THEORY EXAMINATION (SEM–VI) 2016-17 INTRODUCTION TO ELECTRIC DRIVES

INTRODUCTION TO ELECTRIC DRIVES (EEC604)

SECTION – A

(10 × 2 = 20 Marks | Short Answers)

(a) Necessary conditions for turning ON of an SCR

An SCR turns ON when it is forward biased and a gate pulse of sufficient magnitude is applied. The anode current must also exceed the latching current.

(b) Are turn-on and turn-off times constant?

No, turn-on and turn-off times of a thyristor are not constant. They depend on factors such as supply voltage, load current, junction temperature, gate current, and circuit inductance.

(c) Why is power flow control called phase-controlled converter method?

Because the average output voltage and power flow are controlled by varying the firing angle (α) of the thyristor with respect to the AC supply phase.

(d) Advantages of single-phase bridge converter over mid-point converter

A bridge converter:

Utilizes the transformer better                      Requires no center-tapped transformer

Provides higher output voltage                     Has lower PIV requirement per thyristor

(e) What is meant by TRC?

TRC (Turn-off Time Requirement / Circuit) refers to the minimum time required after current zero for a thyristor to regain its forward blocking capability.

(f) Why diodes are connected anti-parallel with thyristors in inverter circuits?

Anti-parallel diodes provide a path for reverse or reactive current, protect thyristors from voltage spikes, and ensure continuity of current in inductive loads.

(g) Advantages of cycloconverter for synchronous motor speed control

Cycloconverters provide:                                 Direct AC-to-AC conversion

Smooth low-speed operation                         High efficiency

Regenerative braking capability

(h) Main disadvantages of armature resistance control

This method causes:                                         Large power loss in resistance

Poor efficiency                                                  Poor speed regulation

Heating of resistors

(i) Define dynamic braking

Dynamic braking is a method in which a motor is disconnected from the supply and its kinetic energy is dissipated in an external resistance.

(j) Methods of starting of synchronous motor

Damper winding (induction motor starting)   Pony motor method

Variable frequency supply method

SECTION – B

(Attempt Any Five | 5 × 10 = 50 Marks)

(a) Turn-off characteristics and triggering methods of thyristor

Turn-off characteristics:
After current zero, excess charge carriers must recombine before the SCR can block forward voltage. This time is called turn-off time (tₒff).

Triggering methods:

Gate triggering                                Forward voltage triggering

dv/dt triggering                               Temperature triggering

Light triggering

(b) V-I characteristics of thyristor, latching & holding current

SCR has three regions:                      Reverse blocking

Forward blocking                              Forward conduction

Latching current: Minimum current to keep SCR ON immediately after triggering.
Holding current: Minimum current below which SCR turns OFF.

(c) Why freewheeling diode improves power factor

A freewheeling diode provides an alternate path for inductive current during negative half cycle, reducing reactive power drawn from supply and improving power factor.

(d) 1-ϕ full wave fully controlled rectifier with R-L-E load

Continuous conduction mode:

Vdc=2Vmπcos⁡αV_{dc} = \frac{2V_m}{\pi}\cos\alphaVdc​=π2Vm​​cosα

Discontinuous conduction mode:
Output voltage depends on conduction angle and load parameters.

Freewheeling improves waveform smoothness. Relevant voltage and current waveforms should be drawn in exam.

(e) Operation of 3-ϕ half-wave controlled rectifier

Each thyristor conducts for 120°. Output voltage is obtained by phase-controlled switching. The rectifier gives lower ripple compared to single-phase rectifier. Waveforms of phase voltage, output voltage, and current are drawn.

(f) Chopper – working and control strategies

A chopper converts fixed DC into variable DC.

Control strategies:                                       Time ratio control

Current limit control

Choppers are widely used in DC motor speed control.

(g) Speed control of 3-ϕ induction motor using semiconductor devices

Methods include:                                            Stator voltage control

V/f control                                                       Slip power recovery

Rotor resistance control                                  Cycloconverter and inverter control

(h) Slip recovery control scheme

Slip power from rotor circuit is recovered and fed back to the supply, improving efficiency. Kramer and Scherbius drives are common slip recovery methods.

SECTION – C

(Attempt Any Two | 2 × 15 = 30 Marks)

3) 120° Mode of 3-ϕ Inverter and Comparison with 180° Mode

In 120° conduction mode, each thyristor conducts for 120°. Only two switches conduct at any time.
Advantages: lower switching loss, simpler commutation.

In 180° mode, three switches conduct simultaneously, giving higher output voltage and smoother waveform.

Comparison:

120° mode → lower output voltage, less harmonics

180° mode → higher output voltage, better utilization

Waveforms of phase and line voltages should be drawn.

4) Principle and Working of Cycloconverter

A cycloconverter converts fixed AC frequency to variable AC frequency directly.

(i) Step-down cycloconverter

Output frequency is less than input frequency. Widely used in low-speed high-power drives.

(ii) Step-up cycloconverter

Output frequency is higher than input frequency but rarely used due to complexity and poor waveform quality.

5) Numerical Problem – DC Motor Fed from 1-ϕ Full Converter

Given:
V = 220 V,
Rₐ = 0.5 Ω,
K = 0.4 V-s/rad,
Speed = 1500 rpm = 157 rad/s,
Torque = 20 Nm

Back emf:

E=Kω=0.4×157=62.8 VE = K\omega = 0.4 \times 157 = 62.8 \text{ V}E=Kω=0.4×157=62.8 V

Armature current:

Ia=TK=200.4=50 AI_a = \frac{T}{K} = \frac{20}{0.4} = 50 \text{ A}Ia​=KT​=0.420​=50 A

Average output voltage:

Vdc=E+IaRa=62.8+25=87.8 VV_{dc} = E + I_a R_a = 62.8 + 25 = 87.8 \text{ V}Vdc​=E+Ia​Ra​=62.8+25=87.8 V

(i) Firing angle

Vdc=2Vmπcos⁡αV_{dc} = \frac{2V_m}{\pi}\cos\alphaVdc​=π2Vm​​cosα α=cos⁡−1(87.8π2×311)\alpha = \cos^{-1}\left(\frac{87.8\pi}{2 \times 311}\right)α=cos−1(2×31187.8π​)

(ii) RMS thyristor current

IT(rms)=Ia2I_{T(rms)} = \frac{I_a}{\sqrt{2}}IT(rms)​=2​Ia​​

(iii) Input power factor

PF=cos⁡α×displacement factor\text{PF} = \cos\alpha \times \text{displacement factor}PF=cosα×displacement factor

(All steps written clearly for exam.)