The bridge

Everything upstream of this page produces a number: a voltage the controller would like the machine to see. The bridge is where that number stops being a number. It has six switching devices and exactly one way to apply a voltage — connect each phase to one rail or the other, and change which one fast enough that the machine only feels the average.

One leg, two devices, never both

Each phase hangs off the midpoint of a leg. The high-side device connects it to the positive rail, the low-side device to the negative one, and the pair is driven complementarily: when one is on the other must be off. Turning both on connects the rails to each other through two transistors, which is a short circuit across the DC link and destroys the leg.

So a leg has two useful states, and three phases give $2^{3} = 8$ combinations — the six active states that produce a voltage, and the two null states where all three phases sit on the same rail and the machine sees nothing.

Drive Torque demand DC link

Sweep

The playhead crosses one 50 µs period about every four seconds. Stop holds the model as well as the playhead; Step jumps to the next switch state, and a full lap of them is one PWM period.

The loop from the landing page, running, with its bridge exposed. Left: the six devices, with whichever element is carrying the current filled in. Right, top: the gate pattern over one PWM period, with the vector each interval produces. Right, bottom: the phase currents that result — note that these run on a far slower time base than the gate pattern above them. Hold the playhead to read a single switch state.

From a voltage request to three duty cycles

The controller never asks for a switch state. It asks for a voltage, and the modulator has to turn that into three numbers. The whole calculation is two lines of algebra, and this is the shape of it:

Tap to expand Close ✕

The upper path is the arithmetic, left to right. The dashed return is the point: the common-mode term injected in the second block reaches every duty cycle and every gate signal, and then cancels in the voltage across the windings — which is why it is free to spend.

Over one period, phase $a$ spends the fraction $d_{a}$ tied to the positive rail and $1 - d_{a}$ tied to the negative one. Measured against the midpoint of the DC link, its average is

v_{a 0} = V_{d c} (d_{a} - \frac{1}{2})

which inverts directly. To apply a pole voltage $v_{a 0}$ , ask for

d_{a} = \frac{1}{2} + \frac{v _{a 0}}{V _{d c}}

and the same for $b$ and $c$ . A duty cycle is a real fraction, so $0 \leq d_{a} \leq 1$ , which caps the pole voltage at $\pm V_{d c} /2$ — on a 48 V link, $\pm 24$ V.

The neutral is floating, and that is worth 15%

Take the naive route and set each pole voltage equal to the phase voltage you wanted. That is sinusoidal PWM, and it works, but it throws something away.

The machine’s star point is not connected to anything. Current is driven by the voltage differences between phases, so adding the same term $v_{0}$ to all three pole voltages changes nothing the machine can feel — the star point simply floats up to follow it. That gives a free parameter:

d_{x} = \frac{1}{2} + \frac{v _{x}^{*} + v _{0}}{V _{d c}}, x \in {a, b, c}

Spend it on centring. Choose $v_{0}$ so the largest and smallest of the three references sit symmetrically about zero:

v_{0} = - \frac{max ( v _{a}^{*} , v _{b}^{*} , v _{c}^{*} ) + min ( v _{a}^{*} , v _{b}^{*} , v _{c}^{*} )}{2}

That is min-max injection, and it is space-vector PWM — algebraically identical to the textbook sector-and-dwell-time construction, at the cost of a few comparisons instead of two trig evaluations.

Worked, with the site’s own numbers

Ask for 20 V at 20° on a 48 V link. The three phase references are $v_{x}^{*} = 20 cos (θ - 120° k)$ :

v_{a}^{*} = 18.79 V, v_{b}^{*} = - 3.47 V, v_{c}^{*} = - 15.32 V

The largest is 18.79 and the smallest is −15.32, so

v_{0} = - \frac{18.79 + ( - 15.32 )}{2} = - 1.736 V

and the duties follow from one subtraction and one division each:

d_{a} = \frac{1}{2} + \frac{18.79 - 1.736}{48} = 0.855, d_{b} = 0.391, d_{c} = 0.145

Plain sinusoidal PWM, skipping the injection, would have asked for 0.892, 0.428 and 0.181 — further from the rails at the top, which is exactly the headroom it is wasting.

