The whole Power Electronics formula sheet, to print

The SCR is modelled as a $p$-$n$-$p$ transistor $Q_1$ (common-base current gain ${\alpha }_1$) and an $n$-$p$-$n$ transistor $Q_2$ (gain ${\alpha }_2$), with the gate current feeding the base of $Q_2$; $I_{CBO1}$ and $I_{CBO2}$ are their leakage currents. The gains rise with current. As ${\alpha }_1+{\alpha }_2\to 1$ the anode current is limited only by the external circuit: the device has latched.

Latching current $I_L$ is the minimum anode current that must be reached before the gate pulse is removed for the SCR to stay on. Holding current $I_H$ is the anode current below which a conducting SCR (gate open) turns off. Typically $I_L$ is about two to three times $I_H$.

Piecewise-linear on-state model $v_T=V_{T0}+r_T{i}_T$: threshold voltage $V_{T0}$ and dynamic resistance $r_T$, with the average and rms values of the SCR current taken over a full period.

Charging current of the junction capacitance $C_{J2}$ of $J_2$ ($C_{J2}$ taken as constant). If this current reaches the triggering level, the SCR turns on with no gate signal; this sets the device's critical $dv/dt$ rating.

For a rectangular gate pulse train of width $T_p$ at frequency $f$, the average gate power is $P_{G,peak}f{T}_p$. A short pulse may therefore carry a peak power above the average gate power rating $P_{G,av(max)}$, as long as it stays within the device's peak gate power limit.

A dc supply $V_s$ is switched by the SCR onto a series $R$-$L$ load, with the current starting from zero. The current $i=\frac{V_s}{R}\left(1-e^{-Rt/L}\right)$ must reach $I_L$ before the gate pulse ends. The approximation holds when $I_{L}R\ll V_s$.

Supply $v=V_m\sin \omega t$; $\alpha$ is measured from the positive-going zero crossing of the supply. The SCR fires when the supply reaches the value that drives $I_{GT}$ through the total series gate-circuit resistance $R$ (including the minimum fixed resistor) with $V_{GT}$ across the gate junction (the series diode drop neglected). Since a sine peaks at $90^{\circ }$, the range is about $0^{\circ }$ to $90^{\circ }$. Increasing $R$ delays firing.

$\eta =R_{B1}/(R_{B1}+R_{B2})$ is the intrinsic stand-off ratio. $C$ charges through $R$ from $V_{BB}$ until $v_C$ reaches the peak-point voltage $V_P=\eta V_{BB}+V_D$; the UJT then fires and $C$ discharges quickly through $R_{B1}$ (or a pulse transformer), giving a gate pulse. This form neglects $V_D$, the valley voltage $V_V$ and the discharge time.

A dc supply $V$ feeds a purely resistive load $R$ through the SCR. A capacitor $C$ charged to $V$ in the reverse-biasing polarity is connected directly across the conducting SCR (auxiliary switch drop neglected). The SCR turns off and $C$ recharges through $R$, so the SCR voltage is $v=V-2V{e}^{-t/RC}$. It passes through zero at this time, which is how long the SCR is held reverse biased.

At the instant of turn-on the whole supply $V_s$ appears across the series inductor $L$ (other resistances neglected), so $di/dt=V_s/L$. Conduction starts near the gate and spreads slowly, so a small inductor in series with the SCR protects it from local hot-spot burn-out.

$Q_g$ is the total gate charge for the gate voltage swing $\Delta V_{GS}$ (from the off level to the on level; equal to $V_{GS}$ when the device is turned off at 0 V), and $f_s$ is the switching frequency. $P_G$ is the power drawn from the driver, and $I_{G,avg}$ the average current drawn from the driver supply. With a constant gate current $I_G$, the time to deliver a charge $Q$ is $t=Q/I_G$.

$I_{D(rms)}$ is the rms drain current taken over the whole switching period, and $R_{DS(on)}$ is taken at the operating junction temperature.

The device switches current $I$ against a blocked voltage $V$ with an inductive load clamped by a diode. At turn-on the current ramps linearly with the voltage held at $V$, then the voltage falls linearly with the current held at $I$; turn-off is the reverse. $t_{on}$ and $t_{off}$ are the total transition times. The energy per cycle is $\frac{1}{2}VI(t_{on}+t_{off})$, and multiplying by $f_s$ gives the power.

With a resistive load the voltage and current ramp linearly at the same time, in opposite directions. The energy per transition is ${\int }_0^{t}V\left(1-\frac{\tau }{t}\right)I\frac{\tau }{t}d\tau =\frac{1}{6}VIt$.

