14.2 Electrical Apparatus & Devices — Industrial Electronics & Power Converters

14.2 Electrical Apparatus & Devices — Industrial Electronics & Power Converters

Topic E of the Electrical Engineering subject in the PRC Enhanced Table of Specifications (PRBEE Resolution No. 40, s. 2024) is Electrical Apparatus & Devices, Industrial Electronics, weighted 2.25% of the whole examination and 5 of the 100 Electrical Engineering items. The 2011 syllabus carried the same material as Electronic Power Equipment and Components & Devices. This is the electronics that actually sits in an electrical engineer's switchroom: drives, soft starters, UPS systems and rectifier banks.

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1. The Thyristor Family

The defining SCR behaviour: a gate pulse latches the device on, after which the gate loses all control. Turn-off requires the anode current to fall below the holding current. In AC circuits this happens naturally at each current zero (natural or line commutation); in DC circuits an auxiliary circuit must force the current to zero (forced commutation). This single fact explains most SCR examination items.

Latching current is the minimum anode current needed to latch the device on at turn-on; holding current is the minimum needed to keep it on. Latching current is always the larger of the two.

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2. Controlled Rectifiers (AC to DC)

Delaying the SCR firing angle $\alpha$ from the natural commutation point reduces the average output.

Single-phase half-wave controlled:

$V_{dc} = \frac{V_m}{2\pi}\left(1 + \cos\alpha\right)$

Single-phase fully controlled bridge (continuous conduction):

$V_{dc} = \frac{2V_m}{\pi}\cos\alpha$

Three-phase full-wave (six-pulse) fully controlled bridge:

$V_{dc} = \frac{3\sqrt{2}\,V_{LL}}{\pi}\cos\alpha = 1.35\,V_{LL}\cos\alpha$

Three consequences are examinable:

1. At $\alpha = 0$ the converter behaves as an uncontrolled diode bridge;
2. For $\alpha > 90°$ with an inductive source of EMF, $V_{dc}$ goes negative — power flows from the DC side back to the AC line and the converter operates in the inversion mode, which is how a regenerative drive returns braking energy;
3. Displacement power factor for a fully controlled converter is approximately $\cos\alpha$, so deep phase control means poor power factor — the reason large rectifier installations need harmonic filters and reactive compensation.

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3. Choppers, Inverters and AC Controllers

DC Chopper (DC to DC)

With duty cycle $D = t_{on}/T$:

$V_o = D\,V_s \quad \text{(buck)}, \qquad V_o = \frac{V_s}{1-D} \quad \text{(boost)}, \qquad V_o = \frac{D\,V_s}{1-D} \quad \text{(buck-boost)}$

Inverter (DC to AC)

Converts DC to AC of controllable magnitude and frequency. Voltage source inverters dominate motor drives. Output waveform quality ranks square wave < modified sine < sinusoidal PWM, and the modulation index sets the fundamental amplitude:

$m_a = \frac{V_{\text{control}}}{V_{\text{carrier}}}, \qquad V_{o1(\text{peak})} = m_a\frac{V_{dc}}{2} \quad (m_a \le 1)$

Raising the switching frequency pushes harmonics higher, where they are easier to filter, at the cost of greater switching loss.

AC Voltage Controller and Cycloconverter

An AC voltage controller uses back-to-back SCRs (or a TRIAC) to vary rms voltage at constant frequency — the basis of the soft starter and of resistive heating control. A cycloconverter converts directly from a fixed AC frequency to a lower variable frequency without a DC link, used in very large low-speed drives such as cement mills.

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4. Variable-Frequency Drives and Soft Starters

A VFD is a rectifier, a DC link and a PWM inverter in series. Below base speed it holds the volts-per-hertz ratio constant so that air-gap flux — and hence available torque — stays constant:

$\frac{V}{f} = \text{constant} \qquad \Longrightarrow \qquad \phi \approx \frac{V}{4.44\,f\,N} = \text{constant}$

For a 460 V, 60 Hz motor the ratio is $460/60 = 7.67$ V/Hz, so 30 Hz operation requires 230 V. Above base speed the voltage cannot rise further, so $V/f$ falls, flux weakens, and the drive enters the constant-power field-weakening region with torque falling as $1/f$.

Synchronous speed follows the frequency directly:

$N_s = \frac{120f}{P}$

A soft starter, by contrast, only ramps voltage at fixed 60 Hz. It limits starting current but reduces starting torque with the square of the voltage, so it cannot start a high-inertia load that a VFD handles easily.

$\frac{T_{st}}{T_{st,\text{full}}} = \left(\frac{V_{\text{applied}}}{V_{\text{rated}}}\right)^{2}$

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5. Harmonics from Converter Loads

Non-linear converter loads draw non-sinusoidal current. A $p$-pulse converter generates characteristic harmonics of order

$h = kp \pm 1, \qquad k = 1, 2, 3, \dots$

so a six-pulse drive produces the 5th, 7th, 11th and 13th harmonics, while a twelve-pulse arrangement cancels the 5th and 7th and starts at the 11th. Total harmonic distortion and true power factor follow:

$THD_I = \frac{\sqrt{\sum_{h \ge 2} I_h^{2}}}{I_1}, \qquad PF_{\text{true}} = \frac{\cos\theta_1}{\sqrt{1 + THD_I^{2}}}$

The distortion factor term is why a drive with near-unity displacement factor can still present a true power factor well below unity, and why triplen (3rd, 9th, 15th) harmonics from single-phase electronic loads add arithmetically in the neutral of a four-wire system, sometimes forcing an oversized neutral conductor.

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Solved Board Exam Examples

Example 1: Three-Phase Controlled Rectifier Output

A six-pulse fully controlled bridge is supplied from a 460 V line-to-line, 60 Hz source and fired at $\alpha = 30°$. Find the average DC output voltage.

Solution.

$V_{dc} = 1.35\,V_{LL}\cos\alpha = 1.35(460)\cos 30° = 621(0.8660) = \boxed{537.8 \text{ V}}$

At $\alpha = 0$ the same bridge would deliver 621 V; at $\alpha = 90°$ the average output is zero; beyond 90° it goes negative and the converter inverts.

Example 2: Constant Volts-per-Hertz Operation

A 460 V, 60 Hz, 4-pole induction motor is driven by a VFD at 45 Hz. Find the applied voltage under constant V/f control and the new synchronous speed.

Solution. The rated ratio is

$\frac{V}{f} = \frac{460}{60} = 7.667 \text{ V/Hz}$

$V_{45} = 7.667 \times 45 = \boxed{345 \text{ V}}$

$N_s = \frac{120(45)}{4} = \boxed{1350 \text{ rpm}}$

Because flux is preserved, the motor develops full rated torque at 1350 rpm — the property that distinguishes a VFD from a soft starter.

Example 3: Soft Starter Torque Penalty

A soft starter limits the applied voltage to 65% of rated during starting. What fraction of full-voltage starting torque remains?

Solution. Starting torque varies with the square of the applied voltage:

$\frac{T_{st}}{T_{st,\text{full}}} = (0.65)^{2} = \boxed{0.4225 \text{ or } 42.25\%}$

Starting current falls to about 65% of its full-voltage value, but torque collapses to 42%. If the load requires more than 42% of full-voltage starting torque, the motor will stall and the soft starter is the wrong solution.