Chapter 8 of 30
- How replacing diodes with thyristors turns a fixed rectifier into a controlled DC source, with the firing angle \( \alpha \) as the only control variable.
- The four circuits — half-wave, full converter, semiconverter, and the freewheeling variant — and the different formula each one needs.
- Why an inductive load changes \( V_{dc} \) from \( \frac{V_m}{\pi}(1+\cos\alpha) \) to \( \frac{2V_m}{\pi}\cos\alpha \), and why that is the most consequential distinction in the chapter.
- Continuous versus discontinuous conduction, and the critical inductance that separates them.
- Inverter mode: how a rectifier sends power back to the AC line when \( \alpha > 90^\circ \), and the two conditions that make it possible.
- Why the input power factor is \( 0.9\cos\alpha \) — and why the semiconverter beats the full converter on that measure.
- How all of it comes together in a DC motor drive, the application this circuit was built for.
One Change, and the Rectifier Becomes Adjustable
Every circuit in Chapters 6 and 7 delivered whatever DC voltage the supply happened to give. Useful — but a battery charger needs to taper its current, a motor needs to change speed, and a plating bath needs to hold a set current as the electrode area changes. None of that is possible with diodes.
Replace the diodes with thyristors and exactly one thing changes: conduction no longer begins at the crossover. It begins when you fire the gate.
Three families of circuit follow, and choosing between them is a real design decision rather than a matter of taste:
| Circuit | Devices | Output range | Quadrants | Can invert? |
|---|---|---|---|---|
| Half-wave | 1 SCR | \(0 \to 0.318V_m\) | One | No |
| Semiconverter (half-controlled bridge) | 2 SCR + 2 diodes | \(0 \to 0.637V_m\) | One | No |
| Full converter (fully controlled bridge) | 4 SCR | \(-0.637V_m \to +0.637V_m\) | Two | Yes |
| Dual converter | 8 SCR (two bridges) | \(\pm 0.637V_m\), either current direction | Four | Yes — Ch. 10 |
The Half-Wave Controlled Rectifier
One thyristor, one load. It is a poor converter and nobody builds it for power — but it is the only circuit in which every formula can be derived in two lines, so it is where the method is learned.
With a resistive load the thyristor conducts from \( \alpha \) to \( \pi \). Current follows voltage, so it reaches zero at \( \pi \) and the device commutates there.
- Average: integrate over what actually reaches the load
Note the \(2\pi\): only one half cycle is used, so the average is taken over the full period, exactly as in Chapter 6.Working\[ V_{dc} = \frac{1}{2\pi}\int_{\alpha}^{\pi} V_m\sin\theta\,d\theta = \frac{V_m}{2\pi}\bigl[-\cos\theta\bigr]_{\alpha}^{\pi} = \frac{V_m}{2\pi}(1 + \cos\alpha) \]
- RMS: the same integral with the square
Working\[ V_{rms} = \frac{V_m}{2}\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} \]
- Check the limits At \( \alpha = 0 \) these give \( V_m/\pi \) and \( V_m/2 \) — the uncontrolled half-wave results of Chapter 6, as they must. At \( \alpha = \pi \) both vanish. Always check that a controlled formula collapses to its uncontrolled version at \( \alpha = 0 \); it catches most algebra errors instantly.
Now make the load inductive, and the behaviour changes exactly as it did in Chapter 6. The current cannot stop at \( \pi \); it carries the thyristor into the negative half cycle until it extinguishes at the angle \( \beta \). The output goes negative over \( (\pi, \beta) \), and the average falls:
Add a freewheeling diode across the load and the negative excursion is clamped away, restoring \( V_{dc} = \frac{V_m}{2\pi}(1+\cos\alpha) \) and smoothing the current — the three-for-one bargain of Chapter 6, unchanged by the addition of gate control.
The Full Converter
Four thyristors in a bridge, fired in diagonal pairs: \( T_1T_2 \) for one half cycle, \( T_3T_4 \) for the next. This is the circuit that matters — the front end of every DC drive built between about 1960 and 2000, and still the standard at high power.
Everything now depends on one question, and getting it wrong is the single largest source of error in this chapter.
If yes — a highly inductive load, a motor armature, a battery — conduction is continuous and \( V_{dc} = \frac{2V_m}{\pi}\cos\alpha \). If no — a resistive or lightly inductive load — conduction is discontinuous and \( V_{dc} = \frac{V_m}{\pi}(1+\cos\alpha) \). The two agree only at \( \alpha = 0 \), and diverge completely by \( \alpha = 90^\circ \), where one gives zero and the other gives \( 0.318V_m \).