Steady state. Thermal resistances in ${}^{\circ }$C/W: junction-to-case, case-to-sink and sink-to-ambient; $P$ is the total device loss.

In steady state the average inductor voltage over one period is zero. This gives the voltage gain of every converter.

In steady state the average capacitor current is zero, so the average current delivered to the load equals the average current of the element feeding the output node: the inductor in the buck, the diode in the boost and buck-boost.

CCM, $I_L$ = average inductor current, $\Delta i_L$ = peak-to-peak ripple. Used for rms ratings and copper loss; with negligible ripple $I_{L,rms}\approx I_L$.

CCM. Output is positive, same polarity as the input, and $V_o\le V_s$.

Peak-to-peak ripple, CCM. Average inductor current equals $I_o$; average input current is $I_s=D{I}_o$.

For fixed $V_s$, $L$ and $f$ in CCM. The same result holds for a step-down chopper feeding an inductive load (with or without back emf) when the load resistance is neglected.

Assumes the whole ripple component of the inductor current flows into the capacitor (the load takes only the average).

CCM if $L>L_{crit}$ (boundary where $I_{min}=0$, i.e. $I_o=\Delta i_L/2$).

CCM. Same polarity as input, $V_o\ge V_s$. In the ideal circuit the gain rises without limit as $D\to 1$; in a real converter the inductor and switch resistances make the gain reach a maximum and then fall as $D\to 1$.

The diode conducts only for $(1-D)T$ and carries $I_L$, so its average current is $(1-D)I_L=I_o$; the switch average current is $D{I}_L$.

CCM, peak-to-peak.

During the ON interval the capacitor alone supplies $I_o$ for $DT$.

CCM if $L>L_{crit}$ (from $I_L=\Delta i_L/2$).

CCM. Output is negative with respect to the input common; $|V_o|<V_s$ for $D<0.5$, $|V_o|>V_s$ for $D>0.5$.

Here $I_o=|V_o|/R$. Diode average current is $(1-D)I_L=I_o$; average input (switch) current is $I_s=D{I}_L=D{I}_o/(1-D)$.

CCM, peak-to-peak. The capacitor alone feeds the load during the ON interval.

CCM if $L>L_{crit}$.

From volt-second balance with the inductor voltage zero during $D_{3}T=(1-D-D_2)T$. $D_2$ is the fraction of the period for which the diode conducts.

Valid for $K<K_{crit}$ (DCM) only. At $K=K_{crit}$ they agree with the CCM gains. For the buck-boost the diode interval is $D_2=\sqrt{K}$; in every case a DCM answer must satisfy $D+D_2<1$.

Separately excited DC motor fed by a step-down chopper, continuous armature current, constant flux; average values.

Step-up (Class B) chopper with the machine generating: $D$ is the duty ratio of the switch that shorts the armature, $I_a$ is the magnitude of the continuous armature current flowing from the machine to the supply, and the average voltage across the machine terminals is $(1-D)V_s$. Average values, ripple neglected.

Holds for any output waveform; $V_{rms}$ is the total rms including the DC part.

One diode conducts for 180 degrees of each cycle.

For the centre-tap type $V_m$ is the peak of each half of the secondary. PIV is $V_m$ for the bridge and $2V_m$ for the centre-tap.

$\varphi ={\tan }^{-1}(\omega L/R)$. The current is $i=\frac{V_m}{Z}\left[\sin (\omega t-\varphi )+\sin \varphi e^{-\omega t/\tan \varphi }\right]$ with $Z=\sqrt{R^2+(\omega L{)}^2}$. The current continues after the supply goes negative, up to the extinction angle $\beta >\pi$ where $i=0$, so the output voltage has a negative portion from $\pi$ to $\beta$.

The diode conducts only while the instantaneous supply voltage exceeds $E$, from ${\theta }_1$ to $\pi -{\theta }_1$. In a full-wave bridge the same interval repeats in every half cycle, so the average current doubles.

Each diode conducts for 120 degrees; its average current is $I_{dc}/3$ and its rms current is $I_{dc}/\sqrt{3}$. The output ripple frequency is $3f$.

The output is made of six line-voltage segments per cycle, so the ripple frequency is $6f$. PIV $=V_{LL,m}$.

Each diode conducts 120 degrees. The line current has one positive and one negative 120-degree block per cycle, and its fundamental is $0.78I_{dc}$.

Applies to an R load and also to an RL load with a freewheeling diode. Range $0\le \alpha \le \pi$.

Pure R load, so the output follows the supply from $\alpha$ to $\pi$ in each conducting half cycle. The bridge result holds for both the fully controlled and the half-controlled bridge.

