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Chapter 3 โ€ข Theory & Derivations

Power Electronics: Thyristors, UJT & Phase Control

Comprehensive treatment of power semiconductor devices: the four-layer pnpn Silicon Controlled Rectifier (SCR) and its two-transistor regenerative latching model; Unijunction Transistors (UJT) and negative-resistance relaxation oscillators; bidirectional Triac and Diac switches in AC phase control systems; complete mathematical derivations and solved university exam problems.

ยง3.1 Silicon Controlled Rectifier (SCR): Structure & Two-Transistor Analogy

1. Physical Architecture and Operating Regimes

The Silicon Controlled Rectifier (SCR) is a four-layer, three-junction ($p_1-n_1-p_2-n_2$) unidirectional semiconductor switching device with three terminals: Anode (A), Cathode (K), and Gate (G). The outer layers are heavily doped $p_1$ (anode) and $n_2$ (cathode), while the inner $n_1$ and $p_2$ layers are moderately/lightly doped.

When an SCR is connected with an external anode-to-cathode voltage $V_{AK}$:

  • Reverse Blocking State ($V_{AK} < 0$): Junctions $J_1$ and $J_3$ are reverse-biased, while $J_2$ is forward-biased. Only a minuscule reverse leakage current flows until the reverse breakdown voltage $V_{BR}$ is reached, at which point avalanche breakdown occurs.
  • Forward Blocking State ($V_{AK} > 0$, $I_G = 0$): Outer junctions $J_1$ and $J_3$ are forward-biased, but central junction $J_2$ is reverse-biased, creating a wide depletion layer. The device supports high forward voltage with only a tiny forward leakage current flowing until the forward breakover voltage $V_{BO}$ is exceeded.
  • Forward Conduction State ($V_{AK} > 0$, $I_G > I_{GT}$): Injection of a gate trigger pulse causes instantaneous regenerative switching (turn-on). Junction $J_2$ collapses, and the SCR drops into a low-impedance forward conduction state with a forward voltage drop of only $V_F \approx 1.0\text{ to }1.5\,\text{V}$.

2. Mathematical Derivation of Two-Transistor Regenerative Analogy

To mathematically understand regenerative latching, consider splitting the four-layer $p_1-n_1-p_2-n_2$ structure into two interconnected bipolar transistors: a PNP transistor $Q_1$ ($p_1-n_1-p_2$) and an NPN transistor $Q_2$ ($n_1-p_2-n_2$).

The collector of $Q_1$ feeds the base of $Q_2$, and the collector of $Q_2$ feeds the base of $Q_1$. Let $\alpha_1$ and $\alpha_2$ denote the common-base current gains of $Q_1$ and $Q_2$ respectively, and let $I_{CO1}, I_{CO2}$ be their reverse saturation currents.

$$I_{C1} = \alpha_1 I_A + I_{CO1}$$
$$I_{C2} = \alpha_2 I_K + I_{CO2}$$

Applying Kirchhoff's Current Law at the cathode: $I_K = I_A + I_G$. The total current crossing central junction $J_2$ is the sum of collector currents and reverse leakage:

$$I_A = I_{C1} + I_{C2} = \alpha_1 I_A + I_{CO1} + \alpha_2 (I_A + I_G) + I_{CO2}$$

Collecting terms in $I_A$:

$$I_A (1 - (\alpha_1 + \alpha_2)) = \alpha_2 I_G + (I_{CO1} + I_{CO2})$$
$$I_A = \frac{\alpha_2 I_G + I_{CO1} + I_{CO2}}{1 - (\alpha_1 + \alpha_2)}$$

Latching Condition: At very low current levels, transistor gains $\alpha_1$ and $\alpha_2$ are very small ($\ll 0.5$), so $(\alpha_1 + \alpha_2) \ll 1$, keeping $I_A$ negligible (off-state). However, when gate current $I_G$ is injected into the base of $Q_2$, $I_A$ increases, causing emitter currents to rise. Because transistor $\alpha$ increases nonlinearly with emitter current, the loop gain sum reaches unity:

$$\alpha_1 + \alpha_2 \to 1 \implies [1 - (\alpha_1 + \alpha_2)] \to 0 \implies I_A \to \infty \quad \text{(limited only by external load } R_L\text{)}$$

Once $\alpha_1 + \alpha_2 = 1$, the internal regenerative loop feeds itself: $Q_1$ drives $Q_2$, which in turn drives $Q_1$. At this stage, the external gate signal $I_G$ loses control and can be safely removed; the device remains latched in the ON state.

