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

Ch.4 Single Phase Full Wave Voltage Controller
ioN  i A1N  sin( t     sin(    )e

Z  R 2  L 
2

1
 
( R / L )  t 
 
 L 

 R 
       tan 1 
2
Thyristor Norm Currents:
 
 1   2

1
iA1N d (t ) I RN  

 iA1N d (t )
2 
 2 

RMS Norm Output Current:


2
2
I RoN  I RN
 I RN
2
1
cn an2  bn2
an 
1
 
2

2V sin t d t  


 n  tan 1 
12
 
v

o
cos nt d t 
1
2
 1  
VR   
 

bn 
2V sin t
Ripple Voltage:
A ;
VRo  2V VRoN
V 


Voltage Ripple Factor:
bn 
1
cos nt d t  
i

o
0
2
i

o
sin nt d t  
0
2

2

i
o
cos nt d t 

i

o
sin nt d t 

1 2
a n  bn2
2
 H n I R   0
I nR

dn 
2
2
R
2
o
Kv 

VnR  I nR R  nL 
2
R
V  Vc
Io  o
R

I R
2
Po  Ro I Ro
p. f . 
VI Ro
Resistive:     
Inductive:   2   
 L 
 min    tan 1  
RL:
 R 
when    treat as complete sinusoidal wave
V
Z
2
 nL 
2

1
rms output current: I Ro
Ch. 5 Single Phase Full-Wave Controlled Rectifier

Determine what modes its operating in

If discontinuous use Ch.3 graphs and theory (the
formulas are for only one thyristor so for, where
I  I off graph  I  I and
R
QR
I N  I B off graph  I Q
 I Ro  2 I QR and I o  2 I Q )
1
2
for p.f @ source
VB  2V , Z B  Z ,
Ch.2 Circuits with Switches and Diodes
First Principles:
1 t
i dt  vc 0
C 0
di
vL  L
dt
vc 
nR

dn
2
A
A

 A
1
  firing angle
  extinction angle
      conduction angel
V
 Vc 
 m  c
2V
 2V 
  sin 1 m  sin 1 
Vo 
an 
1
2
2
1
 vo d t  
0

1

2
o
cos nt d t 
o
sin nt d t 
v

v
nth rms harmonic voltage:
Thyristor Ratings
 1  2

VR    vo d t 
 2 



Io
A
IR
A
2
max. forward or rev. voltage across
thyristor: v
 2 2V V
I QR 
rms ripple voltage:

12

12
1 2
an  bn2
2
VnR 

 Vo  VnR
2

VRI  VR2  Vo2

2 12

12
voltage ripple factor: K  VRI
v
 
s
I 2 R  Vc I o
p. f .  Ro
Vsec I Ro
Vo
i  I o   cn cosnt  n   d n sin nt  n 
(-Vo )then D.C. delivers power
(Vo ) then A.C. delivers power
V0
R
a
cn  n
Zn
if current totally constant(no ripple):
nth rms harmonic current:
I0 
dn 
I srms  I o
with Freewheeling Diode circuit

1
vo d t 
2 

current ripple factor: K  I RI
i
because
2V
Z
IB 
Xo  X N  XB
bn 
VI
p. f .  o o
Vsec I o
2V sin t d t 


because the source current is not constant must calculate
with:
just calculate the second harmonic for IRI to get the rms
output calculation
rms thyristor current:


vo  Vo   an cos nt  bn sin nt 
A
 I o2  I RI2
avg thyristor current: I  I o
Q
2
1
v  2V sin t
A

I RoN  2  I RN
Vo  vo 
Half-Wave Rectifier:
turn off time available: t    
q
I QR  I RN  I B
B
 8 cos 2  
 V 1 
 2 