Continuous conduction: where the negative area comes from
Suppose the load inductance is large enough that \( i_o \) never falls to zero. Then some pair of thyristors must always be conducting, because the current needs a path. So when \( T_1T_2 \) are fired at \( \alpha \), they must keep conducting until \( T_3T_4 \) are fired at \( \pi + \alpha \) — even after the supply voltage has reversed at \( \pi \).
During \( (\pi,\ \pi+\alpha) \) the load is therefore connected to a negative supply. The output voltage goes negative, and that area subtracts from the average:
- Integrate over one conduction interval of width \( \pi \)
Working\[ V_{dc} = \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha} V_m\sin\theta\,d\theta = \frac{V_m}{\pi}\bigl[-\cos\theta\bigr]_{\alpha}^{\pi+\alpha} \]
- Use \( \cos(\pi+\alpha) = -\cos\alpha \)
Full converter, continuous conduction\[ V_{dc} = \frac{V_m}{\pi}\bigl(\cos\alpha + \cos\alpha\bigr) = \frac{2V_m}{\pi}\cos\alpha \]
- Read what the cosine is telling you At \( \alpha = 0 \), \( V_{dc} = 2V_m/\pi \) — the uncontrolled bridge. At \( \alpha = 90^\circ \) the positive and negative areas are exactly equal and \( V_{dc} = 0 \). Beyond \( 90^\circ \), \( V_{dc} \) is negative, and the converter is no longer rectifying. That is Section 8-5.
A fully controlled bridge feeding a highly inductive load, so conduction never stops. Move \( \alpha \) and watch the negative area grow. At \( \alpha = 90^\circ \) it exactly balances the positive area; beyond that the converter starts returning power to the supply.
—
Continuous and Discontinuous Conduction
Since the entire analysis hinges on which mode the converter is in, it is worth knowing what decides it — and it is not simply "is there an inductor".
The current ripples about its average value. Conduction stays continuous as long as the trough of that ripple stays above zero. Three things therefore push the converter toward discontinuous operation:
- Light load — the average current falls toward the ripple amplitude. This is the usual cause.
- Small inductance — the ripple amplitude grows.
- Large firing angle — the applied voltage waveform is more distorted, so the ripple is larger for the same average.
Inverter Mode: Sending Power Back
Set \( \alpha > 90^\circ \) in a continuously conducting full converter and \( V_{dc} \) is negative. That deserves a careful reading, because it is the most useful and most misunderstood result in the chapter.
The thyristors are unidirectional, so the current cannot reverse. If the voltage has reversed and the current has not, then the product \( V_{dc}I_o \) is negative — power is flowing out of the DC side and into the AC supply. The converter has become a line-commutated inverter.
The second condition is the one people forget. A passive load cannot supply power, so a resistor, or an inductor once its stored energy is gone, can never sustain inversion. The DC side must contain a battery, a rotating machine acting as a generator, or another converter — something that can push current against the reversed voltage.
| Firing angle \(\alpha\) | Output \(V_{dc}\) | Current \(I_o\) | Power flow | Machine is |
|---|---|---|---|---|
| \(0 \to 90^\circ\) | Positive | Positive | AC → DC | Motoring |
| \(90^\circ\) | Zero | Positive | None (net) | Coasting at the boundary |
| \(90^\circ \to 180^\circ\) | Negative | Positive | DC → AC | Generating — regenerative braking |
The Semiconverter and the Freewheeling Path
Replace two of the four thyristors with diodes — the two in the lower half of the bridge — and you have a semiconverter, or half-controlled bridge. It gives up inversion, and gets something valuable in return.
The key behavioural change is that the bridge now has an inherent freewheeling path. When the supply reverses at \( \pi \), the conducting thyristor and the diode of the same leg can carry the load current in a local loop, disconnecting the load from the supply. The output is clamped at zero instead of going negative.
