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ELEC4240/9240 Power Lecture 8 - Effect of source on operation

Ideal VS real rectifier with source inductance

The output DC of the rectifier circuits discussed so far have been found by assuming that currents transfer (commutate) from one diode to another instantaneously. However this can not happen when the AC source has some inductance Ls. (Change of current through any inductance must take some time!). This source inductance is associated with the of the supply and the inductance of the AC supply network to the input transformer. The commutation process (or the overlap process) forces more than one diode or a pair of (in a bridge rectifier) to conduct simultaneously, resulting in a drop from the output terminals which is proportional to the load current.

The output dc voltage Vd of a rectifier falls with load current Id, by an amount which is much larger than additional voltage drop across the conducting diodes when the current through the diodes increases. The AC source inductance, which consists of the AC line and the input transformer leakage , is mostly responsible for the additional voltage drop. Consider the half-wave diode rectifier shown below.

D is Ls

vs Id IDf D Load vs = Vmaxsinωt ∼ vi f

Figure 8.1. Half-wave diode rectifier with source inductance.

Let us assume that the load current Id is smooth and -free (i.e., of constant, due to the highly inductive load). Assume also that for ωt > 0, the load current flows through the rectifier diode and that for ωt > π, it commutates to the free-wheeling diode Df. This transfer of the load current between the rectifier and the freewheeling diodes can not however be instantaneous, because of the source inductance Ls. This transfer takes place over a small commutation or overlap angle µ, during which time, the current gradually falls to zero in one circuit and it rises to Id in the other circuit at the same rate. Clearly, the two diodes simultaneously conduct during the commutation process (µ).

Lecture 8 Effect of overlap on rectifier 8-1 F. Rahman ELEC4240/9240 Power Electronics

vs

iD Id

i Df

µ vo

v i

µ µ µ µ

Figure 8.2 Waveforms in the rectifier circuit of figure 8.1

Lecture 8 Effect of overlap on rectifier 8-2 F. Rahman ELEC4240/9240 Power Electronics

Because of the prolonged conduction of Df, the load voltage is clamped to zero for 0 < ωt <µ, resulting in some loss of positive voltage in the vo waveform. Consequently Vd is reduced, the extent of which depends on µ, which in depends on Ls and Id.

During the process of overlap, all of the ac source voltage drops across Ls, so that for 0 < ωt < µ,

di vV== sintLω 8.1 max s dt

Integrating,

µ Id Vsintd(t)Lmaxωω== ω s diLI ω s d 8.2 ∫∫00 or, V(1cos)max−=µ ω LI s d 8.3

ωLs and cosµ = 1- Id 8.4 Vmax

The overlap, or commutation angle, µ can the found from (4) given Id and Ls.

11πµ Vdmax=− V sin(ω t )d(ωωω t ) V max sin td( t ) 22π ∫∫00π

V1max V max⎡ ω L s ⎤ = −=−ωLs I1Idd⎢ ⎥ 8.5 ππ22V π⎣ max ⎦

Vmax π

Vd

Id

Figure 8.3 Voltage regulation characteristic of the rectifier of figure 8.1 due to source inductance

Lecture 8 Effect of overlap on rectifier 8-3 F. Rahman ELEC4240/9240 Power Electronics Overlap in a bridge rectifier due to source inductance

During the positive half cycle, diodes D1 and D4 carries the load current Id. During the negative half cycle, diodes D3 and D2 carry the load current. During overlap all four diodes carry the load current. The output voltage during overlap is zero and all of the supply voltage applies across the source inductor Ls.

vo

Id D1 D3 ip Ls is

Vd Load Vmaxsinωt vi

N:1 D2 D4

Figure 8.4. A diode bridge rectifier with source inductance

vs

vo µ µ

Id i s

- Id

vi

µ µ µ µ

Figure 8.6 Waveforms in the rectifier of figure 8.4

Lecture 8 Effect of overlap on rectifier 8-4 F. Rahman ELEC4240/9240 Power Electronics Thus, during commutation overlap,

di VsintLω = 8.6 max s dt

µ Id Vmax sinωω td( t )== ω L s di 2 ω L s I d ∫∫0I− d

2Lω s ∴ cosµ =− 1 Id 8.7 Vmax

The dc output voltage of the converter is given by

111ππµ Vd==− V max sinω td(ωωωωω t ) V max sin td( t ) V max sin t( d t ) πππ∫∫∫µ 00

2Vmax⎛⎞ω L s =−⎜⎟1Id 8.8 π ⎝⎠Vmax

2Vmax⎛⎞ω L s =−⎜⎟1Id 8.9 π ⎝⎠Vmax

2Vmax

π

Vd

Id

Figure 8.5 Regulation characteristic of a 1-phase bridge rectifier due to source inductance

Lecture 8 Effect of overlap on rectifier 8-5 F. Rahman ELEC4240/9240 Power Electronics Effect of overlap on three-phase center-tap rectifier

In the three-phase, center-tap rectifier of figure below, the load current starts to

commutate to diode D2 from ωt = 0+ when vb starts to become more positive than va. During overlap, both diodes D1 and D2 carry the load current which is assumed to remain constant during the process.

