CH3 Half-wave rectifiers (The basics of analysis) 3-1 Introduction ˙Used often in low-power applications ( average current in the supply will not be zero,

CH3 Half-wave rectifiers (The basics of analysis)

3-1 Introduction

˙Used often in low-power applications ( average current in the supply will not be zero , and cause

problems in transformer performance.)

˙Enable the student to advance to the analysis of more complicated circuits with a minimum of effort.

3-2 Resistive load : Fig.3-1

R

V

R

VI

V

t)td(sinV2

1VV

m0

m

0

mavg0

Average power absorbed by the resistor:

2R

V

R

VI

2

Vt)d(t)sin(V

2

1V

R

VRIP

mrmsrms

m

0

2mrms

2rms2

rms

3-3 Resistive-Inductive load (R-L load): Fig.3-2

Diode on，dt

(t)idL(t)iRtsinVm

response natural(t)i

response forced(t)i(t)i(t)i(t)i

n

f

nf

The force response is the steady-state sinusoidal current.

)R

L(Tan L)(RZ

)-tsin(Z

V(t)i

1-22

mf

，

Natural response is the solution to the homogeneous differential eq.

constant) (Time R

L Ae(t)i

0dt

(t)idL(t)iR

t-

n

，

Some numerical method is required to solve

.

angle. extinction

])e(sin)-t[sin(Z

V

esinZ

V)-tsin(

Z

V(t)i

sinZ

V)sin(-

Z

V-A

0Ae)-sin(0Z

V(0)i

Ae)-tsin(Z

V

(t)i(t)i(t)i

t/-m

t/-mm

mm

0m

t/-m

nf

0)e(sin)-sin(

0)e(sin)-[sin(Z

V)(i

t

])e(sin)-t[sin(Z

V

)e(sinZ

V)-tsin(

Z

Vt)(i

/-

/-m

t/-m

t/-mm

Average power absorbed by the load :

0

22

0

2rms

2rmsL

t)t)d((i2

1t)t)d((i

2

1I

RIP

Average current.

0

t)t)d((i2

1I

RL )

R

L(Tan L)(RZ

2t 0

t0 )e(sinZ

V)-tsin(

Z

Vt)(i

1-22

t/-mm

，，

，

，

3-5 R-L source load : Fig.3-5

(t)i(t)i(t)i

Vdt

(t)idL(t)iRtsinV

ON Diode ,)V

V(sin VsinV

nf

dcm

m

dc1-dcm

,

(t)i f is determined using superposition for the two source ( dcm V ,tsinV )

t-

n

dcmf

Ae(t)i

R

V-)-tsin(

Z

V(t)i

? 0)(i

)eR

V)-sin(

Z

V(-A 0)(i

otherwise 0

t AeR

V-)-tsin(

Z

Vt)(i

dcm

t-dcm

，

，

Average power absorbed by R, P= RI2rms

Average power absorbed by dcV , dcdc IVP

t)t)d((it)sin(V2

1

t)t)d((it)(2

1IVRIP

t)t)d((i2

1I t)t)d((i

2

1I

m

2

0dc2rmsac

2rms ，

3-6 Inductor-Source load : Fig.3-6

At )V

V(sint

m

dc1- ， Diode conducting.

otherwise , 0

t t)-(L

Vt)cos-(cos

L

V

dVL

1-)dsin(V

L

1t)(i

L

V-tsinV

t)d(

t)(id

Vt)d(

t)(idL

Vdt

(t)idLtsinV

dcm

t

dc

t

m

dcm

dc

dcm

,

Power supplied by the AC source=Power absorbed by the dc source.

3-7 The freewheeling diode Fig.3-7 ， 2D : freewheeling diode

The voltage across the load is a half-wave rectified sine wave and

6... 4 2n

o2m

omm

o t)cos(n1)-(n

2V-t)sin(

2

VV(t)

，，

Using superposition, ?(t)io

If L is infinitely large，

R

L

R)(

V

R

VI(t)i mo

oo ,Reducing load current harmonics.

