GATE EE · IN

DC-DC converters

Buck, boost and buck-boost

DC-DC converters change a DC voltage level by switching an inductor and a capacitor, and the buck, boost and buck-boost circuits are the three basic ones. Almost every GATE question reduces to volt-second balance on the inductor (the gain), the ripple formulas, the continuous-versus-discontinuous boundary and the device stresses, so these relations must be instant. Chopper quadrants and DC motor control follow from the same switching idea.

9 previous-year questions · 9 solved · 5 practice questions

Method: balance equations and assumptions

A DC-DC converter turns a fixed input $V_{s}$ into a different average output $V_{o}$ by switching at frequency $f = 1/ T$ . Convention: duty ratio $D = t_{o n} / T$ ; the switch, diode, inductor and capacitor are ideal; the output capacitor is large enough that $V_{o}$ is practically constant; the converter is in continuous conduction (CCM) and in steady state unless stated; $R$ is a resistive load and $I_{o} = ∣ V_{o} ∣/ R$ is the magnitude of the average load current.

VOLT-SECOND BALANCE (INDUCTOR)

\int_{0}^{T} v_{L} d t = 0 \Rightarrow V_{L, o n} D T + V_{L, o f f} (1 - D) T = 0

In steady state the average inductor voltage over one period is zero. This gives the voltage gain of every converter.

CHARGE BALANCE (CAPACITOR)

\overline{i_{C}} = 0 \Rightarrow \overline{i_{L}} = I_{o} (buck), \overline{i_{D}} = I_{o} (boost, buck-boost)

In steady state the average capacitor current is zero, so the average current delivered to the load equals the average current of the element feeding the output node: the inductor in the buck, the diode in the boost and buck-boost.

RMS INDUCTOR CURRENT (TRIANGULAR RIPPLE)

I_{L, r m s} = I_{L}^{2} + \frac{( Δ i _{L} ) ^{2}}{12}

CCM, $I_{L}$ = average inductor current, $Δ i_{L}$ = peak-to-peak ripple. Used for rms ratings and copper loss; with negligible ripple $I_{L, r m s} \approx I_{L}$ .

Procedure for any converter

Draw the circuit for the switch ON interval and for the OFF interval.
Write the inductor voltage in each interval.
Apply volt-second balance to get $V_{o} / V_{s}$ in terms of $D$ .
Use $Δ i_{L} = V_{L} t / L$ in either interval to get the current ripple.
Use charge balance on the capacitor to get the average inductor current, then $I_{ma x} = I_{L} + Δ i_{L} /2$ and $I_{min} = I_{L} - Δ i_{L} /2$ .
CCM holds if $I_{min} > 0$ ; the boundary is $I_{min} = 0$ .

EXAM TIP

Ideal lossless converters: input power equals output power, so $V_{s} I_{s} = ∣ V_{o} ∣ I_{o}$ . The gain therefore also gives the input-to-output current ratio. This is a quick check on every answer.

Buck (step-down) converter

The switch is in series with the supply; the diode freewheels the inductor current when the switch is off, and the inductor and capacitor sit in an LC filter. When the switch is on, $v_{L} = V_{s} - V_{o}$ ; when it is off, $v_{L} = - V_{o}$ . Output is always lower than input.

BUCK: VOLTAGE GAIN

V_{o} = D V_{s}

CCM. Output is positive, same polarity as the input, and $V_{o} \leq V_{s}$ .

BUCK: INDUCTOR CURRENT RIPPLE

Δ i_{L} = \frac{( V _{s} - V _{o} ) D T}{L} = \frac{V _{o} ( 1 - D )}{L f}

Peak-to-peak ripple, CCM. Average inductor current equals $I_{o}$ ; average input current is $I_{s} = D I_{o}$ .

BUCK: MAXIMUM RIPPLE

Δ i_{L} = \frac{V _{s} D ( 1 - D )}{L f} \leq Δ i_{L, ma x} = \frac{V _{s}}{4 L f} at D = 0.5

For fixed $V_{s}$ , $L$ and $f$ in CCM. The same result holds for a step-down chopper feeding an inductive load (with or without back emf) when the load resistance is neglected.

