The One-Sentence Definition: An IGBT (Insulated-Gate Bipolar Transistor) is a high-power, three-terminal semiconductor switch that merges the fast, voltage-controlled gate of a MOSFET with the high-current, low-loss output of a bipolar junction transistor (BJT).
When makers and engineers ask 'what is a IGBT', they are usually trying to solve a specific thermal or drive problem in a high-voltage circuit. In a real installation, the IGBT changes the game for high-voltage switching (typically 600V to 3300V). It allows you to switch massive currents without the massive base-drive current required by a BJT, and without the severe conduction losses ($I^2R$) that plague high-voltage power MOSFETs.
People commonly confuse IGBTs with standard power MOSFETs, assuming they are drop-in replacements. They are not. A MOSFET is a purely unipolar device (majority carriers only), while an IGBT is a bipolar device that injects minority carriers into the drift region to lower on-state resistance. This physical difference dictates entirely different switching frequency limits and failure modes.
The Math: IGBT vs. MOSFET Conduction Losses
To understand why the IGBT exists, we need to look at the silicon physics at high voltages. As the breakdown voltage rating of a MOSFET increases, its silicon drift region must become thicker and more lightly doped. This causes the on-resistance ($R_{DS(on)}$) to skyrocket. An IGBT bypasses this by injecting holes (minority carriers) from a p+ substrate, effectively 'flooding' the drift region with charge carriers and drastically reducing the voltage drop across the device when it is turned on.
Let us run a worked numeric example on a 400V DC bus driving a 20A continuous load, which is typical for a 5kW solar string inverter or a 3-phase Variable Frequency Drive (VFD). We will assume a 25°C junction temperature for baseline comparison.
- Option A: 600V Power MOSFET (e.g., Vishay IRFP460). The $R_{DS(on)}$ is roughly 0.27Ω. Conduction loss is calculated as $I^2 \times R$. Therefore, $20^2 \times 0.27 = 108W$. Dissipating 108W per switch requires a massive extruded aluminum heatsink and forced-air cooling.
- Option B: 1200V IGBT (e.g., onsemi FGA25N120). The on-state voltage drop ($V_{CE(sat)}$) is typically 2.0V at 25A. Conduction loss is calculated as $V_{CE(sat)} \times I$. Therefore, $2.0 \times 20 = 40W$.
Safety Note: Working with 400V DC buses is lethal. Always de-energize, lock out the DC disconnect, and verify the bus capacitors are discharged to < 5V using a tested CAT III/IV multimeter before probing IGBT terminals.
Where You Meet IGBTs in Practice
You will rarely find an IGBT in low-voltage hobbyist circuits like 12V LED drivers or 5V logic level shifters. They live in the high-power, high-voltage domain. According to Infineon Technologies, IGBTs are the foundational building blocks of modern power conversion. Here is where you will physically encounter them:
- Variable Frequency Drives (VFDs): Inside the 3-phase inverter bridge of a motor controller, switching at 2kHz to 16kHz to synthesize a sine wave for industrial AC motors.
- Induction Cooktops: High-frequency resonant half-bridges (switching at 20kHz to 40kHz) use specialized co-packaged IGBTs with built-in fast-recovery diodes to drive the induction coil.
- Solar String Inverters: The DC-to-AC conversion stage relies on 650V or 1200V trench-gate IGBTs to push power back to the grid.
- MIG/TIG Welders: Inverter-based welders use IGBTs to chop rectified mains voltage into high-frequency AC, allowing for a drastically smaller step-down transformer.
