Power Electronics

DC-DC converter steady-state analysis

Steady-state analysis of a DC-DC converter rests on two balance principles. The average inductor voltage over one switching period must be zero (volt-second balance), and the average capacitor current must be zero (charge balance). For an ideal converter in continuous conduction mode (CCM), the buck output is V_O = DV_{in}, the boost output is V_O = V_{in}/(1-D), and the buck-boost output is V_O = -DV_{in}/(1-D), where D is the duty ratio of the switch. Because the ideal converter has no loss, input power equals output power, V_{in}I_{in} = V_O^2/R. From this, the average resistance seen by the source in a boost converter is R_{in} = R(1-D)^2.

In a buck-boost converter the diode conducts only during the off-interval. Its average current is therefore (1-D)I_L, and this must equal the load current I_O, so the average inductor current is I_L = I_O/(1-D).

The output capacitor ripple in a buck-boost converter comes from the on-interval, when the diode is off and the capacitor alone supplies the load. The peak-to-peak ripple is \Delta V_O = I_O DT/C, so the capacitor value follows from C = I_O DT/\Delta V_O.

The inductor ripple is set by the voltage across the inductor and the time it is applied, \Delta I_L = V_L\,\Delta t/L. For a boost converter, \Delta I_L = V_{in}D/(Lf), and the peak inductor current is I_L + \Delta I_L/2.

A useful relation is that the capacitor current equals the inductor current minus the constant load current. When the output ripple is neglected, the capacitor current ripple equals the inductor current ripple, which allows L to be found from the measured capacitor waveform via L = (V_{in}-V_O)t_{on}/\Delta I.

For a triangular inductor current, the average equals the midpoint of the peak values, so the average input current is half the peak when the rise and fall times are equal. Power balance then links the input and output quantities, which is how the input voltage can be found from the waveform.

In a chopper that connects a source to a load through a switch, the average voltage across the load equals the source voltage multiplied by the fraction of time the switch connects it, \langle v\rangle = VD. The average power delivered in such a switched interval is likewise the duty ratio times the instantaneous power during the on-state.

When the load is a constant-current sink in a switched circuit, the average power equals duty ratio times voltage times current during the interval that the source supplies current.

The main sources of error are confusing the polarity and voltage transfer of the three basic topologies, and mixing the ripple formulas for the inductor and the capacitor. Always check the CCM assumption before using the simple transfer formulas.

Full-bridge inverter output waveform

A single-phase full-bridge voltage source inverter (VSI) uses two diagonal pairs of switches. When the diagonal pairs are switched alternately with a 50% duty cycle, the output voltage is a two-level square wave of amplitude V_{dc} with zero average value. Its fundamental has peak value 4V_{dc}/\pi and rms value 2\sqrt{2}V_{dc}/\pi, and the fundamental is in phase with the square wave.

When the load is sinusoidal, the active power is determined only by the fundamental component of the voltage, P = \frac{V_{1m}I_m}{2}\cos\phi, where \phi is the displacement angle between the fundamental voltage and current. Harmonic voltages in the output do not contribute to average power unless they coincide in frequency with a current component.

In 180° conduction mode, the output voltage is \pm V_{dc} and reverses every half period. For a purely inductive load, v = L\,di/dt, so the current changes linearly with slope V_{dc}/L, and the peak-to-peak current is \Delta I = V_{dc}\,(T/2)/L. This gives a triangular load current.

In a complementary leg, each switch and its anti-parallel diode carry the inductor current during their own interval. Current flows through the switch when the load current is positive and through the diode when it is negative, so both paths must be counted. The rms current of a device equals the square root of its conduction fraction times the mean square of the current over its on-interval. For a linear ramp from -I to +I, the mean square is I^2/3.

• Anti-parallel diode — conducts the load current in the direction the switch cannot carry. Example: a triangular current crossing zero uses the switch in one half and the diode in the other.
• Triangular ripple — the peak-to-peak current in a square-wave-fed inductor is set by the volt-seconds applied per half period.

Thyristor rectifier fundamental current

In a controlled rectifier that delivers a ripple-free DC load current, the line current is a quasi-square wave. Its magnitude is I_{dc}, and its conduction starts at the firing angle \alpha, with the thyristors and diodes taking turns during the freewheeling intervals. The fundamental is obtained from the Fourier coefficients a_1 and b_1, with the peak magnitude \sqrt{a_1^2+b_1^2}. For a single-phase half-controlled (semi-converter) bridge, this simplifies to I_{s1} = \frac{4I_{dc}}{\pi}\cos\frac{\alpha}{2}.