The integral runs only to \( \pi \), because after that the load is freewheeling and contributes nothing. \( V_{dc} \) is therefore always positive, falling from \( 2V_m/\pi \) at \( \alpha = 0 \) to zero at \( \alpha = 180^\circ \) — but never reversing. One quadrant only.
| Property | Full converter | Semiconverter |
|---|---|---|
| Devices | 4 thyristors | 2 thyristors + 2 diodes |
| \(V_{dc}\) | \(\frac{2V_m}{\pi}\cos\alpha\) | \(\frac{V_m}{\pi}(1+\cos\alpha)\) |
| Output at \(\alpha=90°\) | 0 | \(0.318V_m\) |
| Quadrants | Two | One |
| Inversion | Yes | No |
| Input displacement factor | \(\cos\alpha\) | \(\cos(\alpha/2)\) |
| Input PF at \(\alpha=60°\) | 0.45 | 0.83 |
| Cost | Higher — 4 SCRs, 4 gate drives | Lower — 2 SCRs, 2 gate drives |
What the Supply Sees
Chapter 6 found a diode bridge drawing a square-wave current in phase with the voltage: unity displacement, 0.9 power factor, all of the deficit being distortion. Phase control keeps the distortion and adds a displacement problem.
With continuous conduction the full converter's supply current is the same square wave of amplitude \( I_o \) — but shifted bodily by \( \alpha \), because that is when conduction now starts.
The distortion factor is fixed at \( 2\sqrt2/\pi = 0.900 \) by the square-wave shape and cannot be improved by phase control. The displacement factor is \( \cos\alpha \), and it falls fast. At \( \alpha = 60^\circ \) the power factor is already 0.45, and the converter is drawing more than twice the current a unity-power-factor load would need for the same real power.
Worked Examples
Problem. A single-phase full converter runs from 230 V, 50 Hz at \( \alpha = 60^\circ \). Find \( V_{dc} \) (a) with a purely resistive load, (b) with a highly inductive load.
Solution. \( V_m = \sqrt2(230) = 325 \) V. The load decides the formula:
A 50% difference, from nothing but the load. The inductive case is lower because the current is forced to keep flowing after the supply reverses, so a negative area is subtracted.
Sanity check both at \( \alpha = 0 \): (a) gives \( 2V_m/\pi = 207 \) V, (b) gives \( 2V_m/\pi = 207 \) V. They agree, as they must — with no firing delay there is no negative excursion to subtract.
Problem. A 220 V, 15 A DC motor with \( R_a = 0.5\ \Omega \) is fed from a single-phase full converter on a 230 V, 50 Hz supply. The armature is highly inductive. Find the firing angle for rated operation, and the back-EMF at that point.
Solution. The converter must supply the armature circuit equation \( V_{dc} = E_a + I_aR_a \). At rated conditions \( V_{dc} = 220 \) V, so:
The cosine exceeds one, so this is impossible. The converter's maximum output is \( 2V_m/\pi = 207 \) V at \( \alpha = 0 \), which is less than the motor's 220 V rating.
This is a genuine and common design outcome, not an arithmetic slip — a 230 V AC supply cannot drive a 220 V DC motor through a single-phase bridge. Three ways out:
- Step up the AC supply with a transformer: \( V_L \ge \pi(220)/(2\sqrt2) = 244 \) V, so use 250 V with margin.
- Accept reduced ratings — run the motor at 207 V and correspondingly lower base speed.
- Use a three-phase bridge (Chapter 9), which gives 560 V from a 415 V line and has ample headroom.
Continuing with a 250 V supply (\( V_m = 354 \) V): \( \cos\alpha = \pi(220)/(2 \times 354) = 0.976 \), so \( \alpha = 12.6^\circ \), and \( E_a = 220 - (15)(0.5) = 212.5 \) V.
Problem. The motor of Example 2 (on the 250 V supply) is to be braked regeneratively at 10 A while its back-EMF is 150 V. Find the required firing angle and the power returned to the supply.
Solution. To brake, the machine's field connections are reversed so that its EMF opposes the current — equivalently, take \( E_a = -150 \) V in the armature equation. The converter must then produce:
Note \( \alpha = 130^\circ \) is comfortably inside the practical limit of about 150–165°, so the margin angle is respected.
Note also the sign convention that makes this work: \( V_{dc} \) is negative and \( I_a \) is positive, so \( P = V_{dc}I_a \) is negative — power leaving the DC side. The 1.45 kW comes from the rotating mass and goes back into the utility, instead of into a braking resistor.
Problem. Both circuits deliver 120 V DC at 20 A from a 230 V, 50 Hz supply into a highly inductive load. Find the firing angle, the supply RMS current and the input power factor for each.
Solution. \( V_m = 325 \) V. Solve each for \( \alpha \), then evaluate its own supply-current relations:
The semiconverter delivers the same 2.4 kW while drawing 15.2 A instead of 20 A — a 24% reduction in supply current — with a power factor of 0.685 against 0.522.
