Question 4 of 6: Half-Wave Controlled Rectifier — Sensitivity of the D.C. Output

Problem 4: Half-Wave Controlled Rectifier — Sensitivity of the D.C. Output (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.

Given. Single-phase half-wave controlled rectifier into a purely resistive load; initial delay angle $\alpha_0 = 18^{\circ}$ with peak input $V_{max,0}$; the peak input is then reduced to $0.85\,V_{max,0}$ while the delay angle is reset to $\alpha_1 = 8^{\circ}$, $\alpha_2 = 12^{\circ}$ or $\alpha_3 = 30^{\circ}$. Part (d) repeats the same three cases for a three-phase controlled rectifier in continuous conduction.

Find. (a) and (b) the operating principle with a resistive load and the change produced by adding inductance and a d.c. source; (c) and (d) the percentage change in mean output voltage relative to the original operating point, for the single-phase and three-phase converters respectively.

Approach. Write the mean output voltage of each converter as a product of an input-amplitude factor and a delay-angle factor, then form the ratio of the new operating point to the original — every constant cancels, so no absolute voltage is needed.

Part (a) — resistive load. In a single-phase half-wave controlled rectifier a single thyristor is placed between the a.c. source and the load. During the positive half cycle the device is forward biased but blocks until a gate pulse is applied at ωt = α; it then latches on and the full source voltage appears across the resistor, so the load current is v_s/R and is exactly in phase with the voltage. At ωt = 180° the source passes through zero, the anode current falls below the holding value and the thyristor turns off naturally (line commutation). Throughout the negative half cycle it is reverse biased and blocks, so the load sees nothing. The output is therefore a single chopped positive lobe per cycle, and its mean value is obtained by integrating from α to 180°: V_dc = (V_max/2π)(1 + cosα), which falls monotonically from V_max/π at α = 0 to zero at α = 180°. Both the conduction angle and the mean voltage are controlled by the gate timing alone; the price is a large d.c. component drawn from the a.c. supply, heavy low-order harmonic content and a poor input power factor.

Part (b) — resistance plus inductance in series with a d.c. source. Three things change. First, the inductance stores energy while the current rises and returns it as the current falls, so the current no longer follows the voltage and does not reach zero at ωt = 180°. The thyristor continues to conduct into the negative half cycle until the volt-second areas balance, extinguishing at an extinction angle β > 180°. During that interval the load terminals are held at a negative voltage, which reduces the mean output below the resistive-load value; the mean becomes V_dc = (V_max/2π)(cosα − cosβ), and β must itself be found from a transcendental extinction condition. Adding a free-wheeling diode across the load prevents this by clamping the terminals at zero, restoring the resistive-load expression and improving the power factor.

Second, the d.c. source E in the load imposes a firing window. The thyristor cannot conduct until the instantaneous supply exceeds E, so the earliest useful firing angle is α_min = sin⁻¹(E/V_max) and conduction ceases once the supply falls back through E (modified by the inductive delay). Firing before α_min simply produces no current. Third, because the current is now continuous within its conduction window and smoothed by L, the current ripple is much reduced and the load current can be treated as approximately constant when ωL ≫ R — the assumption that underlies the constant-armature-current d.c. drive analysis of Problem 6. The mean current follows the loop equation V_dc = E + I_dcR, so the converter now controls current (and hence torque, in a motor load) rather than merely voltage.

1. Part (c) — set up the ratio for the single-phase converter. With a resistive load, $V_{dc} = (V_{max}/2\pi)(1+\cos\alpha)$, so both the peak-voltage scaling and the delay-angle factor appear as simple multipliers: $$\frac{\Delta V_{dc}}{V_{dc,0}} = \frac{V_{dc,i}}{V_{dc,0}} - 1 = 0.85\,\frac{1+\cos\alpha_i}{1+\cos 18^{\circ}} - 1 ,$$ with $1 + \cos 18^{\circ} = 1.95106$. The constant $1/2\pi$ and the reference peak $V_{max,0}$ cancel, so no absolute voltage is required.
2. Part (c) — evaluate the three delay angles. For $\alpha_1 = 8^{\circ}$: $1+\cos 8^{\circ} = 1.99027$, ratio $= 0.85(1.99027)/1.95106 = 0.86708$. For $\alpha_2 = 12^{\circ}$: $1+\cos 12^{\circ} = 1.97815$, ratio $= 0.86180$. For $\alpha_3 = 30^{\circ}$: $1+\cos 30^{\circ} = 1.86603$, ratio $= 0.81296$. Therefore $$\boxed{\Delta V_{dc}/V_{dc,0} = -13.29\ \%,\; -13.82\ \%,\; -18.70\ \% \text{ for } \alpha = 8^{\circ},\,12^{\circ},\,30^{\circ}}$$ Even advancing the firing from 18 degrees to 8 degrees recovers only about two percentage points of the 15 per cent lost with the supply, because near zero delay the cosine is flat.
3. Part (d) — set up the ratio for the three-phase converter. In continuous conduction a three-phase controlled rectifier has a mean output directly proportional to the input amplitude and to $\cos\alpha$ — $V_{dc} = (3\sqrt{3}/\pi)V_{max}\cos\alpha$ for the three-pulse (half-wave) connection and $(3\sqrt{3}/\pi)V_{max}\cos\alpha$ doubled for the six-pulse bridge. Whichever connection is meant, the constant cancels in the ratio, so $$\frac{\Delta V_{dc}}{V_{dc,0}} = 0.85\,\frac{\cos\alpha_i}{\cos 18^{\circ}} - 1, \qquad \cos 18^{\circ} = 0.95106 .$$
4. Part (d) — evaluate and compare with the single-phase case. For $\alpha_1 = 8^{\circ}$: $0.85(0.99027)/0.95106 = 0.88504$. For $\alpha_2 = 12^{\circ}$: $0.85(0.97815)/0.95106 = 0.87421$. For $\alpha_3 = 30^{\circ}$: $0.85(0.86603)/0.95106 = 0.77400$. Hence $$\boxed{\Delta V_{dc}/V_{dc,0} = -11.50\ \%,\; -12.58\ \%,\; -22.60\ \% \text{ for } \alpha = 8^{\circ},\,12^{\circ},\,30^{\circ}}$$ The three-phase converter is the less sensitive of the two at small delay angles and the more sensitive at large ones, because $\cos\alpha$ falls away faster than $(1+\cos\alpha)/2$ once $\alpha$ exceeds roughly 20 degrees.