BUCK: CAPACITOR VOLTAGE RIPPLE

Δ v_{C} = \frac{Δ i _{L}}{8 f C} = \frac{V _{o} ( 1 - D )}{8 L C f ^{2}}

Assumes the whole ripple component of the inductor current flows into the capacitor (the load takes only the average).

BUCK: CRITICAL INDUCTANCE

L_{cr i t} = \frac{( 1 - D ) R}{2 f}

CCM if $L > L_{cr i t}$ (boundary where $I_{min} = 0$ , i.e. $I_{o} = Δ i_{L} /2$ ).

WORKED EXAMPLE

A buck converter has $V_{s} = 60$ V, $D = 0.4$ , $L = 100 μ$ H, $f = 20$ kHz and $R = 4 Ω$ . Find $V_{o}$ , the peak-to-peak inductor ripple, $I_{ma x}$ , $I_{min}$ , and check whether conduction is continuous.

Solution

$V_{o} = D V_{s} = 0.4 \times 60 = 24$ V, so $I_{o} = 24/4 = 6$ A and $I_{L} = 6$ A.
$T = 50 μ$ s. $Δ i_{L} = V_{o} (1 - D) T / L = 24 \times 0.6 \times 50 \times 1 0^{- 6} / (100 \times 1 0^{- 6}) = 7.2$ A.
$I_{ma x} = 6 + 3.6 = 9.6$ A and $I_{min} = 6 - 3.6 = 2.4$ A.
$I_{min} > 0$ , so CCM. Check: $L_{cr i t} = (1 - D) R / (2 f) = 0.6 \times 4/40000 = 60 μ$ H $< 100 μ$ H.

Answer: $V_{o} = 24$ V, $Δ i_{L} = 7.2$ A, $I_{ma x} = 9.6$ A, $I_{min} = 2.4$ A, continuous conduction.

Boost (step-up) converter

The inductor is in series with the supply; the switch shunts the inductor to ground and the diode connects it to the output. Switch on: $v_{L} = V_{s}$ and the capacitor alone feeds the load. Switch off: $v_{L} = V_{s} - V_{o}$ (negative) and the inductor current flows through the diode to the load. Output is higher than the input.

BOOST: VOLTAGE GAIN

V_{o} = \frac{V _{s}}{1 - D}

CCM. Same polarity as input, $V_{o} \geq V_{s}$ . In the ideal circuit the gain rises without limit as $D \to 1$ ; in a real converter the inductor and switch resistances make the gain reach a maximum and then fall as $D \to 1$ .

BOOST: AVERAGE INDUCTOR (INPUT) CURRENT

I_{L} = I_{s} = \frac{I _{o}}{1 - D} = \frac{V _{s}}{( 1 - D ) ^{2} R}

The diode conducts only for $(1 - D) T$ and carries $I_{L}$ , so its average current is $(1 - D) I_{L} = I_{o}$ ; the switch average current is $D I_{L}$ .

BOOST: INDUCTOR CURRENT RIPPLE

Δ i_{L} = \frac{V _{s} D T}{L} = \frac{V _{s} D}{L f}

CCM, peak-to-peak.

BOOST: CAPACITOR VOLTAGE RIPPLE

Δ v_{C} = \frac{I _{o} D T}{C} = \frac{V _{o} D}{R C f}

During the ON interval the capacitor alone supplies $I_{o}$ for $D T$ .

BOOST: CRITICAL INDUCTANCE

L_{cr i t} = \frac{D ( 1 - D ) ^{2} R}{2 f}

CCM if $L > L_{cr i t}$ (from $I_{L} = Δ i_{L} /2$ ).

WORKED EXAMPLE

A boost converter has $V_{s} = 12$ V, $V_{o} = 36$ V, $R = 18 Ω$ , $L = 50 μ$ H, $C = 100 μ$ F and $f = 50$ kHz. Find $D$ , the average input current, the peak inductor current and the peak-to-peak output voltage ripple. Assume CCM and verify it.

