Power Electronics- Chapter 3

BEE4223 Power Electronics & Drives Systems Chapter 3: AC TO DC CONVERSION (RECTIFIER)

LEARNING OBJECTIVES

• Upon completion of the chapter the student should be able to: – State the operation and characteristics of diode rectifier. – Discuss the performance parameters and use different technique for analyzing and design of diode rectifier circuits. – Simulate different arrangement of diode rectifiers by using PSpice.

Overview • Single-phase, half wave rectifier – – – – – –

Uncontrolled R load R-L load R-C load Controlled Free wheeling diode

• Single-phase, full wave rectifier – R load – R-L load, – Controlled R, R-L Load – continuous and discontinuous current mode •Three-phase rectifier – uncontrolled – controlled

Rectifiers • DEFINITION: Converting AC (from mains or other AC source) to DC power by using power diodes or by controlling the firing angles of thyristors/controllable switches. • Basic block diagram

Rectifiers • Input can be single or multi-phase (e.g. 3-phase). • Output can be made fixed or variable • Applications: – DC welder, DC motor drive, Battery charger, DC power supply, HVDC

Root-Mean-Squares (RMS)

(.) 1 2π

2π

∫ 0

. dω t

2

Root Mean Squares of f

1 Step 2: 2π

2π

∫

( f ) dω t 2

0

Step 1: Step 3:

1 2π

2π

( f ) d ω t ∫ 2

0

(f)

2

Concept of RMS

v

2

Square root of the average of v2

Average of v2

ωt v

Average of v=0

Ideal Rectifier: Single-Phase, Half-Wave R-Load

• Considering the diode is Given Voltage Source, vs (t ) = Vs sin(ωt ), ideal, the voltage at R- " DC" output voltage, load during forward π 1 biased is the positive Vo = Vave = VDC = Vm sin(ωt ) dωt ∫ cycle of voltage source, 2π 0 while for negative biased, Vm V = = 0.318Vm the voltage is zero. o π

Ideal Rectifier: Single-Phase, Half-Wave R-Load • We observe that: – DC voltage is fixed at 0.318 or Average power absorbed by resistor, 31.8% of V the peak value P = I 2 rms R =

" DC" output current , Vo  Vm  1  Vm Io = =    = R  π  R  πR

2

rms

R π

1 V – RMS where, V voltage = [V sin( ωreduced t )] d (ωt ) = is from ∫ 2π 2 0.707 (normal V V sinusoidal I = = R 2or R 50% of peak RMS) to 0.5 value. 2

o , rms

s

m

0

o , rms

o , rms

m

– Half wave is not practical , becauseEfficiency of high distortion P V I η= = supply current. P V I The supply current contains DC component that may saturate the input transformer dc

dc dc

ac

rms rms

Example 1 • Consider the half-wave rectifier circuit with a resistive load of 25Ω and a 60 Hz ac source of 110Vrms. – Calculate the average values of Vo and Io. Justify the significant value of Vo and Io. – Calculate the rms values of Vo and Io. – Calculate the average power delivered to the load.

Example 1 (Cont) • Solution • (i) The average values of Vo and Io are given by

Vm 2 (110) Vo = = = 49.52V π π and , Vo 49.52 Io = = = 1.98 A R 25

In this case, for the particular circuit, possible dc output voltage obtained from the circuit is 49.52V and dc output current is 1.98A. That means, for any dc application within this value, this circuit can be used.

Example 1 (Cont) • (ii) The rms value of the of Vo and Io Vm 2 (110) Vrms = = = 77.78V 2 2 and , Vo ,rms 77.78 I o ,rms = = = 3.11A R 25

(iii) average power delivered to the load over one cycle

Po =

Vo2 ,rms R OR

(77.78) 2 = = 242W 25

2 Po = I rms R = (3.11) 2 25 = 242W

Example 2 • For the half-wave rectifier, the source is a sinusoid of 120Vrms at a frequency of 60Hz. The load resistor is 5Ω. Determine (i) the average load current, (ii) the average power absorbed by the load, and (iii) the power factor of the circuit.

