Inverting Buck-Boost Converter Calculator

Why the output of an inverting buck-boost sits below ground

An inverting buck-boost turns a positive input into a negative output without a transformer and without a second inductor. Most of them start life as an ordinary buck regulator that has been rewired. Texas Instruments lays out the change in three steps in its application report on inverting buck-boost converters (SNVA856B, revised 2022): the buck output becomes system ground, the buck ground pins become the negative output, and the positive input stays where it was.

That move explains the sign. With the switch on, the inductor sits between the input and system ground and stores energy, while the output capacitor alone feeds the load. When the switch turns off, the inductor current has to keep flowing, so it takes the only path left, through the catch diode or the low-side switch, into the output. The far end of that path is the negative rail, so the output capacitor ends up charged with its positive plate at ground and its negative plate below ground.

Two things follow. No power travels straight from input to output, since every watt crosses the inductor, so the inductor current is always larger than the load current. And the regulator ground pin now floats at the negative output, so the part sees the input voltage plus the output voltage across its supply pins, not the input alone. Both facts drive every number on this page.

Analog Devices describes the same circuit in its note AN-1083 on inverting buck-boost design with the ADP2300 and ADP2301, which is a useful second check on the wiring and on the peak current formula.

Duty cycle sets the conversion ratio

In steady state the inductor charges and discharges by the same volt-second product. During the on time it sees Vin, and during the off time it sees the output, so Vin x D = |Vout| x (1 - D), which rearranges to |Vout| / Vin = D / (1 - D). Solved for the duty cycle, an ideal inverting buck-boost runs at D = |Vout| / (Vin + |Vout|).

Real parts lose energy, and Texas Instruments folds that into the same equation (SNVA856B, Equation 2): D = |Vout| / (|Vout| + efficiency x Vin). The efficiency term raises the duty cycle, because the converter has to pass more charge to cover its own losses.

Take 12 V in and 5 V out at 85 percent efficiency. The ideal answer is 5 / (12 + 5) = 5 / 17 = 0.294. With the loss term it becomes 5 / (5 + 0.85 x 12) = 5 / 15.2 = 0.329, about 12 percent higher. That second figure is what a real design runs at.

The ratio climbs past 1 when the output is bigger than the input. A 5 V input making a 5 V negative rail runs at 5 / (5 + 4.25) = 0.541, so it spends more than half of each cycle charging the inductor.

One check ties the two forms together: the conversion ratio should equal D / (1 - D) times the efficiency. For the 12 V case, 0.329 / (1 - 0.329) = 0.490, and 0.490 x 0.85 = 0.417, which matches 5 / 12 = 0.417.

The table holds the output at 5 V, the load at 2 A and the efficiency at 85 percent. A higher input lowers the duty cycle but does not lower the inductor current much, because that current depends on the ratio of output to input rather than on the input alone.

Average and peak inductor current

The load draws current only while the switch is off, so the inductor has to carry more than the load. Charge balance gives the average as Iout / (1 - D) for an ideal part, or Iout x (1 + |Vout| / (efficiency x Vin)) once losses are counted. Texas Instruments writes the second form as Equation 4 in the same report.

For 2.5 A at D = 0.329 the ideal average is 2.5 / (1 - 0.329) = 2.5 / 0.671 = 3.73 A. The efficiency form lands on the same figure from the other direction: 2.5 x (1 + 5 / 10.2) = 2.5 x 1.490 = 3.73 A.

Peak current adds half the ripple. With a 30 percent ripple target the swing is 0.75 A, so the peak is 3.73 + 0.375 = 4.10 A and the valley is 3.73 - 0.375 = 3.35 A. A 10 µH inductor at 400 kHz widens the swing to 0.99 A, which pushes the peak to 3.73 + 0.493 = 4.22 A.

Those peaks decide which regulator you can use, because the switch carries the inductor current rather than the load current. The TPS5430 note (SLVA257A, 2007) makes the trade plain: at D = 0.25 the usable load is 3 x (1 - 0.25) = 2.25 A. A 2.5 A load at D = 0.329 needs a peak limit above 4.2 A, so a part sold as a 3 A buck with a 4.8 A minimum peak limit is the practical floor.

Sizing the inductor

The inductor sets the ripple. During the on time the current rises at Vin / L, so the swing is the on time multiplied by that slope, and solving for L gives L = Vin x D / (frequency x ripple current). Texas Instruments suggests a swing between 20 and 40 percent of the load current, with 30 percent a common choice.