2
AK max
IQ  I N  I B
RN
2
cn off graph and then solve for cn
2
1
cn
A ; n  tan n L R  rad 
2
2
Ro
 I Ro 

1
Vo I o
Vsec I SR
vR  Ri
V 
V
VRI
Vo
12
I
RMS value of n th harmonic at angle 


ripple current:
 nR
2
I

I
 nR 
RI
RMS value of line current at angle   0 I R  0


1
2
sin nt d t 
12
nth harmonic component of current: I
I nRpeak  2 I nR
H n 
avg output current:
get this:
io  iA1
I nR 
o

io  I o   d n cosnt   n  n 
0
0
1
v
V 
Load circuit Current
n 1
an 
 
V cos 
 sin n  1 sin n  1 


n  1 
 n  1

d t 

2

io   an cos nt  bn sin nt
2
2

VRI  V  V
Harmonic Analysis
1

RMS Voltage
1
1


  2 sin 2  2 sin 2    

2V
I RoN
Z
2 2

p. f . 
1 

2
I SR    I o  d t 

 

◄BASE VALUES►
 bn 

 an 

 cosn  1 cosn  1 


n  1 
 n  1
2
RMS Ouputs:
I Ro 
1
Vo 
 2I RN
 1  

   sin 2 t  d t 
  

 1

 2
vo  Vo   cn cosnt   n 

2
RMS Norm Voltage:
VRoN

n 1
IN 
1
gamma from graph then find beta
else continue with following formulae:
Vc
 L 
m
;   tan 1 
 ; i N  sin t   
2Vsec
 R 
Fourier Analysis
rms ripple current:
 nL 
 R 
bn
Zn
n  tan 1 
I nR 



12
1 2
cn  d n2
2
I RI  I R2  I o2

12
current ripple factor:
i N  sin t  m
I RI
Io
Iv 
Ch.3 Single Phase Half Wave Controlled Rectifier

mainly done through graphs
RL Load Circuit
di
L  Ri  2V sin t
dt

voltage ripple factor:
  (   )  

 i d t   avg norm


1

cos   1  m 2  m     
2
 

d t 

2
 R L   t  
i
sin t     sin    e


Z 
I RN
Z  R  L 
L Load with EMF Circuit

2

1
 L 
  tan  
 R 
1
2
 in d t 

2
 iN d t 

1
 RMS norm
1
I RN  
 2

1
1  cos 
sin t d t  
2 
2
Z  L  

2
2
 1 
sin 2 
     

2 
 4 

1
2
1

  2  
 cos   cos t  d t 

 1
VRN  
 2
2 

2
 sin t d t 
2
 Vc 
1
  sin m
 2V 
max  2  sin 1
2V
 i d t   avg. norm
N

 i d t 

2
N

2
max  2  sin
1
1
or
2
which leads to:
V
 nR
Zn
 2 
I RI   I nR

 n 1 
Ki 
1

2
 I R2  I o2

1
2
I RI
Io
Discontinuous Current Mode

 V  Vc
t x   ln etON  1 
1  e tON 
V
c



T  t x V
t
Vo  ON V 
c
T
T
 

V  Vc
Io  o
R
current at moment of commutation: Imax
V  Vc
1  e tON 
R
0

I max 


V  Vc
Io  o
R


V e

R e


 1 V

 1 R
c
 Vo   cn sin( nt   n )

c n  a n2  bn2
n 1


VRI   VnR2 
 n 1 

1
2
12
1

rms nth harmonic voltage: V  1 a 2  b 2
nR
n
n
2
rms output voltage: V  V 2 
R
 o




 n  tan a n bn 
the preceding 4 formulas good for C.C.M. and D.C.M. but
C.C.M. has simplified ones in its section
V
an 
sin ntON
n
V
1  cos ntON 
bn 
n
2V
1  cos ntON 1 2
cn 
n
 sin ntON 
 n  tan 1 

1  cos ntON 
the preceding 4 formulas only for C.C.M.
rms “ripple voltage”:

 Vo   cn sin( nt   n )
n 1
V
V 1  e tON 
 c
R 1  e T 
R
T
n 1
2 T
a n   v0 cos nt dt
T 0
T
2  tON
2nt
2nt 
   V cos
dt   Vc cos
dt 
tx
T0
T
T

Vc
V

sin nt ON 
sin nt x
n
n
2 T
bn   v0 sin nt dt
T 0
V
1  cos nt ON   Vc 1  cos nt x 

n
n
2 
rms output voltage: V  V 2 
R
 o  VnR 
n 1


2V

2 12
vo  Vo   an cos nt  bn sin nt

Vc

2
look on graph with all the variables, continue with
respective formulae:
Continuous Current Mode
2
 rms norm