It reaches that at a larger firing angle, which seems paradoxical until you remember why: during the freewheeling interval the supply carries no current at all, so the current it does draw is both smaller and better aligned with the voltage. If this load never needs to regenerate, the semiconverter is the better circuit — and it uses two fewer thyristors.
Problem. A full converter with continuous conduction runs from 230 V. Find \( V_{rms} \) and the RMS ripple content of the output at \( \alpha = 0^\circ \), \( 60^\circ \) and \( 90^\circ \).
Solution. Each conduction interval is a complete half cycle of the supply, merely shifted, so \( V_{rms} = V_m/\sqrt2 = 230 \) V at every firing angle. The DC content changes, so the ripple is what is left:
The ripple factor \( V_r/V_{dc} \) goes 0.48 → 2.00 → ∞. Phase control buys adjustability and pays for it in waveform quality, and the bill grows steeply with \( \alpha \).
This is why a phase-controlled DC drive needs a smoothing reactor sized for its lowest speed, not its rated one — and why a converter operated deep into phase control causes far more motor heating, from harmonic currents, than its DC output would suggest.
Problem. A full converter on 230 V, 50 Hz at \( \alpha = 45^\circ \) feeds a load of \( R = 10\ \Omega \) in series with \( L = 50 \) mH and a 100 V battery. Estimate whether conduction is continuous.
Solution. First the operating point, assuming continuous conduction:
Now compare the load's AC impedance at the dominant ripple frequency (\( 2f = 100 \) Hz) with its resistance:
The inductive reactance at the ripple frequency is about three times the resistance, so the ripple current is strongly attenuated — its peak-to-peak value works out at roughly 2 A against a 4.6 A average. The trough stays comfortably above zero, so conduction is continuous and the assumption used to find \( I_o \) is self-consistent.
The general test: assume continuous conduction, compute \( I_o \), estimate the ripple amplitude from \( V_r/(\omega_r L) \), and check that \( I_o \) exceeds half of it. If not, the discontinuous equations apply and the whole calculation must be redone — which is exactly why designers fit enough inductance to make the question disappear.
Summary & Formula Sheet
\( \alpha \) moves the instant conduction starts. Lossless control, and turn-off is free by line commutation.
Continuous: \( V_{dc} = \frac{2V_m}{\pi}\cos\alpha \). Discontinuous: \( \frac{V_m}{\pi}(1+\cos\alpha) \). They differ in sign beyond 90°.
With continuous conduction \( V_{rms} = V_m/\sqrt2 \) whatever \( \alpha \) is. All the change is in the DC content.
\( \alpha > 90° \) reverses \( V_{dc} \) but not \( I_o \), so power returns to the line — if a DC-side EMF sustains the current.
Freewheeling clamps the output positive: \( \frac{V_m}{\pi}(1+\cos\alpha) \), one quadrant, displacement \( \cos(\alpha/2) \).
\( \mathrm{PF} = 0.9\cos\alpha \). The lag comes from the firing delay itself, in both rectifier and inverter modes.
Key terms
- Firing angle, \(\alpha\)
- The delay from the instant a device becomes forward biased to the instant it is gated. The control variable of every phase-controlled converter.
- Conduction angle, \(\gamma\)
- How long a device actually conducts. Decided by the circuit and the load, not commanded. \(\pi\) for a continuously-conducting full converter.
- Extinction angle, \(\beta\)
- The angle at which current in an RL load finally reaches zero, beyond \(\pi\). Found from a transcendental equation.
- Continuous conduction
- Load current never reaches zero, so some device is always on. Requires enough inductance and enough load current.
- Full converter
- Fully controlled bridge, four thyristors. Two-quadrant: output voltage can reverse, output current cannot.
- Semiconverter
- Half-controlled bridge, two thyristors and two diodes. One quadrant, inherent freewheeling, better power factor.
- Line-commutated inverter
- A full converter at \(\alpha > 90^\circ\) returning power to the AC supply. Needs a DC-side EMF to sustain the current.
- Margin angle
- The angle left before \(180^\circ\), covering the thyristor's \(t_q\) and the commutation overlap. Too small means commutation failure.
- Displacement factor
- \(\cos\phi_1\), the phase shift of the fundamental supply current. Equals \(\cos\alpha\) for a full converter — caused by the firing delay itself.
- Smoothing reactor
- Inductance added to the DC circuit to keep conduction continuous down to the minimum load, giving the control loop one consistent plant.
