DC-DC converters: practice problems with step-by-step solutions

Free DC-DC converters practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: ideal conversion ratios, inductor ripple and currents, output ripple and the CCM boundary.

Free practice problemEasy
+−24 VD = 0.6LCR−+Vo​ = ?

An ideal inverting buck-boost converter runs from 24 V with a duty cycle D = 0.6 in continuous conduction. Find the magnitude of the output voltage |Vo|.

Vin​ = 24 VD = 0.6
V
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What you practise

  1. Easyideal conversion ratios
  2. Mediuminductor ripple and currents
  3. Hardoutput ripple and the CCM boundary

How to solve DC-DC converters problems

  1. Identify the converter (buck, boost or buck-boost) and the duty cycle D from the voltage ratio.
  2. Use volt-second balance on the inductor: the average inductor voltage over a period is zero.
  3. Find the inductor voltage while the switch is on, then ΔIL = VL·D/(L·f).
  4. For the output ripple, find the charge that flows in and out of the capacitor; for the CCM boundary, set the average inductor current equal to ΔIL/2.

Common mistakes

  • Using the buck relation for a boost converter, or D for 1 − D.
  • Using Vin instead of Vin − Vo as the inductor voltage in a buck converter.
  • Forgetting that the boost inductor carries the input current Io/(1 − D), not the load current.
  • Mixing kHz, µH and µF without converting to SI units.

Key formulas

Buck
Vo = D·Vin
Boost
Vo = Vin/(1 − D)
Inverting buck-boost
|Vo| = Vin·D/(1 − D)
Ideal power balance
VinIin = VoIo
Inductor ripple
ΔIL = VL,on·D/(L·f); buck Vin − Vo, boost and buck-boost Vin
Average inductor current
buck Io; boost and buck-boost Io/(1 − D)
Output ripple
buck ΔIL/(8fC); boost and buck-boost Io·D/(fC)
CCM boundary
buck (1 − D)R/(2f); boost D(1 − D)²R/(2f); buck-boost (1 − D)²R/(2f)

Learn DC-DC converters from scratch

Before this: RC & RL transients

1What a switching converter does

A DC-DC converter turns one DC voltage into another, for example 12 V from a battery into 5 V for a processor. Instead of burning the difference in a resistor, it switches the supply on and off very fast and smooths the result.

The switch S is on for a fraction D of each period T = 1/f. D is the duty cycle. An inductor L and a capacitor C filter the pulses into a steady output.

+−Vin​SLCR+−Vo​
  • The buck converter steps the voltage down.
  • The boost converter steps it up.
  • The inverting buck-boost converter can do both, with a negative output.

With ideal parts no energy is lost, so Vin·Iin = Vo·Io.

Self-check An ideal buck converter runs from 24 V with D = 0.25. What is its output voltage?
  1. 18 V
  2. 32 V
  3. 6 V
  4. 96 V

Answer: 6 V. Vo = D·Vin = 0.25·24 = 6 V. Using 1 − D gives 18 V, the boost formula gives 32 V, and dividing by D gives 96 V.

2Volt-second balance and the conversion ratios

In steady state the inductor current returns to the same value after each period. So the average inductor voltage over a period must be zero: the volt-seconds while the switch is on cancel those while it is off.

For the buck converter the inductor sees Vin − Vo for the time DT and −Vo for (1 − D)T:

(Vin − Vo)·D = Vo·(1 − D) → Vo = D·Vin

The same balance gives the other two converters:

boost: Vo = Vin/(1 − D) buck-boost: |Vo| = Vin·D/(1 − D)

These hold in continuous conduction (CCM), when the inductor current never drops to zero.

Self-check An ideal boost converter must turn 12 V into 48 V. What duty cycle does it need?
  1. 0.25
  2. 0.75
  3. 0.8
  4. 4

Answer: 0.75. 48 = 12/(1 − D), so 1 − D = 0.25 and D = 0.75. The value 0.25 is the off-time fraction, and 4 is the voltage ratio itself.

