4. diode rectifiers

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


Single-phase diode rectifiers • Basic rectifier, single diode, purely resistive load •  Large ripple in vd


Single-phase diode rectifiers

• Basic rectifier, single diode, inductive load

RivdtdiLv sL −=⋅=

03

0, == ∫

t

LavL dtvV


Single-phase diode rectifiers •  Single diode, load with internal dc voltage, load with large C

03

1

, == ∫t

tLavL dtvV


Single-phase bridge rectifier

+ rail: Diode with highest anode potential is conducting) - rail: Diode with lowest cathode potential is conducting


Single-phase bridge rectifier • Resistive load


•  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


Single-phase bridge rectifier •  Large inductive load - Effect of Ls on current commutation •  is (inductor current) can not change from Id to –Id instantaneously


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∫


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


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


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


Single-phase bridge rectifier

• Constant dc-side voltage (Approximation of RC-load with RC>> 10 ms)


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


Practical bridge rectifier

vd

id ic

t1 t2 t3

t1<t<t2

t2<t<t3

vd<vs → diodes conducting

Ls=0


Practical bridge rectifier

vd<vs → diodes conducting

Ls>0


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∞

== + ∑


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


( ) ( ) 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:


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

= = = =


Commutation

Three phase full-bridge rectifiers

Commutation starts when vac = vcn Driving voltage in a mesh: vcomm = van - vcn


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


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.

4. diode rectifiers

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