Decision Tree: Choosing Between IGBT, MOSFET, and SiC
Selecting the right switch requires balancing bus voltage, switching frequency, and thermal budget. Use this decision matrix to terminate your selection process with a concrete part number.
| Bus Voltage | Switching Frequency | Best Technology | Default Part Pick |
|---|---|---|---|
| < 150V | > 50kHz | Si Power MOSFET | IRF3205 (55V, 110A) |
| 200V - 400V | 50kHz - 200kHz | GaN FET | EPC2212 (100V, 75A) |
| 600V - 1200V | 5kHz - 30kHz | Trench IGBT | FGA25N120ANTD (1200V, 25A) |
| 600V - 1200V | > 50kHz | SiC MOSFET | SCTW100N120G2AG (1200V, 38A) |
Bench Testing and Common Failure Modes
When troubleshooting a blown VFD or welder, you need to know how to test an IGBT on the bench and understand how it died. As detailed in All About Circuits, IGBTs have unique failure signatures compared to MOSFETs.
How to Test an IGBT with a Multimeter
- Discharge the Gate: Short the Gate (G) to the Emitter (E) with a resistor or a screwdriver to ensure the internal Miller capacitance is fully discharged.
- Check the Body Diode: Set your DMM to diode test mode. Place the red probe on E and black on C. You should read a standard diode drop (0.4V to 0.6V). Reverse the probes (red on C, black on E); it should read open (OL).
- Turn it On: Use a 9V battery or a bench supply to apply +10V to +15V between G and E (positive to G).
- Verify Conduction: With the gate still charged, measure C to E with the DMM in diode mode or resistance mode. The meter should now show a low voltage drop or near-zero resistance in both directions, indicating the channel is fully enhanced.
- Turn it Off: Short G to E again. The C-E path should immediately return to the open (OL) state in the C-to-E direction.
Common Failure Modes
- Shoot-Through (Cross Conduction): In a half-bridge, if the dead-time between the high-side IGBT turning off and the low-side turning on is too short, both conduct simultaneously. This shorts the DC bus, resulting in a catastrophic, explosive die failure.
- Collector-Emitter Overvoltage (Turn-off Snubbing): Because IGBTs switch large currents, the parasitic inductance of the busbars ($V = L \frac{di}{dt}$) creates massive voltage spikes during turn-off. If this spike exceeds the 1200V rating, the IGBT avalanches and shorts out. Always verify your RC snubber networks are intact.
- Gate Oxide Rupture: The gate threshold voltage ($V_{GE(th)}$) is typically 5V to 6V, but the absolute maximum gate voltage is usually ±20V. A noisy gate driver lacking a proper negative turn-off bias (e.g., -5V to -8V) can allow Miller capacitive coupling to spike the gate voltage, punching through the oxide layer.
Frequently Asked Questions
Can I replace a power MOSFET with an IGBT in my existing circuit?
Generally, no. While the pinout (Gate, Drain/Collector, Source/Emitter) and gate-drive voltage requirements (10V-15V) are similar, an IGBT has a fixed minimum voltage drop ($V_{CE(sat)}$) of about 1.5V to 2.5V, even at very low currents. A MOSFET's voltage drop scales linearly with current ($I \times R_{DS(on)}$) and can be in the millivolt range at light loads. Swapping a MOSFET for an IGBT in a low-current or high-frequency circuit will result in severe overheating.
Do I need a negative gate voltage to turn off an IGBT?
For high-power industrial applications (above 50A), yes. Applying a negative voltage (typically -5V to -8V) to the gate during the off-state prevents the Miller effect from capacitively coupling high $dv/dt$ transients from the collector back into the gate, which could cause false turn-on and shoot-through. For small hobbyist induction heaters or sub-20A welders, pulling the gate hard to 0V (ground) with a low-impedance push-pull driver is usually sufficient.
What is the 'tail current' in an IGBT?
Tail current is the lingering flow of minority carriers (holes) in the drift region after the gate voltage has been removed and the MOS channel has closed. Because these carriers cannot be swept out by the gate, they must recombine naturally. This causes the collector current to 'tail off' slowly during turn-off, creating a period where both high voltage and high current exist simultaneously. This results in high turn-off switching losses and is the primary reason IGBTs are limited to switching frequencies below 30kHz to 50kHz.