The ratio of the rms fundamental current to the total rms input current is the distortion factor. For a half-controlled bridge with ripple-free current, the total rms is I_{dc}\sqrt{(\pi-\alpha)/\pi}, and the rms fundamental is \frac{2I_{dc}}{\pi}\sqrt{1+\cos\alpha} divided by \sqrt{2}. Their ratio simplifies to \frac{2}{\pi}\sqrt{\frac{\pi(1+\cos\alpha)}{\pi-\alpha}}.

When the firing angle is measured from the positive zero crossing and some devices are fired beyond 180°, the output voltage follows the input only while a conducting path exists. Between intervals where the output is blocked the output is taken as zero, and the freewheeling diode reverses the output in the intervals where it conducts. The average output voltage is \frac{1}{2\pi}\int v_o\,d\theta over one full period, so each conducting segment must be integrated with its correct sign.

Do not confuse the fundamental (the first harmonic of the line current, which carries the real power) with the total current (which includes all harmonics and sets the rms and apparent power). Also do not confuse the average output voltage of a rectifier with the rms value of its line current.

Power device selection for switching frequency

The choice of a power semiconductor depends on its voltage and current rating, its on-state (conduction) loss and its switching loss. Switching loss grows in proportion to the switching frequency f_s, so as f_s rises the device must switch quickly to keep the total loss low. A device's switching speed is governed by how stored charge is removed. Majority-carrier devices have no minority-carrier storage time and switch much faster than bipolar devices.

• Power MOSFET — unipolar, majority-carrier, voltage-controlled; very fast and well suited to low-voltage, high-frequency converters.
• IGBT — combines a MOS gate with a bipolar output; medium frequency and high voltage, with a current tail at turn-off.
• SCR / GTO — bipolar thyristors; high power and low frequency, with slow turn-off because stored charge must be removed.

At comparable power ratings, the power MOSFET is the fastest device in this list, because its turn-off does not depend on recombination of stored minority charge. The SCR and GTO are the slowest.

The typical mistake is to pick a high-power device for a high-frequency application. A low-voltage, moderate-power, high-frequency converter is better served by a MOSFET than by a thyristor family device, even though thyristors handle much higher power.

Thyristor commutation circuit

Forced commutation turns off a thyristor by making its current fall to zero with a reverse-biased path, using an external resonant or capacitive circuit. In an LC commutation circuit, a pre-charged capacitor and an inductor form a resonant loop with \omega_0 = 1/\sqrt{LC}. The thyristor current is the sum of the constant load current and the oscillatory current, i_T = I_L + I_p\sin\omega_0 t. The thyristor turns off when i_T first falls to zero, which occurs when \sin\omega_0 t = -I_L/I_p. The conduction time follows from this condition.

A commutation capacitor charged to a voltage V_C can also be used to turn off a main thyristor. Once the auxiliary device is triggered, the capacitor discharges through the main thyristor path and supplies the constant load current, so the main device current reaches zero when the capacitor has delivered the charge CV_C at the load current. The available turn-off time is therefore t_{off} = CV_C/I_L.

The available turn-off time must exceed the device's turn-off time t_q, and the reverse voltage must be held long enough for the device to recover its blocking capability. A holding current below the load current cannot be reached by the commutation current alone, so the circuit must drive the current below the holding level.

Quasi-square wave inverter modulation index

A quasi-square wave output is a symmetric waveform whose pulses have an adjustable width. The modulation index is defined as the ratio of the peak fundamental of the output voltage to the applied DC value, M = V_{1m}/V_d. When each pulse of height V_d occupies the interval from \theta to \pi - \theta and is centred on \pi/2, the fundamental sine coefficient is b_1 = \frac{2}{\pi}\int_\theta^{\pi-\theta}V_d\sin\omega t\,d(\omega t) = \frac{4V_d}{\pi}\cos\theta. Therefore the modulation index is M = \frac{4}{\pi}\cos\theta.

The maximum value is 4/\pi at \theta = 0, which corresponds to the full square wave. As the pulse narrows, \theta increases and M falls to zero at \theta = \pi/2. To obtain a required M, the angle is found from \cos\theta = \pi M/4.

Do not confuse this index with the sinusoidal PWM index m_a, which is defined as the ratio of the peak reference to the carrier peak and lies in the linear range for values up to 1.

Diode rectifier and regulator circuits

An uncontrolled diode bridge converts an AC input into a pulsating DC output. For an ideal full-wave bridge, the output is the rectified sine |v_s(t)|. Its average value is 2V_m/\pi, but for power calculations the rms value matters. The rms value of |v_s| equals the rms value of the input, V_{rms}, so the power delivered to a purely resistive load is P = V_{rms}^2/R, regardless of the rectified waveshape.

The average and rms values of a rectified waveform are different quantities. The average gives the DC component used to size filters and set the operating point of a regulator. The rms value gives the resistive heating and the load power.