pushing material beyond their limits: breakdown voltage

Breakdown voltage is a critical parameter in power electronics - it determines whether a device survives high voltage stress or fails catastrophically. Here’s a breakdown of the physics and practical implications.

what is breakdown voltage?

The breakdown voltage (also called withstand voltage) is the maximum reverse voltage a semiconductor device - such as a diode, triac, MOSFET or IGBT power transistor - can withstand across its terminals before it’s blown.

It is fundamentally tied to the material’s intrinsic properties: bandgap and dielectric strength and is strongly temperature-dependent.

why temperature matters so much

increased intrinsic carrier generation
Higher temperatures exponentially increase intrinsic carrier concentration. In reverse-biased p-n junctions, these carriers contribute to leakage current. More carriers → higher leakage → lower voltage needed to reach the critical field for avalanche breakdown.

reduced mean free path
Hotter lattice vibrations scatter charge carriers more frequently, shortening their mean free path and reducing energy gain from the electric field.

avalanche multiplication weakens
Although avalanche rates decrease slightly with temperature due to scattering, the dominant effect is the higher leakage current, which causes the device to hit the critical field at a lower applied voltage.

temperature’s toll: pushing materials beyond limits

As temperature rises, withstand voltage decreases due to increased carrier generation and reduced resistivity. This is critical in power electronics, where devices face high thermal stress or limited cooling. Overheating can trigger thermal runaway.

The physics is governed by intrinsic carrier concentration:

$$ n_i = \sqrt{N_C N_V} \cdot e^{-\frac{E_g}{2kT}}$$

where:

• $% n_i $% - intrinsic carrier concentration [cm⁻³]
• $%N_C $% - effective density of states in conduction band [cm⁻³]
• $% N_V $% - effective density of states in valence band [cm⁻³]
• $% E_g $% - bandgap energy [eV]
• $% k $% - Boltzmann constant
• $% T $% - absolute temperature [K]

The exponent $% e^{-\frac{E_g}{2kT}} $% is extremely sensitive to temperature.

Example for silicon:

• At 300 K (~27°C): $% n_i \approx 1 \times 10^{10} $% cm⁻³
• At 400 K (~127°C): $% n_i \approx 2 \times 10^{12} $% cm⁻³
  → 200× increase in carriers

In practice, for silicon devices, breakdown voltage drops by 0.1–0.2% per °C above 25°C.
For a 1200 V IGBT, a temperature rise from 27°C to 127°C can reduce $% V_{BR} $% by 120–240 V.

Datasheets typically show dramatic derating at higher temperatures. Example from a popular IGBT (e.g., IHW30N120R5):

Half the capability - just from temperature rise.

choosing the right material is a game-changer

Silicon has a relatively low breakdown voltage compared to wide-bandgap materials.

At 400 K, GaN has only ~$% 1 \times 10^4 $% cm⁻³ intrinsic carriers (vs. silicon’s $% 2 \times 10^{12} $%).
→ $% V_{BR} $% barely changes with temperature - one reason GaN excels in high-voltage, high-temperature applications.

GaN also offers better speed and (often) lower cost than SiC for mid-range voltages, though processing is more challenging.

where does the heat come from?

Temperature rise is a consequence - the root cause is power dissipation in semiconductors (IGBTs, MOSFETs, diodes). Key interacting parameters are:

• $% V_{CE(sat)} $% - saturation voltage (conduction losses)
• $% I^2t $% - energy absorbed during overcurrent events
• junction temperature rise $% \Delta T_j $%
• breakdown voltage $% V_{BR} $%

Together they define the safe operating area (SOA). Let’s break them down.

$% V_{CE(sat)} $% – expected heat generation

Saturation voltage $% V_{CE(sat)} $% causes conduction losses:

$$ E = V_{CE(sat)} \cdot I_C \cdot t $$

This energy becomes heat.

$% I^2t $% – the thermal chain reaction trigger

$% I^2t $% [A²s] quantifies how much overcurrent energy the device can absorb before thermal damage:

$$ E = I^2 \cdot R \cdot t $$

Example: IGBT with $% I^2t = 500 $% A²s
→ can handle 50 A for 0.2 s ($% 50^2 \times 0.2 = 500 $%)
With $% R_{CE(on)} = 0.05 \ \Omega $%:
→ $% E = 25 $% J
With given transient thermal impedance $% Z_{th} = 0.5 \ ^\circ\mathrm{C}/\mathrm{J} $%:
→ $% \Delta T_j = 12.5 \ ^\circ\mathrm{C} $% jump

case study: total thermal collapse (or when the inductor saturates…)

Inductor saturation is a common killer. Here’s the most common, yet spectacular scenario:

• magnetic core saturates → permeability drops dramatically
• inductance collapses (e.g., 10 µH → 1 µH)
• current rise rate $% di/dt $% explodes (e.g., 30 A/µs → 300 A/µs), current spikes far beyond ratings (50 A → 80 A)
• voltage overshoots exceed $% V_{BR} $%
• conduction losses quadruple ($% P = I^2 R $%)
• switching losses surge
• final act: junction temperature rockets (e.g., 100°C → 160°C in seconds)

Classic fiasco scenario. Accompnied by classic questions: how’s that even possible that the 30A rated MOSFET blew while the current was just a few amps?

aftermath

tip #1 choose components made from semiconductors with appropriate parameters.

tip #2 source parts from suppliers that maintain high quality control (crystal dislocations, point defects, surface irregularities, process variability).

tip #3 avoid counterfeits - they often use inferior materials and poor fabrication processes or misrepresented specs.

tip #4 old proverb says “trust but verify”. Buy a cheap breakdown voltage tester for a few bucks and test your MOSFETs. You will be surprised by the results - and you’ll quickly realize the money was better spent than on a six-pack.

waiting your feedback

Did this post finally explain why things burn when they shouldn’t? Have you ever tracked down the real reason? Or do you already have a breakdown voltage tester - and the six-pack anecdote is just part of the story..?