= H-bridge
{c}
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.<Brushed DC Motor 2: The H-Bridge by Hector Hernandez>[Slides 2–7]{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 $V_d$. Then
$$
v_A=q_AV_d,\qquad v_B=q_BV_d,\qquad v_{AB}=(q_A-q_B)V_d.
$$
Opposite 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 $v_{AB}=-12\,\mathrm V$ gives
$$
\frac{di}{dt}=\frac{v_{AB}}L=-1200\,\mathrm{A\,s^{-1}}.
$$
Current 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 $Li^2/2=0.020\,\mathrm J$.
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.
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