \text{V},\ I_L = \frac{1}{0.4} = 2.5\ \text{A},\ L_{crit} = \frac{0.6 \times 0.16 \times 30}{2 \times 50000} = 28.8\ \mu\text{H}$.

Trap: The three gains belong to continuous conduction only. If the inductor current falls to zero in each cycle, discard them and write volt-second and power balance from the waveform you are given.

Many questions run backwards: they give a current waveform and ask for $L$, $C$ or the source voltage. Read each slope as $V = L\,\Delta I/\Delta t$.

Uncontrolled rectifiers: averages from the peak voltage

A diode rectifier is a thyristor rectifier fired at $\alpha = 0$. Each average is the area under the output voltage divided by the period.

Worked example. A 50 V battery charges through $10\ \Omega$ from a half-wave rectifier with $V_m = 100$ V: $\displaystyle \theta_1 = 30^\circ,\ I_{avg} = \frac{2 \times 100 \times 0.866 - 50 \times 2\pi/3}{2\pi \times 10} = \frac{173.2 - 104.7}{62.83} = 1.09\ \text{A}$.

The trap is the peak in the three-phase bridge. On 400 V, $3\sqrt{2} \times 400/\pi = 540.2$ V; using the phase peak gives a figure $\sqrt{3}$ times too small.

Controlled rectifiers: firing angle, inversion and overlap

Delaying the firing by $\alpha$ removes a slice of each output pulse. With continuous current, a fully controlled converter follows $\cos\alpha$; a semi-converter follows $(1 + \cos\alpha)$ because its freewheeling path clamps the output at zero.

Worked example. A single-phase full converter on 230 V at $\alpha = 60^\circ$, with $\omega L_s = 1\ \Omega$ and $I_d = 10$ A: $\displaystyle V_{dc} = 207.1 \times 0.5 - \frac{2 \times 1 \times 10}{\pi} = 103.5 - 6.4 = 97.2\ \text{V},\ \cos(60^\circ + \mu) = 0.5 - \frac{20}{325.3} = 0.4385,\ \mu \approx 4.0^\circ$.

Remember: Regeneration needs a negative average voltage at positive current. A fully controlled converter gives it above 90 degrees with a reversed DC source; a semi-converter never can.

Commutation: turning a thyristor off

In line (voltage) commutation the AC supply reverses the thyristor voltage, as in every rectifier above. In forced (current) commutation a charged capacitor or an LC ring drives the current to zero.

Worked example. With $L = 40\ \mu$H, $C = 10\ \mu$F and 100 V on an uncharged capacitor: $\pi\sqrt{LC} = \pi \times 20\ \mu\text{s} = 62.8\ \mu\text{s},\ I_{peak} = 100\sqrt{10/40} = 50\ \text{A}$.

Worked example. A $20\ \mu$F capacitor at 100 V commutating 40 A: $\displaystyle t_c = \frac{20 \times 10^{-6} \times 100}{40} = 50\ \mu\text{s} > t_q = 30\ \mu\text{s}$.

The AC controller trap is firing below the load angle. With $R = 10\ \Omega$ and $\omega L = 17.32\ \Omega$, $\phi = 60^\circ$, so firing at $30^\circ$ acts like firing at $60^\circ$.

Line-current harmonics, distortion factor and displacement factor

A rectifier with a large inductive load draws a block-shaped line current. Fourier analysis splits it into a fundamental and harmonics. With a sinusoidal supply only the fundamental does work; harmonics add RMS current but no power.

Worked example. An 8 A fundamental lagging $36.87^\circ$ plus a 6 A fifth harmonic: $I_s = \sqrt{64 + 36} = 10\ \text{A},\ \text{DF} = 0.8,\ \text{THD} = 0.75,\ \text{PF} = 0.8 \times 0.8 = 0.64$.

Worked example. A semi-converter with $I_d = 10$ A at $\alpha = 90^\circ$: $\displaystyle I_s = 10\sqrt{0.5} = 7.07\ \text{A},\ I_{s1} = 0.9003 \times 10 \times 0.7071 = 6.37\ \text{A},\ \text{PF} = \frac{6.37}{7.07} \times 0.7071 = 0.637$.

In one line: Add the squared harmonic RMS values for the total, divide the fundamental by it, then multiply by the cosine of the fundamental's own phase angle.

Inverters and sinusoidal PWM: fundamentals, phase and line

An inverter turns a DC bus into AC. In square-wave mode you need the Fourier series of a fixed pattern; in sinusoidal PWM a modulation index scales the fundamental.

Worked example. Three-phase SPWM with $V_{dc} = 400$ V and $m_a = 0.9$: $\displaystyle \hat{V}_{ph} = \frac{0.9 \times 400}{2} = 180\ \text{V},\ V_{ph} = \frac{180}{\sqrt{2}} = 127.3\ \text{V},\ V_L = \sqrt{3} \times 127.3 = 220.5\ \text{V}$.

Worked example. A 120 V full bridge feeding 40 mH at 50 Hz: $\displaystyle \Delta I = \frac{120 \times 0.02}{2 \times 0.04} = 30\ \text{A}$ peak to peak, from $-15$ A to $+15$ A.

The trap is mixing waveform RMS with fundamental RMS. A 200 V full bridge has 200 V output RMS but only 180.1 V fundamental, and only that fundamental delivers power to a sinusoidal current.

How to use this sheet in the exam

Practise these examples on the GATE virtual calculator, since many are numerical answer questions. The MCQ, MSQ and NAT marking scheme, common to every paper, puts no negative mark on a NAT, so carry full precision and round only at the end. Check your session in the GATE 2027 exam dates.

The last-minute sheet in the GATE EE 2027 book covers every syllabus topic with the conditions each formula needs. The book has 643 pages, 914 questions with worked solutions and 10 full mock tests, built on all 5 GATE EE papers from 2022 to 2026 counted question by question.

More formula sheets: all of GATE EE · Control Systems · Electrical Machines · Power Systems

Quick revision

1. Buck $D$, boost $1/(1 - D)$, buck-boost $-D/(1 - D)$, in continuous conduction only.
2. Buck ripple $\Delta I_L/(8Cf)$; boost and buck-boost ripple $I_o D/(Cf)$.
3. Diode bridges: single-phase $0.9\,V$, three-phase $1.35\,V_L$, always from the peak.
4. Full converter $\cos\alpha$ inverts above 90 degrees; semi-converter $(1 + \cos\alpha)$ never does.
5. Source inductance subtracts $2\omega L_s I_d/\pi$ or $3\omega L_s I_d/\pi$.
6. Forced commutation: pulse $\pi\sqrt{LC}$, turn-off time $CV/I$ above $t_q$.
7. Power factor $(I_{s1}/I_s)\cos\phi_1$; semi-converter displacement angle $\alpha/2$.
8. SPWM leg peak $m_a V_{dc}/2$; line RMS $0.612\,m_a V_{dc}$ in the linear range.