Converters and Rectifiers

LessonAbout 1 h

Converters and Rectifiers

Converter (or rectifier) is an electrical machine or static electronic device used to convert alternating current (AC) electrical power into direct current (DC) power or to alter the frequency and phase characteristics of AC power.

1. Relationship Between AC and DC Voltages

  • Synchronous Converter: rotary machine with AC slip rings and DC commutator
  • Polyphase Taps: AC voltage measured between adjacent slip rings depends on phase count \(n\)

AC to DC Voltage Ratio

\[ V_{\text{ac}} = \dfrac{V_{\text{dc}}}{\sqrt{2}} \sin\left(\dfrac{180^\circ}{n}\right) \]

Where: \(V_{\text{ac}}\) = rms AC line-to-line voltage between adjacent slip rings (V), \(V_{\text{dc}}\) = DC voltage at commutator brushes (V), \(n\) = number of slip rings (\(n=2\) for \(1\phi\), \(n=3\) for \(3\phi\), \(n=4\) for \(4\phi\), \(n=6\) for \(6\phi\))

EE Board Exam April 1994

Example. In a three-phase synchronous converter, determine the ratio of the AC line voltage to DC voltage.

Solution.

\[ \dfrac{V_{\text{ac}}}{V_{\text{dc}}} = \dfrac{1}{\sqrt{2}} \sin\left(\dfrac{180^\circ}{3}\right) = \dfrac{\sin 60^\circ}{\sqrt{2}} = 0.612 \]

Answer: \(0.612\)

2. Efficiency of Converters

  • DC Output Power: total active power delivered to DC load
  • AC Input Power: active electrical power drawn from AC supply line
  • Power Factor Dependency: input current varies inversely with power factor

Converter Power and Efficiency Equations

\[ P_{\text{dc}} = V_{\text{dc}} I_{\text{dc}} \qquad\qquad P_{\text{ac}} = \dfrac{P_{\text{dc}}}{\eta} \]
\[ P_{\text{ac}} = V_{\text{ac}} I_{\text{ac}} \cos\theta \qquad\qquad (\text{Single-Phase}) \]
\[ P_{\text{ac}} = \sqrt{3} V_{\text{ac}} I_{\text{ac}} \cos\theta \qquad\qquad (\text{Three-Phase}) \]

Where: \(P_{\text{dc}}\) = DC power output (W), \(P_{\text{ac}}\) = AC power input (W), \(\eta\) = efficiency (per unit), \(V_{\text{dc}}\) = DC voltage (V), \(I_{\text{dc}}\) = DC current (A), \(V_{\text{ac}}\) = AC line voltage (V), \(I_{\text{ac}}\) = AC line current (A), \(\cos\theta\) = power factor

Example. A three-phase rotary converter delivers \(150\ \text{A}\) at \(500\ \text{V DC}\). Determine the current on the AC side if the operating power factor of the machine is \(0.80\) and the efficiency is \(90\%\).

Solution.

\[ P_{\text{dc}} = 500(150) = 75000\ \text{W} \qquad\qquad P_{\text{ac}} = \dfrac{75000}{0.90} = 83333.33\ \text{W} \]
\[ V_{\text{ac}} = \dfrac{500}{\sqrt{2}} \sin 60^\circ = 306.19\ \text{V} \implies I_{\text{ac}} = \dfrac{83333.33}{\sqrt{3}(306.19)(0.80)} = 196.4\ \text{A} \]

Answer: \(196.4\ \text{A}\)

3. Synchronous-Synchronous Frequency Converters

  • Direct Shaft Coupling: two synchronous machines coupled on one shaft
  • Pole Ratio Principle: output frequency dictated by ratio of pole counts
  • Constant Speed Operation: both machines turn at identical synchronous speed

Frequency Conversion Ratio

\[ \dfrac{f_1}{f_2} = \dfrac{P_1}{P_2} \]

Where: \(f_1, f_2\) = frequencies of first and second synchronous machines (Hz), \(P_1, P_2\) = number of poles of first and second synchronous machines