v an L s D 1

v o D2 L s v bn Id

Load v cn L s D3 V d

n

Figure 8.7 Three-phase center-tap rectifier with source inductance

van vbn vcn

vo

v abi

µ

i a

i b

i c Figure 8.7

µ

Figure 8.8 Waveforms in the rectifier of figure 8.7

Lecture 8 Effect of overlap on rectifier 8-6 F. Rahman ELEC4240/9240 Power Electronics During overlap,

di vL=+a v 8.10 an sdt o di vL=+b v 8.11 bn sdt o

Assuming that Id remains constant during the overlap time, and noting that iiab+= I d, so that

di di ab=− . 8.12 dt dt

Adding the voltage equations and canceling the equal but opposite terms,

vv+ v = an bn , during the overlap process. 8.13 o 2

Thus, during the commutation overlap, the converter output voltage vo is the average of the voltages of the lines undergoing commutation. Once the load current is fully commutated, vo jumps up to the potential vb. Form the ideal output voltage waveform, the area bounded by vb and (va +vb)/2 is lost due to overlap of two conducting diodes.

In the following analysis, the line-neutral voltages are:

vVsintan= max ω ; vVsint2/3bn=− max (ω π ) ; vVsint4/3cn= max (ω − π )

The part of the positive voltage pulse lost due to overlap starting from angle ωt = π/6 is given by

vvbn+− an vv bn an di vL−== 8.14 bn22dt s

The area (shaded) inside the voltage pulse lost due to overlap is given by

π + µ Id 6 ⎛⎞vvbn− an π ⎜⎟d(ωω t )== Ls di ω Lsd I 8.15 ∫∫⎝⎠2 0 6

Note that (vb - va) is the line-line voltage vba. The integral on the right hand side by shifting the origin by π/6 to the left. Thus

Lecture 8 Effect of overlap on rectifier 8-7 F. Rahman ELEC4240/9240 Power Electronics

µ 3Vmax sinωω td( t )= ω Ls Id 8.16 ∫0 2

2Lω s ∴1cos−=µ Id , so that 8.17 Vmax ll−

2Lω s cosµ =− 1 Id where Vmax l-l = √3 Vmax 8.18 Vmax l− l

The dc output voltage is

33V 3Lωω3Vmax l− l ⎛⎞ L VI1I=−=max s⎜⎟ − s 8.19 dd⎜⎟ d 22ππ 2 π⎝⎠ Vmax l− l

3Vmax l− l

Vd

Id

Figure 8.9 Regulation characteristic of the rectifier in figure 8.7

Lecture 8 Effect of overlap on rectifier 8-8 F. Rahman ELEC4240/9240 Power Electronics Effect of source inductance on three-phase diode bridge rectifier

vL+ vo = vL+ − vL−

van Ls ia D1 D3 D5 iL

Ls ib vabi R vbn Vd Load L Ls ic vcn D4 D6 D2

v L−

Figure 8.10 Three-phase diode bridge rectifier with source inductance

As for the three-phase CT rectifier, the voltage equations are

di vL=+a v 8.20 asdt L+ di vL=+b v 8.21 bsdt L+ when D1 and D3 are in overlap due to the source inductance Ls and where all voltages are with respect to the fictitious neutral point. vL+ is the potential of the positive voltage bus (cathodes of the upper diodes) of the rectifier with respect to the neutral point.

As before, during each overlap, the positive and negative dc buses have voltages which are average values of the commutating line-line potentials.

During the commutation overlap of diodes D1 and D3, the positive rail voltage is (vb + va)/2, and the positive voltage lost from VL+ as a result of the overlap is

vv+− vv di vv−=− vba = ba = L 8.21 bL+ b22dt s

Integrating for the duration of the overlap

π +µ Id 6 ⎛⎞vvba− π ⎜⎟d(ωω t)== Ls di ω Lsd I 8.22 ∫∫⎝⎠2 0 6

Lecture 8 Effect of overlap on rectifier 8-9 F. Rahman ELEC4240/9240 Power Electronics

va vb vc

vo

v ABi

ia

ib

ic

Commutation notches in vabi

Figure 8.11 Waveforms in the rectifier of figure 8.10

Lecture 8 Effect of overlap on rectifier 8-10 F. Rahman ELEC4240/9240 Power Electronics

Note again that (vb - va) is the line-line voltage. The integral in the right hand side by shifting the origin by π/6 to the left. Thus

µ 3Vmax sinωω td( t )= ω Ls Id 8.23 ∫0 2

2Lω s ∴ 1cos−=µ Id , so that 8.24 Vmax ll−

2Lω s cosµ =− 1 Id where Vmax l-l = √3 Vmax 8.25 Vmaxl− l

The dc output voltage Vd is given by

µ 3Vmax l−− l13L V max l l 3V max l − l ω s Vd =− sinωω td() t =− Id 8.26 π πππ/3∫0 2

3V ⎛⎞ωL V1I=−max l− l s 8.27 dd⎜⎟ π ⎝⎠Vmax l− l

3V max l− l π

Vd

Id

Figure 8.12. Voltage regulation characteristic of the three-phase diode bridge rectifier due to source inductance

Lecture 8 Effect of overlap on rectifier 8-11 F. Rahman