The Dual full bridge Converters with suitable examples and sketches are discuss here advanced. Dual full bridge Converters is needed to read.

Dual full-bridge converter uses two nos. of full-bridge converters in series or parallel. These converters are called “12-pulse converters” because there are 12 thyristors, each requiring a separate trigger pulses.

Series-converter gives a higher output voltage and the parallel converter gives higher output current. Per unit voltage-ripple at the output is lesser for both types.

Dual converter produces an input current having 12-steps, bringing it more towards the sinusoidal shape. So, THD of input current is lesser. Because of these reasons, most high power converters prefer dual-bridge arrangement.

## 3-phase dual series-bridge converter with inductive load

Dual series converter gives higher DC voltage output with still higher ripple frequency. Two full-bridges are supplied with three-phase voltages of same RMS value but 30° phase-shift in between. High power applications, eg: HVDC, prefer this converter.

In general, dual series converter gives, Higher DC voltage output Higher ripple-frequency in output voltage Lesser THD at the supply side AC current

Required 2-sets of three-phase voltages with 30° phase-shift in between is obtained using a single 3-winding transformer (or two separate transformers). The same transformer does the current addition to result in a low THD in the net input current.

Turn-ratio of the Ddy transformer should be 𝑁1:𝑁2:𝑁3 ≡ 1:𝑚:𝑚 3 to give equal voltage to two bridges. Value of 𝑚 is decided according to the utility voltage and the required converter voltage. Vector connection Dd0y1 gives necessary phase-shift, i.e. bridge-2 voltage 30° leading ahead of bridge-1 voltage.

To operate dual-bridge we can select one of two alternative controls. • Concurrent control • Sequential control

(i) Concurrent Control of dual series-bridges

This is a popular mode of control. Here, both bridges are operated at the same delay-angle, i.e. α1 = α2. Then,

𝑉𝑜2and 𝑉𝑜1of individual bridges each has 6 ripples per AC cycle. The net output 𝑉 𝑜 has 12 ripples per AC cycle. This doubled ripple frequency is beneficial for reducing the size of filter components at output and input, both.

Adding up of bridge output voltages to give the net output voltage

Using Ampere’s Law, we can easily show that Input current at the utility side is given by,

Currents 𝐼𝑎 2, 𝐼𝑎 1 and 𝐼𝑏 1 are quasi-square currents having 120°wide single pulse in each half-cycle. 𝐼𝑎 2 is 30° leading ahead of 𝐼𝑎 1because bridge-2 voltage is 30°leading ahead of bridge-1. 𝐼𝑏 1 is 120° lagging behind 𝐼𝑎 1 as usual.

Utility side input current waveform has significantly improved in the form of a stepped sinusoidal waveform. Using Fourier expansion,

Orders of harmonics present at input current are 12𝑘 ±1 , where 𝑘 = 1,2,3,4,……

Contributions to 𝐼𝐴 𝐹𝑢𝑛𝑑 by bridge-2 and bridge-1 can be given separately, with respect to waveforms of 𝐼𝑎 2, 𝐼𝑎 1 and 𝐼𝑏 1. Contribution from bridge-2 is 𝑚 𝐼𝑎 2 and that from bridge-1 is 𝑚 3 𝐼𝑎 1 − 𝐼𝑏 1 . Mathematically, both contributions are equal and inphase, and angle 𝛼 lagging behind utility phase-A voltage.

Taking input phase-voltage as 𝑉𝐴 = 𝑉𝑚 sin𝜔𝑡, and using Fourier expansion,

Order of harmonics in IA 𝑖𝑠 = 12𝑘 ±1 , 𝑘 = 1,2,3,4,….⫽ Input Displacement Angle = 𝛼 ⫽ Input 𝐷𝑖𝑠𝐹 = cos𝛼 ⫽

## Sequential Control of dual series-bridges

1. mode-1, where α2 = 0° and α1 is varied from 0° to 180°.

2. mode-2, where α1 = 180° and α2 is varied from 0° to 180°

Sub-mode 1 gives variable positive voltage and sub-mode 2 gives variable negative voltage. With this control, the displacement-factor at the input gets improved but THD tends to get deteriorated.

## Sub-mode 1 : Sequential Control of dual series-bridges

Let take 𝐼1′ and 𝐼2′ be components of input line-current at utility end, produced by bridge-1 and bridge-2, respectively.

According to operation of Dd0y1 transformer and full-bridge converters, 𝐼2′ is inphase with input voltage (because 𝛼2 = 0) and 𝐼1′ is 𝛼1lagging behind input voltage (because of 𝛼1).

Let

Thus input current 𝐼𝐴 contains harmonics of order 6𝑘 ± 1 , where k = 1, 2, 3, 4, ….⫽

## Sub-mode 2 : Sequential Control of dual series-bridges

𝐼2′ is 𝛼2lagging behind input voltage (because of 𝛼2) and 𝐼1′is out of phase with input voltage (because 𝛼1 = 180°).