4. Motor-Generator Sets

  • AC Motor to DC Generator: converts AC line power to DC power
  • Cascaded Efficiency: overall efficiency equals product of individual efficiencies
  • Electrical Isolation: complete physical isolation between AC and DC circuits

Motor-Generator Efficiency Equations

\[ \eta_m = \dfrac{P_m}{P_{\text{in}}} \qquad\qquad \eta_c = \dfrac{P_c}{P_m} \qquad\qquad \eta_g = \dfrac{P_{\text{out}}}{P_c} \]
\[ \eta_{\text{overall}} = \dfrac{P_{\text{out}}}{P_{\text{in}}} = \eta_m \times \eta_c \times \eta_g \]

Where: \(\eta_m\) = motor efficiency, \(\eta_c\) = coupling efficiency, \(\eta_g\) = generator efficiency, \(P_{\text{in}}\) = motor electrical input (W), \(P_m\) = motor mechanical output (W), \(P_c\) = coupling output power (W), \(P_{\text{out}}\) = generator electrical output (W)

Example. A single-phase motor is connected across a \(230\ \text{V}\), \(60\ \text{Hz}\) supply and draws \(30\ \text{A}\) at \(0.85\) pf lagging. This motor drives a \(120\ \text{V DC}\) generator through direct coupling. If motor, coupling, and generator efficiencies are \(0.85\), \(0.96\), and \(0.88\) respectively, determine current delivered by generator.

Solution.

\[ P_{\text{in}} = (230)(30)(0.85) = 5865\ \text{W} \qquad\qquad P_{\text{out}} = 5865(0.85)(0.96)(0.88) = 4211.54\ \text{W} \]
\[ I_{\text{dc}} = \dfrac{P_{\text{out}}}{V_{\text{dc}}} = \dfrac{4211.54}{120} = 35.09\ \text{A} \]

Answer: \(35.09\ \text{A}\)

5. Relation Between Pulley Diameter and Speed

  • Inverse Proportionality: rotational speed varies inversely with pulley diameter
  • Belt Drive Transmission: speed conversion proportional to diameter ratio

Pulley Speed Ratio

\[ \dfrac{N_1}{N_2} = \dfrac{d_2}{d_1} \]

Where: \(N_1\) = speed of driving pulley (rpm), \(N_2\) = speed of driven pulley (rpm), \(d_1\) = diameter of driving pulley, \(d_2\) = diameter of driven pulley

Example. A \(60\ \text{Hz}\), \(4\text{-pole}\) motor is belt connected to a \(6\text{-pole}\) generator with pulley diameters of \(6\ \text{inches}\) and \(10\ \text{inches}\) respectively. Determine the frequency generated by the generator.

Solution.

\[ N_1 = \dfrac{120(60)}{4} = 1800\ \text{rpm} \qquad\qquad N_2 = 1800 \left(\dfrac{6}{10}\right) = 1080\ \text{rpm} \]
\[ f_2 = \dfrac{P_2 N_2}{120} = \dfrac{6(1080)}{120} = 54\ \text{Hz} \]

Answer: \(54\ \text{Hz}\)

6. Single-Phase AC Input Rectifiers

Single-phase rectifier is a circuit that converts single-phase AC voltage into pulsating DC output voltage.

6.1 Half-Wave

  • Unidirectional Conduction: conducts current during positive half-cycles only
  • Peak AC Voltage: \(V_m = \sqrt{2} V_{\text{ac}}\)

Single-Phase Half-Wave Output Voltage and Current

\[ V_{\text{out,dc}} = \dfrac{V_m}{\pi} = 0.318 V_m \qquad\qquad V_{\text{out,rms}} = 0.5 V_m \]
\[ I_{\text{out,dc}} = \dfrac{I_m}{\pi} = 0.318 I_m \qquad\qquad I_{\text{out,rms}} = 0.5 I_m \]

Where: \(V_m\) = peak AC input voltage (V), \(I_m\) = peak AC input current (A), \(V_{\text{out,dc}}\) = average DC output voltage (V), \(V_{\text{out,rms}}\) = rms output voltage (V), \(I_{\text{out,dc}}\) = average DC output current (A), \(I_{\text{out,rms}}\) = rms output current (A)

Example. A half-wave rectifier draws power from a power transformer at \(24\ \text{V}\) and delivers \(200\ \text{mA}\) to a load. Determine the power drawn by the load.

