Test: AC Voltage Controllers - Electrical Engineering (EE) MCQ

Voltage controllers convert the fixed ac voltage to variable ac by changing the values of the firing angle.

For controlling the load, the minimum value of firing angle α = load phase angle

Range of firing angle for which output voltage is uncontrollable 0 < α ≤ ø
0 < α ≤ 30°

 A tap changing transformer can give variable ac from fixed ac without a change in frequency.

In electric heating, variable ac supply is needed. The devices are fired appropriately to apply enough temperature.

For resistive load, peak power = 

 = 10,580 W

As it is half wave, it consists of one SCR ( the control element) in anti parallel with one diode.

As T1 is triggered at an angle α, the conduction in the positive have cycle will start at α. In the negative half cycle, the diode is forward biased and load is connected as it is to the supply.

Single-phase induction motors work on single-phase AC supply. It has two windings; Main windings and auxiliary winding. For speed control of a single-phase induction motor by an ac voltage controller, and only a fundamental component of current is required to develop a toque. The harmonics in the motor current may cause power losses resulting in heating of the motor.

As it consists of two SCRs, it is a full wave controller.

As it consists of two SCRs, it is a full wave controller and both the half cycles can be controlled by varying their respective firing angles.

The diode conducts from π to 2π in the negative half cycle. If it were a non resistive load, the diode would conduct for less than 90°, as the inductor would force conduct the SCR for some time.

Vo = (1/2π) ∫Vm sinωt d(ωt). Where the integration would be run from α to 2π as the conduction takes place from α to 2π.
Vo = (Vm/2π) (cosα – 1).

As the positive half is chopped off due to some value of firing angle, the firing angle is not equal to zero. As the negative half is a half sine wave, either it is a full wave controller with firing angle for T2 set to zero or it is a half wave controller with a Thyristor (T1) and a diode.

 T2 is triggered at π + α, it conducts from π + α to 2π after which it is line commutated.

As the control is in both the directions or cycles, it is a full wave ac voltage controller circuit.

 In integral cycle control, power is delivered for m cycles and not delivered for n cycles. Hence, the average value of the power delivered is controlled by manipulating m and n.

Over a complete cycle of 2π x (on cycles + off cycles) the power is delivered for n cycles.
Hence, P = n/2π(m+n).

Vrms = [ (k/2π) x ∫Vm² sin⁡² ωt d(ωt) ] ^1/2 
Where the integral runs from 0 to 2π.
Vrms = (Vm/√2) k = Vs x √k.

 Output voltage = Vs x √k
P = (Vs x √k)²/R.

As each SCR conducts for π radians during each cycle of n on cycles, the average value of SCR current is
I = (k/2π) ∫Im sin ωt d(ωt). Where the integration is from 0 to π.

Vo = [(√2 x 230) x (cos45 – 1)]/2π = -15.17 V.
Negative value is due to the fact that the average value in the positive half cycle is less than that in the negative half cycle.

Vo = [(√2 x 230) x (cos45 – 1)]/2π = -15.17 V.
Io = (Vo)/R = -0.7585
Negative value is due to the fact that the average value in the positive half cycle is less than that in the negative half cycle.

 Vrms = Vo x √k
k = (6/6+4) = 6/10 = 0.6
Vrms = √0.6 x 230 = 178.157 V.

k = (6/6+4) = 6/10 = 0.6
input pf = √0.6 = 0.7746.

Vrms = Vo x √k
k = (6/6+4) = 6/10 = 0.6
Vrms = √0.6 x 230 = 178.157 V
P = (Vrms)²/R = 2116 W = 2.116 kW.

Peak current Im = (230 x √2)/15 = 21.681 A
k = (6/6+4) = 6/10 = 0.6
Avg current = (k Im)/π = 4.14 A.

In RL loads with pulse gating, the incoming SCR may be fired during the interval when it is reversed biased by the outgoing SCR, thus it won’t get turn on even after the outgoing SCR is not reveres basing the incoming SCR is the pulse width is over before forward biasing the SCR.

As the gating is applied for a longer duration, the device is heated up.

In high frequency gating a train of pulses are used to overcome the thermal problems due to continuous gating.

As firing angle is 90, there is ideally be no conduction in the positive half. Hence, the average value will be zero.
Vo = (√2 Vs)/2π x (cos90 – 1) = – 51.8 V.