Buck

Buck converter — CCM

A simple step-down converter with continuous inductor current.

Press Run: the example simulates in your browser. Open and simulate takes it to the full app, where you can change it.

1Overview

The simplest switching converter in the gallery: a buck (step-down) converter running open loop at a fixed duty cycle. It turns 5 V into about 0.88 V at 2 MHz, and its inductor current never reaches zero, which is continuous conduction mode (CCM).

Use it to learn the app, and to check the textbook buck equations against a simulation: every result below is worked out by hand first.

2Model

Buck converter schematic — **Fig. 1:** The example as it opens in PWMSim.

S1 connects the input to the switch node sw for a fraction D = 0.18 of each 500 ns period, as PWM1 commands. When S1 opens, D2 carries the inductor current. L1 and C1 filter the switch node into a nearly constant output across the load R4.

The switches and diodes are two-state models (Ideal switched in the app): on, they are a 10 mΩ resistance; off, an open circuit (100 MΩ to 1 GΩ). The diode's forward drop is 1 nV, so it is ideal in practice. R2 and R3 add the inductor's and the trace's resistance.

Part	Kind	Value	Role
V1	DC voltage	5	Input source
S1	Switch	—	Main switch
PWM1	PWM	2 MHz · 18%	Gate signal for S1
D1	Diode	1n	Body diode of S1
D2	Diode	1n	Freewheeling diode
R1	Resistor	100meg	Keeps the switch node defined
R2	Resistor	10m	Inductor resistance
L1	Inductor	10u	Output inductor
R3	Resistor	1m	Trace resistance
C1	Capacitor	22u	Output capacitor
R4	Resistor	1	Load

3How it works

Output voltage

In steady state the inductor's average voltage is zero (volt-second balance). During the on-time it sees V_in − V_out, during the off-time −V_out, so

D \cdot (V_{in} - V_{out}) - (1 - D) \cdot V_{out} = 0 ⟹ V_{out} = D \cdot V_{in}

That gives 0.18 × 5 V = 0.90 V. The load current also flows through 21 mΩ of series resistance (the switch or the diode, 10 mΩ, plus R2 and R3), which takes a little off:

V_{out} = \frac{D \cdot V_{in}}{1 + R_{path} / R}

Inductor ripple

During the on-time the inductor current rises with slope (V_in − V_out)/L for D·T:

{ΔI}_{L} = \frac{(V_{in} - V_{out}) \cdot D}{L \cdot f_{sw}}

Staying in CCM

The current's valley is I_out − ΔI_L/2. It reaches zero, and the converter leaves CCM, when the load resistance rises to

R_{crit} = \frac{2 \cdot L \cdot f_{sw}}{1 - D}

Here R_crit = 48.8 Ω, far above the 1 Ω load.

Start-up

Switched on from zero, L1 and C1 ring at f₀ = 1/(2π√(LC)) = 10.7 kHz, damped by the load with Q = R√(C/L) = 1.48. A second-order step response overshoots by exp(−πζ/√(1−ζ²)) = 32% (ζ = 1/2Q), peaking at t_p = 1/(2f₀√(1−ζ²)) = 49.5 µs.

4Simulation

Open the example and it runs once, for 600 µs (1,200 switching periods). Fig. 2 shows the start-up; Fig. 3 the last four periods, where the inductor current is the triangle the ripple equation describes.

Fig. 2: Start-up: output voltage and inductor current over the whole run.

Fig. 3: Steady state: switch node and inductor current, last four periods.

Measured over the last ten periods, against the hand calculations of section 3:

Quantity	By hand	PWMSim
Output voltage, mean	0.881 V	0.881 V
Load current, mean of i(L1)	0.881 A	0.881 A
Inductor ripple, peak to peak	37.1 mA	36.9 mA
Inductor current, minimum	0.863 A	0.863 A
Start-up peak of v(out)	1.17 V	1.16 V
Time of that peak	49.5 µs	49.2 µs

The steady-state values agree within 0.4%. The start-up estimate is looser: it treats the converter as an LC filter switched onto a step of D·V_in, which ignores the switching itself.

5Try this

Raise the duty cycle

Select PWM1 and change its duty from 0.18 to 0.22, then press Run transient. The output and the ripple both follow the equations of section 3:

Duty 0.22	By hand	PWMSim
Output voltage, mean	1.077 V	1.077 V
Inductor ripple, peak to peak	43.1 mA	42.9 mA

Lighten the load into DCM

Set R4 to 100 Ω, above R_crit = 48.8 Ω, and the inductor current now reaches zero every period: discontinuous conduction (DCM). The output is no longer D·V_in; with K = 2L/(R·T) = 0.40 it is

V_{out} = \frac{2 \cdot V_{in}}{1 + \sqrt{1 + \frac{4 \cdot K}{D^{2}}}}

The output time constant grows to RC = 2.2 ms, so open Settings and set the stop time to 10 ms (type 10m) before you run it.

Fig. 4: R4 = 100 Ω: the inductor current falls to zero and stays there until the switch turns on again. While neither S1 nor D2 conducts, nothing drives the switch node, and it sits at the output voltage.

R4 = 100 Ω	By hand	PWMSim
Output voltage, mean	1.235 V	1.235 V
Inductor current, minimum	0 A	28 nA

6What the model leaves out

Switches and diodes change state instantly. There are no switching transitions, no reverse recovery, no gate drive and no parasitic capacitance, so the only losses are the on-resistances.
The open-loop output follows the input and the load: there is no controller. The PI and peak-current examples close the loop.
A simulation is not a hardware test.

7References

R. W. Erickson and D. Maksimović, Fundamentals of Power Electronics, 3rd ed., Springer, 2020. Chapter 2 (volt-second balance, ripple) and chapter 5 (discontinuous conduction mode).
N. Mohan, T. M. Undeland and W. P. Robbins, Power Electronics: Converters, Applications, and Design, 3rd ed., Wiley, 2003. Chapter 7.

Every waveform and simulated number above comes from this example as the app opens it, run by PWMSim.