Example 2 (Cont) Solution • (i) The average load current Vm = 2 (120) = 169.7V and , Vs 169.7 Io = = = 10.8 A πR π (5)

(ii) The average power absorbed by the load

Vrms

Vm 2 (120) = = = 84.9V 2 2

and 2 Vrms 84.9 2 P= = = 1440W R 5

Example 2 (Cont) (iii) Power Factor

P P pf = = = S Vm , rms I m , rms

P  Vm  Vm , rms    2R 

1440 = = 0.707  2 (120)   (120)  2(5)    Note:

The power factor at the input of the rectifier circuit is poor even for resistive load and decreases as triggering angle for controlled rectifier is delayed.

Half-wave with R-L load • Industrial load typically contain inductance as well as resistance. • By adding an inductor in series with the load resistance causes an increase in the conduction period of the load current, hence resulting the halfwave rectifier circuit working under an inductive load.

Vm = Ri + L •

di dt

That means, the load current flows not only during Vs > 0, 0 but also for a portion of Vs < 0. This is due to

Half-wave with R-L load • Until certain time (<π), Vs>VR (hence VL= VsVR is positive), the current builds up and inductor stored energy increases. increases • At maximum of VR, Vs=VR hence, VL =0V. =0V • Beyond this point, VL becomes negative (means releasing stored energy), and current begins to decrease. •After T=π, the input, Vs becomes negative but current still positive and diode is still conducts due to inductor stored energy. energy The load current is present at certain period, but never for the entire period, regardless of the inductor size. •This will results on reducing the average output voltage due to the negative segment. Mohd Rusllim Mohamed The larger the Inductance, the larger negative

VL = L

di dt

Half-wave with R-L load • The point when the current reaches zero, is when the diode turns off, given by sin( β − θ ) + sin(θ )e

 β  −   ωτ 

The average power absorbed by load , 1 P= 2π

2π

1 = 2π

2π

=0

and

L  ωL  Z = R 2 + (ωL) 2 , θ = tan −1  , τ = R  R 

0

∫ v(ωt )i (ωt ) d (ωt ) 0

OR

where,  −ωt  V   V  m sin( wt − 0) + m sin(θ )e  ωτ  for 0 ≤ ωt ≤ β i ( wt ) =  Z Z 0 for β ≤ ωt ≤ 2π 

∫ p(ωt ) d (ωt )

2 P = I rms R ; Since the average power

absorbed by inductor is zero (0). rms current , I rms

1 = 2π

β

2π

1 2 i ( ω t ) d ( ω t ) = i (ωt ) d (ωt ) ∫0 2π ∫0 2

and , average current , β

1 Io = i (ωt ) d (ωt ) ∫ 2π 0

Example 3 • For half-wave rectifier with R-L load, R=100Ω, L=0.1H, ω=377rad/s, and Vs=100V. Determine • An expression for the current in this circuit • The point where diode turns off • The average current • The rms current • The power absorbed by the R-L load, and • The power factor

Example 3 (cont) Solution

(ii) β (diode stop)

For parameter given Z = R 2 + (ω L) 2 = (100) 2 + [(377)(0.1)]2 = 106.9Ω

 β  −   0.377 

sin( β − 0.361) + sin(0.361)e =0  ωL   (377)(0.1)  o θ = tan −1   = tan −1   = 20.7 = 0.361 rad  R   100  Using numerical root finding, β L  0.1  is found to be 3.50 rads or 201o ωτ = ω = (377)  = 0.377 rad R  100  (i) Current Equation

i (ωt ) = 0.936 sin(ωt − 0.361) + 0.331e for 0 ≤ ωt ≤ β

−

ωt 0.377

A

Example 3 (cont) v) Power absorbed by resistor

iii) Average current β

1 2 Io = i (ωt ) d (ωt ) 2π ∫0 1 = 2π

3.50

∫ [0.936 sin(ωt − 0.361) + 0.331e

2 P = I rms R −

ωt 0.377 2

] d (ωt )

0

= 0.308 A

iv) rms current I rms =

1 2π

3.50

∫ [0.936 sin(ωt − 0.361) + 0.331e

= [0.474]2 (100) = 22.4W

vi) Power factor −

ωt 0.377 2

] d (ωt )

pf =

0

= 0.474 A

P P = S Vm ,rms I m , rms

22.4 100   0.474  2 = 0.67

=

Half-wave with R-C load • In some applications in which a constant output is desirable, a series inductor is replaced by a capacitor. • parallel The purpose of capacitor is to reduce the variation in the output voltage, voltage making it more like dc. dc • The resistance may represent an external load, while the capacitor is a filter of rectifier circuit.