I min
rms nth harmonic voltage: V  1 a 2  b 2
nR
n
n
R Load with EMF Circuit
 V 
  sin 1  c   sin 1 m
 2V 
Vc e   1
tx
T
L
 
,   ,   ,   ON
V
e 1

R
T

Vc

 1
I RN  
 2

Z n  R 2  nL 
current ripple factor:

tON 
 
 

2

1
1 

 

sin 2  vo  Vo   an cos nt  bn sin nt
 2 2 4

n 1
1
 n  tan 1 nL R 
if   tON then C.C.M. else D.C.M, if you don’t have tON
I min
 m

iN  sin t     
 Be  t  tan  
 cos 

m
B
 sin    
cos 
1
IN 
2
m
I max 
12
RL Load with EMF Circuit
  sin 1 
2
12
Ch.6 One Quad Chopper DC to DC Converter

Type A: Vo and Io only positive

2
find:

T
2
1
2V
 cos   cos t  mt    d t 

current at moment of commutation: Imax
    cos   sin  

 cos   cos t  d t 
R 2  nL 
rms ripple current:
a 
t
Vo  ON V
T
2 
2
bn
Bn 
I nR
T
2 
 1
I RN  
 2
Vc
max  2  sin 1
(solve for ρ from m equation)
2V
cos   cos t 
i
L
IN 
R  nL 
2
rms nth harmonic current: I  1 A2  B 2
nR
n
n
1 
m 
IN 
 cos   sin   sin     

2 
2 
2V
i
sin t
Z
1
an
An 
2
Z  R  0  
 1 

I RN    sin 2 t d t 
 2 

L Load Circuit
 rms norm
m  cos   cos(   )
2
R Load Circuit
IN 
12
iN  cos   cos t  mt   
 avg. norm
 
N

 Vc 
1
  sin m
 2V 

 1

 2
i
2
  sin 1 
 
1
IN 
2
I RN
2
n 1
N

 1

 2
VRI
Vo
io  I o   An cosnt  n   Bn sin nt  n 
 
1
IN 
2
Kv 
1

1
 VR2  Vo2
rms “ripple voltage”:


VRI   VnR2 
 n 1 
voltage ripple factor:
Kv 
2
2

VnR2 

n 1


1
2
1


1
2
2
 VR2  Vo2

1
VRI
Vo


1
2
An 
Bn 
an
R 2  nL 
2
bn
R 2  nL 
2
n  tan 1 nL R 
2
io  I o   An cosnt  n   Bn sin 
n 1
rms nth harmonic current:
1
1 2
I 
A  B 2 2 or
nR
2

n
n


Z n  R 2  nL 

2 12
VnR
Zn
I nR 

2 
I RI   I nR

 n 1 
I RI
current ripple factor:

rms ripple current:
Ki 
1
2

 I R2  I o2

1
  n
 sin 
  2
n 1, 3
where  is pulse width
2
Io
I 01  I max  current at commutation
1
r
4Vs
 n   n 
sin 
 sin    VnPeak
n
 2   2 

tc  t2  t1  t1
2
PL  I Drms
RDS (ON )
Ch. 9 DC to DC Converters
Buck
Rise:
2
Vs  Vo  L
V C 
x
 
I 01  V 

tq 2  t2 
r
I
t2

Vs  Vo t1 Vo t 2
I 

L
L
Vo  L
Ch. 8 Power Transistors
BJT
General for Buck:
td :
T

td
V
Pi t   ic (t )  vce (t )  I cs
for Full-Bridge: E
V
2

n 1, 3
n 1, 3

t
Vcc  Vce ( sat)  Vcc  
t
r 

 P t dt
Vo 
tr
i
0
Pi t   ic (t )  vce (t )  Vce ( sat) I cs
Pn 
4E
 n sin nt  V 
4E
t
tr
sin( nt   n )
1
T
 P t dt
tn
0
i
inst . power
avg power
Pi t   ic (t )  vce (t )  Vce ( sat) I cs
n

 nL  1 nC 
n  tan 1 

R

tq 
Ps 
rad 
commutation
load current at instant of commutation sub this into the io
formula:  t   , io  I 01 A