3The method step by step

  1. Identify the converter and find D from the voltage ratio.
  2. Find the inductor voltage while the switch is on: Vin − Vo for the buck, Vin for the boost and the buck-boost.
  3. The ripple current is ΔIL = VL·D/(L·f).
  4. The average inductor current is Io for the buck and Io/(1 − D) for the boost and buck-boost. The peak is the average plus ΔIL/2.
  5. For the output ripple, use ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost.

Convert the units first: µH to H, µF to F and kHz to Hz.

4A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A buck converter turns 24 V into 12 V for a 6 Ω load. It switches at 100 kHz with L = 47 µH and C = 22 µF. Find the inductor ripple current, the peak inductor current and the output ripple.

1

Look at the circuit

+−24 VS47 µH22 µF6 Ω+−12 V

This is a buck converter, so D = 12/24 = 0.5.

2

Inductor ripple current

While S is on, the inductor sees 24 − 12 = 12 V.

ΔIL = 12·0.5/(47·10⁻⁶·100·10³) = 1.28 A
3

Peak inductor current

The average inductor current is the load current, 12/6 = 2 A.

Imax = 2 + 1.28/2 = 2.64 A
4

Output ripple

ΔVo = ΔIL/(8fC) = 1.28/(8·100·10³·22·10⁻⁶) = 0.073 V
5

Check the answer

The lowest inductor current is 2 − 0.64 = 1.36 A, still above zero, so the converter is in continuous conduction and the formulas apply.

Answer

ΔIL = 1.28 A, Imax = 2.64 A and ΔVo = 73 mV.

5Ripple and the conduction boundary

The inductor current rises while the switch is on and falls while it is off. Its average equals the load current in a buck converter.

iL​tImax​Imin​Io​ΔIDTT2T

If the load current falls to ΔIL/2, the triangle just touches zero: the converter is at the CCM boundary. Below that it enters discontinuous conduction and the simple ratios no longer hold.

Setting the average current equal to ΔIL/2 gives the smallest inductance for continuous conduction:

buck: (1 − D)R/(2f) boost: D(1 − D)²R/(2f) buck-boost: (1 − D)²R/(2f)
Self-check A buck converter runs at D = 0.4 into a 10 Ω load at 50 kHz. What is the smallest inductance for continuous conduction?
  1. 120 µH
  2. 30 µH
  3. 40 µH
  4. 60 µH

Answer: 60 µH. L = (1 − D)R/(2f) = 0.6·10/(2·50 000) = 60 µH. Using D instead of 1 − D gives 40 µH, and leaving out the 2 gives 120 µH.

6Common mistakes

  • Using the buck relation for a boost converter, or D where 1 − D belongs.
  • Using Vin instead of Vin − Vo as the inductor voltage of a buck converter.
  • Taking the load current as the boost inductor current: the inductor carries the input current, Io/(1 − D).
  • Mixing kHz, µH and µF without converting them.
Self-check The inductor voltage is 10 V for the on-time, D = 0.5, L = 100 µH and f = 50 kHz. What is ΔIL?
  1. 0.001 A
  2. 1 A
  3. 1000 A
  4. 2 A

Answer: 1 A. ΔIL = 10·0.5/(100·10⁻⁶·50·10³) = 5/5 = 1 A. Leaving L in µH or f in kHz gives answers that are a factor 1000 off.

7How exam questions look

  • Easy: ideal conversion ratios. You find the output voltage or the duty cycle of a buck, boost or buck-boost converter, or the input current from the power balance.
  • Medium: inductor ripple and currents. You find the ripple current, the inductance for a given ripple, the peak inductor current of a buck, or the average inductor current of a boost.
  • Hard: output ripple and the CCM boundary. You find the output ripple or the capacitance it needs, the critical inductance of each converter, or the smallest load current for continuous conduction.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
+−12 VD = ?LCR−+48 V

An ideal inverting buck-boost converter must produce −48 V from 12 V in continuous conduction. Find the duty cycle D.