Mohd Rusllim Mohamed

Half-wave with R-C load • Assume the capacitor is uncharged, and as source positively increased, diode is forward biased • As diode is on, on the output voltage is the same as source voltage, voltage and capacitor charges. charges • Capacitor is charged to Vm as input voltage reaches its positive peak at ωt = π/2. /2 • As source decreases after ωt = π/2, /2 the capacitor discharges into load resistor. resistor As diode is reversed biased, biased the load is isolated from source, source and the output voltage (capacitive Mohd Rusllim Mohamed charge) decaying exponentially with time constant RC. RC

Half-wave with R-C load • The angle ωt = θ is the point when diode turns 2π + α ≤ ω t ≤ 2π + θ Vm sin ω t off. Vo (ω t ) =  − (ω t − θ ) / ω RC θ ≤ ω t ≤ 2π + α Vθ e • The diode will stay off until the capacitor and where, Vθ = Vs sin θ input voltages become equal again.

• The effectiveness of capacitor filter is determined by the variation in output voltage, or expressed as maximum and minimum output voltage, which is peak-to-peak ripple voltage. voltage

Half-wave with R-C load (Ripple Voltage) • The ripple:

∆Vo = Vmax − Vmin = Vm − Vm sin α = Vm (1 − sin α )

• if Vθ≈Vm and θ≈π/2, then ripple can be approximated as  2π   Vm   ∆Vo = Vm   =   ωRC   fRC  •

The output voltage ripple is reduced by increasing the filter capacitor, C. Anyhow, this results in a larger peak diode current. Vm sin α R sin α   = Vm ωC cos α +  R  

I D , peak = ωCVm cos α +

Example 4 • The half-wave rectifier has 120Vrms source at 60Hz, R=500Ω, C=100µF and delay when diode turns on is given 48o. Determine – – – – –

The expression of output voltage Ripple voltage Peak diode current Sketch and label the output waveform Value of C as ripple voltage is 1% of Vm, and hence find new α under this condition.

Example 4 (cont) Solution (ii) Ripple Voltage

For parameter given Vm = 120 2 = 169.7V

ωRC = 2π (60)(500)(1×10 ) = 18.85 rad −6

Angle θ = − tan −1 (−18.85) + π = 1.62 rad Angle α = 48o = 0.843 rad Vm sin θ = (169.7) sin 1.62 = 169.5V

(i) Output Voltage 169.7 sin ωt Vo (ωt ) =  − (ωt −1.62 ) / 18.85 169.5 e

2π + α ≤ ωt ≤ 2π + θ

θ ≤ ωt ≤ 2π + α

∆Vo = Vm (1 − sin α ) = 169.7[1 − sin(0.843)] = 43V (iii) Peak diode current sin(0.843)   I D , peak = 169.7 377(10 −4 ) cos(0.843) +  500   = 4.50 A

Example 4 (cont) (iv) Waveform must be properly labeled according to data (v) Capacitor value For ∆Vo =0.01Vm  Vm C =  fR∆V o  hence,

α =sin

−1

   Vm   =  (60)(500)(0.01V )   =3333µF m   

 ∆Vo  1 −  Vm 

   

 1 =sin −1  1 −  fRC 

   

 1 =sin −1  1 −  (60)(500)(3333 ×10 −6  =81.9 o

   

RL Source Load • To supply a dc source from an ac source • The diode will remain off as long as the voltage of ac source is less than dc voltage. • Diode starts to conduct at ωt=α. Given by,

