Simplified analysis of graetz circuit copy - copy

Chapter III:SimplifiedAnalysis of Graetz Circuit

3.3.1 Without Overlap•At any instant, two valves are conducting in the bridge, one from the upper and one from the lower commutation group.

•As the next valve of a commutation group fires, the preceding valve turns off.

•This assumption, that no overlap between two valves(meaning no two valves are “on” at the same instant), is incorrect.

•However, this assumption provides a simpler analysis into the operation of a converter.

3.3.1 Without Overlap•The firing of valves are numbered in sequence, with 60° intervals, and conducts for 120°.

•Consecutive firing pulse is 60° in steady state.

Circuit diagram of a basic Graetz circuit

3.3.1 Without OverlapTo further simplify analysis, the following assumptions are taken into consideration

1. The DC current is constant

2. The valves can be modelled as ideal switches with zero impedance when on or conducting, and infinite impedance when off or not conducting.

3. The AC voltages at the converter bus are sinusoidal and remain constant.

3.3.1 Without Overlap•A period of an AC voltage supply can be divided into 6 intervals, corresponding to the firing of a pair of valves.

•The DC voltage waveform repeats for each interval, thus simplifying the calculation of the average DC voltage, since we only have to consider one interval.

•Assuming the firing of the 3rd valve is delayed by an angle α (α° after the crossing of the commutation voltage for valve 3 – voltage eba ) the instantaneous DC voltage vd during the interval is given by

vd = eb – ec = ebc, α ≤ ωt ≤ α + 60°

3.3.1 Without OverlapLet eba = ELL sin ωt

Then ebc = ELL sin (ωt + 60°)

2

2

3.3.1 Without Overlap•Eq. 3.8 that different values of α (range of α is from 0 to 180°), vd is variable. More importantly, when α crosses 90°, vd is negative value.

•This means that the same converter can act as an inverter or rectifier depending on the direction of the DC voltage.

3.3.1 Without Overlap: DC Voltage Waveform

•The DC voltage wave contains a ripple whose fundamental frequency is six times the supply frequency.

•It contains harmonics of the order h = np, where p is the pulse number, and n is an integer.

•The rms value of the hth order harmonic is given by:

3.3.1 Without Overlap: DC Voltage Waveform

As seen on the figure, 3 voltage jumps arise due to the commutation from one valve to the next. The jumps have the same magnitude given by eq. 3.10:

3.3.1 Without Overlap: DC Voltage Waveform

The full range of 180° cannot be utilized. For all series thyristors to fire, a minimum limit of alpha is set greater than zero, likewise, for it to go off, an upper limit is set less than 180°. Α is not allowed to go beyond the upper limit (180°-γ), where γ is called the extinction angle or margin angle. The minimum γvalue is 10°. For normal inverter operation, γ is not allowed to go below 15° or °.

3.3.1 Without Overlap: AC Current Waveform

While it is assumed that dc has no ripple(or harmonics) due to smoothing reactors in series with the bridge circuit. The AC currents flowing through the primary and secondary windings of the transformer has.

Fig. 3.6 shows the current waveform in a valve winding.

3.3.1 Without Overlap: AC Current Waveform

• The rms value of the fundamental component of current is given by:

Where n is an integer, and p is the pulse number.

3.3.1 Without Overlap: AC Current Waveform, Power Factor

AC harmonics order(6 pulse bridge converter) : 5, 7, 11, 13 and higher order.

• The first four are filtered out using tuned filters for each.

• The rest requires high pass filters.

The rms value of the hth is given by

The AC power supplied to the converter is

𝐼ℎ =𝐼𝐼ℎ

3.3.1 Without Overlap: Power Factor

•Ignoring the losses in the converter, the DC power will be equal to the AC power. Therefore:

Substituting for Vdo and II from eqs. 3.8 and 3.11 gives : cosΦ = cosα

The reactive power requirement increases as α closes to 90° from 0° or 80°. When α=90°, pf=0 and only reactive power is consumed.

3.3.2 With Overlap• Due to the leakage inductance of the converter transformers and the supply network’s impedance, the current in the valve will not suddenly change. An example is when commutation from valve 1 to 3, there is a finite period of time when both valves are conducting. That is called overlap and its duration is measured by u called overlap(commutation) angle.

3.3.2 With Overlap

•The first subinterval shows three valves are conducting and 2nd

subinterval shows 2 valves. This is based on the assumption that the u<60°. As u goes beyond 60°, there is a finite period where 4 valves are commutating and 3 valves at the rest of the interval.

• Because of this, there are 3 modes of the converter.

3.3.2 With Overlap

The three modes are: ◦ 1. Mode 1 – two and three valve conduction u<60°

◦ 2. Mode 2 – Three valve conduction u=60°

◦ 3. Mode 3 – Three and four valve conduction u<60°

3.3.2 With Overlap: Analysis of Mode 1

For this Mode, the bridge circuit can be simplified to this:

3.3.2 With Overlap: Analysis of Mode 1

For that circuit eb – ea = Lc eq. 3.16

The L.H.S is called the commutating emf and its value is:

Which is also the voltage across valve 3 just before it fires.

