GATE Electrical Engineering Power Electronics Syllabus

What is the body diode of a power MOSFET?

An intrinsic parasitic $P\text{-}N$ diode formed between the source (body) and drain, which conducts in the reverse direction. It is useful as a freewheeling diode but has a relatively slow reverse recovery.

What does IGBT stand for and what two devices does it combine?

Insulated Gate Bipolar Transistor. It combines the voltage-controlled (insulated) gate of a MOSFET with the high-current, low-conduction-loss bipolar output of a BJT.

Name the three terminals of an IGBT.

Gate (G), Collector (C), and Emitter (E).

Is an IGBT voltage-controlled or current-controlled, and is it a majority- or minority-carrier device?

It is voltage-controlled (via gate-emitter voltage $V_{GE}$) like a MOSFET, but it is a minority-carrier (bipolar conduction) device in its output, giving low on-state voltage drop at high currents.

Compare the on-state voltage drop of an IGBT versus a power MOSFET at high current.

The IGBT has a nearly constant, low saturation voltage $V_{CE(sat)}$ (bipolar conduction) so it conducts high current with low loss, whereas the MOSFET's drop $I_D R_{DS(on)}$ rises with current. Hence IGBTs are favored at high voltage/high current.

What is the 'tail current' problem in an IGBT?

During turn-off, stored minority carriers in the drift region recombine slowly, producing a residual collector 'tail current' that increases turn-off switching loss and limits the switching frequency.

Give a typical application domain comparison: MOSFET vs IGBT vs thyristor by frequency/power.

MOSFET: low-to-medium power, high frequency (>100 kHz). IGBT: medium-to-high power, medium frequency (a few kHz to ~50 kHz). Thyristor/SCR: very high power, low frequency (line frequency, DC drives, HVDC).

Rank MOSFET, IGBT, and BJT/thyristor by switching speed (fastest first).

MOSFET (fastest) > IGBT (intermediate) > BJT/Thyristor (slowest). MOSFETs have no minority-carrier storage; IGBTs have a turn-off tail; thyristors are slow latching devices.

What is a buck converter and what is its basic function?

A buck (step-down) converter is a DC-DC switching converter whose average output voltage is less than or equal to the input voltage. It steps the voltage down while ideally stepping the current up.

State the voltage conversion ratio of an ideal buck converter in continuous conduction mode (CCM).

$$\frac{V_o}{V_{in}} = D$$ where $D$ is the duty cycle ($0 \leq D \leq 1$). Since $D \leq 1$, $V_o \leq V_{in}$.

For an ideal (lossless) buck converter, express the input/output current relationship.

From power balance $V_{in}I_{in} = V_o I_o$, with $V_o = D V_{in}$, the average input current is $I_{in} = D\,I_o$, so the output current is stepped up by $\frac{1}{D}$.

Name the four main components of a basic buck converter.

A controlled switch (MOSFET/IGBT), a freewheeling diode, an inductor $L$, and an output filter capacitor $C$ (across the load).

In a buck converter, what is the inductor voltage during the switch-ON interval?

During ON time, $v_L = V_{in} - V_o$ (positive), so the inductor current rises with slope $\frac{di_L}{dt} = \frac{V_{in}-V_o}{L}$.

In a buck converter, what is the inductor voltage during the switch-OFF interval?

During OFF time the inductor freewheels through the diode and $v_L = -V_o$, so the current falls with slope $\frac{di_L}{dt} = -\frac{V_o}{L}$.

What is a boost converter and its basic function?

A boost (step-up) converter is a DC-DC switching converter whose average output voltage is greater than or equal to the input voltage. It steps the voltage up while stepping the current down.

State the voltage conversion ratio of an ideal boost converter in CCM.

$$\frac{V_o}{V_{in}} = \frac{1}{1-D}$$ where $D$ is the duty cycle. As $D \to 1$, $V_o \to \infty$ (ideally), so $V_o \geq V_{in}$.

