Switching Device Load Locus Calculator
Switching Device Load Locus Calculator
The path a hard-switched device traces in the voltage-current plane over one cycle, drawn against the safe operating area you type in from its datasheet. At turn-off the current holds while the voltage rises, then falls at full voltage — out to the top-right corner. At turn-on the opposite diode’s reverse recovery adds a current spike while the voltage is still up. Both corners are where devices die.
Switching load locus against SOA
400 V bus, 30 A load, 60/40 ns turn-off and 30/50 ns turn-on, 25 nH of stray loop inductance, a 300 nC soft-recovery diode, and SOA corners of 120 A, 20 kW and 600 V
The trajectory, and the energy it costs
Eoff = ½·IL·(Vbus·trv + Vpk·tfi)
Eon = ½·(Vbus − LσIL/tri)·IL·tri + Vbus·(IL·t2 + Qrr) + ½·Vbus·IL·tfv
with a snubber: v(t) = ILt² ÷ (2Cstfi), so Eoff = IL²tfi² ÷ (24Cs)
- Lσ
- stray inductance of the commutation loop: device, DC-link capacitor and the copper between them
- S
- recovery softness, tb ÷ ta. The same charge in a snappier diode makes a taller spike
- t2
- the whole reverse-recovery interval, ta + tb. The device carries the LOAD current at full bus voltage for all of it, as well as the recovery current — which is the term the usual Vbus·Qrr shorthand leaves out
- Cs
- snubber capacitance. Above ILtfi ÷ (2Vbus) the device current reaches zero before the voltage reaches the bus
Worked example
400 V bus, 30 A load, 60/40 ns turn-off and 30/50 ns turn-on, 25 nH of stray loop inductance, a 300 nC soft-recovery diode, and SOA corners of 120 A, 20 kW and 600 V
The current falls in 40 ns, so di/dt is 750 MA/s and the stray inductance adds 18.75 V: the device sees 418.8 V while it still has 30 A in it
At turn-on di/dt is 1 GA/s, so the diode's 300 nC comes back as Irr = √(2 × 300 nC × 1 GA/s ÷ 2) = 17.32 A, and the device peaks at 47.32 A with the bus still across it
Instantaneous power: 12.56 kW at the turn-off corner, 18.93 kW at the turn-on spike
Your SOA allows 47.76 A at 418.8 V and 50 A at 400 V. The margins are 1.592× and 1.057×, so the worst point is the turn-on spike at 1.06×
Energy: 611.3 µJ at turn-off and 1.004 mJ at turn-on, which at 20 kHz is 32.31 W. Of the turn-off figure, 11.25 µJ is the energy that was sitting in the stray inductance and had nowhere else to go
A 2.2 nF snubber is above the critical 1.5 nF, so the current reaches zero at 272.7 V — well short of the bus — and the device's turn-off energy drops to 27.27 µJ, moving 11.68 W out of the die. It costs 7.04 W in the snubber resistor and adds 40 A of discharge current at turn-on
Where the locus goes, phase by phase
| Phase | Voltage | Current | Why it matters |
|---|---|---|---|
| Turn-on, current rise | bus minus L·di/dt | 0 up to the load current | the stray inductance helps here: it takes voltage off the device while the current builds |
| Turn-on, reverse recovery | still at the bus | load current plus the recovery spike | the peak of the whole cycle in many designs. It is the opposite diode’s property, it grows with di/dt and roughly doubles from 25 °C to 125 °C |
| Turn-on, voltage fall | bus down to I·RDS(on) | the load current | the last of the turn-on energy |
| On-state | I·RDS(on) | the load current | sits on the SOA’s resistance line by definition. Conduction loss, not an SOA question |
| Turn-off, voltage rise | up to the bus | still the load current | the freewheel diode cannot take the current until the voltage has arrived, so the device holds both |
| Turn-off, current fall | bus plus L·di/dt | load current down to zero | the top-right corner. The stray inductance’s whole stored energy is delivered into the device here |
| Ring-down | decaying back to the bus | zero | the loop inductance and the device’s output capacitance ring. An RC snubber’s resistor is chosen to damp this, which is a separate job from reshaping the trajectory |
Why the locus goes to the corner, and what to do about it
Plot a switch’s voltage on one axis and its current on the other, and the path it traces over a switching cycle is its load locus. For a resistive load that path is a straight line between the on and off points and nothing much happens. For the clamped-inductive cell — which is what every motor bridge, every buck converter and every half bridge is — the path goes out to the top-right corner, and it is worth understanding exactly why.
At turn-off the inductor will not let the current change. The gate goes low, the device starts to come out of conduction, and the voltage across it rises — but the freewheel diode cannot pick the current up until that voltage has reached the bus. So for the whole voltage rise the device carries the full load current, and by the end of it the device is holding the full bus voltage and the full load current at the same time. Only then does the current start to fall, and as it falls the stray loop inductance adds L·di/dt on top of the bus. The trajectory goes right and then down, round the outside of the plot. Every joule that was stored in the stray inductance is delivered into the device on the way.