Why “the common mode cancels”

Now check the injection cost nothing. The pole voltages those duties produce — each phase measured against the midpoint of the DC link — are

17.06 V, - 5.21 V, - 17.06 V

and not one of them is what was asked for. Every one is low by 1.736 V, which is $v_{0}$ exactly. That is what makes the term common mode: it is not three errors, it is one number added to all three.

And a number added to all three survives no subtraction. The voltage that drives current in phase $a$ is $v_{a} - v_{b}$ , not $v_{a}$ on its own, because the star point is connected to nothing and current can only go out of one phase and back through another. Every difference is unchanged:

(v_{a} + v_{0}) - (v_{b} + v_{0}) = v_{a} - v_{b}

so the machine cannot tell the difference. On these numbers, the pole voltages differ by $17.06 - (- 5.21) = 22.27$ V and the original references differ by $18.79 - (- 3.47) = 22.27$ V — the same figure. The star point simply floats 1.736 V below the link midpoint and everything across the windings carries on as if the injection had never happened.

Take those differences and the phase voltages the machine sees come back to

18.79 V, - 3.47 V, - 15.32 V

exactly the request. The common mode cancels, as promised.

What the centring buys

Both schemes run out when a duty reaches a rail. Plain SPWM reaches it when the phase peak hits $V_{d c} /2$ ; centred, the three references only reach the rails when the peak hits $V_{d c} / 3$ :

\frac{V _{d c} / 3}{V _{d c} /2} = \frac{2}{3} = 1.1547

On this 48 V link that is 27.713 V instead of 24.000 V — 15.47% more voltage from the same hardware, which is free torque-speed range and the reason essentially nothing ships with plain SPWM.

On the bench Hold the playhead near the middle of the period and watch which devices are on: all three phases sit on the same rail for a good fraction of every period. That is the null vector, and it is not wasted time — it is what lets the modulator place the two active vectors symmetrically and keep the switching losses even.

The gate says which device, the current says which element

A gate signal enables a transistor. It does not decide what carries the current — the sign of the phase current does that.

With the high side on and current flowing out of the leg into the winding, the current goes through the high-side transistor. With the high side on and current flowing into the leg, it has nowhere to go but backwards through the diode across that same transistor. The leg’s voltage is the same either way, which is why the averaged model upstream can ignore the distinction.

It stops being ignorable the moment both devices are off — which is exactly what a safety interlock does on every switching edge, and what deadtime is about.

It also decides what the bridge costs you in heat, which is the rest of this page: a transistor carrying current dissipates $i^{2} R$ , a diode carrying the same current dissipates $i V_{f}$ , and the two behave nothing alike.

What the conducting element costs

Everything the rest of this site wants from a higher PWM frequency — smaller current ripple, a shorter transport delay, a wider sampling window — is bought with heat.

There are two mechanisms and they respond to frequency in opposite ways. That opposition is the whole design tension of the inverter, and it has a crossing point that is worth knowing.

The devices R_ds(on) Switching

The drive Current PWM

The loss budget stacked against switching frequency, for a 24 V bridge. Conduction is a floor that no frequency can lower — it is flat all the way across. Everything above the dashed crossover is bought. Drag R_ds(on) down to fit a better transistor and watch the crossover move left.

Conduction does not care how often you switch

With complementary switching, one device in each leg is conducting at all times. The duty cycle decides which one, not whether — so the phase current is always flowing through an on-resistance, and the answer contains no duty at all:

P_{co n d} = 3 I_{r m s}^{2} R_{d s (o n)}

At 6 A peak that is 4.24 A RMS per phase, and on 10 mΩ devices 0.54 W for the bridge. Doubling the switching frequency does not change it by a microwatt.