Since i1 = Id – i3 (eq. 3.18), we get:

𝑑𝑖3𝑑𝑡

−𝑑𝑖1ⅆ𝑡

3.3.2 With Overlap: Analysis of Mode 1

Where:

The sol’n from 3.20 is from the initial condition i3 (ωt = a) = 0 (eq. 3.22)

At ωt = α + u, i3 = Id,, this gives Id = Is[cos α – cos (α +u)] (eq. 3.23)

Note: during commutation, the instantaneous dc voltage is:

Instead of (eb – ec)

−3

2𝑒𝑐

3.3.2 With Overlap: Analysis of Mode 1 : Average Direct Voltage

The average direct voltage Vd is obtained as:

3.3.2 With Overlap: Analysis of Mode 1 : Average Direct Voltage

Where Rc is called equivalent commutation resistance. This is similar to armature reaction in DC machines as it only represents voltage drop and not a power loss.

3.3.2 With Overlap: Analysis of Mode 1 : DC voltage and Valve voltage waveforms

Notice the valve voltage has now 8 jumps. The two major jumps occur at the firing and turning off and is given by:

3.3.2 With Overlap: Analysis of Mode 1 : DC voltage and Valve voltage waveforms

•The remaining 6 jumps are divided to 2 groups, the first group composed of jumps that are equal in magnitude to , and the other equals with .

•The jumps result in extra losses in the damper circuit.

𝑉𝑗1

2

𝑉𝑗2

2

3.3.2 With Overlap: Analysis of Mode 1 : Inverter equations

To operate as an inverter, an advance angle β is definedwhere β = π – α, and use opposite polarity for the dc voltage with voltage rise opposite to the direction of current. Thus:

3.3.2 With Overlap: Analysis of Mode 1 : Inverter equations

The subscript “i” refers to the inverter.

3.3.2 With Overlap: Analysis of Mode 1 : AC current and DC voltage harmonics

The waveform of the valve voltage and current are distorted. And so, the expressions derived for the case w/out overlap becomes invalid. Through Fourier Analysis, the actual expression for the current is :

Where Φ is the p.f angle and δ = α + u

3.3.2 With Overlap: Analysis of Mode 1 : AC current and DC voltage harmonics

From those expressions Φ can be obtained by:

The harmonic components in the AC current are also changed. The following is reduced form the values calculated with no overlap and is called the reduction factor:

3.3.2 With Overlap: Analysis of Mode 1 : AC current and DC voltage harmonics

The DC voltage expression are also affected due to overlap. It can be shown that

3.3.2 With Overlap: Analysis of Mode 2

•Mode 1 is the normal mode of operation. Mode 2 occurs during DC line faults or a dip in the AC voltage occurs.

•The overlap angle u may exceed 60°.

•When u>60°, the minimum number of valves conducting are three and there are intervals where four valves.

3.3.2 With Overlap: Analysis of Mode 2

The equivalent circuit of this mode is shown:

3.3.2 With Overlap: Analysis of Mode 2

= Is cos(α + u -60°) + B (3.46)

3.3.2 With Overlap: Analysis of Mode 2

3.3.2 With Overlap: Analysis of Mode 2 : Average Direct Voltage

3.3.2 With Overlap: Analysis of Mode 2 : Average Direct Voltage

In this case, the comparison between eqs. 3.50 and 3.26 shows that Rc is three times that of the case where the u<60°. When α=0, u reaches a value of 60°when Id increases to : Id = Is [1 – cos 60°] = (0.5)Is.

At u=60°, three valves are conducting all the time. For α≥30, it is possible for the current to be increased further, which also increases u further. The reason the overlap angle cannot go beyond 60° for α<30° is shown in the next figure.

3.3.2 With Overlap: Analysis of Mode 2 : Average Direct Voltage

As an example, picture valve 2 and 6 are still conducting, the anode of valve 3 is at a potential of and the 𝑒𝑏 + 𝑒𝑐

2=−𝑒𝑎2

3.3.2 With Overlap: Analysis of Mode 2 : Average Direct Voltage

Cathode is at a potential of ea. Hence, unless ea is negative, valve 3 cannot conduct and this is possible only for α≥30°. Thus there is an intermediate mode when the value of overlap angle remains constant at 60°. In this mode, an increase in direct current is accompanied by automatic increase in the firing angle.

3.3.2 With Overlap: Analysis of Mode 2 : DC and Valve Voltage Waveforms

Instantaneous voltage across the converter bridge (vd)

Valve voltage of #3 and #4 valves.

3.3.2 With Overlap: Analysis of Mode 2 : DC and Valve Voltage Waveforms

The instantaneous voltage across the bridge can have both positive and negative excursions, followed by periods of zero magnitude. The valve voltage has 6 jumps with three of the having the magnitude

Reference:Padiyar, K. R. (1990). HVDC Power Transmission Systems: Technology and System Interactions. New Delhi: New Age International