Vm sin α = Vdc OR V  α = sin −1  dc   Vm 

RL Source Load rms current , 2π

I rms

1 2 = i (ω t ) d (ω t) ∫ 2π α

and , average current , β

Io

1 = i (ω t ) d (ω t) ∫ 2πα

Power ,

Pac =PR +Pdc 2 =I rms R +IVdc

Example 5 • The RL half-wave rectifier has 120Vrms source at 60Hz, R=2Ω, L=20mH, Vdc =100V with extinction angle given by 193o. Determine – – – – – –

The expression of current in the circuit Power absorbed by resistor Power absorbed by dc source Power supplied by ac source Power factor Draw the waveform

Example 5 (cont) Solution

(i) Current Equation

For parameter given

ii) Power absorbed by resistor

Example 5 (cont) iii) Power absorbed by dc source

iv) Power supplied

v) Power factor

v) Waveform - Refer notes

Freewheeling Diode (FWD) • Note that, previously discussed uncontrolled half-wave RL load rectifier allows load current to present at certain period (current decreasing by time since opposing negative cycle of input), hence reducing the average output voltage due to the negative segment. • In other word, for single-phase, half wave rectifier with R-L load, the load (output) current is NOT CONTINUOUS. • A FWD (sometimes known as commutation diode) can be placed in parallel to RL load to make the load (output) current continuous.

Freewheeling Diode (FWD) •

Note that both D1and D2 cannot be turned on at the same time. time

•

For a positive cycle voltage source, – D1 is on, on D2 is off – The voltage across the R-L load is the same as the source voltage.

•

For a negative cycle voltage source, – D1 is off, off D2 is on

Mohd Rusllim Mohamed

– The voltage across the R-L load is zero. zero – However, the inductor contains energy from positive cycle. cycle The load current still circulates through the R-L path.

Freewheeling Diode (FWD) •

negative cycle voltage source (cont), – But in contrast with the normal half wave rectifier, negative cycle of FWD does not consist of supply voltage in its loop. – Hence the “negative part” of Vo as shown in the normal half-wave disappear. disappear

Vm Vo = ; π •

Vo Io = ; R

2 Po = I rms R

Irms is determined from Fourier component of current

V In = n ; Zn

2V Vn = 2 m n −1 π

Z n = R + jnωo L ;

(

)

∞

Hence, I rms = I + ∑ I k2, rms 2 o

k =1

π

Vrms ,n=1

Vm 1 2 = [ V sin( ω t )] d ( ω t ) = s 2π ∫0 2

Mohd Rusllim Mohamed

n =2, 4, 6... - same as uncontrolled RLoad Rectifier

Example 6 • Uncontrolled R-L load rectifier, has a problem of discontinuous load current. Suggest a solution to the problem by justifying your answer through its principles of operation and waveform.

Solution Operation of FWD and its waveform (refer notes)

Example 7 • Determine the average load voltage and current, and determine the power absorbed by the resistor in the FWD circuit, where R=2Ω and L=25mH, Vm=100V; 60Hz. Solution •The average load voltage and current,

Vm 100 Vo = = = 31.8V π π Vo 31.8 Io = = = 15.9 A R 2

Example 7 (cont) Vn = Zn

.......(i )

) = 2 + jn[( 2π )( 60 ) ( 25 ×10 )] .......(ii )

In =

(

2(100) n 2 −1 π

−3

Vn 200 = 2 .......(iii ) Zn n −1 π [ 2 + j 9.425n ]

[(

) ]

n = 2,4,6.....

