4. diode rectifiers
DESCRIPTION
TRANSCRIPT
1 Challenge the future 1 Challenge the future
Electronic Power Conversion Line-Frequency Diode Rectifiers
2 Line-Frequency Diode Rectifiers
5. Line-frequency diode rectifiers • Conversion: line frequency ac → uncontrolled dc (control often in
next converter stage) • Applications: power supplies, ac/dc/ac drives, dc servo drives, ...
J Low cost, popular
L Large capacitor needed for ripple-free dc voltage → distorted line current
L Filters required to comply with standards on allowable line-current harmonics
3 Line-Frequency Diode Rectifiers
Single-phase diode rectifiers • Basic rectifier, single diode, purely resistive load • Large ripple in vd
4 Line-Frequency Diode Rectifiers
Single-phase diode rectifiers
• Basic rectifier, single diode, inductive load
RivdtdiLv sL −=⋅=
03
0, == ∫
t
LavL dtvV
5 Line-Frequency Diode Rectifiers
Single-phase diode rectifiers • Single diode, load with internal dc voltage, load with large C
03
1
, == ∫t
tLavL dtvV
6 Line-Frequency Diode Rectifiers
Single-phase bridge rectifier
+ rail: Diode with highest anode potential is conducting) - rail: Diode with lowest cathode potential is conducting
7 Line-Frequency Diode Rectifiers
Single-phase bridge rectifier • Resistive load
8 Line-Frequency Diode Rectifiers
Single-phase bridge rectifier • Resistive load
9 Line-Frequency Diode Rectifiers
• For inductive load
ss
T
sd VVtdtVT
V 9.022sin22/1 2/
00 === ∫ π
ω
• Output voltage
Fourier analysis of is:
hII ssh /1=dds III 9.022
1 ==π
9.0/1 == ds IIDF
9.0cos)/( 1 =⋅= ϕds IIPF
10 Line-Frequency Diode Rectifiers
Single-phase bridge rectifier • Large inductive load - Effect of Ls on current commutation • is (inductor current) can not change from Id to –Id instantaneously
11 Line-Frequency Diode Rectifiers
sin sL s s
div = 2 V t = L 0 < t < udt
ω ω
u - commutation angle
0( ) sin
t
s s ss
1i t = i (0) + 2 V t d(t)L
ω∫0)0( =si ds Iui =)/( ω
coss d sL I 2 V ( 1 u) ω = −
cos s d
s
L I u = 1 2 V
ω−0
sin ( )u
u sA = 2 V t d tω ω∫
12 Line-Frequency Diode Rectifiers
Voltage drop due to current commutation
0sin ( ) s
d s s2 2 V1V = 2 V t d t = = 0.45 V
2 2π
ω ωπ π∫
sin ( )d su
1V = 2 V t d t2
πω ω
π ∫
0sin ( )
u
d s s
u ss s d
1V = 0.45V 2 V t d t2
area A L = 0.45V = 0.45V I2 2
ω ωπ
ωπ π
−
− −
∫
u sd d
area AV = = I2
L 2 πωπ
Δ
For Ls = 0:
For Ls ≠0:
= apparent resistor (no losses)
Static model of dc-side of rectifier
Every cycle the area Au is lost once
13 Line-Frequency Diode Rectifiers
commutation: 3+4 ==> 1+2 1. describe problem as superposition of
Id and commutation current iu
2. identify commutation meshes and driving voltages
3. note that Id is constant
14 Line-Frequency Diode Rectifiers
0( ) sin
t
s s ss
1i t = i (0) 2 V t d( t)L
ωω ω ω
ω+ ∫
0sin
u
s s s s d L i 2 V t d( t) with i I2ω ω ωΔ = Δ =∫
( cos )s d s2 L I = 2V 1 u ω − cos 2 s d
s
L I u = 1 2 Vω−
0sin
u
u sA 2 V t d( t) ω ω= ∫
0sin
u
s d s L I 2 V t d( t) 2 ω ω ω= ∫