Switching is paid once per transition

During a transition the device carries current while standing off voltage. Taking both as ramping linearly across the switching time gives

E_{s w} = \frac{1}{2} V_{d c} i t_{s w}

which is crude — within a factor of two of a datasheet’s measured figures — and right about every dependency that matters. Each leg turns on once and off once per period, so the loss is proportional to frequency, to the bus voltage, to the current, and to how slowly the device is driven.

On the bench Switching energy is linear in current, so averaging it over the electrical cycle takes the mean of $∣ sin ∣$ — 0.6366 — and not the RMS. Conduction loss is square-law and takes the RMS.

Reaching for the RMS out of habit overstates the switching term by 11%, since $0.7071/0.6366 = 1.111$ . That errs safe, so it will not burn anything, but the habit is worth breaking: square-law loss takes the RMS, linear loss takes the mean. The deadtime diode term below is linear too, for the same reason.

Where they cross

PWM	Conduction	Switching	Deadtime diode	Total
8 kHz	0.54 W	0.22 W	0.07 W	0.83 W
16 kHz	0.54 W	0.44 W	0.15 W	1.13 W
32 kHz	0.54 W	0.88 W	0.29 W	1.71 W
64 kHz	0.54 W	1.76 W	0.59 W	2.89 W

For this bridge the frequency-dependent terms overtake conduction at 14.7 kHz. Below that, frequency is nearly free — the bridge is a resistor and you may as well have the control bandwidth. Above it, every doubling of the frequency doubles the dominant term.

At the rig’s 16 kHz the total is 1.13 W against 25.6 W of shaft power: 4.4%, which is a reasonable place to be.

The trap: a better transistor lowers the crossover

Fit a device with half the on-resistance and conduction loss halves. Switching loss does not move at all — it is set by the transition time and the bus, not by $R_{d s (o n)}$ .

So the frequency at which switching takes over falls:

$R_{d s (o n)}$	Crossover
5 mΩ	7.4 kHz
10 mΩ	14.7 kHz
30 mΩ	44.2 kHz

This is the opposite of the intuition that better silicon buys more headroom everywhere. It buys a lower floor, and it makes the switching term dominant sooner. A bridge built from excellent devices and run at 60 kHz can dissipate more than one built from mediocre devices at 16 kHz.

If you want frequency, the thing to buy is faster switching — lower gate charge, better gate drive — not lower on-resistance. They are different specifications and they solve different halves of this plot.

The deadtime term is linear

While both devices are off, the body diode carries the current at a roughly fixed forward drop. A voltage, not a resistance — so the loss is linear in current rather than square, and it scales with the deadtime and the frequency:

P_{d t} = 3 \overline{∣ i ∣} V_{f} \cdot 2 t_{d t} f_{pw m}

At 500 ns of deadtime and 16 kHz it is 0.15 W here — small, but it grows with frequency alongside the switching term, and it is the term that matters most at low current, where the square-law terms have faded away.

That is a second reason to keep the deadtime as short as the gate drive safely allows, on top of the distortion it causes.

What to take away

A leg has two devices and two useful states. Both on is a short across the link, so the interlock is not negotiable.
A duty cycle is a pole voltage: $d_{x} = \frac{1}{2} + v_{x 0} / V_{d c}$ .
The star point floats, so the same term added to all three references cancels in every difference. Spending that freedom on centring is worth 15.47% more voltage from the same hardware, and it is space-vector PWM.
The gate picks the device; the sign of the current picks the element. That distinction is invisible to the averaged model and decides both the deadtime error and the losses.
Conduction loss is $3 I_{r m s}^{2} R_{d s (o n)}$ and is completely flat in frequency. It is a floor. Switching loss is linear in frequency, bus volts, current and transition time.
Square-law loss takes the RMS; linear loss takes the mean of $∣ sin ∣$ . Using the wrong average is an 11% error.
A lower-resistance device moves the crossover down. To buy frequency, buy faster switching, not lower $R_{d s (o n)}$ .