At turn-on the danger comes from the other device. The current rises into the switch, and when it reaches the load current the freewheel diode should stop conducting — but it cannot, until the stored charge in its junction has been swept out. During that interval the diode is still a short circuit, so the switch carries the load current plus the recovery current while the bus is still across it. That spike is often the worst point of the whole cycle, and it is easy to miss because it is a property of the diode opposite rather than of the switch you are looking at. It grows with di/dt, so a faster gate drive makes it worse, and it roughly doubles between 25 °C and 125 °C.
Reading the trajectory. The long horizontal stretch at the load current is travelled twice: once to the right as the voltage rises at turn-off, and once to the left as it falls at turn-on. The two lie exactly on top of one another, which is why the picture looks like a single line with two excursions off it — the recovery spike up at the bus voltage, and the loop out to the right where the overshoot takes the device past the bus while the current is still falling.
The SOA is the reader’s, not this page’s. Device safe-operating-area charts are copyrighted datasheet material, and they are not interchangeable between parts or even between pulse widths on the same part. So the page asks for your three corners — the current limit, the constant-power line at the pulse width of your transition, and the voltage rating — and draws the boundary those make. Read them off your own chart, at the right pulse width: a switching transition is a sub-microsecond pulse and the SOA there is far above the DC curve. The Infineon application note listed below is the clearest published explanation of what the five legs of a power MOSFET’s SOA each mean.
What a snubber actually does. Put a capacitor across the device and the load current has somewhere to go while the device’s own current falls: the voltage then rises as the integral of that current rather than ahead of it, and if the capacitance is above I·tfi ÷ 2Vbus the device reaches zero current before the voltage reaches the bus. The trajectory never visits the corner — which is the amber curve on the chart. It is not free: the snubber resistor burns C·V²·f every cycle, and the capacitor’s discharge adds V ÷ R to the turn-on current. What the snubber buys is that the loss is moved out of the silicon and into a resistor you can heatsink. For choosing the R, and for measuring the parasitics it should be matched to, use the RC snubber calculator. This page deliberately does not restate the loss arithmetic — conduction, output capacitance, gate drive and the thermal path all belong to the MOSFET loss calculator — and the gate drive that sets these transition times belongs to the gate driver calculator.
Frequently asked questions
Is the switching energy the area enclosed by the locus?
No, and it is a common thing to say. A closed curve in the voltage-current plane encloses an area whose units are volts times amperes — a power, not an energy. The switching energy is the time integral of v(t)·i(t) through the transition, and that is what this page computes: it depends on how long the device spends at each point of the locus, which the shape of the locus alone does not tell you.
Why is my turn-on loss higher than Vbus × Qrr?
Because during the whole reverse-recovery interval the device carries the load current at the full bus voltage as well as the recovery current. Vbus × Qrr accounts only for the recovery charge itself. With a 30 A load, a 300 nC diode and 1000 A/µs, the load-current term is more than three times the recovery-charge term.
How do I reduce the voltage overshoot at turn-off?
It is L·di/dt, so there are exactly two levers. Reduce L — the commutation loop through the device and the DC-link capacitor is a layout problem, and moving the decoupling capacitor next to the device is usually worth more than anything else you can do. Or reduce di/dt with a larger gate resistor, which costs switching energy. A snubber or a clamp is the third answer when neither is enough.
What SOA pulse width should I read the corners at?
The duration of the transition, which is the sum of your voltage and current times — typically 100 ns to a few hundred. Most datasheets do not draw a line that short; extrapolating the shortest published line by the square root of pulse-width ratio is the usual approach for the thermally-limited part of the chart, but the thermal-instability leg of a power MOSFET’s SOA does not extrapolate that way and is the reason linear mode kills devices. When in doubt, ask the manufacturer.
Does a SiC or GaN device change the picture?
Yes, in both directions. There is no reverse recovery in a SiC Schottky or a GaN device, so the turn-on spike largely disappears — enter zero recovery charge and watch the amber corner go. But the transitions are much faster, so the same stray inductance produces a far larger overshoot at turn-off, and loop inductance that was tolerable with silicon is not.
Why does the snubber curve on the chart retrace itself?
The chart positions every series from one x row, which is the blue trajectory’s own voltage. That trajectory doubles back on itself — the voltage goes up, holds, and comes down — so any curve drawn as a function of voltage is drawn twice, exactly on top of itself. The snubbed turn-off current and the SOA boundary are both single-valued functions of voltage, so nothing is distorted; the line is simply traced more than once.
Related calculators
References
- Infineon Technologies. Linear Mode Operation and Safe Operating Diagram of Power-MOSFETs, application note AP99007, V1.1, May 2017 (J. Schoiswohl). Sets out the five limiting lines of a power MOSFET SOA — RDS(on), package current, maximum dissipation, thermal instability and breakdown voltage — and why the thermal-instability leg does not scale with pulse width the way the others do.
- Mohan N, Undeland T M, Robbins W P. Power Electronics: Converters, Applications and Design, 3rd ed. Wiley, 2003. The switching-waveform treatment of the clamped-inductive cell, and the snubber chapter, including the capacitance that makes the voltage reach the bus exactly as the current reaches zero. Edition and publisher verified by search; chapter numbers are not quoted because they were not.
- Erickson R W, Maksimović D. Fundamentals of Power Electronics, 3rd ed. Springer, 2020, chapter 4 ‘Switch Realization’: the switching loss of the clamped-inductive cell, including the diode reverse-recovery contribution and why it appears in the transistor’s loss rather than the diode’s.