Test Yourself
Six questions on the converter that drove industry for forty years. Answer each before revealing it.
A full converter at \( \alpha = 120^\circ \) feeds a 10 Ω resistor. What is the output voltage?
Use the discontinuous formula, because a resistor cannot sustain current past \( \pi \):
\( V_{dc} = \frac{V_m}{\pi}(1+\cos 120°) = \frac{V_m}{\pi}(0.5) = 0.159V_m \) — small, but positive.
The trap: applying \( \frac{2V_m}{\pi}\cos 120° = -0.318V_m \) gives a negative voltage across a resistor. That would mean current flowing backwards through thyristors, which cannot conduct in reverse. The formula has been used outside its assumptions.
The deeper point: a resistive load can never make this converter invert, at any firing angle. Inversion needs stored or generated energy on the DC side, and a resistor has neither. Whenever a controlled-rectifier answer comes out with an impossible sign, the first suspect is the continuous-conduction assumption.
Why does \( V_{rms} \) stay at \( V_m/\sqrt2 \) for a continuously-conducting full converter, when \( V_{dc} \) changes from \( 0.637V_m \) to zero?
Because squaring destroys the sign, and the conduction interval is always a full half cycle.
Each pair conducts for exactly \( \pi \), so the output over that interval is a complete half sine — just starting at a different place. Firing later moves which part of the sine appears, but not how much of it. Mean-square is therefore \( V_m^2/2 \) regardless.
The average, however, cares very much about sign: the portion after \( \pi \) is negative and subtracts.
The consequence to remember: the ripple content \( \sqrt{V_{rms}^2 - V_{dc}^2} \) grows as \( \alpha \) increases, reaching 100% of the RMS at \( \alpha = 90^\circ \). A phase-controlled drive running slowly is running with an output that is mostly ripple — which is why its smoothing reactor must be sized at the lowest speed, and why the motor heats more than its shaft power suggests.
A drive's speed controller is tuned at full load and behaves perfectly. At 10% load it oscillates. What has changed?
Almost certainly the converter has entered discontinuous conduction, and the controller is now facing a different plant.
- In continuous conduction \( V_{dc} = \frac{2V_m}{\pi}\cos\alpha \) — output depends only on \( \alpha \), and the gain \( dV_{dc}/d\alpha \) is predictable.
- In discontinuous conduction the output also depends on the load current, and the incremental gain rises steeply — often several-fold.
A loop tuned for the lower gain becomes under-damped when the gain jumps, and oscillates. The symptom is characteristic: stable at load, unstable at light load.
Three fixes, in order of preference:
- Add a smoothing reactor sized to keep conduction continuous down to the minimum expected current. This removes the problem rather than compensating for it.
- Gain-schedule the controller — detect discontinuous operation and reduce the loop gain there.
- Add a small permanent load so the current never falls that far. Crude, and it wastes power, but it is sometimes the cheapest retrofit.
A full converter is inverting at \( \alpha = 150^\circ \) when the AC supply voltage dips by 15%. What happens, and why is it dangerous?
A commutation failure is very likely, and it is one of the classic destructive faults in line-commutated converters.
Commutation relies on the incoming thyristor having a more positive anode voltage than the outgoing one, for long enough to sweep out the outgoing device's stored charge and let it regain blocking capability — the turn-off time \( t_q \) of Chapter 3. Two things now go wrong at once:
- The reduced supply voltage slows the current transfer, so the commutation overlap \( \mu \) grows.
- At \( \alpha = 150^\circ \) the margin angle was only \( 30^\circ \) — about 1.7 ms at 50 Hz — and the growing overlap eats into it.
If the margin is consumed, the outgoing thyristor is still conducting when forward voltage returns. It stays on, the incoming device is also on, and the pair short-circuits the supply through the DC-side EMF.
Why it is dangerous: the fault current is limited only by the supply and transformer impedance, and it rises within a half cycle. The DC-side EMF keeps feeding it. This is why inverter-mode controllers include a hard \( \alpha \) end-stop, supply-voltage monitoring that retards firing on a dip, and fast \( I^2t \)-coordinated fuses — the protection chain of Chapter 4.
Both a full converter and a semiconverter can deliver 0.318\(V_m\). Which uses the smaller firing angle, and which is the better choice?
Full converter: \( 0.318V_m = \frac{2V_m}{\pi}\cos\alpha \Rightarrow \cos\alpha = 0.5 \Rightarrow \alpha = 60^\circ \).
