Problem 1 — controlled half-wave rectifier feeding a dc motor armature: 22-Elec-B8 Power Electronics and Drives, May 2018, Q2

Question 2: Problem 1 — controlled half-wave rectifier feeding a dc motor armature (15 points)

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. A single-thyristor half-wave converter is fed from a 220 V rms supply and loaded by an armature resistance in series with a back EMF, and the paper states the measured mean current together with the conduction and minimum delay angles.

Given data
Quantity	Symbol	Value
Supply (rms)	$V_{s}$	220 V
Supply peak	$V_{m}=\sqrt{2}V_{s}$	311.13 V
Mean output current	$I_{dc}$	32 A
Conduction angle	$\gamma$	145°
Minimum delay angle applied	$\alpha_{min}$	21°
Delay angle in part (b)	$\alpha_{2}$	30°

Find. (a) the back EMF $E_{c}$, the delay angle $\alpha$ and the armature-circuit resistance $R$; (b) with the firing advanced to $\alpha_{2}=30^\circ$, the mean power absorbed by $E_{c}$ and the corresponding shaft output in horsepower.

The thyristor is gated at α and stops conducting at β, where the supply falls back through the back EMF. The shaded area is the voltage available to drive current through R, and it is what the mean-current integral measures.

Approach. Because there is no inductance, the load itself fixes the extinction angle — conduction stops the instant the supply falls back through $E_{c}$ — so $\beta$ follows from $\alpha$ and $\gamma$, $E_{c}$ follows from $\beta$, and the one mean-current integral then delivers $R$.

Convert the rms supply to a peak value. The waveform algebra is all written in peak volts: $$V_{m}=\sqrt{2}\,V_{s}=\sqrt{2}\times 220=311.13\ \text{V}$$
Identify the delay angle actually applied. The paper quotes $\alpha_{min}=21^\circ$ as the smallest firing delay the gate controller will issue, and part (b) then speaks of “the delay angle” being adjusted to a larger value. Part (a) is therefore the converter running at its minimum delay, $\boxed{\alpha=21^\circ}$.
Get the extinction angle from the conduction angle. Conduction runs from the gate pulse to extinction, so $$\beta=\alpha+\gamma=21^\circ+145^\circ=166^\circ$$
Read the back EMF off the extinction condition. With a purely resistive load in series with $E_{c}$ there is no stored magnetic energy, so current ceases the moment the supply voltage falls back to the back EMF, $v_{s}(\beta)=E_{c}$: $$E_{c}=V_{m}\sin\beta=311.13\sin 166^\circ=\boxed{75.27\ \text{V}}$$

Before using that value, it is worth checking that the firing instant is physically reachable. The supply first rises through $E_{c}$ at $\arcsin(E_{c}/V_{m})=\arcsin(75.27/311.13)=14.00^\circ$, so a gate pulse at 21° arrives while the thyristor is already forward-biased and fires it immediately. The 21° minimum is a controller limit sitting above the 14° the load would permit, which is exactly the situation the phrase “minimum delay angle” describes.

Write the mean output current over the whole period. The converter is half-wave, so the average is taken over $2\pi$ and the load sits at $E_{c}$ outside the conduction window: $$I_{dc}=\frac{1}{2\pi R}\left[V_{m}(\cos\alpha-\cos\beta)-E_{c}\,(\beta-\alpha)\right]$$ with $(\beta-\alpha)$ in radians.
Substitute and solve for the resistance. $V_{m}(\cos 21^\circ-\cos 166^\circ)=311.13\times1.90388=592.34$ and $E_{c}\gamma=75.27\times2.53073=190.48$, so $$R=\frac{592.34-190.48}{2\pi\times 32}=\frac{401.86}{201.06}=\boxed{2.00\ \Omega}$$

The round 2 Ω is a strong sign that the reading of the data is the one the examiner intended. A loop check confirms it: the mean load voltage must be $E_{c}+I_{dc}R=75.27+32\times1.9987=139.23$ V, and integrating the actual output waveform — the supply during conduction, $E_{c}$ during the dead band — returns 139.23 V as well.

