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Dead-time output-voltage error is the difference between the intended switching-cycle average electric potential difference and the actual average produced when nonoverlap intervals alter the terminal waveform. Bridge switching dead time reduces H-bridge shoot-through, but during it an inductive load still carries electric current. Its recirculation path helps determine which rail clamps the output. The primary manufacturer discussion shows the bipolar/unipolar switching states and diode paths during that interval.^[ref]

An original event-accounting model avoids assuming a universal sign. Let an ideal half-bridge output take values 0 and $V_d$ over period $T$. During disjoint affected intervals of lengths $\Delta t_k$, suppose the intended normalized level is $q_k$ and the actual clamped level is $c_k$, each either 0 or 1. ThenAn interval replaced by the same rail contributes no error. An intended high interval replaced by low contributes negatively; the reverse contributes positively. Device drops require additional voltage terms.

For a stipulated single-pulse switching model, assume the load current keeps one sign throughout a cycle and that ideal recirculation clamps the midpoint to the low rail for current leaving the midpoint, or to the high rail for current entering it. With one effective high interval shortened or extended by $t_d$,For $V_d=24\mathrm{V}$, $T=50\mu \mathrm{s}$ and $t_d=100\mathrm{ns}$, the magnitude is $48\mathrm{mV}$. The corresponding ideal duty error magnitude is $t_d/T=0.002$. In a full H-bridge, subtract the two measured midpoint averages; their errors can reinforce or partly cancel according to their states.

Current-polarity dead-time compensation would add an opposing correction in that specified model. Near a current zero crossing, however, the sign may change within the interval, and the assumed clamping model can fail. Blindly applying the same signed offset throughout a crossing need not correct the actual waveform.

A proposed low-voltage test records both switch commands, actual midpoint potential and load electric current, integrates the difference from the intended waveform over a cycle, and compares it with the event sum. Repeat across current directions and pulse widths. Separate switching delay, diode drops and uncertain timing rather than fitting every discrepancy to one nominal dead-time number. The formulas above are illustrative switching models, not specifications for a particular driver.

An H-bridge is a four-switch arrangement that can apply either polarity of a DC supply across a two-terminal electrical load. Two switches form each side, with the load joining their midpoint nodes. Power inverters and bidirectional electric motor drives use this switching geometry. The primary TI lesson illustrates its components, forward/reverse states and inductive-current paths.^[ref]

In an idealized switching model, let $q_A,q_B\in \{0,1\}$ select the lower or upper rail at midpoint A and B, with supply magnitude $V_d$. ThenOpposite choices give $+V_d$ or $-V_d$ across the load. Equal choices give zero terminal difference even though both midpoints can be at the upper rail. This zero-output state does not mean every switch is open.

Each leg's upper and lower devices must not simultaneously provide a low-resistance path between supply rails: that is H-bridge shoot-through. Bridge switching dead time introduces a nonoverlap interval during a transition. Inductive load electric current still needs a path during that interval, often involving diodes or controlled conducting devices. The actual current path depends on the bridge state and current direction.

An original inductor example shows why reversing the commanded terminal polarity need not instantly reverse electric current. Neglect winding electrical resistance and opposing motor electromotive force, and take $L=10\mathrm{mH}$ with initial current $i=2\mathrm{A}$. Applying $v_{AB}=-12\mathrm{V}$ givesCurrent reaches zero only after about $1.667\mathrm{ms}$ if this voltage is maintained. Initially $v_{AB}i=-24\mathrm{W}$: the inductive load is returning stored energy through the bridge rather than absorbing it. Its initial magnetic energy is $L{i}^2/2=0.020\mathrm{J}$.

During rapid ideal bipolar switching, applying $+V_d$ for fraction $d$ and $-V_d$ for the rest gives interval-average output $(2d-1)V_d$. With $V_d=12\mathrm{V}$ and $d=0.75$, the average is $6\mathrm{V}$. This average does not specify instantaneous current or heating without a load model.

A proposed low-voltage observation records both midpoint potentials, differential load electric potential difference and load electric current. Comparing them through reversal distinguishes a changed switch command from changed current direction. Driver delays, device drops, recirculation and the supply's ability to accept returned energy determine how closely a real bridge follows the ideal states.

H-bridge shoot-through is unintended conduction directly between the supply rails through devices in the same bridge leg. It differs from the intended load-current path through opposite legs of an H-bridge. Overlapping device conduction during switching can create that rail-to-rail path even when the intended logical states never command it. The manufacturer application note explicitly identifies cross-conduction as shorting the supply and introduces nonoverlap timing to prevent it.^[ref]

An original lumped fault-loop model shows why a short overlap can matter. Let a previously unenergized parasitic loop have electrical resistance $R$, inductance $L$ and constant supply electric potential difference $V_d$. During overlap,soFor $t\ll \tau$, the initial ramp is $i\simeq V_{d}t/L$; the steady resistive value $V_d/R$ is not reached instantaneously.

