RTUEE / EC / EEEYr 2020 · Sem 62020

Q1Advanced Power Electronics

Question

16 marks

Q.1. (a) A single phase half wave ac voltage controller feeds a load of R = 20Ω, with an input voltage of 230V, 50Hz, firing angle of thyristor is 45°, determine: (i) RMS value of voltage (output) (ii) Power delivered to load & input PF (iii) Average input current [8]

(b) For a single phase full wave controller with RL load, draw the waveform of output voltage, output current and voltage across thyristors. Determine the expression for rms output voltage. [8]

Answer

For the given single-phase half-wave AC voltage controller (R=20Ω, 230V/50Hz, α=45°), RMS output voltage ≈ 155.07V, power delivered ≈ 1202.4W with input PF ≈ 0.674, and average input current ≈ 4.42A; for a full-wave RL-load controller, the output voltage/current waveforms exhibit an extended conduction period beyond the voltage zero crossing (due to load inductance), with RMS output voltage given by a standard integral expression in terms of α and the extinction angle β.

(a) Single-Phase Half-Wave AC Voltage Controller (R load)

Given: R = 20Ω, Vs = 230V (rms), f = 50Hz, firing angle α = 45° = π/4 rad. Peak supply voltage Vm = √2×230 = 325.27V.

(i) RMS output voltage: for a half-wave controller with a purely resistive load, the output voltage exists only from firing angle α to π (conducting for the positive half-cycle only, from α to π, since only one thyristor is used and no current flows during the negative half-cycle at all):

Substituting α = π/4 (45°): (π-α) = 2.356, sin(2×45°)/2 = sin(90°)/2 = 0.5, sum = 2.856, divided by π = 0.9091, square root = 0.9535:

(ii) Power delivered to load and input power factor:

Input (source) apparent power S = Vs×Irms = 230×7.754 = 1783.3 VA (using the full source rms voltage of 230V, since the source itself always supplies its full rated rms voltage regardless of the controller's firing angle):

(iii) Average input current: for a half-wave controller, the average (DC) component of the output voltage is:

Summary of results: RMS output voltage ≈ 155.07 V, power delivered ≈ 1202.4 W, input power factor ≈ 0.674 (lagging, due to the phase-controlled thyristor delaying current conduction relative to the source voltage), and average input current ≈ 4.42 A. Note that a half-wave controller feeding a purely resistive load produces a non-zero DC component in both output voltage and current, since only one half-cycle is controlled — a practical drawback of half-wave configurations, since this DC component can cause core saturation problems if the load were instead transformer- or inductor-based, which is one reason full-wave (bidirectional) AC voltage controllers are generally preferred in practice despite requiring twice the number of thyristors.

(b) Single-Phase Full-Wave AC Voltage Controller with RL Load

In a full-wave AC voltage controller supplying an RL (resistive-inductive) load, two thyristors (or a triac) are connected in anti-parallel, each conducting for one half-cycle. Due to the load inductance, the load current does not fall to zero exactly at the point where the supply voltage crosses zero — the inductor's stored energy sustains current flow for some additional angle β beyond the natural voltage zero crossing, called the extinction angle, which must be found by solving the load's differential equation i(t) = 0 at the conduction end point (a transcendental equation, generally requiring iterative/numerical solution for a given α, R, L, ω).

Waveforms: the output voltage waveform follows the supply sinusoid from firing angle α until the extinction angle β (both measured from the start of the respective half-cycle), and is zero for the remaining portion of each half-cycle until the next thyristor is fired at the start of the following half-cycle plus α; the output current waveform, governed by the RL circuit's exponential response, rises from zero at α, peaks, and decays back to zero at β (extending beyond π if β>π, i.e., if the conduction angle γ=β-α exceeds the natural π radians conduction that would occur for pure resistive load); the voltage across each non-conducting thyristor equals the full supply voltage during the interval it is blocking (reverse or forward blocking, depending on firing conditions), while the voltage across a conducting thyristor is approximately zero (a small forward drop).

Full-wave AC Voltage Controller with RL load (waveforms)VoαβioVthy

Expression for RMS output voltage: the RMS output voltage is found by integrating the squared instantaneous output voltage over one full cycle (accounting for conduction only between α and β in each half-cycle, by symmetry the same expression applies to both halves):

Evaluating this standard integral gives:

where β (extinction angle) must be determined for the specific R, L, ω and α values by solving the transcendental load current equation i(ωt)=0 at ωt=β (typically via iterative numerical methods, since β cannot generally be expressed in closed form). This general expression reduces to the earlier pure-resistive-load formula when L=0 (in which case β=π+α is not applicable; rather β=π exactly, since current follows voltage instantaneously with no lag for a pure resistance, giving the standard R-load-only RMS formula as a special case).

Derivation of the transcendental extinction-angle equation: the instantaneous load current for the RL branch, valid from ωt=α until the current decays back to zero at ωt=β, is obtained by solving the first-order differential equation governing the circuit during conduction:

with the standard particular-plus-complementary solution giving:

where Z = √(R²+(ωL)²) is the load impedance magnitude and φ = tan⁻¹(ωL/R) is the load impedance angle. Setting i(β)=0 in this expression yields the transcendental equation that must be solved (graphically or by Newton-Raphson iteration, since β appears both inside a sine term and inside an exponential term) to determine β for any given α, R, L and ω:

A key practical consequence of this relationship is that the conduction angle γ = β - α always lies between π (the value it would take for a purely resistive load, φ=0) and π+φ (its asymptotic upper bound as α approaches φ from above); when the firing angle α is reduced below the load impedance angle φ, the circuit enters continuous conduction, where the current in one thyristor has not yet reached zero by the time the next thyristor's gate pulse would normally be applied — in this condition, one thyristor conducts continuously and the firing pulses to the alternate device have no effect until α is increased back above φ, so the practical firing-angle control range for an RL-load full-wave AC voltage controller is effectively restricted to α ≥ φ for meaningful phase control to occur.

Power and power factor implications: because the load current for an RL load is not simply proportional to instantaneous supply voltage (unlike the resistive case), the real power delivered is:

which must be evaluated by direct numerical integration of the voltage-current product over the actual conduction interval, since Vrms and Irms alone (without knowledge of their relative phase/waveshape) do not fully capture the true power delivered when both waveforms are non-sinusoidal segments; this is an important distinction from the pure-resistive case, where v(t) and i(t) are always in phase and simple RMS-based power computation is valid throughout.

Practical design note: in commercial AC voltage controllers feeding inductive loads (induction motor soft-starters, transformer-coupled loads), the firing circuit must incorporate a means of detecting or estimating φ (or conservatively assume the worst-case, highest expected φ) to ensure the minimum firing angle command never falls below φ, otherwise the intended phase-control action is lost and the controller effectively behaves as an uncontrolled full-conduction bridge for that portion of the control range — a subtlety frequently tested in RTU examination numericals involving RL-load voltage controllers.

Back to Paper