A buck converter is a DC/DC converter that steps down voltage.
This post gathers useful design equations.
Assumptions:
-Synchronous converter (i.e. no diode switches), therefore always continuous conduction
-Small output voltage ripple
Capital letters indicate DC values and lower case are AC.
AC quantities are derived using the small-ripple approximation.
Note that the rms values will be very different from the DC values unless the inductor ripple is very small!
Contents
Ideal Converter
DC Quantities
Voltages
The output voltage is reduced linearly by the duty cycle.
$$\frac{V_o}{V_i} = D$$
The capacitor is the output.
$$V_C = V_o$$
The switches see the input voltage.
$$V_{Sw1} = V_{Sw2} = V_i$$
Currents
For a resistor load:
$$I_o = \frac{V_o}{R_o}$$
The inductor current is equal to the output current.
$$I_L = I_o$$
The input current is reduced compared to the output current, and is equal to the high-side switch current.
$$I_i = D \cdot I_o$$
$$I_{Sw1} = I_i = D \cdot I_o$$
$$I_{Sw2} = (1 – D) \cdot I_o$$
Losses
Ideal converter.
$$\eta = 1$$
$$P_{loss} = 0$$
AC Quantities
Voltages
Output voltage ripple (with ESR = 0):
$$\Delta v_C = \frac{\Delta i_{L}}{8 \cdot C \cdot f_s}$$
$$v_{C,pk-pk} = 2\Delta v_C$$
$$v_{C,max} = V_C + \Delta v_C$$
Currents
Peak inductor ripple current (one-sided):
$$\Delta i_{L} = \frac{(V_i – V_o)D}{2 \cdot L \cdot f_s}$$
$$i_{L,max} = I_L + \Delta i_{L}$$
$$i_{L,pk-pk} = 2\Delta i_{L}$$
Component rms currents.
$$i_{L,rms} = I_L \sqrt{1 + \frac{1}{3}(\frac{\Delta i_{L}}{I_L})^2}$$
$$i_{C,rms} = \frac{\Delta i_{L}}{\sqrt{3}}$$
$$i_{i,rms} = i_{Sw1,rms} = I_o \sqrt{D}\sqrt{1 + \frac{1}{3}(\frac{\Delta i_{L}}{I_L})^2}$$
$$i_{Sw2,rms} = I_o \sqrt{1-D}\sqrt{1 + \frac{1}{3}(\frac{\Delta i_{L}}{I_L})^2}$$
Non-Ideal Converter
Coming soon…