3. Latching Current ($I_L$) vs. Holding Current ($I_H$)

  • Latching Current ($I_L$): The minimum anode current that must be attained immediately after gate triggering for the SCR to sustain conduction when the gate trigger pulse is terminated. If $I_A < I_L$ when $I_G$ ceases, the SCR falls back into the forward blocking state.
  • Holding Current ($I_H$): The minimum anode current below which the SCR automatically turns OFF (re-enters the forward blocking mode) during commutation. If an external circuit reduces $I_A$ below $I_H$, the regenerative loop collapses.
  • Crucial Relationship: For all thyristors, $I_L > I_H$, typically $I_L \approx 2\text{ to }3 \times I_H$.

ยง3.2 SCR Turn-On Mechanisms, Firing Angles & Phase-Controlled Rectification

1. Firing Angle $\alpha$ and Conduction Angle $\gamma$

In AC power circuits, the point on the sinusoidal voltage waveform at which the SCR is triggered into conduction is quantified by the firing angle (or delay angle) $\alpha$ measured in electrical degrees or radians from the zero-crossing of the positive half-cycle ($0 \le \alpha \le \pi$).

The conduction angle $\gamma$ represents the duration over which the SCR conducts before natural line commutation turns it off:

$$\gamma = \pi - \alpha$$

2. Half-Wave Phase-Controlled Rectifier with Resistive Load

Consider an AC source $v_s(t) = V_m \sin(\omega t)$ connected to an SCR in series with a resistive load $R$. During the negative half-cycle ($\pi \le \omega t \le 2\pi$), the SCR is reverse-biased and $v_o(t) = 0$. During the positive half-cycle, conduction begins only when triggered at $\omega t = \alpha$ and extinguishes naturally at $\omega t = \pi$.

The average (DC) load voltage is derived via definite integration:

$$V_{dc} = \frac{1}{2\pi} \int_{\alpha}^{\pi} V_m \sin(\omega t)\, d(\omega t) = \frac{V_m}{2\pi} \left[ -\cos(\omega t) \right]_{\alpha}^{\pi}$$
$$V_{dc} = \frac{V_m}{2\pi} [ -(-1) - (-\cos\alpha) ] = \frac{V_m}{2\pi} (1 + \cos\alpha)$$

The RMS load voltage is calculated by evaluating the root-mean-square integral:

$$V_{rms} = \sqrt{ \frac{1}{2\pi} \int_{\alpha}^{\pi} V_m^2 \sin^2(\omega t)\, d(\omega t) } = \frac{V_m}{\sqrt{2\pi}} \sqrt{ \int_{\alpha}^{\pi} \frac{1 - \cos(2\omega t)}{2}\, d(\omega t) }$$
$$V_{rms} = \frac{V_m}{2\sqrt{\pi}} \sqrt{ (\pi - \alpha) + \frac{1}{2} \sin(2\alpha) }$$

3. Full-Wave Mid-Point & Bridge Phase-Controlled Rectifier

For a full-wave controlled bridge rectifier where SCR pairs are fired symmetrically at $\alpha$ and $\pi + \alpha$:

$$V_{dc} = \frac{1}{\pi} \int_{\alpha}^{\pi} V_m \sin(\omega t)\, d(\omega t) = \frac{V_m}{\pi} (1 + \cos\alpha)$$
$$V_{rms} = \frac{V_m}{\sqrt{2}} \sqrt{ 1 - \frac{\alpha}{\pi} + \frac{\sin(2\alpha)}{2\pi} }$$

When $\alpha = 0$, $V_{dc} = \frac{2V_m}{\pi}$, which corresponds exactly to an uncontrolled full-wave diode rectifier. By modulating $\alpha$ continuously from $0$ to $\pi$, $V_{dc}$ smoothly varies from $\frac{2V_m}{\pi}$ down to $0\,\text{V}$, providing continuous DC motor speed and heating control.

ยง3.3 Unijunction Transistor (UJT): Intrinsic Standoff Ratio & Relaxation Oscillator

1. Architecture and Equivalent Circuit of UJT

The Unijunction Transistor (UJT) is a three-terminal single-junction device consisting of a lightly doped $N$-type silicon bar with ohmic contacts at both ends, denoted as Base-1 ($B_1$) and Base-2 ($B_2$). Near $B_2$, a heavily doped $P$-type emitter region is alloyed, forming a single $P-N$ junction with the bar.

