Test: Single Phase Full wave mid point converters - Electrical Engineering (EE) MCQ

The maximum voltage across SCRs in mod point connection is 2 Vm. So the maximum voltage of SCR = = 400 V

The maximum average power that the midpoint converter can handle is 

= 12.732 kW

Concept:

Peak inverse voltage: It is a voltage across the diode when it is reversed bias.

The following circuit is the center tapped full-wave rectifier.

Consider for positive half cycle, D₁ is FB and D₂ is reversed biased.

In loop 1, Apply KVL

⇒ V₀ = V_in

Now, apply KVL in loop 2

V_in + V₀ = V_D2

⇒ V_D2 = V_in + V_in = 2V_in

PIV of the centre tapped full wave rectifier = 2V_in

Calculation:

Given that, the center-tapped full-wave rectifier and V_in = 200 V (Peak) between the centre tap and one end of secondary.

Peak inverse voltage = 2V_in = 400 V

Concept:

Center tapped full wave rectifier:

• The Center tapped full-wave rectifier is a device used to convert the AC input voltage into DC voltage at the output terminals.
• It employs a transformer with the secondary winding tapped at the center point. And it uses only two diodes, which are connected to the opposite ends of a center-tapped transformer as shown in the figure below.
• The center tap is usually considered as the ground point or the zero voltage reference point.

Analysis: The DC output voltage or average output voltage can be calculated as follows,

 

V₀ = 2V_m / π 

Now we can calculate the average or mean current of load by dividing the average load voltage by load resistance RL. Therefore mean load current is given by

I₀ = V₀ / RL

If the internal resistance of the diode is given in that case mean load current I₀ = V₀ / (R_L + r)

Where r = internal resistance of the diode.

Calculation:

Given that 

Rms value of supply voltage V = 50 V

The internal resistance of diode r = 20 Ω 

The load resistance R_L = 980 Ω 

Maximum voltage on the secondary side Vm = √2 V = √2 × 50 = 70.7 V

Average or DC output voltage V0 = (2 × 70.7) / π = 45 V

Average or mean load current is

I₀ = V₀ / (R_L + r) = 45 /(980 + 20) = 45 mA

Concept: 

Center tapped full wave rectifier:

• The Center tapped full-wave rectifier is a device used to convert the AC input voltage into DC voltage at the output terminals.
• It employs a transformer with the secondary winding tapped at the center point. And it uses only two diodes, which are connected to the opposite ends of a center-tapped transformer as shown in the figure below.
• The center tap is usually considered as the ground point or the zero voltage reference point.
  

Analysis:

The DC output voltage or average output voltage can be calculated as follows,

V₀ = 2V_m / π 

Now we can calculate the average or mean current of load by dividing the average load voltage by load resistance RL. Therefore mean load current is given by

I₀ = V₀ / RL

Current through each diode will be half of the load current, as each diode conducts only for the half cycle.

Calculation:

Load current = I₀ = 4.2 A

Current through each diode = I₀ / 2 = 4.2 / 2 = 2.1 A

∴The current ratings of the each diode is 2.1 A.

A single-phase mid-point full-wave SCR converter:

In this circuit, two thyristors are used and are connected as shown below

Where,

V_s is the source voltage

V_m is the maximum voltage of the midpoint 

Let α be the firing angle 

V₀ is the output voltage

When the thyristors are triggered with a firing angle 0 and π/2 

Then the output waveform will be as follows

 

Let the average output voltage of the circuit be V_a

The value of source voltage is V_m sin ωt 

Calculation:

Given,

α₁ = 0, α₂ = π/2

For the first thyristor, the average output voltage is

V_a = 2V_m / π 

For the second thyristor, the average output voltage is

V_a = V_m / π 

We know that 

Ripple frequency = 2nf = 2 × 12 × 30 = 720 Hz 

Concept:

In a full wave rectifier though a transformer,

Average output voltage, V₀ = 2V_m/π

Average output current, I₀ = V₀/R

RMS value of output voltage V_or = V_s

RMS value of load current, I_or = V_s/R

Average value of diode current, I_d = I_m/2

RMS value of diode current, I_dr = I_or

Peak value of diode current, I_dm = √2 I_or

Power delivered to the load = V_or I_or

Input voltamperes = V_s I_or

Calculation:

Given that, DC output voltage (V_DC) = 400 V

Average output voltage of rectifier (V₀) = 400 V

⇒ V_s = 444.28 V

Load resistance (R) = 10 Ω

RMS value of load current,  I_or = 444.2810 = 44.428A

kVA rating of transformer = 444.28 × 44.428 = 19.738 kVA

Concept:

The average output voltage of the single-phase full-wave rectifier with a resistive load is given by
V₀ = 2V_m/π cos α

Where,

V_m = peak value of source voltage

α = firing angle

Calculation:

Given that:

Source voltage Vs = 120 V rms

firing angle α = 60⁰

From the source voltage peak value V_m = √2 × 120 V

The average output value of a single-phase full-wave rectifier is given by

Concept:

The average output voltage of a 1ϕ full wave converter is given by:

Case 1: α ≤ 90° 

V_o(avg) = I_oR + E

This is a motoring mode of operation of a rectifier. 

Case 2: α ≥ 90° 

V_o(avg) = I_oR - E

This is a generating mode of operation of a rectifier. 

In generating mode, the polarity of battery voltage gets reversed.

Calculation:

Given, α = 120° 

 
I_o = 4.64 A

Concept:

Single-phase fully controlled bridge circuit:

 

• The bridge rectifier circuit is made of four SCRs T₁, T₂, T₃, T₄, and a load resistor RL.
• The four SCRs are connected in a closed-loop configuration to effectively convert the alternating current (AC) into Direct current (DC).
• The input signed is applied across terminals A and B and output DC signal is obtained across the load resistor RL connected across the terminals C and D.
• The four SCRs are arranged in such a way that only two SCRs conduct electricity during each half cycle.
• SCRs T₁ and T₃ are pairs that conduct electric current during the positive half cycle.
• Similarly, SCRs T₂ and T₄ conduct electric current during a negative half cycle.
• During the positive half cycle, SCRs T₁ and T₃ become forward biased while SCRs T₂ and T₄ become reverse biased.
• During the negative half-cycle, SCRs T₂ and T₄ become forward biased while SCRs T₁ and T₃ become reverse biased.
• The current flow across load resistor RL is the same during positing half-cycles and the negative half cycles.
• The output DC signal polarity may be either completely positive or negative.
• The bridge rectifier allows electric current during both positive and negative half cycles of the input AC signal.
• Peak inverse voltage is the maximum voltage that an SCR can withstand in a reverse bias condition.
• During the positive half cycle, SCRs T₁ and T₃ are conducting.
• Similarly, during the negative half cycle, SCR T₂ and T₃ are in the conducting stage while SCRs T₁ and T₃ are in a non-conducting state.

Calculation:

Given that

Vm = 100 V (AC signal)

PIV of bridge Rectifier =  PIV of SCR T₁ and T₃ = Vm = 100 V

Similarly, PIV of SCR T₂ and T₂ is Vm = 100 V.

i.e. PIV of each SCR T₁, T₂, T₃, T₄ is Vm = 100 V