Take stipulated $V_d=24\mathrm{V}$, $R=0.10\Omega$ and $L=100\mathrm{nH}$, giving $\tau =1\mu \mathrm{s}$. An overlap lasting $100\mathrm{ns}$ reaches about $22.84\mathrm{A}$. The supply deliversSome of this energy remains in the loop's magnetic field at turn-off; it should not all be labeled heat dissipated in one switch. A $500\mathrm{ns}$ overlap in the same ideal model reaches about $94.43\mathrm{A}$, illustrating the nonlinear dependence on duration. Device and supply behavior can invalidate these constant-parameter assumptions.

Bridge switching dead time addresses timing overlap by allowing one conduction path to cease before the complementary one starts. An inductive load still needs a recirculation path while the controlled devices are off. TI's primary bridge lesson illustrates current paths during that interval.^[ref]

A proposed diagnostic compares actual device control waveforms and supply-current transients with load current during transitions, using a limited-energy test arrangement. A large load current by itself does not identify shoot-through; the timing and path establish the mechanism. The calculated fault-loop pulse is an original model, not an asserted device rating.

Hector Hernandez is the credited presenter and preparer of the TI Precision Labs training material Brushed DC Motor 2: The H-Bridge by Hector Hernandez. Its scientific subject is the H-bridge: a switching network that applies either polarity to a motor and provides paths for the electric current already stored in the inductive load. The primary title slide gives his name exactly as used here.^[ref]

The material connects a controller's commands to MOSFET gate drive, then distinguishes forward/reverse conduction from recirculation. That distinction matters because reversing a command does not instantaneously reverse the electric current in a motor's inductance. Slides 5–7 show electric current decay paths through diodes or conducting switches.^[ref]

For an original bridge-loss example, take a $12\mathrm{V}$ supply and a stated operating electric current of $2\mathrm{A}$ through two conducting switches, each with electrical resistance $0.05\Omega$. Their combined conduction loss isand their combined potential drop is $2IR=0.20\mathrm{V}$. The motor terminals receive $11.8\mathrm{V}$ in the chosen conducting state, giving $23.6\mathrm{W}$ of electrical input from $24\mathrm{W}$ supplied. This is not a calculation of mechanical output: motor winding loss, induced electric potential difference and mechanical losses remain separate.

A bench comparison records terminal electric potential difference and motor electric current together while changing the bridge state. The waveforms separate voltage reversal from electric current reversal and expose the recirculation interval discussed in the training material. The numerical values above are illustrative calculations, not performance measurements attributed to Hernandez.

A power inverter converts electrical power from a direct current input to an alternating current output. Its switches reverse or modulate the connection to the DC supply, while its control and filtering determine the output waveform. It transfers energy from its input; the switching operation itself is not an energy source. The DOE overview explains the switching-and-filtering principle and distinguishes equipment designed for independent operation from equipment relying on a grid reference.^[ref]

An ideal H-bridge uses four controlled switches around a two-terminal load. One diagonal pair applies $+V_d$ across the load, and the opposite pair applies $-V_d$. In a stipulated square-wave mode with equal half-periods,Its root mean square magnitude is $V_d$, but it is not sinusoidal. Its Fourier series isThe fundamental RMS magnitude is $2\sqrt{2}{V}_d/\pi$. Thus a $48\mathrm{V}$ input gives a $48\mathrm{V}$ RMS ideal square-wave output, but only about $43.22\mathrm{V}$ RMS in the fundamental. Equal total RMS does not imply equal waveform or equal response of every electrical load.

For a $12\Omega$ purely resistive load directly across this ideal square-wave output, mean active power is $V_d^2/R=192\mathrm{W}$. The fundamental contributes $8V_d^2/({\pi }^{2}R)\approx 155.63\mathrm{W}$ and the other harmonic components supply the remainder. The harmonic active power sum accounts for that difference. Adding a filter changes the circuit; the directly connected resistor calculation should not then be reused unchanged.

Rapid switching also permits averaged control. If a short switching interval applies $+V_d$ for fraction $d$ and $-V_d$ for the rest, its average output is $(2d-1)V_d$. Choosing $d(t)=(1+m\sin \omega t)/2$, with $0\le m\le 1$, makes that local average sinusoidal. The actual waveform still contains switching components that the load and filter must handle.

A proposed low-voltage demonstration records DC input electric potential difference and electric current, output waveform and load electric current, comparing average input and output power. Output frequency, total RMS, fundamental RMS and waveform distortion answer different questions. For islanded microgrid operation, control of a usable local supply is an additional capability, not a consequence of DC-to-AC conversion alone.