The total resistance of the silicon bar between $B_1$ and $B_2$ when the emitter is open-circuited is the interbase resistance $R_{BB}$:

$$R_{BB} = R_{B1} + R_{B2} \quad (\text{typically } 4\,\text{k}\Omega \text{ to } 10\,\text{k}\Omega)$$

The internal voltage division between Base 1 and Base 2 defines the fundamental device parameter, the Intrinsic Standoff Ratio $\eta$:

$$\eta = \frac{R_{B1}}{R_{B1} + R_{B2}} = \frac{R_{B1}}{R_{BB}} \quad (0.51 \le \eta \le 0.82)$$

2. Dynamic Conduction & Negative Resistance Characteristic

When an interbase bias $V_{BB}$ is applied with $B_2$ positive relative to $B_1$, the potential at the internal junction point $A$ inside the silicon bar is $V_A = \eta V_{BB}$.

If emitter voltage $V_E < V_A + V_D$ (where $V_D \approx 0.6\,\text{V}$ is the silicon diode forward barrier drop), the emitter junction is reverse-biased, permitting only a tiny reverse leakage current $I_{EO}$ to flow (Cutoff Region).

When $V_E$ reaches the Peak-Point Voltage $V_P$:

$$V_P = \eta V_{BB} + V_D$$

The emitter junction becomes forward-biased, injecting a burst of holes into the $N$-type channel between the emitter and $B_1$. This massive carrier injection drastically lowers the resistance $R_{B1}$ via conductivity modulation. As $I_E$ increases, $V_E$ rapidly plummets, establishing a dynamic Negative Differential Resistance region ($dV_E/dI_E < 0$) until reaching the Valley Point $(V_V, I_V)$, beyond which the device enters saturation.

3. UJT Relaxation Oscillator Circuit & Derivation of Period

In a relaxation oscillator, a timing resistor $R$ charges an external capacitor $C$ from $V_{BB}$. As $C$ charges, capacitor voltage rises exponentially:

$$v_C(t) = V_{BB} \left( 1 - e^{-t / (RC)} \right)$$

When $v_C(t)$ reaches the peak point $V_P \approx \eta V_{BB}$ (neglecting small $V_D$ and assuming discharge to $V_V \approx 0$):

$$\eta V_{BB} = V_{BB} \left( 1 - e^{-T / (RC)} \right) \implies 1 - \eta = e^{-T / (RC)}$$

Taking the natural logarithm of both sides yields the oscillation period $T$ and frequency $f$:

$$T = R C \ln\left( \frac{1}{1 - \eta} \right)$$
$$f = \frac{1}{T} = \frac{1}{R C \ln\left( \frac{1}{1 - \eta} \right)}$$

When the capacitor discharges rapidly through the low resistance of $R_{B1}$ and an external resistor $R_1$, a sharp, high-energy positive voltage pulse is developed across $R_1$, perfectly matched for reliably triggering SCR and Triac gates.

ยง3.4 Triac and Diac: Bidirectional AC Switching & Light Dimmer Circuits

1. Diac (Diode AC Switch): Symmetrical Bi-directional Trigger

A Diac is a three-layer, two-terminal ($A_1$ and $A_2$) bidirectional semiconductor diode. It does not have a control gate. The device remains in forward or reverse blocking mode until the applied voltage across its terminals reaches the symmetrical breakover voltage $V_{BO}$ (typically $\pm 30\text{ to }35\,\text{V}$).

Once $|V| \ge V_{BO}$, the Diac undergoes negative-resistance breakdown, dropping terminal voltage suddenly to $\sim 20\,\text{V}$ and discharging a sharp spike of current in either positive or negative polarity. This makes Diacs the primary trigger element for driving Triac gates symmetrically on both alternating half-cycles.

2. Triac (Triode AC Switch): Architecture and Quadrant Modes

A Triac is functionally equivalent to two inverse-parallel (back-to-back) SCRs integrated monolithically on a single silicon chip with a single common gate terminal. The main power terminals are designated Main Terminal 1 (MT1) and Main Terminal 2 (MT2), referenced to MT1.