1
2

4E
sin nt  n d t 
n Z n
 1  4 E  2

 sin 2 nt   n d t 
I nRo    
 n Z 
2



n 


1
4E
I nD 
sin nt  n d t 
2  n Z n
Io 
Po  RI

I
n 1, 3,..
2
1R
on
I Ro 
V
2 I1R
Ps  
2

 P t dt
ts
0
i
avg power

I
n 1, 3,..
2
on
1
2

t   t
Pi t   ics (t )  vce (t )  I cs 1  Vcc 
 t  t
f 

 f
1 tf
Pf   Pi t dt avg power
T 0

 inst . power


to  1  k T  t s  t f :
Pi t   ics (t )  vce (t )  I ceoVcc
Po 
1 to
Pi t dt
T 0
inst . power
avg power
General Equations:
tON  td  tr
tOFF  ts  tr
Ptotal  PT  Pd  Pr  Pn  Ps  Pf  Po
dPi t 
 0 for max power
dt
Thermal Evaluation
TJUNCT  TAMB  PT  JC  CS   SA 
if resistive:
the only difference use this formula for rms output current: Secondary Breakdown Derating
Factor: TC  TJUNCT  PT JC
Vo
q 1 I

C 8 fC
Vo
Dk
1
T
f
T  t1  t 2 
I  L
Vs
t2 
_

1 D
8 fLC
I
I
L
L
Vs  Vo
Vo
I  L
V0  Vs
t 2  1  D T
V
V DT Vo D
Vo  s  o

1 D
RC
RCf
tf :
T
 t x  t off : has to be true for guaranteed
2
I on 
1
T
D
t1  DT
inst . power
12
2

1  

Z n   R 2   nL 
 
nC  


V
t1
 k  o  duty cycle
T
Vs
Boost:
T  Ts
t1 
ts :
A
I
I
L
L
Vs  Vo
Vo
t1
t
D  2
T
T
Vo Vs  Vo  kVs 1  k 
I 

fLVs
fL
t n  kT  t d  t r :
using the following 2 equations, the nth harmonic voltage
and current equations can be found:
 nZ
1
T
T  t1  t 2 
kD
tr :
Pr 
1
f
Vo  kVs
Pi (t ) dt
0
Ch. 7 Single Phase Voltage-Source Inverter
for Half-Bridge: E
I 2  I1
I
L
t1
t1
Fall:
VnPeak
sin t  n 
Zn
t angle desired,
1
Pd 
T
12
1 
1
tq1     2 sin 1 
r 
x

Vds
 miller effect
Vgs
Pi (t )  ic (t )  vce (t )  I ceo Vcc

io 
1
i .e. commutation


2 r
vo 

 sin nt 

V ABn 
I on 
 CV
1  cos  r t 2   x 1  cos  r t 2 
t1 
I 01
r
t1
 n
2
4Vs
 n sin 
V
ion  nPeak sin t  n 
Zn
tq  t2  t1


v AB 
V ABnR  V ABn
1
L1C
1
 LI
  LI  
sin 1   r 1 01  
 sin 1 r 1 01
V  r r
V

2 1
2
 3
1  r L1 I 01 
t2 
 sin 
 t1 

t

r r
r r 1
 V  r
t1 
Qx  VxCx
Q
Q
I1  gd , I 2  gs , IT  I1  I 2
t
t
Vgd  Vgs   Vds 
Ch 7 (cont)
single pulse width:
Current Commutation of Type ‘A’ Chopper
r 
MOSFET
I o max io1 peak 
I

, I QR  Ro
2
2
2
I Ro 
which leads to:
I 
Vo Vs  Vo  Vs D

fLVo
fL
t  DTs
Ic  Io
1 DTs
DT
I D
I o dt  I o s  o

C 0
C
fC
Vc 
2
Po 
Vo
R
 Vs 


2
Vo2  1  D 
Vs
Vs I s 


R
R
1  D 2 R
IL 
Vs
1  D 2 R
I max  I L 
i L
2
I min  I L 
i L
2
i L Vs DT

2
2L
Vs
V DT

 s
1  D 2 R 2 L
Vs
V DT

 s
1  D 2 R 2 L
D1  D  R
2f
V
 
Q   o  DT  CVo
R
2
Lmin 
Vo
D

 given ripple percent
Vo
RCf