# Een 100 Electrical Circuits I POWER ELECTRONICS DIODE RECTIFIERS (UNCONTROLLED R ECTIFIERS) Mustafa A. Fadel Assistant Lecture Electrical Engineering Department College of Engineering, Mustansiriyah University Rectifier is a circuit that converts an AC signal to a DC signal Rectifier AC supply Transformer Block diagram of an uncontrolled diode rectifier circuit Note: In a diode rectifiers, the power flows only from the AC source to the DC side. Applications of Uncontrolled Rectifiers DC power supply for consumer electronic products such as radios, TVs, DVD players, mobile phone chargers, computers, laptops and so on (low power) DC motor drives (high power)

2 Depending on the type of input source, rectifiers are classified into two main groups: Half-wave Single phase Full-wave Uncontrolled rectifier Half-wave Three phase Full-wave Important equations 1

= ( ) 0 = 1 2 () 0 3 Single-phase half-wave rectifier (R load) Circuit diagram

Waveforms 4 The average value of output voltage The average value of load current = = = The rms value of output voltage

= 2 The rms value of load current = Peak inverse voltage across the diode = The efficiency of rectification = =

5 The effective (rms) value of the ac component of output voltage 2 = 2 The form factor (a measure for the shape of output voltage) FF= The ripple factor (a measure for the ripple content) 2 2 2

RF= = = 1 6 The harmonic factor or total harmonic distortion (a measure for distortion of a waveform] of the input current h 2 2 1 = = = = 1 1 ( 2 1 1

) Where Is1 is the rms value of the fundamental component of the input current. And I s is the rms value of the input current. 7 If is the angle between the fundamental component of the input current and the voltage, the displacement factor =cos The input power factor 1 cos 1 cos = = 8

Single-phase half-wave rectifier (RL load) without freewheeling diode Due to inductive load, the conduction period of the diode D 1 will extend beyond 180 degree until the current becomes zero. Circuit diagram Waveforms 9 The average value of the output voltage = [ 1 cos ( + ) ] 2 Where the angle can be calculated as: = 1

, =2 [ ] The average value of the load current = The average value of the output voltage (and hence the current) can be increased by making =0, which is possible by adding a freewheeling diode Dm across the load. 10 11 Single-phase half-wave rectifier (RL load) with freewheeling diode

Circuit diagram Waveforms The effect of the freewheeling diode Dm is to prevent the negative voltage appearing across the load, and as a result, the magnetic stored energy is increased. At t=/, the current from D1 is transferred to Dm and this process is called commutation of diodes. Depending on the load time constant, the load current may be discontinuous. 12 Single-phase center-tap full-wave rectifier (R load) Circuit diagram Waveforms 13 The average value of output voltage The average value of load current

2 = 2 = = The rms value of output voltage = The rms value of load current =

Peak inverse voltage across each diode =2 2 14 Single-phase bridge full-wave rectifier (R load) Circuit diagram 15 The average value of output voltage 2 = The average value of load current

2 = = The rms value of output voltage = 2 The rms value of load current = Peak inverse voltage across each diode

= Waveforms 16 Single-phase bridge full-wave rectifier (RL load) With a resistive load, the load current is identical in shape to the load voltage. In practical applications, most loads are inductive. The load current shape and magnitude depend on both the load resistance R and inductance L. Circuit diagram Waveforms 17 Single-phase bridge rectifier with very large inductive load 18

E#1 If a single-phase bridge rectifier supplies a very high inductive load such as a dc motor. the turns ratio of the transformer is unity. Determine a) the HF of the input current, and b) the input PF of the rectifier. Note the output (load) current is constant and ripple free due to the highly inductive load. 19 Using Fourier series, the input current is can be analyzed as ()= + =1 ,2 , 3 , 1 = 2

1 = ( cos+ ) 2 ( ) ( )=0 0 2 ( ) cos ( )=0 0 2 4 1 = ( ) sin ( )= 0

( ) = 4 3 5 + + + 1 3 5 ( ) Therefore, the rms value of the input current is 4

1 = 1+ 3 2 2 1 + 5 2 ( () () ) + = 20 Therefore, the rms value of the fundamental component of the input current is

4 1= =0 . 9 2 Therefore, the harmonic factor is = = ( 2 1= 1 ) ( 2 1

1=0 . 4843 48 . 43 % 0.9 ) Since =0, the displacement factor is =cos =1 Therefore, the input power factor 1 cos = =0 . 9 21 Multiphase rectifier 22

Three phase bridge rectifier Circuit diagram 23 Waveforms 24 The average value of output voltage 3 3 = =1 . 654 The rms value of output voltage 1/ 2 3 9 3 = +

=1 .6554 2 4 Peak inverse voltage across each diode = 3 25 E#2 A three-phase bridge rectifier supplies a high inductive load such that the average load current Idc= 60 A and the ripple content is negligible. Determine the ratings of the diodes if the line-toneutral voltage of the supply is 120V, 60Hz. As shown, the average value of the diode current ( ) = 60 = =20 A 3 3

The rms value of the diode current The peak inverse voltage across the diode PIV = 3 = 3 2 120=294 Currents through diodes 26 E#3 For the following current waveform, determine the average, rms, peak value of the current. = 1 + 2 + 3 3 400 1 =150 =150 =12 5000

2 2 1 2 150 100 2 = = =1 . 91 5000 5 4 200 3 =100 =100 =4 5000 =1 2+1. 91+4=17 . 91

27 = 2 1 + 2 2 + 2 3 3 400 1 =150 =150 =42 .43

5000 2 1 150 1 00 2 = = =15 2 2 5000

5 4 2 00 3 =100 =100 =20 5000 = 42 . 43 2 +152 +20 2=49 . 25 =300 28