Solution

$D = 1 - V_{s} / V_{o} = 1 - 12/36 = 2/3$ .
$I_{o} = 36/18 = 2$ A. $I_{L} = I_{o} / (1 - D) = 2 \times 3 = 6$ A (check: $P_{o} = 72$ W, $72/12 = 6$ A).
$T = 20 μ$ s. $Δ i_{L} = V_{s} D T / L = 12 \times (2/3) \times 20 \times 1 0^{- 6} / (50 \times 1 0^{- 6}) = 3.2$ A.
$I_{ma x} = 6 + 1.6 = 7.6$ A, $I_{min} = 4.4$ A $> 0$ , so CCM.
$Δ v_{C} = I_{o} D T / C = 2 \times (2/3) \times 20 \times 1 0^{- 6} / (100 \times 1 0^{- 6}) \approx 0.267$ V.

Answer: $D = 2/3$ , $I_{s} = 6$ A, $I_{ma x} = 7.6$ A, $Δ v_{C} \approx 0.267$ V.

Buck-boost and Cuk converters

In the buck-boost converter the switch is in series with the supply, the inductor is across the supply while the switch is on, and the diode is connected so that the inductor discharges into the capacitor and load when the switch is off. Switch on: $v_{L} = V_{s}$ . Switch off: $v_{L} = V_{o}$ (negative). The output polarity is inverted and its magnitude can be below or above $V_{s}$ depending on $D$ . Energy reaches the load only during the OFF interval.

BUCK-BOOST: VOLTAGE GAIN

V_{o} = - \frac{D}{1 - D} V_{s}

CCM. Output is negative with respect to the input common; $∣ V_{o} ∣ < V_{s}$ for $D < 0.5$ , $∣ V_{o} ∣ > V_{s}$ for $D > 0.5$ .

BUCK-BOOST: AVERAGE INDUCTOR CURRENT

I_{L} = \frac{I _{o}}{1 - D} = \frac{V _{s} D}{( 1 - D ) ^{2} R}

Here $I_{o} = ∣ V_{o} ∣/ R$ . Diode average current is $(1 - D) I_{L} = I_{o}$ ; average input (switch) current is $I_{s} = D I_{L} = D I_{o} / (1 - D)$ .

BUCK-BOOST: RIPPLES

Δ i_{L} = \frac{V _{s} D T}{L}, Δ v_{C} = \frac{∣ V _{o} ∣ D}{R C f}

CCM, peak-to-peak. The capacitor alone feeds the load during the ON interval.

BUCK-BOOST: CRITICAL INDUCTANCE

L_{cr i t} = \frac{( 1 - D ) ^{2} R}{2 f}

CCM if $L > L_{cr i t}$ .

Cuk converter (in brief). It has an inductor at the input, an inductor at the output and a series coupling capacitor that carries the energy from input to output. Both inductor currents are continuous (low ripple at input and output) and the output is inverted. In CCM its gain is the same as the buck-boost: $V_{o} = - \frac{D}{1 - D} V_{s}$ . The switch carries the sum of the two inductor currents, and the coupling-capacitor voltage is $V_{s} + ∣ V_{o} ∣$ .

Comparison and device stresses

The switch (when off) and the diode (when reverse biased) must block a voltage, and each must carry the peak inductor current when it conducts. The peak current of the switch and of the diode is $I_{L} + Δ i_{L} /2$ , with the ripple not neglected. Average currents below follow from charge balance and $V_{s} I_{s} = ∣ V_{o} ∣ I_{o}$ .