Solution.

\[ V_m = \sqrt{2}(24) = 33.94\ \text{V} \qquad\qquad V_{\text{out,dc}} = 0.318(33.94) = 10.79\ \text{V} \]
\[ P_{\text{dc}} = V_{\text{out,dc}} I_{\text{out,dc}} = 10.79(0.200) = 2.16\ \text{W} \]

Answer: \(2.16\ \text{W}\)

6.2 Full-Wave

  • Bidirectional Rectification: converts both positive and negative half-cycles to DC
  • Higher Output Average: double the DC output voltage of half-wave rectifier

Single-Phase Full-Wave Output Voltage and Current

\[ V_{\text{out,dc}} = \dfrac{2 V_m}{\pi} = 0.636 V_m \qquad\qquad V_{\text{out,rms}} = \dfrac{V_m}{\sqrt{2}} = 0.707 V_m \]
\[ I_{\text{out,dc}} = 0.636 I_m \qquad\qquad I_{\text{out,rms}} = 0.707 I_m \]

Where: \(V_m\) = peak AC input voltage (V), \(I_m\) = peak AC input current (A), \(V_{\text{out,dc}}\) = average DC output voltage (V), \(V_{\text{out,rms}}\) = rms output voltage (V)

Example. An AC ammeter reads \(8\ \text{A}\) in a single-phase full-wave rectifier circuit. What is the reading on a DC ammeter connected in the same load circuit?

Solution.

\[ I_m = \dfrac{I_{\text{out,rms}}}{0.707} = \dfrac{8}{0.707} = 11.315\ \text{A} \]
\[ I_{\text{out,dc}} = 0.636 I_m = 0.636(11.315) = 7.2\ \text{A} \]

Answer: \(7.2\ \text{A}\)

7. Three-Phase AC Input Rectifiers

Three-phase rectifier is a polyphase rectifier that converts three-phase AC input into smooth DC output voltage with low ripple factor.

7.1 Half-Wave

  • Three-Pulse Output: three conduction pulses per AC cycle
  • Output Voltage Average: \(0.827 V_m\) where \(V_m\) is peak input phase voltage

Three-Phase Half-Wave Output Voltage and Current

\[ V_{\text{out,dc}} = \dfrac{3\sqrt{3}}{2\pi} V_m = 0.827 V_m \qquad\qquad V_{\text{out,rms}} = 0.841 V_m \]
\[ I_{\text{out,dc}} = 0.827 I_m \qquad\qquad I_{\text{out,rms}} = 0.841 I_m \]

Where: \(V_m = \sqrt{2} V_{\text{ac}}\) = peak AC input phase voltage (V), \(I_m = \sqrt{2} I_{\text{ac}}\) = peak AC input phase current (A)

Example. A three-phase half-wave rectifier delivers a DC current of \(1000\ \text{A}\) to a load. Determine the rms value of the output current.

Solution.

\[ I_m = \dfrac{I_{\text{out,dc}}}{0.827} = \dfrac{1000}{0.827} = 1209.19\ \text{A} \]
\[ I_{\text{out,rms}} = 0.841 I_m = 0.841(1209.19) = 1017\ \text{A} \]

Answer: \(1017\ \text{A}\)

7.2 Full-Wave

  • Six-Pulse Output: six conduction pulses per AC cycle providing very low ripple
  • High DC Output Ratio: output DC voltage equals \(0.955 V_m\)

Three-Phase Full-Wave Output Voltage and Current

\[ V_{\text{out,dc}} = \dfrac{3}{\pi} V_m = 0.955 V_m \qquad\qquad V_{\text{out,rms}} = 0.95577 V_m \]
\[ I_{\text{out,dc}} = 0.955 I_m \qquad\qquad I_{\text{out,rms}} = 0.96577 I_m \]