Positive for $\alpha <90^{\circ }$ (rectification) and negative for $\alpha >90^{\circ }$ (inversion, which needs an active load such as a reversed battery or a regenerating machine). With an RLE load and continuous current, $I_{dc}=(V_{dc}-E)/R$.

The output cannot go negative, so only the rectifying quadrant is available.

Square-wave source current; each thyristor conducts for 180 degrees, with average current $I_{dc}/2$ and rms current $I_{dc}/\sqrt{2}$. The source current has odd harmonics $n=3,5,7,\dots$ with $I_n=I_{s1}/n$, and THD $=48.3\%$. Input powers: $P=\frac{2V_m}{\pi }{I}_{dc}\cos \alpha =V_{dc}{I}_{dc}$ and $Q=\frac{2V_m}{\pi }{I}_{dc}\sin \alpha$.

The source current is $+I_{dc}$ from $\alpha$ to $\pi$, $-I_{dc}$ from $\pi +\alpha$ to $2\pi$, and zero while the load freewheels. Its displacement factor $\cos (\alpha /2)$ is better than the fully controlled bridge's $\cos \alpha$.

Negative for $\alpha >90^{\circ }$, which is inversion. The load current stays continuous for all $\alpha$ when $L$ is large enough.

The output never goes negative, so $0\le \alpha \le 180^{\circ }$ gives rectification only.

Valid for $60^{\circ }\le \alpha \le 120^{\circ }$; the output reaches zero at $\alpha =120^{\circ }$. For $\alpha \le 60^{\circ }$ with an R load the output stays positive and the cosine law $\frac{3V_{LL,m}}{\pi }\cos \alpha$ still holds.

The line current has one positive and one negative 120-degree block per cycle. Its harmonics are of order $n=6k\pm 1$ ($5,7,11,13,\dots$) with $I_n=I_{s1}/n$, and THD $=31.1\%$. Input powers: $P=\frac{3V_{LL,m}}{\pi }{I}_{dc}\cos \alpha$ and $Q=\frac{3V_{LL,m}}{\pi }{I}_{dc}\sin \alpha$.

Large L, continuous conduction. The drop equals $4f{L}_s{I}_{dc}$ and does not depend on $\alpha$.

Constant $I_{dc}$; $L_s$ is per phase. The drop equals $6f{L}_s{I}_{dc}$.

An auxiliary thyristor switches a capacitor $C$, charged to $V_C$, across the conducting thyristor, reverse-biasing it at once. With the load current $I_0$ constant during commutation, the capacitor voltage changes linearly from $-V_C$ to zero in time $t_c$.

An LC circuit drives a resonant current pulse of peak $I_{cp}$ through the thyristor, opposite to its load current $I_0$, assumed constant during commutation. The thyristor current reaches zero when the pulse reaches $I_0$. A diode across the thyristor then carries the excess, and its forward drop reverse-biases the thyristor until the pulse falls back to $I_0$. Commutation needs $I_{cp}>I_0$.

Two antiparallel thyristors (or a triac), fired symmetrically. $V_s$ is the rms supply voltage and $\alpha$ is measured from the voltage zero crossing. Control range $0\le \alpha \le 180^{\circ }$. The output has no DC component. Input power factor $=V_{o,rms}/V_s$.

Steady state with continuous armature current. Speed is raised by increasing $V_{dc}$ (reducing $\alpha$).

Continuous conduction, zero source inductance.

Continuous conduction, zero source inductance. At $\alpha =0$ this is the diode bridge value $V_{d0}=1.35V_{LL}$.

In each cycle a line carries $+I_d$ for $120^{\circ }$, nothing for $60^{\circ }$, $-I_d$ for $120^{\circ }$, and nothing for $60^{\circ }$. With an ideal ripple-free $I_d$ this does not depend on $\alpha$.

Reference $v_{an}=V_m\cos \omega t$. At $\alpha =0$ the positive $120^{\circ }$ block is centred on the positive peak of $v_{an}$. For $\alpha >0$ the whole waveform is the $\alpha =0$ waveform delayed by $\alpha /\omega$. The peak of the $n$-th term is $\frac{4I_d}{n\pi }\sin \frac{n\pi }{3}$. Zero overlap, ripple-free $I_d$.

The peak of the fundamental is $\frac{2\sqrt{3}}{\pi }{I}_d$. Divide by $\sqrt{2}$ for the rms value.

Six-pulse bridge, zero overlap, ripple-free $I_d$. Relative to the fundamental the magnitude falls as $1/n$, and it does not depend on $\alpha$.

Reference $v_s=V_m\sin \omega t$. The current is a square wave of $\pm I_d$ delayed by $\alpha$, so every odd order $n$ is present. Ripple-free $I_d$, zero overlap. The magnitudes do not depend on $\alpha$.