Property (CCM)	Buck	Boost	Buck-boost
Gain $V_{o} / V_{s}$	$D$	$1/ (1 - D)$	$- D / (1 - D)$
Polarity of $V_{o}$	Same	Same	Inverted
Average inductor current $I_{L}$	$I_{o}$	$I_{o} / (1 - D)$	$I_{o} / (1 - D)$
Average input current $I_{s}$	$D I_{o}$	$I_{o} / (1 - D)$	$D I_{o} / (1 - D)$
Average switch current	$D I_{o}$	$D I_{o} / (1 - D)$	$D I_{o} / (1 - D)$
Average diode current	$(1 - D) I_{o}$	$I_{o}$	$I_{o}$
Peak switch and diode current	$I_{L} + Δ i_{L} /2$	$I_{L} + Δ i_{L} /2$	$I_{L} + Δ i_{L} /2$
Inductor ripple $Δ i_{L}$	$V_{o} (1 - D) / (L f)$	$V_{s} D / (L f)$	$V_{s} D / (L f)$
Capacitor ripple $Δ v_{C}$	$Δ i_{L} / (8 f C)$	$V_{o} D / (R C f)$	$∣ V_{o} ∣ D / (R C f)$
$L_{cr i t}$	$(1 - D) R / (2 f)$	$D (1 - D)^{2} R / (2 f)$	$(1 - D)^{2} R / (2 f)$
Switch off-state voltage	$V_{s}$	$V_{o}$	$V_{s} + ∣ V_{o} ∣$
Diode reverse voltage	$V_{s}$	$V_{o}$	$V_{s} + ∣ V_{o} ∣$
Input current	Pulsed	Continuous (= $I_{L}$ )	Pulsed

EXAM TIP

Capacitor ripple: only the buck uses the triangular-ripple formula $Δ i_{L} / (8 f C)$ . In boost and buck-boost the output capacitor supplies the full load current for $D T$ , so $Δ v_{C} = I_{o} D T / C$ . Using the wrong one is a common error.

Discontinuous conduction

If $L < L_{cr i t}$ (or the load current is light) the inductor current falls to zero before the period ends: a third interval $D_{3} T$ with the switch and diode both off appears, so $D + D_{2} < 1$ , where $D_{2} T$ is the diode conduction interval. The gain then depends on the load as well as on $D$ . Define $K = \frac{2 L}{R T} = \frac{2 L f}{R}$ . The converter is in DCM when $K < K_{cr i t}$ , where $K_{cr i t} = 1 - D$ for the buck, $D (1 - D)^{2}$ for the boost and $(1 - D)^{2}$ for the buck-boost.

DCM: GAIN IN TERMS OF THE DIODE INTERVAL $D_2$

Buck: \frac{V _{o}}{V _{s}} = \frac{D}{D + D _{2}}; Boost: \frac{V _{o}}{V _{s}} = \frac{D + D _{2}}{D _{2}}; Buck-boost: \frac{V _{o}}{V _{s}} = \frac{D}{D _{2}}

From volt-second balance with the inductor voltage zero during $D_{3} T = (1 - D - D_{2}) T$ . $D_{2}$ is the fraction of the period for which the diode conducts.

DCM: GAIN IN TERMS OF $D$ AND $K$

Buck: \frac{V _{o}}{V _{s}} = \frac{2}{1 + 1 + 4 K / D ^{2}}; Boost: \frac{V _{o}}{V _{s}} = \frac{1 + 1 + 4 D ^{2} / K}{2}; Buck-boost: \frac{V _{o}}{V _{s}} = \frac{D}{K}

Valid for $K < K_{cr i t}$ (DCM) only. At $K = K_{cr i t}$ they agree with the CCM gains. For the buck-boost the diode interval is $D_{2} = K$ ; in every case a DCM answer must satisfy $D + D_{2} < 1$ .

EXAM TIP

First decide CCM or DCM by comparing $L$ with $L_{cr i t}$ (or $K$ with $K_{cr i t}$ ). Applying the CCM gain to a lightly loaded or small-inductor converter gives a wrong $V_{o}$ ; in DCM the output of a boost rises as the load is lightened.

Choppers and DC motor control

A chopper is a DC-DC converter used to vary the DC voltage applied to a DC load or the armature of a DC motor. Quadrants refer to the $(V_{o}, I_{o})$ plane. For a separately excited DC motor with armature voltage $V_{a} = D V_{s}$ , in steady state $V_{a} = E_{b} + I_{a} R_{a}$ with $E_{b} = K ϕ ω$ , so speed is controlled by varying $D$ (below base speed, with constant field).