Vin​ = 12 VVo​ = −48 V

Find D

Show hints
  1. Start from the ideal conversion ratio: |Vo| = Vin·D/(1 − D).
  2. Solve it for D.
  3. D = |Vo|/(Vin + |Vo|).
Show the answer and the step-by-step solution

DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.

Conversion ratio

|Vo| = Vin·D/(1 − D)

Solve for D

D = |Vo|/(Vin + |Vo|) = 0.8

Answer

D = 0.8
Example 2Medium
tiL​Imax​Imin​ΔI = ?DTT2T

An ideal boost converter turns 15 V into 60 V at 25 kHz with L = 330 µH. Find the peak-to-peak inductor ripple current ΔIL.

Vin​ = 15 VVo​ = 60 Vf = 25 kHzL = 330 µH

Find ΔIL (A)

Show hints
  1. While S is on, the inductor sits directly across Vin.
  2. ΔIL = VL·D·T/L, T = 1/f.
  3. VL = 15 V.
Show the answer and the step-by-step solution

DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.

Duty cycle

D = 1 − 15/60 = 0.75

Inductor voltage during the on-time

While S is on, the inductor sits directly across Vin.

VL = 15 V

Ripple current

ΔIL = VL·D/(L·f) = 15·0.75/(330·10⁻⁶·25·10³) = 1.364 A

Answer

ΔIL = 1.364 A
Example 3Hard
+−12 VSLC = 22 µFR = 10 Ω−+−36 V

An ideal buck-boost converter turns 12 V into −36 V for a 10 Ω load at 100 kHz. The output capacitor is 22 µF. Find the peak-to-peak output voltage ripple ΔVo in mV.

Vin​ = 12 VVo​ = −36 Vf = 100 kHzC = 22 µFR = 10 Ω

Find ΔVo (mV)

Show hints
  1. While the switch is on, the diode blocks and the capacitor feeds the load alone.
  2. ΔVo = Io·D·T/C.
  3. Io = 36/10 = 3.6 A.
Show the answer and the step-by-step solution

DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.

Duty cycle

D = 36/(12 + 36) = 0.75

Load current

Io = 36/10 = 3.6 A

Output ripple

ΔVo = 3.6·0.75/(100·10³·22·10⁻⁶) = 1227 mV

Answer

ΔVo = 1227 mV

Questions and answers

What is DC-DC converters?

A DC-DC converter turns one DC voltage into another, for example 12 V from a battery into 5 V for a processor. Instead of burning the difference in a resistor, it switches the supply on and off very fast and smooths the result.

The switch S is on for a fraction D of each period T = 1/f. D is the duty cycle. An inductor L and a capacitor C filter the pulses into a steady output.

How do I solve DC-DC converters problems step by step?
  1. Identify the converter (buck, boost or buck-boost) and the duty cycle D from the voltage ratio.
  2. Use volt-second balance on the inductor: the average inductor voltage over a period is zero.
  3. Find the inductor voltage while the switch is on, then ΔIL = VL·D/(L·f).
  4. For the output ripple, find the charge that flows in and out of the capacitor; for the CCM boundary, set the average inductor current equal to ΔIL/2.
What are common mistakes in DC-DC converters?
  • Using the buck relation for a boost converter, or D for 1 − D.
  • Using Vin instead of Vin − Vo as the inductor voltage in a buck converter.
  • Forgetting that the boost inductor carries the input current Io/(1 − D), not the load current.
  • Mixing kHz, µH and µF without converting to SI units.
What do the DC-DC converters exercises cover?
  • Easy: ideal conversion ratios
  • Medium: inductor ripple and currents
  • Hard: output ripple and the CCM boundary

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

RC & RL transients

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Printable worksheet

DC-DC converters worksheet with answers
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