Zero-to-peak fluctuation in load current can be estimated as being equal to the amplitude of the first ac term in the Fourier series ( )

1I

The peak-to-peak ripple is then 1o 2II

3-8 Half-wave rectifier with a capacitor filter: Fig.3-11

t sinVV

off diode eV

on diode tsinV(t)

m

RC)(-t

m

o

,

,

,

At ， the slopes of the voltage function are equal. t

RC)(-t-

RC)(-t-

e)1

(sinV

]esin[V)(

cosV)sinV()(

RC

td

d

tttd

dm

ｍ

ｍ

ｍ

)()(

11

1

sinV

cosV

sinVsinVcosV

11

)/()(m

RCTanRCTan

RCTanRCTan

RC

RCe

RC

m

m

mRCm

，

In practical circuits where the time constant is large.

mmand VsinV2

At ONDiodet 2 ，

0)(sinsin

)sinV()2sin(V)/()2(

)/()2(

RC

RCmm

e

e

Solved numerically for ?

on diode t )tcos(Vc

off diode t eR

sinV

)t(d

)t(dct)(i

dt

)t(dvc(t)i

R/)t(v(t)i

m

)RC/()t(m

oC

oC

oR

，22，

，2，

CRDS iiii

Since the diode is on for a short time in each cycle, the peak diode current is generally much larger than the average diode current. Peak capacitor current occurs when the diode turn on at 2t ,

Peak diode current is

)sin

cos(V

sinVcosVI ,

Rc

Rc

m

mmpeakD

Peak-to-peak ripple voltage

)sin(VsinVVV mmmo 1

If R-C time constant is large compared to the period of the sine wave

2 ,V V ,

2 m

2

,2

......!3!2!1

1

，)(，

)2

(

]2

11[

]1[

)2(

32

,

)(2

)(2

2222

TRC

RC

x

xxxe

IrippleVC

fRC

V

RCV

RCV

eVeVVV

eVeVv

x

peakDo

mm

m

RCm

RCmmo

RCm

RCmo

3-9 The controlled half-wave rectifier

Resistive load (R): Fig.3-13

.angle delay)cos(

V)t(tdsinVV

,resistor load the across voltage Average

mmo

1

22

1

Power absorbed by R is R

VP rms

2

2

21

2

2

1

2

1 22

0

2

)sin(V

)t(d]tsinV[)t(d)t(V

m

morms

R-L load : Fig.3-14

angle ction?....condu-

angle nction?.....extiy ,numericall solved

]e)sin())[sin(Z

V()(i

otherwise , 0

t , ]e)sin()t)[sin(Z

V(

)t(i

e)]sin()Z

V([A

Ae)sin()Z

V()(i

Ae)tsin()Z

V(

)t(i)t(i)t(i

)/()(m

)/()t(m

)(m

)(m

)(t

m

nf

0

0

Average output voltage

]cos[cos2

V)(sinV

2

1V

m

mo tdt

Average current :

)t(d)t(iI

2

1

Power absorbed by the load is RIP rms2

)t(d)t(iI rms

2

2

1

.R-L source load : Fig.3-15

The gate signal may be applied to the SCR at any time that the ac source is larger than the dc source.

)/(dcm

)t/(-dcm

m

dc-1min

]eR

V-sin

Z

V[-A 0)(i

otherwise , 0

t ,AeR

V-)-tsin(

Z

Vt)(i

)V

V(sin

3-11 Commutation

Effect of source inductance: Load current is constant.

Fig.3-17

1D 2DThe interval when both and are on is called commutation time (or angle). Commutation is the process of turning off an electronic switch, which usually involves transferring the load current from one switch to another.

When both and are on, the voltage across is , and current in and source is

1D 2D SL tsinVv mLS

SL

t

0S

mm

S

t

0 SLSS

S t)cos-(1L

V0t)td(sin

L

1(0)it)d(

L

1i

)2V

XI-(1

V

cosu)(12

Vt)os(-c

2

Vt)td(sinV

2

1V

LX , )V

XI-(1cos)

V

LI-(1cosu

0cosu)-(1L

V-I(u)i

t)cos-(1L

V-Ii-Ii

m

SLm

mu

m

u mo

SSm

SL1-

m

SL1-

S

mLD2

S

mLSLD2

SL reduces average load voltage ( oV )