Buck-Boost Converter Designer

Buck-Boost Converter Designer

Design an inverting buck-boost converter, which makes a negative output from a positive input: duty cycle, inductance, peak and RMS currents, switch voltage stress, output capacitance and ESR, the right-half-plane zero and standard parts.

Inverting buck-boost power stage

Vin, −Vout, Iout, fsw → L, C, stress
Ripple is largest at the maximum input, peak current at the minimum — see the chart.
Enter 5 for a −5 V rail. The output of this topology is always negative with respect to ground.
The full-load current (magnitude).
The inductor carries input plus output current. 20–40% is typical.
0 = ideal or synchronous. Use the figures from your part’s datasheet; typical values vary widely between manufacturers.
0 = ideal. For a MOSFET, current × RDS(on).
Inverting buck-boost: with the switch on, the inductor charges from Vin; with it off, the inductor pulls current up from the output through the diode, so the output sits below ground. The dots show average currents. The real switch chops at the switching frequency, so each current is a ripple around these averages, not a steady flow.
16.61µHExample

12 V in, −5 V out at 1 A, 500 kHz, 30% ripple, 25 mV output ripple, ideal switches

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Inverting buck-boost power stage (CCM)

D = (|Vo| + Vd) ÷ (Vin − Vsw + |Vo| + Vd); IL = Io ÷ (1 − D); L = (Vin − Vsw)·D ÷ (f·ΔIL); Vstress = Vin + |Vo|; Cmin = Io·D ÷ (f·ΔV); fRHPZ = (1 − D)²·R ÷ (2π·D·L)
|Vo|
output voltage magnitude; the output itself is −|Vo|
D
duty cycle; with both drops at 0 it is |Vo| ÷ (Vin + |Vo|)
IL
average inductor current = input current + output current
R
load resistance at full load, |Vo| ÷ Io

Worked example

12 V in, −5 V out at 1 A, 500 kHz, 30% ripple, 25 mV output ripple, ideal switches
D = 5 ÷ (12 + 5) = 29.41%
IL = 1 ÷ (1 − 0.2941) = 1.417 A (of which 0.417 A from the input)
ΔIL = 0.30 × 1.417 = 0.425 A
L = 12 × 0.2941 ÷ (500,000 × 0.425) = 16.61 µH → 18 µH (E12, rounded up)
Switch and diode each block 12 + 5 = 17 V; Cmin = 23.53 µF → 33 µF; RHP zero 74.9 kHz

The inverting buck-boost converter

This page designs the classic inverting buck-boost: one switch, one diode, one inductor. While the switch is on the inductor charges from the input; when it turns off, the inductor’s current flows up out of the output capacitor through the diode, pulling the output below ground. The magnitude of the output can be above or below the input — D = |Vout|/(Vin + |Vout|) ideally, so 50% duty gives −Vin. A step-down IC used this way (TI SNVA856A) has its ground pin on the negative rail, so it sees Vin + |Vout| across it: 17 V in the example, not 12 V. Set the diode and switch drops to 0 for an ideal (synchronous, lossless) design; enter them to see how a Schottky or a MOSFET’s I·RDS(on) drop stretches the duty cycle.

Signs. Enter the output as a magnitude; the page prints the output voltage as negative and every current as a magnitude. Feedback for a negative rail needs care: a controller that regulates a positive feedback voltage against its own ground either has to sit on the negative rail or needs an inverting level shift.

Not the four-switch buck-boost. The non-inverting four-switch buck-boost used in battery-powered equipment has a positive output and runs as a buck, a boost or a mix of both depending on the input; its inductor current, duty cycles and component stresses are different, and these formulas do not apply to it. SEPIC and Ćuk converters also give either polarity, with other equations.

Currents. The inductor carries the input and output currents together, Iout/(1 − D). Both terminal currents are pulsed — the input current flows only while the switch is on, the output current only while the diode conducts — so both capacitors carry substantial RMS current, shown above. The chart shows the trade-off across input voltage with the E12 inductor: ripple rises with the input, peak current rises as the input falls.

Right-half-plane zero. Like the boost, this converter has an RHP zero, at (1 − D)²R/(2πDL) (Erickson & Maksimović, Table 8.2); it limits the loop bandwidth. The value here was checked against the zero of the linearised averaged circuit, not only the textbook formula. For positive outputs see the buck converter designer and the boost converter designer; for switch losses, the MOSFET loss calculator.

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Frequently asked questions

Why is the output of a buck-boost converter negative?

In the inverting buck-boost the inductor, charged from the input, releases its current through the diode out of the output capacitor, so the output is driven below ground. The non-inverting four-switch buck-boost is a different circuit.

What is the duty cycle of an inverting buck-boost converter?

|Vout| ÷ (Vin + |Vout|) for an ideal converter. 12 V to −5 V needs 29.41%; 50% gives an output equal to −Vin.

What voltage rating do the switch and diode need?

At least Vin + |Vout| — 17 V for 12 V in and −5 V out — plus margin for ringing.

What is the right-half-plane zero of a buck-boost converter?

(1 − D)² × R ÷ (2π × D × L), with R the load resistance. For the example with the 18 µH inductor at full load, 74.9 kHz.

Related calculators

References

  1. Erickson RW, Maksimović D. Fundamentals of Power Electronics, 3rd ed. Springer, 2020. Ch. 2 (inductor volt-second and capacitor charge balance, ripple), Ch. 5 (the CCM–DCM boundary), Ch. 8, Table 8.2 (right-half-plane zero: D′²R/L for the boost, D′²R/(DL) for the buck-boost).
  2. De Stasi F. Working With Inverting Buck-Boost Converters. Texas Instruments application report SNVA856A, revised May 2020. Duty cycle, IL = IIN + IOUT and the VIN + |VOUT| stress on the IC.
  3. Ridley R. Buck-Boost Converter with Voltage-Mode Control (Design Center article 019). Ridley Engineering. Loop bandwidth limited to about a fifth of the RHP-zero frequency.
  4. IEC 60063:2015. Preferred number series for resistors and capacitors. The E6, E12 and E24 series used for the suggested standard parts.