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Chapter 2 AC to DC CONVERSION (RECTIFIER) Single-phase, half wave rectifier Uncontrolled: R load, R-L load, R-C load Controlled Free wheeling diode Single-phase, full wave rectifier Uncontrolled: R load, R-L load, Controlled Continuous and discontinuous current mode

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chapter 2 ac to dc conversion rectifier
Chapter 2AC to DC CONVERSION (RECTIFIER)
  • Single-phase, half wave rectifier
    • Uncontrolled: R load, R-L load, R-C load
    • Controlled
    • Free wheeling diode
  • Single-phase, full wave rectifier
    • Uncontrolled: R load, R-L load,
    • Controlled
    • Continuous and discontinuous current mode
  • Three-phase rectifier
    • uncontrolled
    • controlled

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

rectifiers

AC input

DC output

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
  • 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

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

single phase half wave r load

+

vs

_

+

vo

_

w

t

vs

p

2p

vo

io

Single-phase, half-wave, R-load

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

half wave with r l load

i

+

vR

_

+

vo

_

+

vs

_

+

vL

_

Half-wave with R-L load

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

r l load
R-L load

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

r l waveform

vs,

io

b

w

t

vo

vR

vL

2p

0

4p

p

3p

R-L waveform

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

extinction angle
Extinction angle

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

rms current power
RMS current, Power

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

half wave rectifier r c load

Vm

+

vs

_

iD

+

vo

_

vs

Vo

2p

3p

3p /2

4p

D

p /2

p

vo

Vmax

Vmin

iD

Half wave rectifier, R-C Load

a

q

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

operation
Let C initially uncharged. Circuit is energised at wt=0

Diode becomes forward biased as the source become positive

When diode is ON the output is the same as source voltage. C charges until Vm

After wt=p/2, C discharges into load (R).

The source becomes less than the output voltage

Diode reverse biased; isolating the load from source.

The output voltage decays exponentially.

Operation

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

estimation of q
Estimation of q

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

estimation of a
Estimation of a

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

ripple voltage
Ripple Voltage

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

capacitor current
Capacitor Current

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

peak diode current
Peak Diode Current

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

example

Vm

vs

Vo

2p

3p

3p /2

4p

D

p /2

p

vo

Vmax

Vmin

iD

a

q

Example

A half-wave rectifier has a 120V rms source at 60Hz. The load is =500 Ohm, C=100uF. Assume a and q are calculated as 48 and 93 degrees respectively. Determine (a) Expression for output voltage (b) peak-to peak ripple (c) capacitor current (d) peak diode current.

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

example cont
Example (cont’)

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

controlled half wave

ig

v

ia

s

+

vs

_

w

w

w

t

t

t

vo

v

i

g

a

Controlled half-wave

+

vo

_

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

controlled h w r l load

i

+

vR

_

+

vo

_

+

vs

_

w

+

vL

_

t

vs

p

2p

vo

io

b

a

Controlled h/w, R-L load

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

controlled r l load
Controlled R-L load

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

examples
Examples
  • A half wave rectifier has a source of 120V RMS at 60Hz. R=20 ohm, L=0.04H, and the delay angle is 45 degrees. Determine: (a) the expression for i(wt), (b) average current, (c) the power absorbed by the load.

2. Design a circuit to produce an average voltage of 40V across a 100 ohm load from a 120V RMS, 60Hz supply. Determine the power factor absorbed by the resistance.

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

freewheeling diode fwd

+

vR

_

+

vL

_

io

+

vo

_

+

vs

_

io

io

vo= 0

+

vo

_

vo= vs

+

vs

_

+

vo

_

io

D2 is on, D1 is off

D1 is on, D2 is off

Freewheeling diode (FWD)
  • Note that for single-phase, half wave rectifier with R-L load, the load (output) current is NOT continuos.
  • A FWD (sometimes known as commutation diode) can be placed as shown below to make it continuos

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

operation of fwd
Operation of FWD
  • Note that both D1 and D2 cannot be turned on at the same time.
  • For a positive cycle voltage source,
    • D1 is on, D2 is off
    • The equivalent circuit is shown in Figure (b)
    • The voltage across the R-L load is the same as the source voltage.
  • For a negative cycle voltage source,
    • D1 is off, D2 is on
    • The equivalent circuit is shown in Figure (c)
    • The voltage across the R-L load is zero.
    • However, the inductor contains energy from positive cycle. The load current still circulates through the R-L path.
    • But in contrast with the normal half wave rectifier, the circuit in Figure (c) does not consist of supply voltage in its loop.
    • Hence the “negative part” of vo as shown in the normal half-wave disappear.

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

fwd continuous load current

output

vo

io

iD1

Diode

current

iD2

t

w

0

2p

3p

p

4p

FWD- Continuous load current
  • The inclusion of FWD results in continuos load current, as shown below.
  • Note also the output voltage has no negative part.

