Buck-boost regulator output and ripple derivation

A Buck-Boost regulator gives an output voltage that can be lower or higher than the input and is of opposite polarity: Va = −k·Vs/(1 − k).

Circuit

        Q (switch)        Dm
 +Vs ---/ ---+-----------|<---+------+
             |                |      |
             L  iL            C    LOAD   Va
             |                |   (− at top)
 −   --------+----------------+------+

Operation

Mode 1 (Q ON, 0 < t < t1 = kT): Current flows from Vs through Q and L. Inductor current rises linearly from I1 to I2; Dm is reverse biased; C supplies the load.

Vs = L·ΔI/t1   →  t1 = L·ΔI/Vs

Mode 2 (Q OFF, t1 < t < T): The inductor keeps the current flowing through C, the load and Dm. Energy stored in L is transferred to the output, which becomes negative. Inductor voltage = Va (negative); current falls from I2 to I1.

Va = −L·ΔI/t2   →  t2 = −L·ΔI/Va

Waveforms

Q   |‾‾‾‾|____|‾‾‾‾|____
    0   kT    T
vL  |+Vs |    |+Vs |
    |    |____|    |____ Va (negative)
iL  |   /\     /\        I2
    |__/  \___/  \___    I1
iD  |     |\_   |\_      (when Q OFF)

Average output voltage

Rise and fall of inductor current are equal in steady state:

Vs·t1/L = −Va·t2/L
Vs·kT = −Va(1 − k)T
Va = −Vs·k/(1 − k)

• k < 0.5: |Va| < Vs (buck)
• k = 0.5: |Va| = Vs
• k > 0.5: |Va| > Vs (boost)

Peak-to-peak ripple current of inductor

ΔI = Vs·t1/L = Vs·k·T/L
ΔI = Vs·k/(f·L)

Also, using t1 + t2 = T:

T = L·ΔI/Vs − L·ΔI/Va = L·ΔI(Va − Vs)/(Vs·Va)
ΔI = Vs·Va/(f·L(Va − Vs))

(Both forms give the same value since Va = −kVs/(1 − k).)

Capacitor ripple (C supplies Ia during t1): ΔVc = Ia·k/(f·C).

Average input current Is = Ia·k/(1 − k) (ideal). The circuit needs only one switch and gives polarity reversal, but the switch and diode carry high peak currents and input current is discontinuous.