The ac voltage amplitudes,

100 = 50V 2 2(100) V2 = 2 = 21.2V ( 2 − 1)π 2(100) V4 = 2 = 4.24V ( 4 − 1)π 2(100) V6 = 2 = 1.82V ( 6 − 1)π V1 =

Fourier Impedance

Z 0 = 2 + j ( 0 )[9.425] = 2Ω

Z1 = 2 + j (1)[9.425] = 9.63Ω

Z 2 = 2 + j ( 2 )[9.425] = 18.96Ω

Z 4 = 2 + j ( 4 )[9.425] = 37.75Ω Z 6 = 2 + j ( 6 )[9.425] = 56.58Ω Note: angle note included in calculation

Example 7 (cont) Resulting Fourier Terms are as follows:

• Power Absorbed

n

Vn (V)

Zn (Ω)

In (A)

0

31.8

2

15.9

1

50

9.63

5.19

2

21.2

18.96

1.12

4

4.24

37.75

0.11

6

1.82

56.58

0.03

2 Po = I rms R

= (16.34 ) (2) = 534W 2

• rms current ∞

I rms = I o2 + ∑ I k2,rms k =1

2

2

 5.19   1.12   0.11  = 15.9 2 +   +  +   2   2   2  = 16.34 A

2

The Controlled Half-wave Rectifier • Previously discussed are classified as uncontrolled rectifiers. • Once the source and load parameters are established, the dc level of the output and power transferred to the load are fixed quantities. quantities • A way to control the output is to use SCR instead of diode. Two condition must be met before SCR can conduct: – The SCR must be forward biased (VSCR>0) – Current must be applied to the gate of SCR

Controlled, Half-wave R load • A gate signal is applied at ωt = α, where α is the delay/firing angle. Average " DC" output voltage, Vo = Vave = VDC

π

1 = Vm sin(ωt ) dωt Average power absorbed by resistor, ∫ 2π α =

Vs [1 + cos α ] 2π

I o , rms =

Vo , rms R

Vs = 2R

V 2 rms P = I rms R = R 2

π

where, Vo , rms

1 2 = [ V sin( ω t )] d (ωt ) m ∫ 2π 0 =

Vm α sin(2α ) 1− + 2 π 2π

Example 8 • Design a circuit to produce an average voltage of 40V across 100Ω load resistor from a 120Vrms 60 Hz ac source. Determine the power absorbed by the resistor and the power factor. Briefly describe what happen if the circuit is replaced by diode to produce the same average output.

Example 8 (Cont) • Solution In such that to achieved 40V average voltage, the delay angle must be Vs [1 + cos α ] 2π 120 2 40 = [1 + cos α ] 2π α = 61.2 o = 1.07 rad

Vo =

Vo , rms =

Vm α sin( 2α ) 1− + 2 π 2π

120 2 1.07 sin[ 2(1.07)] 1− + 2 π 2π = 75.6V =

V 2 rms 75.6 2 P= = = 57.1W R 100 pf =

57.1 = 0.63  75.6  (120)   100 

• If an uncontrolled diode is used, the average voltage would be

Vs 2 (120) Vo = = = 54V π π • That means, some reducing average resistor to the design must be made. A series resistor or inductor could be added to an uncontrolled rectifier, rectifier while controlled rectifier has advantage of not altering the load or introducing the losses

Controlled, Half-wave R-L load • The analysis of the circuit is very much similar to that of uncontrolled rectifier.  α −ωt   V    m  [sin( wt − θ ) − sin(α − θ )e  ωτ  for α ≤ ωt ≤ β i ( wt ) =  Z  0 otherwise 

and

L  ωL  Z = R 2 + (ωL) 2 , θ = tan −1  , τ = R  R 

rms current , I rms

1 = 2π

β

2π

1 2 i ( ω t ) d ( ω t ) = i (ωt ) d (ωt ) ∫α ∫ 2π α 2

and , average current , β

1 Io = i (ωt ) d (ωt ) ∫ 2π α

Controlled, Half-wave R-L load The average output voltage, Vm [cos α − cos β ] 1 Vm sin(ωt )dωt = Vo = 2π ∫ 2π α β

The average power absorbed by load , 2 P = I rms R ;

Example 9 •

For controlled RL rectifier, the source is 120Vrms at 60Hz, R=20Ω, L=0.04H, delay angle is 45o and extinction angle is 217o. Determine i. ii. iii. iv.

An expression for i(ωt) Average current and voltage Power absorbed by load Power factor

Example 9 (cont) Solution For parameter given

(i) Current Equation

Example 9 (cont) ii) Average current and voltage

iv) Power absorbed by resistor

v) Power factor

iii) rms current