ARMATURE-VOLTAGE SPEED CONTROL

ω = \frac{D V _{s} - I _{a} R _{a}}{K ϕ}

Separately excited DC motor fed by a step-down chopper, continuous armature current, constant flux; average values.

REGENERATIVE BRAKING (CLASS B CHOPPER)

E_{b} = (1 - D) V_{s} + I_{a} R_{a}, P_{returned} = (1 - D) V_{s} I_{a}

Step-up (Class B) chopper with the machine generating: $D$ is the duty ratio of the switch that shorts the armature, $I_{a}$ is the magnitude of the continuous armature current flowing from the machine to the supply, and the average voltage across the machine terminals is $(1 - D) V_{s}$ . Average values, ripple neglected.

Class	Quadrant(s)	Output voltage / current	Use
A	I	$V_{o} > 0$ , $I_{o} > 0$	Step-down chopper, motoring
B	II	$V_{o} > 0$ , $I_{o} < 0$	Regenerative braking (power returned to supply)
C	I and II	$V_{o} > 0$ ; $I_{o}$ of either sign	Motoring and regenerative braking, forward
D	I and IV	$I_{o} > 0$ ; $V_{o}$ of either sign	Forward motoring and reverse regenerative braking (power returned to supply)
E	I, II, III, IV	Both of either sign	Four-quadrant drive (H-bridge)

EXAM TIP

Average motor current is set by the torque ( $T \propto I_{a}$ at constant flux), not by $D$ directly: raising $D$ raises $V_{a}$ , the speed then rises until $E_{b} + I_{a} R_{a} = D V_{s}$ again. A Class A chopper cannot return energy to the supply, because neither its average output voltage nor its current can reverse.

PREVIOUS-YEAR QUESTIONS

9 questions from the EE papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE EE 2026 · GATE EE 2025 · GATE EE 2024 · GATE EE 2023 · GATE EE 2022 · GATE EE 2020 · GATE EE 2018

GATE 2026 EE Q.48

GATE 2026 · EEQ.48 · 2 marks · MCQ

Consider the boost converter circuit shown. Assume that the semiconductor devices are ideal. In steady state, the inductor current rises linearly from 0 A to 6 A in the first 10 µs and then falls linearly from 6 A to 0 A in the next 10 µs of every switching cycle as shown. The load resistance R is 10 Ω and the capacitance C is 500 µF. Neglect the ripple in the output voltage. What is the input voltage V_dc?

GATE 2026 EE question 48: Left: boost converter. — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
10.0 V
(B)
15.0 V
(C)
7.5 V
(D)
12.5 V

Answer and solution

GATE 2026 EE Q.48: answer and solution

GATE 2025 EE Q.34

GATE 2025 · EEQ.34 · 1 mark · NAT

In the circuit with ideal devices, the power MOSFET is operated with a duty cycle of 0.4 in a switching cycle with $I = 10$ A and $V = 15$ V. The power delivered by the current source, in W, is ____________ (round off to the nearest integer).

GATE 2025 EE question 34: A DC current source I (arrow pointing up) sits on the left branch between the bottom rail and the top rail. — Figure from the GATE 2025 EE paper. It is part of the question.

Answer and solution

GATE 2025 EE Q.34: answer and solution

GATE 2025 EE Q.60

GATE 2025 · EEQ.60 · 2 marks · NAT

The steady state capacitor current of a conventional DC-DC buck converter, working in CCM, is shown in one switching cycle. If the input voltage is 30 V, the value of the inductor used, in mH, is ____________ (round off to one decimal place).

GATE 2025 EE question 60: Plot of capacitor current (A) versus time (µs). — Figure from the GATE 2025 EE paper. It is part of the question.

Answer and solution

GATE 2025 EE Q.60: answer and solution

GATE 2024 EE Q.64

GATE 2024 · EEQ.64 · 2 marks · NAT

In the DC-DC converter shown in the figure, the current through the inductor is continuous. The switching frequency is 500 Hz. The voltage ( $V_{o}$ ) across the load is assumed to be constant and ripple free. The peak inductor current in amperes is __________ (rounded off to the nearest integer).

