A boost converter is a DC/DC converter that steps up 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 increased non-linearly by the duty cycle.
$$\frac{V_o}{V_i} = \frac{1}{1-D}$$
$$D = 1 – \frac{Vi}{Vo}$$
The capacitor is the output.
$$V_C = V_o$$
The switches see the output voltage.
$$V_{Sw1} = V_{Sw2} = V_o$$
Currents
For a resistor load:
$$I_o = \frac{V_o}{R_o}$$
The inductor current is increased compared to the output current, and is equal to the input current.
$$I_L = I_i = I_o \frac{1}{1-D}$$
The switches see some combination of the input and output current.
$$I_{Sw1} = D \cdot I_L$$
$$I_{Sw2} = I_o$$
Losses
Ideal converter.
$$\eta = 1$$
$$P_{loss} = 0$$
AC Quantities
Voltages
Output voltage ripple (with ESR = 0):
$$\Delta v_C = \frac{I_o \cdot D}{2 \cdot C \cdot f_s}$$
$$v_{C,pk-pk} = 2\Delta v_C$$
$$v_{C,max} = V_C + \Delta v_C$$
Currents
Peak inductor ripple current:
$$\Delta i_{L} = \frac{D \cdot V_i}{2 \cdot L \cdot f_s}$$
$$i_{L,max} = I_L + \Delta i_{L}$$
$$i_{L,pk-pk} = 2\Delta i_{L}$$
Component rms currents. Note that all of these are higher than the output current.
$$i_{L,rms} = i_{i,rms} = I_L \sqrt{1 + \frac{1}{3}(\frac{\Delta i_{L}}{I_L})^2}$$
$$i_{C,rms} = \sqrt{\frac{D \cdot I_o^2}{1 – D} + \frac{1}{3}\Delta i_{L}^2}$$
$$i_{Sw1,rms} = I_L \sqrt{D}\sqrt{1 + \frac{1}{3}(\frac{\Delta i_{L}}{I_L})^2}$$
$$i_{Sw2,rms} = I_L \sqrt{1-D}\sqrt{1 + \frac{1}{3}(\frac{\Delta i_{L}}{I_L})^2}$$
Non-Ideal Converter
Coming soon…