Where: \(V_m = \sqrt{2} V_{\text{ac}}\) = peak AC line-to-line input voltage (V), \(I_m\) = peak AC line current (A)

Formula Summary

Formula Name Formula Variables
AC to DC Voltage Ratio \(V_{\text{ac}} = \dfrac{V_{\text{dc}}}{\sqrt{2}} \sin\left(\dfrac{180^\circ}{n}\right)\) \(V_{\text{ac}}\) = AC voltage between adjacent slip rings (V)
\(V_{\text{dc}}\) = DC voltage (V)
\(n\) = number of slip rings
Converter DC Power \(P_{\text{dc}} = V_{\text{dc}} I_{\text{dc}}\) \(P_{\text{dc}}\) = DC power output (W)
\(V_{\text{dc}}\) = DC voltage (V)
\(I_{\text{dc}}\) = DC current (A)
Converter AC Input Power (3-Phase) \(P_{\text{ac}} = \sqrt{3} V_{\text{ac}} I_{\text{ac}} \cos\theta\) \(P_{\text{ac}}\) = AC input power (W)
\(V_{\text{ac}}\) = AC line voltage (V)
\(I_{\text{ac}}\) = AC line current (A)
\(\cos\theta\) = power factor
Synchronous Frequency Ratio \(\dfrac{f_1}{f_2} = \dfrac{P_1}{P_2}\) \(f_1, f_2\) = frequencies (Hz)
\(P_1, P_2\) = pole counts
Motor-Generator Overall Efficiency \(\eta_{\text{overall}} = \eta_m \times \eta_c \times \eta_g\) \(\eta_{\text{overall}}\) = combined efficiency
\(\eta_m\) = motor efficiency
\(\eta_c\) = coupling efficiency
\(\eta_g\) = generator efficiency
Pulley Speed Ratio \(\dfrac{N_1}{N_2} = \dfrac{d_2}{d_1}\) \(N_1, N_2\) = pulley speeds (rpm)
\(d_1, d_2\) = pulley diameters
Single-Phase Half-Wave DC Voltage \(V_{\text{out,dc}} = 0.318 V_m\) \(V_{\text{out,dc}}\) = average DC voltage (V)
\(V_m\) = peak AC voltage (V)
Single-Phase Half-Wave RMS Voltage \(V_{\text{out,rms}} = 0.5 V_m\) \(V_{\text{out,rms}}\) = rms output voltage (V)
\(V_m\) = peak AC voltage (V)
Single-Phase Full-Wave DC Voltage \(V_{\text{out,dc}} = 0.636 V_m\) \(V_{\text{out,dc}}\) = average DC voltage (V)
\(V_m\) = peak AC voltage (V)
Single-Phase Full-Wave RMS Voltage \(V_{\text{out,rms}} = 0.707 V_m\) \(V_{\text{out,rms}}\) = rms output voltage (V)
\(V_m\) = peak AC voltage (V)
Three-Phase Half-Wave DC Voltage \(V_{\text{out,dc}} = 0.827 V_m\) \(V_{\text{out,dc}}\) = average DC voltage (V)
\(V_m\) = peak AC voltage (V)
Three-Phase Half-Wave RMS Voltage \(V_{\text{out,rms}} = 0.841 V_m\) \(V_{\text{out,rms}}\) = rms output voltage (V)
\(V_m\) = peak AC voltage (V)
Three-Phase Full-Wave DC Voltage \(V_{\text{out,dc}} = 0.955 V_m\) \(V_{\text{out,dc}}\) = average DC voltage (V)
\(V_m\) = peak AC voltage (V)
Three-Phase Full-Wave RMS Voltage \(V_{\text{out,rms}} = 0.95577 V_m\) \(V_{\text{out,rms}}\) = rms output voltage (V)
\(V_m\) = peak AC voltage (V)