Back to the 12 V to 5 V design at 2.5 A and 400 kHz. A 30 percent swing is 0.75 A, and 12 x 0.329 = 3.95 V. Then 3.95 / (400,000 x 0.75) = 0.0000132, which is 13.2 µH. The nearest standard value is 10 µH, which raises the swing to 12 x 0.329 / (400,000 x 0.00001) = 0.99 A and the peak to 4.22 A.

Choose the part for saturation current, not just inductance. The inductor has to hold the peak without saturating, and the peak moves with input voltage, so size it at the lowest input in the range. Resistance matters too: 3.73 x 3.73 x 0.05 = 0.70 W, which is a large share of a 12.5 W output. The RMS current rating has to clear the same figure with margin.

The same 12 V to 5 V design at 2.5 A and 400 kHz, with only the ripple target moved, gives this short list of inductor options:

Every row keeps the average current at 3.73 A, so the peak column is that figure plus half the ripple. A looser ripple target buys a smaller inductor but asks the switch and the diode for more peak current, and a 40 percent swing already sits at 4.23 A.

Voltage stress on the switch, diode and capacitors

The regulator supply pins see the input plus the output, because the part ground sits at the negative rail. That stress is Vin + |Vout|: 17 V for 12 V to 5 V, and 24 + 15 = 39 V for 24 V to 15 V. Texas Instruments uses that second case to warn that a 36 V part is not enough and a 40 V part is the entry point.

Analog Devices shows the same limit from the data sheet side: the ADP230x is rated for 20 V between its input and ground pins, so a 12 V to 5 V design sits at 17 V with 3 V to spare, while a 15 V input making a 5 V rail would land exactly on the ceiling (AN-1083). Pick the part from that sum, not from the input alone.

The catch diode in a non-synchronous design blocks the same voltage, and the input capacitor between the input and system ground is charged to that sum, so its rating has to clear it with room for ceramic derating. The output capacitor only sees the output voltage, but it carries the chopped output current, so its ripple current rating matters as much as its capacitance.

The efficiency cost of inverting

Inverting costs efficiency. Texas Instruments reports that the LMR33630 reaches close to 95 percent converting 12 V to 5 V at 2 A, a loss near 0.53 W, and that the same part used as an inverting buck-boost can lose three to four times as much, landing near 85 percent.

Run the budget for 5 V at 2.5 A. The load takes 5 x 2.5 = 12.5 W. At 85 percent the input side supplies 12.5 / 0.85 = 14.7 W, so 14.7 - 12.5 = 2.2 W becomes heat, and the supply current is 12.5 / (12 x 0.85) = 12.5 / 10.2 = 1.23 A.

That heat comes from switch conduction, diode forward drop, inductor resistance and the switching edges. Since efficiency appears in both the duty cycle and the inductor current, the first pass at 0.85 feeds back into both, and one more pass with a measured figure is enough to close the design.

Stability changes as well. The topology brings a right-half-plane zero into the loop, which adds phase lag instead of phase lead. Texas Instruments notes that a smaller inductor pushes that zero higher and that a current mode part is easier to compensate.

Two designs from start to finish

The 5 V to 5 V design at 1 A shows the boost side of the family. The duty cycle is 5 / (5 + 4.25) = 0.541, the average inductor current is 1 x (1 + 5 / 4.25) = 2.18 A, and a 30 percent swing of 0.3 A puts the peak at 2.18 + 0.15 = 2.33 A. At 400 kHz the inductor works out to 5 x 0.541 / (400,000 x 0.3) = 0.0000225 H, so a 22 µH part, which is the value the TI example picks, along with a regulator whose minimum peak limit is 2.5 A.

Set the two cases side by side and the pattern shows. As the output grows toward the size of the input, the duty cycle moves toward 0.5, the inductor current climbs to roughly twice the load, and the switch stress stays at Vin + |Vout|. The 12 V to 5 V case needs 13.2 µH and a 4.8 A limit, while the 5 V to 5 V case needs 22 µH and 2.5 A.

The calculator on this page does not model thermal resistance, loop stability or the diode forward drop in detail. It gives the first-pass numbers that a datasheet check and a bench measurement then confirm.

Set the two worked examples against each other and the shared arithmetic shows up row by row:

The lower duty cycle of the 12 V case is the only line that looks easier. Its inductor current is 71 percent higher and its switch stress is 7 V higher, so the part that suits the 5 V design will not suit the 12 V one.