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

full wave rectifier

D1

D3

is

is

iD2

iD1

iD1

io

+

vs

_

+

vo

_

D4

Full Bridge

D2

D1

+

vs1

_

+ vD1 -

+

vs

_

- vo +

+

vs2

_

io

+ vD2 -

D2

Center-tapped

Full wave rectifier
  • Center-tapped (CT) rectifier requires center-tap transformer. Full Bridge (FB) does not.
  • CT: 2 diodes
  • FB: 4 diodes. Hence, CT experienced only one diode volt-drop per half-cycle
  • Conduction losses for CT is half.
  • Diodes ratings for CT is twice than FB

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

bridge waveforms

D1

D3

is

iD1

io

+

vs

_

+

vo

_

D4

Full Bridge

D2

Vm

vs

4p

3p

2p

p

Vm

vo

vD1 vD2

-Vm

vD3 vD4

-Vm

io

iD1 iD2

iD3 iD4

is

Bridge waveforms

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

center tapped waveforms

is

iD1

iD2

Vm

vs

D1

4p

+

vs1

_

3p

+ vD1 -

2p

p

Vm

+

vs

_

- vo +

vo

+

vs2

_

io

vD1

+ vD2 -

-2Vm

D2

Center-tapped

vD2

-2Vm

io

iD1

iD2

is

Center-tapped waveforms

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

full wave bridge r l load

io

is

iD1

+

vR

_

+

vo

_

+

vs

_

w

t

+

vL

_

vs

p

2p

iD1

, iD2

iD3

,iD4

io

vo

is

Full wave bridge, R-L load

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

approximation with large l
Approximation with large L

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

r l load approximation

vs

w

t

p

2p

io

iD1

, iD2

vo

iD3

,iD4

is

R-L load approximation

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

examples31
Examples

Given a bridge rectifier has an AC source Vm=100V at

50Hz, and R-L load with R=100ohm, L=10mH

  • determine the average current in the load
  • determine the first two higher order harmonics of the load current
  • determine the power absorbed by the load

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

controlled full wave r load

is

T1

iD1

io

T3

+

vs

_

+

vo

_

T2

T4

Controlled full wave, R load

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

controlled r l load33

io

is

iD1

+

vR

_

+

vo

_

+

vs

_

+

vL

_

io

b

p

a

2p

+a

p

vo

Discontinuous mode

p+a

io

b

p

2p

a

vo

Continuous mode

Controlled, R-L load

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

discontinuous mode
Discontinuous mode

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

continuous mode
Continuous mode

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

single phase diode groups

D1

io

D3

vp

+

vs

_

+

vo

_

D4

vn

D2

vo =vp -vn

Single-phase diode groups
  • In the top group (D1, D3), the cathodes (-) of the two diodes are at a common potential. Therefore, the diode with its anode (+) at the highest potential will conduct (carry) id.
  • For example, when vs is ( +), D1 conducts id and D3 reverses (by taking loop around vs, D1 and D3). When vs is (-), D3 conducts, D1 reverses.
  • In the bottom group, the anodes of the two diodes are at common potential. Therefore the diode with its cathode at the lowest potential conducts id.
  • For example, when vs (+), D2 carry id. D4 reverses. When vs is (-), D4 carry id. D2 reverses.

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

three phase rectifiers

Vm

Vm

van

vbn

vcn

vp

vn

vo =vp - vn

3p

4p

p

2p

0

Three-phase rectifiers

D1

io

+ van -

D3

D5

vpn

+ vbn -

n

+

vo

_

D2

+ vcn -

vo =vp -vn

vnn

D6

D4

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

three phase waveforms
Three-phase waveforms
  • Top group: diode with its anode at the highest potential will conduct. The other two will be reversed.
  • Bottom group: diode with the its cathode at the lowest potential will conduct. The other two will be reversed.
  • For example, if D1 (of the top group) conducts, vp is connected to van.. If D6 (of the bottom group) conducts, vn connects to vbn . All other diodes are off.
  • The resulting output waveform is given as: vo=vp-vn
  • For peak of the output voltage is equal to the peak of the line to line voltage vab .

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

three phase average voltage

vo

vo

p/3

Vm, L-L

0

2p/3

p/3

Three-phase, average voltage

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

controlled three phase

io

Vm

+ van -

T3

T5

vpn

+ vbn -

n

+

vo

_

T2

+ vcn -

vnn

T6

T4

Controlled, three-phase

T1

a

vbn

van

vcn

vo

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003

output voltage of controlled three phase rectifier
Output voltage of controlled three phase rectifier
  • EXAMPLE: A three-phase controlled rectifier has an input voltage of 415V RMS at 50Hz. The load R=10 ohm. Determine the delay angle required to produce current of 50A.

Power Electronics and Drives (Version 3-2003), Dr. Zainal Salam, 2003