Dual Active Bridge Basics: Phase Shift, Power and Soft Switching

Each bridge produces a 50 % duty square wave. The primary square wave has amplitude V1; the secondary one, referred to the primary through the turns ratio n = N2/N1, has amplitude V2/n. The two square waves are shifted by a phase d, expressed here as a fraction of half the switching period (0 ≤ d ≤ 0.5). The series inductance L sees the difference of the two voltages, and the resulting current carries power from the leading bridge to the lagging one.

Voltage conversion ratio

M = V2 / (n · V1)

Transferred power (L referred to the primary)

P = d · (1 − d) · V1 · V2 / (2 · n · f_s · L)

Maximum power (at d = 0.5)

P_max = V1 · V2 / (8 · n · f_s · L)

Power rises with d up to d = 0.5 and falls beyond it. In practice designs stop well short of 0.5, because near the top of the curve a small change in d gives almost no extra power while the circulating current keeps growing.

Over one half period the inductor voltage takes only two values: V1 + V2/n while the bridges are in opposite states (duration d) and V1 − V2/n while they agree (duration 1 − d). Using half-wave symmetry, i(T/2) = −i(0), gives the current at the two switching instants directly.

Current when the primary switches

i(0) = −[V1 + (V2/n)·(2d − 1)] / (4 · f_s · L)

Current when the secondary switches

i(d) = [V1·(2d − 1) + V2/n] / (4 · f_s · L)

These two numbers are worth computing for every operating point. Their magnitudes are the peak currents the switches must turn off, and their signs decide whether each bridge achieves zero-voltage switching.

A bridge switches at zero voltage when the current at its switching instant flows in the direction that discharges the incoming switch, so its body diode conducts first. From the expressions above, ignoring switch capacitance and dead time:

Primary bridge ZVS

d > (1 − 1/M) / 2 (limits when M > 1)

Secondary bridge ZVS

d > (1 − M) / 2 (limits when M < 1)

At M = 1 both conditions hold for any positive phase shift. As M moves away from 1, one bridge needs a minimum phase shift, which means a minimum power, to keep ZVS. At light load the phase shift is small, so a DAB with M far from 1 loses ZVS on one side exactly where efficiency is already hardest to maintain. In a real converter the switch output capacitance must also be charged during the dead time, so the true ZVS boundary is somewhat narrower than these ideal limits.

These conditions are derived here for ideal switches. The DAB calculator on this site additionally accounts for the switches' parasitic capacitance when it checks ZVS.

When the two square-wave amplitudes differ, part of the inductor current only circulates between the bridges without carrying net power. The worked example below shows the effect: at the same 200 W, the peak current is almost four times higher when M = 0.75 than when M = 1. That reactive current costs conduction loss in the switches, the transformer and the inductor.

This is the main motivation for dual and triple phase shift modulation (DPS, TPS). By adding a phase shift inside one or both bridges, they shape the voltages so that less current circulates and ZVS is kept over a wider range. They add control complexity, and definitions differ between papers, so compare formulas carefully before mixing sources.

1. 1Fix the operating range: V1, the minimum and maximum of V2, the power range and the switching frequency.
2. 2Choose n so that M is close to 1 over the operating range where the converter spends most of its time. For a battery, that is usually the middle of the state-of-charge range rather than the extremes.
3. 3Choose the maximum phase shift d_max at full power. Values around 0.25 to 0.35 leave control headroom without excessive circulating current.
4. 4Compute L from the power equation at the corner that needs the most power with the lowest V2, because P_max falls in proportion to V2.
5. 5At every corner of the range, compute i(0) and i(d) to get peak currents and check both ZVS conditions. If light-load ZVS or RMS current is unacceptable, consider DPS or TPS.

1. 1With n = 400/48 ≈ 8.33, M = 1 at V2 = 400 V. Choosing d_max = 0.35 gives L = 0.35 · 0.65 · 48 · 400 / (2 · 8.33 · 100 kHz · 1 kW) ≈ 2.6 µH on the 48 V side.
2. 2At full power and M = 1, |i(0)| = |i(d)| ≈ 32 A. Both bridges have ZVS.
3. 3At 200 W and M = 1, d ≈ 0.048 and the peak current falls to about 4.4 A. Both bridges still have ZVS (in the ideal model).
4. 4Now let V2 drop to 300 V (M = 0.75). The maximum possible power falls to about 825 W, so 1 kW is no longer reachable with this inductance. At 200 W, d ≈ 0.065, which is below the secondary ZVS limit of (1 − 0.75)/2 = 0.125: the secondary bridge loses ZVS, and the primary peak current rises to about 16 A.

The last step is the important one. A design that looks comfortable at its nominal point can lose both power capability and soft switching at the edge of a battery's voltage range. The DAB calculator on this site lets you repeat these checks for your own operating points.