22 u s
d s s dArea A 2 LV = 0.9 V = 0.9 V Iω
π π− −
or
Shaded area forces the current from -Id to Id :
Output voltage:
Every cycle the area Au is lost twice
Current is(t) goes from -Id to Id in interval (0,u)
è
15 Line-Frequency Diode Rectifiers
Single-phase bridge rectifier
• Constant dc-side voltage (Approximation of RC-load with RC>> 10 ms)
16 Line-Frequency Diode Rectifiers
sin bd sV = 2 V θ
( ) ( sin ) ( )b
d s ds
1i = 2 V t - V d tL
θ
θθ ω ω
ω ∫
0 ( sin ) ( )f
bs d
s
1 = 2 V t V d tL
θ
θω ω
ω−∫
• id has maximum when vs crosses Vd again ê
• θf follows from:
• high peak line current
17 Line-Frequency Diode Rectifiers
Practical bridge rectifier
vd
id ic
t1 t2 t3
t1<t<t2
t2<t<t3
vd<vs → diodes conducting
Ls=0
18 Line-Frequency Diode Rectifiers
Practical bridge rectifier
vd<vs → diodes conducting
Ls>0
19 Line-Frequency Diode Rectifiers
Line voltage distortion 1
spcc s s
div = v L dt
−
11 1 2
shpcc s s s h
di div = v L L dt dt
∞
=− − ∑
1,1 1
spcc s s
div = v L dt
−
, 1 2sh
pcc dis s h
div = L dt
∞
=− ∑
Fundamental component of vpcc
Voltage distortion component
is contains harmonics:
1 2s s shhi i i∞
== + ∑
20 Line-Frequency Diode Rectifiers
d Pn Nnv = v v−
How many diodes are conducting? Which diodes are conducting? Diode with highest anode potential is conducting (+rail) Diode with lowest cathode potential is conducting (-rail)
ia = +Id when diode 1 is conducting = - Id when diode 4 is conducting = 0 when neither diode 1 or 4 is conducting
Assume: id constant and equal to Id and Ls = 0
Fig. (b):
Three phase full-bridge rectifiers
21 Line-Frequency Diode Rectifiers
( ) ( ) cosd ab LLv t = v t = 2 V tω
6
62 cosdo LL
LL LL
1V = V t /3
3= 2 V = 1.35 V
π
π ωπ
π
−∫
1 1 < t < 6 6π ω π−For:
22 Line-Frequency Diode Rectifiers
Line current
s d d2I = I = 0.816 I3
1s d d6I = I = 0.78 Iπ
1ssh
II =h
1 1s
s
I 3PF = DF DPF = = 0.955I π
⋅ = ⋅
From Fourier analysis:
h = 5, 7, 11, 13, 17, ...
with
1 3 0.955 coss
s
IDF and DPFI
απ
= = = =
23 Line-Frequency Diode Rectifiers
Commutation
Three phase full-bridge rectifiers
Commutation starts when vac = vcn Driving voltage in a mesh: vcomm = van - vcn
24 Line-Frequency Diode Rectifiers
comm an cn La Lc = v v v vv − = − ucomm s
div = 2L dt
0
(t
an cnu u
s
v v1i t) = i (0) + d(t)L 2
−∫
0
sinuLL
s d2 V tL i = dt
2ωω Δ ∫
( co2 s )s d LLL I = 2 V 1 uω − cos 2 s d
s
L I u = 1 2 Vω−
2u an cn
Pn an sdi v vv = v L dt
+− =
0
1 sin2
u
u LLA 2 V t d( t) ω ω= ∫ u s dA L Iω=
6 u sdo d
area A 3 LV = = I2
ωπ π
⋅Δs
d do d LL d3 LV = V V = 1.35V Iωπ
− Δ −
è
During commutation:
Shaded area forces the current from 0 to Id in one inductor:
or:
Every period the area Au is lost six times Voltage drop:
so:
è
25 Line-Frequency Diode Rectifiers
Image credits
• All uncredited diagrams are from the book “Power Electronics: Converters, Applications, and Design” by N. Mohan, T.M. Undeland and W.P. Robbins.
• All other uncredited images are from research done at the EWI faculty.