Re-fire later and note that the extinction angle does not move. The load, not the gate, fixes $\beta$, so with $\alpha_{2}=30^\circ$ the window becomes $$\gamma_{2}=\beta-\alpha_{2}=166^\circ-30^\circ=136^\circ$$
Recompute the mean current with the new lower limit. $V_{m}(\cos 30^\circ-\cos 166^\circ)=311.13\times1.73632=571.33$ and $E_{c}\gamma_{2}=75.27\times2.37365=178.67$: $$I_{dc,2}=\frac{571.33-178.67}{2\pi\times1.9987}=\boxed{31.27\ \text{A}}$$
Take the power absorbed by the back EMF. The resistance carries the copper loss and the back EMF absorbs the electromechanical conversion, so $$P_{E_{c}}=E_{c}I_{dc,2}=75.27\times31.268=\boxed{2353.5\ \text{W}}$$
Convert to horsepower. The power absorbed by $E_{c}$ is the developed mechanical power, because the armature copper loss has already been accounted for in $R$: $$P_{out}=\frac{2353.5}{745.7}=\boxed{3.16\ \text{hp}}$$

Final results
Quantity	Symbol	Result
Delay angle, part (a)	$\alpha$	21.0°
Extinction angle	$\beta$	166.0°
Back EMF	$E_{c}$	75.27 V
Armature-circuit resistance	$R$	2.00 Ω
Mean load voltage, part (a)	$V_{dc}$	139.23 V
Conduction angle, part (b)	$\gamma_{2}$	136.0°
Mean current, part (b)	$I_{dc,2}$	31.27 A
Power absorbed by $E_{c}$	$P_{E_{c}}$	2353.5 W
Shaft output, part (b)	$P_{out}$	3.16 hp

Check: which reading of “α_min = 21°” the answer uses. A second reading is arithmetically available — take $\alpha_{min}=\arcsin(E_{c}/V_{m})$, so that $E_{c}=311.13\sin 21^\circ=111.50$ V, $\beta=180^\circ-21^\circ=159^\circ$ and $\alpha=\beta-\gamma=14^\circ$. It is rejected for two reasons. It requires the gate pulse at 14°, which is 7° before the supply has risen through $E_{c}$, so the thyristor is still reverse-biased and cannot fire; the conduction angle would then be $159^\circ-21^\circ=138^\circ$, not the 145° stated. And it returns $R=1.543\ \Omega$ against the 1.9987 Ω — a round 2 Ω to four figures — that the reading used above produces. Worth noting that the two readings differ by only 0.7 per cent in $R$, because near the natural start the integrand is almost zero; the choice matters for $E_{c}$ and $\alpha$, not for the resistance.

Topic. Half-wave phase control into a resistive load in series with a back EMF.

Key relations. $V_{m}=\sqrt{2}V_{s}$; extinction where $v_{s}(\beta)=E_{c}$, i.e. $E_{c}=V_{m}\sin\beta$; mean current $I_{dc}=[V_{m}(\cos\alpha-\cos\beta)-E_{c}(\beta-\alpha)]/(2\pi R)$; loop check $V_{dc}=E_{c}+I_{dc}R$; developed power $P=E_{c}I_{dc}$.

Why this works. With no inductance in the circuit the current is proportional to the instantaneous excess of supply over back EMF, so it reaches zero at the same instant that excess does. The extinction angle is therefore a property of the load, and advancing or retarding the gate moves only the start of conduction. That single fact converts a two-unknown problem into a chain of one-line substitutions, and it is why part (b) needs no new data.

Common pitfall. Averaging over $\pi$ instead of $2\pi$. The full-wave version of this problem does divide by $\pi$; using it here halves the deduced resistance and doubles every downstream answer. The second pitfall is subtracting an efficiency or a friction allowance from $E_{c}I_{dc}$ to “get the shaft power” — the paper gives no such data, and the armature copper loss is already inside $R$.

Source. M. H. Rashid, Power Electronics: Circuits, Devices and Applications, 4th ed., Ch. 10 §10.2 (single-phase half-wave controlled rectifiers) and Ch. 15 §15.2 (single-phase dc drives).

Question 2 of 6: Problem 1 — controlled half-wave rectifier feeding a dc motor armature

Question 2: Problem 1 — controlled half-wave rectifier feeding a dc motor armature (15 points)