Semiconverter: \( 0.318V_m = \frac{V_m}{\pi}(1+\cos\alpha) \Rightarrow \cos\alpha = 0 \Rightarrow \alpha = 90^\circ \).
The full converter uses the smaller angle — but that is not the figure of merit. Compare what the supply sees:
- Full converter: \( \mathrm{PF} = 0.9\cos 60° = 0.45 \), and the supply carries \( I_o \) for the whole cycle.
- Semiconverter: displacement \( \cos 45° = 0.707 \), supply current only \( I_o\sqrt{(\pi - \pi/2)/\pi} = 0.707I_o \), giving \( \mathrm{PF} \approx 0.64 \).
So the semiconverter is better on power factor, supply current, device count and gate-drive complexity — unless the load must regenerate. That single requirement is what the extra two thyristors buy, and it is the whole basis of the choice.
Why does a phase-controlled converter draw lagging reactive power even when it is exporting real power to the grid?
Because the reactive demand comes from the firing delay, and the firing delay is what makes inversion possible in the first place.
The supply current is a square wave shifted by \( \alpha \) from the voltage. Its fundamental therefore lags by \( \alpha \), and lagging fundamental current is absorbed reactive power — regardless of which way the real power happens to be flowing. In inverter mode \( \alpha > 90^\circ \), so the lag is larger than in rectifier mode, and the reactive demand is worse.
Why it cannot be designed away. A line-commutated converter needs the supply to turn its thyristors off, which requires the device current to lag the supply voltage. Absorbing reactive power is the price of getting free commutation — the two are the same phenomenon.
Where this bites hardest: an HVDC terminal transmitting 1000 MW may need 500–600 MVAr of compensation — banks of capacitors and filters occupying a large part of the site. And it is the central argument for voltage-source HVDC using IGBTs, which self-commutates and can therefore control its reactive power independently, or even supply it to a weak grid.
Problems
Before any algebra, settle three things — they select the formula for you:
- Which circuit? Half-wave, full converter, or semiconverter.
- Continuous or discontinuous? Look at the load, not the bridge.
- Rectifying or inverting? If \( \alpha > 90^\circ \), check there is a DC-side EMF to sustain the current.
Problems 1–5 are direct application; 6–10 need judgement; 11–12 are design questions worth discussing in a tutorial.
- A single-phase full converter runs from 230 V, 50 Hz. Find \( V_{dc} \) at \( \alpha = 0°, 30°, 60°, 90°, 120° \) for (a) a highly inductive load, (b) a resistive load. Tabulate and comment on where they diverge.
- A half-wave controlled rectifier on 230 V feeds a resistive load at \( \alpha = 45° \). Find \( V_{dc} \), \( V_{rms} \), the form factor and the ripple factor.
- A semiconverter on 230 V feeds a highly inductive load. Find the firing angle needed for a 150 V output, and the supply RMS current if the load draws 12 A.
- For the converter of Problem 3, find the input power factor, and compare it with a full converter delivering the same 150 V.
- A full converter with continuous conduction runs from 415 V single-phase at \( \alpha = 75° \). Find \( V_{dc} \), \( V_{rms} \) and the RMS ripple voltage.
- A 180 V, 20 A DC motor with \( R_a = 0.4\ \Omega \) is fed from a single-phase full converter. Determine the minimum AC supply voltage that allows rated operation with at least 10° of firing-angle margin.
- The motor of Problem 6 is braked regeneratively at 15 A with a back-EMF of 120 V. Find the firing angle and the power returned to the supply.
- Explain, with reference to the negative area of the output waveform, why a resistive load can never make a full converter invert — at any firing angle.
- A full converter delivers 100 V at 25 A. Find the input power factor, the apparent power drawn, and the reactive power. Repeat at 200 V and comment on how the supply loading changes with speed in a drive.
- A converter feeds an RL load with \( R = 8\ \Omega \), \( L = 20 \) mH from 230 V at \( \alpha = 60° \). Estimate whether conduction is continuous, and state what you would change if it is not.
- A drive must run a fan in one direction only, never regenerating, from 415 V single-phase. Choose between a full converter and a semiconverter, justify the choice on at least four grounds, and state what you would lose by choosing the other.
- An inverter-mode converter is limited to \( \alpha_{max} = 155° \). Given a thyristor with \( t_q = 100\ \mu\text{s} \) on a 50 Hz supply, find the margin angle in degrees and the overlap angle that would just consume it. Discuss what happens during a supply-voltage dip.