GATE 2024 EE question 64: Boost-type circuit. — Figure from the GATE 2024 EE paper. It is part of the question.

Answer and solution

GATE 2024 EE Q.64: answer and solution

GATE 2023 EE Q.23

GATE 2023 · EEQ.23 · 1 mark · MCQ

The chopper circuit shown in figure (i) feeds power to a 5 A DC constant current source. The switching frequency of the chopper is 100 kHz. All the components can be assumed to be ideal. The gate signals of switches S1 and S2 are shown in figure (ii). Average voltage across the 5 A current source is

GATE 2023 EE question 23: Figure (i): a 20 V DC source (+ terminal at top) feeds a half-bridge leg. — Figure from the GATE 2023 EE paper. It is part of the question.

(A)
10 V
(B)
6 V
(C)
12 V
(D)
20 V

Answer and solution

GATE 2023 EE Q.23: answer and solution

GATE 2022 EE Q.61

GATE 2022 · EEQ.61 · 2 marks · NAT

The steady state current flowing through the inductor of a DC-DC buck boost converter is given in the figure below. If the peak-to-peak ripple in the output voltage of the converter is 1 V, then the value of the output capacitor, in µF, is ______________. (round off to nearest integer)

GATE 2022 EE question 61: Plot of inductor current (A, vertical axis) against time (µsec, horizontal axis). — Figure from the GATE 2022 EE paper. It is part of the question.

Answer and solution

GATE 2022 EE Q.61: answer and solution

GATE 2020 EE Q.50

GATE 2020 · EEQ.50 · 2 marks · NAT

In the dc-dc converter circuit shown, switch Q is switched at a frequency of 10 kHz with a duty ratio of 0.6. All components of the circuit are ideal, and the initial current in the inductor is zero. Energy stored in the inductor in mJ (rounded off to 2 decimal places) at the end of 10 complete switching cycles is ______.

GATE 2020 EE question 50: A 50 V source on the left with its positive terminal at the top. — Figure from the GATE 2020 EE paper. It is part of the question.

Answer and solution

GATE 2020 EE Q.50: answer and solution

GATE 2018 EE Q.50

GATE 2018 · EEQ.50 · 2 marks · NAT

The figure shows two buck converters connected in parallel. The common input dc voltage for the converters has a value of 100 V. The converters have inductors of identical value. The load resistance is 1 Ω. The capacitor voltage has negligible ripple. Both converters operate in the continuous conduction mode. The switching frequency is 1 kHz, and the switch control signals are as shown. The circuit operates in the steady state. Assuming that the converters share the load equally, the average value of i_S1, the current of switch S1 (in Ampere), is _____ (up to 2 decimal places).

GATE 2018 EE question 50: A 100 V dc source (+ at top) feeds two buck stages sharing the same input rails. — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

GATE 2018 EE Q.50: answer and solution

GATE 2018 EE Q.53

GATE 2018 · EEQ.53 · 2 marks · NAT

A dc to dc converter shown in the figure is charging a battery bank, B2 whose voltage is constant at 150 V. B1 is another battery bank whose voltage is constant at 50 V. The value of the inductor, L is 5 mH and the ideal switch, S is operated with a switching frequency of 5 kHz with a duty ratio of 0.4. Once the circuit has attained steady state and assuming the diode D to be ideal, the power transferred from B1 to B2 (in Watt) is ____________ (up to 2 decimal places).

GATE 2018 EE question 53: Battery B1 of 50 V (positive terminal up) at the left, in series with inductor L = 5 mH carrying current i_L to the right… — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

GATE 2018 EE Q.53: answer and solution

PRACTICE

5 questions to test yourself

PRACTICE 11 mark · NAT

An ideal buck converter operates in continuous conduction with $V_{s} = 48$ V, output $V_{o} = 12$ V, $L = 60 μ$ H and switching frequency $f = 50$ kHz. Find the peak-to-peak inductor current ripple in amperes.

