The DC link

Every other page here looks at the inverter from the machine’s side: three phases, a rotating vector, smooth torque. Turn round and look at the same bridge from the supply’s side and it is a switch that connects the link to a winding for part of each period and disconnects it for the rest.

So the supply does not see a current. It sees a train of pulses whose average is the power — and every departure from that average has to come from somewhere closer than the supply.

The operating point Depth Current

Where in the revolution Angle

Top: the link current across one PWM period, with the null vectors shaded — during those the machine is shorted to a rail and freewheeling, and the link supplies nothing at all. Bottom: the RMS current the capacitor has to carry, swept across modulation depth, with the operating point marked. Drag Depth and watch the dot climb away from the peak in both directions.

The mean is the power, and nothing else is

Integrating the switching sequence over an electrical revolution at 60% modulation depth and 6 A gives a mean link current of 3.12 A. The same operating point put through $p = \frac{3}{2} (v_{d} i_{d} + v_{q} i_{q})$ gives 74.8 W, and 74.8 W over 24 V is 3.12 A.

Those two calculations share nothing. One knows about duty cycles and dwell times; the other knows about dq quantities and a factor of 3/2. That they agree is what makes the rest of the page worth reading.

What the capacitor carries

The supply provides the average. It cannot provide anything faster — the leads alone have too much inductance — so every excursion above or below is sourced or sunk by the link capacitor. Since the mean and the ripple are orthogonal, the subtraction is exact rather than approximate:

I_{c a p} = I_{r m s}^{2} - I_{m e an}^{2}

At that same operating point the capacitor carries 2.71 A RMS while the supply provides 3.12 A. The component nobody specifies is doing nearly as much work as the one on the front panel.

And it is worst in the middle

Modulation depth	Supply mean	Capacitor RMS	Ratio
20%	1.04 A	2.09 A	2.01
40%	2.08 A	2.64 A	1.27
60%	3.12 A	2.71 A	0.87
80%	4.16 A	2.36 A	0.57
95%	4.94 A	1.67 A	0.34

The absolute RMS peaks around half output and falls away either side, which is not what anyone guesses. The reason is at both ends: near full modulation the bridge is connected to the link for most of the period, so the draw is nearly continuous and there is little for the capacitor to smooth. Near zero it is barely connected at all, so there is little current to smooth either. The worst case is in between, where the pulses are both large and intermittent.

A drive specified only at full load is specified at the wrong point.

Sag is a modulation limit

The volts on the front panel are not the volts at the bridge. At 3.12 A through half an ohm of leads, connectors and supply impedance, 1.56 V disappears — 6.5% of a 24 V link, and 6.5% straight off the modulator’s linear range.

That is a real reason a drive that behaves on the bench misbehaves on a long cable, and it is why the bus voltage belongs in the measurement path rather than as a constant in the firmware. The modulator scales its duties by whatever $V_{d c}$ actually is; feed it a number from a datasheet and every duty is wrong by the sag.

Ripple, and which term leads

The capacitor supplies the ripple current for roughly half a switching period between refills, so it dips by $I Δ t / C$ , and its ESR adds a step of $I R$ on top. On a 470 µF part with 40 mΩ of ESR at 16 kHz, carrying that 2.71 A:

Term	Contribution
charge, $I /2 f_{pw m} C$	180 mV
ESR, $I R$	109 mV

The charge term leads here — but the ESR term is 38% of the total, which is enough that choosing a capacitor on capacitance alone gets it wrong. Worth computing rather than assuming; the ranking flips with part size and switching frequency.

Braking has to put the energy somewhere

Slowing a spinning rotor means removing its kinetic energy. It comes out through the inverter into the link, and the link has exactly three places to put it: back into the supply, into the capacitor as a voltage rise, or into a resistance as heat.

Most bench supplies will not take current in. That leaves two.

The reference machine at 800 rpm holds 42.1 mJ. A 470 µF link at 24 V has 76.1 mJ of room before 30 V, so one stop is comfortable — the bus settles at 27.5 V.

Now double the speed. Kinetic energy goes as $ω^{2}$ , so 1600 rpm holds 168 mJ against the same 76 mJ of headroom, and the link reaches the trip before the rotor reaches standstill.

On the bench The constraint is usually the supply, not the over-voltage trip. A laboratory PSU is a source, not a sink: push current back into it and the link simply rises until something objects. That reframes the fix — there is no braking without somewhere to put the joules, and adding a bigger capacitor buys a fixed number of millijoules rather than a capability.

The cheap answer on a drive with no brake chopper is to make the machine itself the resistance. Injecting d-axis current adds copper loss without adding torque: 6 A of $i_{d}$ burns $\frac{3}{2} R_{s} i_{d}^{2}$ = 18.9 W, which consumes the whole 42 mJ in 2.2 ms. The energy never reaches the link at all.

It costs heat in the windings and it is the difference between a controlled stop and a fault.

What to take away

The mean link current is the power over the bus volts, and nothing else about the switching changes that.
The capacitor carries $I_{r m s}^{2} - I_{m e an}^{2}$ , exactly. At a typical operating point that is the same order as the supply current.
It peaks near half modulation depth. Specifying the capacitor at full load under-specifies it.
Sag is a modulation limit. Measure $V_{d c}$ ; do not assume it.
Ripple is a charge term plus an ESR term, and which leads depends on the part. Compute both.
Braking needs a destination for the energy. The supply usually will not take it, the capacitor holds only millijoules, and kinetic energy goes as speed squared. Injected loss turns the machine into the brake resistor.