Half-controlled bridge (semi-converter), or a full bridge with a freewheeling diode, with ripple-free $I_d$. The line current is $+I_d$ from $\alpha$ to $\pi$, zero while the load freewheels, then $-I_d$ from $\pi +\alpha$ to $2\pi$. Here the harmonic magnitudes do depend on $\alpha$, and the fundamental lags the supply voltage by $\alpha /2$.

${\varphi }_1$ is the phase angle between the supply voltage and the fundamental line current. With zero overlap, ${\varphi }_1=\alpha$ for a fully controlled bridge and ${\varphi }_1=\alpha /2$ for a single-phase half-controlled bridge.

The ratio of the fundamental rms current to the total rms line current. DF $\le 1$.

Sinusoidal supply voltage. For the three-phase case $P=\sqrt{3}{V}_{LL}{I}_{s1}\cos {\varphi }_1$ and $S=\sqrt{3}{V}_{LL}{I}_s$. For the single-phase case $P=V_s{I}_{s1}\cos {\varphi }_1$ and $S=V_s{I}_s$.

Harmonics are measured relative to the fundamental. Equivalently, $\text{DF}=1/\sqrt{1+{\text{THD}}^2}$.

Fully controlled bridge, ripple-free $I_d$, zero overlap. DF and THD do not change with $\alpha$. Only DPF does. A diode bridge ($\alpha =0$) has PF $=0.955$.

Ripple-free $I_d$, zero overlap. At $\alpha =0$ this gives PF $=0.900$, the same as the full bridge. For $\alpha >0$ its PF is higher than the full bridge's $0.9\cos \alpha$.

Zero overlap, lossless converter. $Q_1$ is the fundamental reactive power. With $V_{d0}=\frac{3\sqrt{2}}{\pi }{V}_{LL}$, the fundamental apparent power is $\sqrt{P^2+Q_1^2}=V_{d0}{I}_d$.

$L_s$ is the source inductance per phase, with ripple-free $I_d$. For the single-phase fully controlled bridge the matching result is $V_d=0.9V_s\cos \alpha -\frac{2\omega L_s}{\pi }{I}_d$.

Half-bridge: the output is $\pm V_s/2$, so the rms is $V_s/2$.

Time origin at a rising edge. Only odd harmonics; the nth has peak $4V_s/(n\pi )$. For the half-bridge use $V_s/2$ in place of $V_s$.

Half-bridge: $0.45V_s$.

About 48.3 percent, the same for half- and full-bridge. The harmonic factor of the nth harmonic is ${\mathrm{HF}}_n=V_{n,\mathrm{rms}}/V_{1,\mathrm{rms}}$.

Steady state. Each harmonic is treated separately, with the impedance at its own frequency $n\omega$. Higher harmonics see a larger reactance, so the current is closer to a sine wave than the voltage. For a lossless inverter the average dc source current is $I_s=P/V_s$.

$\sigma$ is the width of each output pulse in radians per half-cycle. Even harmonics are absent.

Full-bridge output of $p$ equal pulses of height $V_s$ per half-cycle, each $\delta$ radians wide ($p\delta \le \pi$). With $p=1$ and $\delta =\sigma$ this is the quasi-square result.

For 180° conduction. The nth harmonic ($n=6k\pm 1$) has rms $1/n$ of the fundamental, so $V_{LLn,\mathrm{rms}}=0.78V_s/n$. Triplen harmonics are absent from both phase and line voltages.

Balanced load. Holds for the total rms and for the fundamental when the phase voltage has no triplen harmonics, as in both conduction modes here.

$m_a$ is the amplitude modulation ratio; $m_f$ is the frequency modulation ratio.

Large $m_f$; the full-bridge result is the same for bipolar and unipolar switching. The fundamental grows linearly with $m_a$ and its frequency is $f_m$.

Each leg gives a pole voltage of peak $m_a{V}_s/2$, and the line voltage is $\sqrt{3}$ times its rms. $V_s$ is the dc bus voltage, $m_a\le 1$, sinusoidal references with no third-harmonic injection.

Single-phase: a square wave of current $\pm I_d$. Three-phase: each line current is a 120°-wide block of height $I_d$, the same shape as the 180°-mode VSI line voltage, with harmonics of order $6k\pm 1$.

Stator impedance drop neglected. Constant $V/f$ gives constant flux and roughly constant breakdown torque below base speed. $P$ is the number of poles and $f$ is in Hz.

$\Delta c$ is the count in time $\Delta t$ seconds and $K$ is the counts per revolution ($N$, or $4N$ with 4x decoding). Counting pulses in a fixed time suits high speed, while timing between pulses suits low speed.