A Triac can be triggered into conduction by either a positive or negative gate pulse for either polarity of terminal voltage $V_{MT2-MT1}$, operating across four quadrants:

  • Mode I+ ($MT2+, G+$): Most sensitive mode. Both MT2 and Gate are positive with respect to MT1. Normal transistor junction action.
  • Mode I- ($MT2+, G-$): MT2 positive, Gate negative with respect to MT1. Gate junction junction injection initiates turn-on.
  • Mode III+ ($MT2-, G+$): MT2 negative, Gate positive with respect to MT1. Least sensitive mode requiring highest gate trigger current $I_{GT}$. Often avoided in symmetrical trigger circuits.
  • Mode III- ($MT2-, G-$): Highly sensitive mode. MT2 negative, Gate negative. Remote gate injection triggering.

3. Full-Wave AC Phase Control (Light Dimmer Circuit)

In a standard phase-controlled light dimmer or AC fan motor speed controller:

  • A potentiometer $R_v$ and fixed resistor $R_1$ charge capacitor $C$ from the AC mains line.
  • The voltage across $C$ lags behind the line voltage by phase angle $\theta = \arctan(\omega R_{total} C)$.
  • When capacitor voltage $|v_C(t)|$ reaches the Diac breakdown voltage $V_{BO} \approx 32\,\text{V}$, the Diac breaks down and rapidly dumps capacitor charge into the Triac gate.
  • The Triac fires into full conduction, applying the remaining portion of the mains half-cycle to the lamp or motor load.
  • At the end of the half-cycle, when AC load current crosses zero, the Triac commutates off naturally until the Diac fires in the opposite polarity during the subsequent half-cycle.
Adjusting $R_v$ varies the firing delay angle $\alpha$ continuously from $\approx 15^\circ$ to $\approx 165^\circ$, smoothly controlling the RMS power delivered to the load from $\approx 98\%$ down to $0\%$.

Solved Problem Example 3.1: SCR Half-Wave Phase-Controlled Rectifier Analysis

An SCR is used in a half-wave phase-controlled rectifier circuit supplied from a $230\text{ V}$ RMS, $50\text{ Hz}$ sinusoidal source, feeding a resistive load of $R = 50\ \Omega$. The firing angle is set to $\alpha = 60^\circ$. (a) Derive and calculate the average DC load voltage $V_{\text{dc}}$. (b) Calculate the average load current $I_{\text{dc}}$ and total power delivered to the load. (c) Compute the circuit rectification efficiency.

Step 1: Calculate Peak Supply Voltage $V_m$

$$V_m = \sqrt{2} \times V_{rms,in} = \sqrt{2} \times 230\,\text{V} \approx 325.27\,\text{V}$$

Step 2: Calculate Average (DC) Load Voltage $V_{dc}$

Using the half-wave phase-controlled equation for firing angle $\alpha = 60^\circ$:

$$V_{dc} = \frac{V_m}{2\pi} (1 + \cos\alpha) = \frac{325.27}{2\pi} (1 + \cos 60^\circ) = \frac{325.27}{2\pi} (1 + 0.5) = \frac{325.27 \times 1.5}{6.28318} \approx 77.65\,\text{V}$$

Step 3: Calculate Average DC Load Current $I_{dc}$

$$I_{dc} = \frac{V_{dc}}{R} = \frac{77.65\,\text{V}}{40\,\Omega} \approx 1.941\,\text{A}$$

Step 4: Calculate RMS Load Voltage $V_{rms}$

Recall the RMS formula for half-wave phase control:

$$V_{rms} = \frac{V_m}{2\sqrt{\pi}} \sqrt{ (\pi - \alpha) + \frac{1}{2} \sin(2\alpha) }$$

Here $\alpha = \pi/3 \approx 1.0472\,\text{rad}$, so $\pi - \alpha = 2\pi/3 \approx 2.0944\,\text{rad}$, and $\sin(2\alpha) = \sin(120^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660$:

$$\text{Term inside root} = 2.0944 + \frac{0.8660}{2} = 2.0944 + 0.4330 = 2.5274$$
$$V_{rms} = \frac{325.27}{2 \times \sqrt{3.14159}} \times \sqrt{2.5274} = \frac{325.27}{3.5449} \times 1.5898 \approx 145.86\,\text{V}$$

Step 5: Total Power Delivered to Load $P_L$

$$P_L = \frac{V_{rms}^2}{R} = \frac{(145.86\,\text{V})^2}{40\,\Omega} = \frac{21275.14}{40} \approx 531.88\,\text{W}$$
Solved Problem Example 3.2: UJT Relaxation Oscillator Frequency & Peak Point Calculation