Answer and solution

Answer: $3$ (accepted range $\pm 0.02$ )

$D = V_{o} / V_{s} = 12/48 = 0.25$ .
$T = 1/ f = 20 μ$ s.
$Δ i_{L} = V_{o} (1 - D) T / L = 12 \times 0.75 \times 20 \times 1 0^{- 6} / (60 \times 1 0^{- 6}) = 3$ A.

PRACTICE 21 mark · MCQ

An ideal buck-boost converter in continuous conduction is supplied from $V_{s} = 12$ V with duty ratio $D = 0.6$ . The average output voltage is

(A)
$- 7.2$ V
(B)
$- 18$ V
(C)
$- 30$ V
(D)
$+ 18$ V

Answer and solution

Answer: (B)

$V_{o} = - \frac{D}{1 - D} V_{s} = - \frac{0.6}{0.4} \times 12 = - 18$ V.
The buck-boost output is inverted, so options with a positive sign are wrong.

PRACTICE 32 marks · NAT

An ideal boost converter has $V_{s} = 20$ V and delivers $V_{o} = 50$ V into a resistive load $R = 25 Ω$ at switching frequency $f = 25$ kHz. Find the minimum inductance, in $μ$ H, for which the inductor current is just continuous (on the boundary of continuous conduction).

Answer and solution

Answer: $48$ (accepted range $\pm 0.5$ )

$D = 1 - V_{s} / V_{o} = 1 - 20/50 = 0.6$ .
$L_{cr i t} = \frac{D ( 1 - D ) ^{2} R}{2 f} = \frac{0.6 \times 0.16 \times 25}{2 \times 25000} = 48 μ$ H.
Check: $I_{o} = 2$ A, $I_{L} = I_{o} / (1 - D) = 5$ A, $Δ i_{L} = V_{s} D T / L = 20 \times 0.6 \times 40 μ s /48 μ H = 10$ A $= 2 I_{L}$ , so $I_{min} = 0$ .

PRACTICE 42 marks · MSQ

For ideal buck, boost and buck-boost converters operating in continuous conduction, which of the following statements are correct?

(A)
The buck-boost converter produces an output of polarity opposite to the input, whose magnitude can be smaller or larger than $V_{s}$ .
(B)
In the boost converter, the reverse voltage across the diode while the switch is on equals $V_{o}$ .
(C)
In the buck converter, the average inductor current equals the average load current.
(D)
In the boost converter, the average inductor current equals the average load current.

Answer and solution

Answer: (A), (B), (C)

A is true: $∣ V_{o} ∣ = D V_{s} / (1 - D)$ , below $V_{s}$ for $D < 0.5$ and above for $D > 0.5$ , with inverted polarity.
B is true: with the switch on, the diode anode is at ground and its cathode at $V_{o}$ , so it blocks $V_{o}$ .
C is true: average capacitor current is zero, so $I_{L} = I_{o}$ in the buck.
D is false: in the boost $I_{L} = I_{o} / (1 - D)$ , which exceeds $I_{o}$ .

PRACTICE 52 marks · NAT

An ideal buck-boost converter has $V_{s} = 24$ V, $D = 0.3$ , $L = 25 μ$ H, $f = 40$ kHz and load $R = 10 Ω$ . The inductor current is found to be discontinuous. Find the magnitude of the average output voltage in volts.

Answer and solution

Answer: $16.1$ (accepted range $\pm 0.1$ )

$T = 25 μ$ s, so $K = \frac{2 L}{R T} = \frac{2 \times 25}{10 \times 25} = 0.2$ .
$K_{cr i t} = (1 - D)^{2} = 0.49 > K$ , so the converter is indeed in DCM.
$∣ V_{o} ∣ = \frac{D V _{s}}{K} = \frac{0.3 \times 24}{0.4472} \approx 16.1$ V.
Check: $D_{2} = K = 0.447$ and $D + D_{2} = 0.747 < 1$ , consistent.

OFFICIAL GATE 2027 SYLLABUS WORDING

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DC to DC conversion: Buck, Boost and Buck-Boost Converters

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buck and boost converters

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