Question 1 of 6: SCR Characteristics and a Single-Phase A.C. Voltage Controller

Question 1: SCR Characteristics and a Single-Phase A.C. Voltage Controller (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — reading the thyristor characteristic

The static anode characteristic of a silicon-controlled rectifier is a four-region curve in the $I_{A}$–$V_{AK}$ plane, and each region corresponds to a distinct internal state of the three-junction $p\text{-}n\text{-}p\text{-}n$ structure.

[Figure not reproduced: Figure 1 — SCR static anode characteristic, redrawn from the examination figure. Increasing gate current moves the forward break-over knee to the left; the on-state branch is common to the whole family. See the official exam paper.]

Region 2 (forward blocking). With $V_{AK}>0$ and no gate drive, the centre junction $J_{2}$ is reverse-biased and only a small forward leakage flows. Region 1 (forward conduction, or on-state). Once the device has been triggered, all three junctions are forward-biased, the anode–cathode drop collapses to roughly one to two volts and the current is set entirely by the external circuit — the near-vertical branch. Region 3 (reverse blocking). With $V_{AK}<0$, junctions $J_{1}$ and $J_{3}$ are reverse-biased and only the reverse leakage $I_{R}$ flows. Region 4 (reverse avalanche). Beyond the reverse repetitive peak voltage the device breaks down destructively; this region is a rating limit, not an operating mode.

The family of curves is the answer to the A-or-B part. The three forward curves are drawn for gate currents in the order $I_{G2}>I_{G1}>I_{G0}$, and the break-over knee moves steadily towards the origin as the gate current is raised. The statement that is correct is therefore the one asserting that the forward break-over voltage decreases as gate current increases; the alternative — that a larger gate current raises the break-over voltage, or that the gate has no influence on it — contradicts the figure. Physically, gate current injects carriers into the $p$-base, so the internal current gains $\alpha_{1}+\alpha_{2}$ reach unity at a lower anode voltage. In the limit of a large gate pulse the knee reaches the origin and the SCR behaves like a plain rectifier diode.

The two marked points are the current thresholds that bound turn-on and turn-off, and they are the only points on the forward characteristic that carry a name:

• X1 — the latching current $I_{L}$: the minimum anode current that must be established while the gate pulse is still present for the device to stay on after the pulse is removed. It fixes the minimum gate-pulse width a firing circuit must deliver into an inductive load, where the current rises slowly.
• X2 — the holding current $I_{H}$: the minimum anode current that will sustain conduction once the device is latched. Fall below it and the SCR reverts to the forward-blocking state. It is the lower knee of the on-state branch, and always $I_{H}<I_{L}$ — typically $I_{L}\approx 2$ to $3\,I_{H}$.

The practical consequence is the one this examination returns to in Problems 1 and 4: an SCR is a latching device with no gate turn-off capability, so in an a.c. controller it commutates naturally when the line current falls through $I_{H}$, and in a d.c. chopper it needs forced commutation or a self-commutating switch.

Parts (b) and (c) — delay-angle range and voltage ratio

Given.

Find. The delay angle $\alpha$ that produces a conduction angle of $130^{\circ}$ at each of the two load angles, and the corresponding ratio $V_{o}/V_{s}$ of output to input rms voltage.

Approach. Solve the transcendental extinction condition of the inductive-load a.c. controller backwards — the conduction angle is given and the delay angle is the unknown — then integrate $v_{s}^{2}$ over the real conduction window to get the rms output.

1. Write the current expression for one conducting half cycle. With $v_{s}=\sqrt{2}\,V_{s}\sin\omega t$ applied to $Z\angle\phi$ from firing at $\alpha$, the series R–L current is the forced sinusoid plus the transient that cancels it at $\omega t=\alpha$:$$i(\omega t)=\frac{\sqrt{2}\,V_{s}}{Z}\left[\sin(\omega t-\phi)-\sin(\alpha-\phi)\,e^{(\alpha-\omega t)/\tan\phi}\right]$$Conduction stops at the extinction angle $\beta$ where this current returns to zero, not where the supply voltage does. Setting $i(\beta)=0$ and writing $\beta=\alpha+\gamma$:$$\sin(\alpha+\gamma-\phi)=\sin(\alpha-\phi)\,e^{-\gamma/\tan\phi}$$
2. Recognise which quantity is unknown. The usual textbook exercise gives $\alpha$ and hunts for $\beta$. Here the paper fixes $\gamma=130^{\circ}$ and asks for $\alpha$, so the same equation is solved as a root problem in $\alpha$ for each load angle. The resistive-load shortcut $\gamma=180^{\circ}-\alpha$ does not apply: it would give $\alpha=50^{\circ}$ for both power factors and lose the whole point of the question.
3. Evaluate the load angles. $\phi=\arccos(0.50)=60.000^{\circ}$ at starting and $\phi=\arccos(0.85)=31.788^{\circ}$ at full load. The decay factor $e^{-\gamma/\tan\phi}$, with $\gamma=2.2689$ rad, is 0.2698 at starting but only 0.0257 at full load — the more resistive the load, the faster the transient dies and the closer the behaviour to a resistive controller.
4. Solve for the two delay angles. Bracketing the root between $\phi$ and $180^{\circ}$ and iterating:$$\boxed{\alpha_{\text{start}}=100.0^{\circ}\quad\text{(pf }0.50\text{)},\qquad\alpha_{\text{full}}=80.7^{\circ}\quad \text{(pf }0.85\text{)}}$$The delay angle therefore ranges over roughly $80.7^{\circ}\le\alpha\le 100.0^{\circ}$ as the motor accelerates, with the extinction angle moving from $210.7^{\circ}$ to $230.0^{\circ}$.
5. Check against the closed-form estimate. When the exponential term is small the extinction condition collapses to $\alpha\approx 180^{\circ}+\phi-\gamma$, giving $81.8^{\circ}$ at pf 0.85 (error $1.1^{\circ}$) and $110.0^{\circ}$ at pf 0.50 (error $10^{\circ}$). The approximation tracks the decay factor, so it is a useful sanity check at high power factor and useless at low power factor — quote it, but never substitute it for the root.
6. Integrate for the rms output voltage. The output is the supply sinusoid over the real conduction window only, twice per cycle:$$V_{o}=V_{s}\sqrt{\frac{1}{\pi}\left[\gamma-\frac{\sin 2\beta-\sin 2\alpha}{2}\right]}$$with all angles in radians. Substituting the starting case ($\alpha=100.0^{\circ}$, $\beta=230.0^{\circ}$) gives $V_{o}/V_{s}=0.7148$; the full-load case ($\alpha=80.7^{\circ}$, $\beta=210.7^{\circ}$) gives 0.7959.
7. State the ratio range.$$\boxed{0.715\le\frac{V_{o}}{V_{s}}\le 0.796}$$i.e. the motor sees between 85.8 V and 95.5 V rms from the 120 V line. The ratio rises as the machine accelerates even though the conduction angle is held constant, because the conduction window slides towards the peak of the supply wave as $\alpha$ falls.