A UJT has an intrinsic standoff ratio $\eta = 0.65$, interbase resistance $R_{BB} = 8\text{ k}\Omega$, and forward diode drop $V_D = 0.70\text{ V}$. It is used in a relaxation oscillator with $V_{BB} = 15\text{ V}$, charging resistor $R = 100\text{ k}\Omega$, and capacitor $C = 0.1\ \mu\text{F}$. Assume valley voltage $V_V = 1.5\text{ V}$. (a) Calculate the peak point firing voltage $V_P$. (b) Determine the oscillation frequency $f$ of the generated sawtooth waveform. (c) Check whether $R$ satisfies the firing condition $R_{\text{min}} < R < R_{\text{max}}$ if peak current $I_P = 2\ \mu\text{A}$ and valley current $I_V = 3\text{ mA}$.

Step 1: Calculate Internal Base Resistances $R_{B1}$ and $R_{B2}$

$$R_{B1} = \eta \cdot R_{BB} = 0.65 \times 8\,\text{k}\Omega = 5.2\,\text{k}\Omega$$
$$R_{B2} = R_{BB} - R_{B1} = 8.0\,\text{k}\Omega - 5.2\,\text{k}\Omega = 2.8\,\text{k}\Omega$$

Step 2: Calculate Peak Point Voltage $V_P$

$$V_P = \eta V_{BB} + V_D = (0.65 \times 15\,\text{V}) + 0.7\,\text{V} = 9.75\,\text{V} + 0.7\,\text{V} = 10.45\,\text{V}$$

Step 3: Calculate Period of Oscillation $T$

Using the theoretical UJT charging equation with $V_V \approx 0$:

$$T = R C \ln\left( \frac{1}{1 - \eta} \right) = (47 \times 10^3\,\Omega) \times (0.1 \times 10^{-6}\,\text{F}) \times \ln\left( \frac{1}{1 - 0.65} \right)$$
$$R C = 4.7 \times 10^{-3}\,\text{s} = 4.7\,\text{ms}$$
$$\ln\left(\frac{1}{0.35}\right) = \ln(2.8571) \approx 1.0498$$
$$T = 4.7\,\text{ms} \times 1.0498 \approx 4.934\,\text{ms}$$

Step 4: Calculate Oscillation Frequency $f$

$$f = \frac{1}{T} = \frac{1}{4.934 \times 10^{-3}\,\text{s}} \approx 202.67\,\text{Hz}$$
Solved Problem Example 3.3: Full-Wave Bridge Controlled Rectifier with Inductive Load

A single-phase full-wave controlled SCR bridge rectifier feeds a highly inductive load such that load current is continuous and ripple-free at $I_{dc} = 15\text{ A}$. The AC supply is $230\text{ V}$ RMS at $50\text{ Hz}$. If the firing angle is $\alpha = 45^\circ$: (a) Calculate the average output DC voltage $V_{dc}$. (b) Calculate the active power delivered to the load. (c) Compute the input displacement factor and power factor of the converter.

Step 1: Calculate Average Output Voltage for Highly Inductive Load

For a continuous-conduction inductive load, current continues through the zero crossing until the next pair of thyristors is fired at $\pi + \alpha$. Thus the integration limits are from $\alpha$ to $\pi + \alpha$:

$$V_{dc} = \frac{1}{\pi} \int_{\alpha}^{\pi + \alpha} V_m \sin(\omega t)\, d(\omega t) = \frac{V_m}{\pi} [ -\cos(\pi + \alpha) + \cos\alpha ] = \frac{2V_m}{\pi} \cos\alpha$$

With $V_{rms} = 240\,\text{V}$, $V_m = 240\sqrt{2} \approx 339.41\,\text{V}$:

$$V_{dc} = \frac{2 \times 339.41}{\pi} \cos(45^\circ) = 216.08 \times 0.7071 \approx 152.8\,\text{V}$$

Step 2: Calculate Active DC Power Delivered to Load

$$P_{dc} = V_{dc} \times I_o = 152.8\,\text{V} \times 15\,\text{A} \approx 2292\,\text{W} = 2.292\,\text{kW}$$

Step 3: Calculate Input Displacement Power Factor (DPF)

For a phase-controlled converter with continuous current, the fundamental input AC current lags behind fundamental source voltage by exactly the firing angle $\alpha$:

$$\text{DPF} = \cos(\phi_1) = \cos(\alpha) = \cos(45^\circ) \approx 0.7071 \text{ (lagging)}$$
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