Name the four main components of a basic boost converter.

An input inductor $L$ (in series with the source), a controlled switch (MOSFET/IGBT), a diode, and an output capacitor $C$ across the load.

In a boost converter, what happens during the switch-ON interval?

The switch shorts the inductor to ground; the inductor charges with $v_L = V_{in}$ and current rises ($\frac{di_L}{dt} = \frac{V_{in}}{L}$). The diode is reverse-biased, and the load is supplied by the output capacitor.

In a boost converter, what happens during the switch-OFF interval?

The inductor current flows through the diode to the load; the inductor voltage is $v_L = V_{in} - V_o$ (negative), releasing stored energy so that $V_o > V_{in}$.

For an ideal boost converter, express the average input current in terms of output current.

From power balance and $V_o = \frac{V_{in}}{1-D}$: $I_{in} = \frac{I_o}{1-D}$, i.e. the input (inductor) current is larger than the output current.

What is a buck-boost converter and its function?

A buck-boost converter is a DC-DC switching converter that can produce an output voltage either lower or higher than the input, typically with inverted (opposite) polarity.

State the voltage conversion ratio of an ideal inverting buck-boost converter in CCM.

$$\frac{V_o}{V_{in}} = -\frac{D}{1-D}$$ The magnitude is less than $V_{in}$ for $D<0.5$ (buck) and greater for $D>0.5$ (boost); the polarity is inverted.

In a buck-boost converter, at what duty cycle does $|V_o| = V_{in}$?

At $D = 0.5$, since $\left|\frac{V_o}{V_{in}}\right| = \frac{D}{1-D} = \frac{0.5}{0.5} = 1$.

Why is the basic buck-boost converter said to have inverted output polarity?

Because of the way the inductor delivers its stored energy to the output during the OFF interval, the output voltage is negative with respect to the common (input) ground reference.

In a buck-boost converter, what is the inductor voltage during the ON interval?

During ON, the inductor is connected directly across the source: $v_L = V_{in}$, and current rises with slope $\frac{V_{in}}{L}$ (diode reverse-biased).

In a buck-boost converter, what is the inductor voltage during the OFF interval?

During OFF, the inductor discharges into the output: $v_L = V_o$ (negative), and the diode conducts to transfer energy to the load and capacitor.

What distinguishes Continuous Conduction Mode (CCM) from Discontinuous Conduction Mode (DCM) in a DC-DC converter?

In CCM the inductor current never reaches zero during a switching cycle; in DCM the inductor current falls to zero for part of each cycle (typically at light load), changing the voltage conversion ratio so it depends on load.

What is the volt-second balance (inductor) principle used to analyze DC-DC converters in steady state?

In steady state the average inductor voltage over one switching period is zero: $$\int_0^{T_s} v_L\,dt = 0$$ i.e., (volt-seconds during ON) = (volt-seconds during OFF). This yields the converter's $V_o/V_{in}$ ratio.

What is capacitor charge (amp-second) balance in steady-state converter analysis?

In steady state the average capacitor current over one switching period is zero: $$\int_0^{T_s} i_C\,dt = 0$$ so the net charge delivered to the capacitor each cycle is zero, keeping output voltage constant.

Define duty cycle $D$ for a switching converter.

The fraction of the switching period for which the main switch is ON: $$D = \frac{t_{on}}{T_s} = t_{on}\,f_s$$ where $T_s$ is the switching period and $f_s$ the switching frequency. $0 \leq D \leq 1$.

Compare the conversion ratios of buck, boost, and buck-boost converters in CCM.

Buck: $\frac{V_o}{V_{in}} = D$ (step-down). Boost: $\frac{V_o}{V_{in}} = \frac{1}{1-D}$ (step-up). Buck-boost: $\frac{V_o}{V_{in}} = -\frac{D}{1-D}$ (step up or down, inverted polarity).