Thus input current 𝐼𝐴 contains harmonics of order 6𝑘 ± 1 , where k = 1, 2, 3, 4, ….⫽

Variation of (IA)Fund and its Displacement angle for different delay angles in the concurrent and sequential control. (VA is phase-A voltage at Utility)

With respect to this diagram of fundamental current, too, we can identify the operational parameters for the dual series converter, as given below.

For sequential control Mode-1:

For sequential control Mode-2:

For concurrent control:

## Example – Dual full bridge Converters

Three-phase, dual series-bridge thyristor converter is operating on 400 V, 50 Hz utility three-phase supply via 400V/350V/350V, Dd0y1, three-phase transformer. The converter delivers 16 kW at 800 V to an inductive load. Converter is operated with concurrent control. Ignore supply internal inductance and transformer leakage inductances. Determine, (i) Delay angles for the two bridges (ii) Displacement Factor at utility input (iii) RMS value of the fundamental current supplied by the utility (iv) RMS values of the two lowest order harmonics present in the utility input current. If the converter is operated in sequential control, how the values in (I) to (iii) above would be modified.

Ans: 𝑉𝐿 = 350 𝑉 𝑓 𝑠 = 50 𝐻𝑧 𝑉 𝑜 𝑚𝑒𝑎𝑛 = 800 𝑉 𝑃 𝑜 = 16 𝑘𝑊 𝑚 = 350 400 = 0.875 𝐷𝑑0𝑦1 normal connection

## 3-phase dual parallel-bridge converter with inductive load

## (i) Dual Ordinary Parallel Converter Dual full bridge Converters

Two bridges are operated concurrently with 𝛼1 = 𝛼2 = 𝛼. The center-tapped inductor absorbs the instantaneous voltage difference between 𝑉𝑜1 and 𝑉𝑜2. In general, dual parallel converter gives, Higher DC current output Higher ripple-frequency in output voltage Lesser THD at the supply side AC current

Mean inductor-voltage is zero and hence,

Ripple in 𝑉 𝑜 has a frequency of 12𝑓 𝑠 as in the series converter. Amplitude of ripple is 50% of that produced by an individual bridge. Both are desirable outcomes.

𝐼𝑜 = 𝐼𝑜1 +𝐼𝑜2

Thus, output current is greater. Input current 𝐼𝐴 at the utility side is a stepped sine waveform as in the series converter, having only 12𝑘 ±1 𝑡 order of harmonics, where 𝑘 = 1,2,3…. (assuming equal current sharing).

- Dual Inverse Parallel Converter

An alternative version of dual parallel converter is obtained when the two bridges are connected in inverse parallel. In this case, we can give either positive or negative 𝐼𝑜 by making 𝐼𝑜2 greater or lesser than 𝐼𝑜1, respectively. So, the load can absorb or regenerate real power, offering true 4-quadrant operation.

Since the mean voltage across the inductor is zero, using KVL,

𝑉𝑜2 𝑚𝑒𝑎𝑛 + 𝑉𝑜1 𝑚𝑒𝑎𝑛 = 0

This means the dual inverse-parallel converter should be operated complying the condition 𝛼2 +𝛼1 = 180°.

In the steady state,

𝑉 𝑜 𝑚𝑒𝑎𝑛 is positive when 𝛼2 < 90° or negative when 𝛼2 > 90°. (𝛼1 is determined by 𝛼2). 𝐼𝑜 is determined by the load but set by the relative values of 𝐼𝑜2 and 𝐼𝑜1. It is positive when 𝐼𝑜2 > 𝐼𝑜1 or negative when 𝐼𝑜2 < 𝐼𝑜1 or zero when 𝐼𝑜2 = 𝐼𝑜1.

The load can either consume or regenerate real power depending on the product 𝑉 𝑜 𝑚𝑒𝑎𝑛𝐼𝑜. Positive product means consuming, negative product means regenerating or zero product means idling.

Contributions to the fundamental component of 𝐼𝐴 at utility side by bridge-2 and bridge-1, assuming utility phase-A voltage as 𝑉𝐴 = 𝑉𝑚 sin𝜔𝑡, are:

Using this expression of 𝐼𝐴,𝐹𝑢𝑛𝑑 , we can determine the input Displacement Angle and input DisF. For example, when 𝐼𝑜2 = 𝐼𝑜1, 𝐼𝐴,𝐹𝑢𝑛𝑑 is 90° lagging behind 𝑉𝐴, indicating zero DisF.

We can show,

## Assignment: dual parallel bridge converter

Drive mathematical expressions for 𝑉 𝑜 𝑚𝑒𝑎𝑛 , 𝐷𝑖𝑠𝐹 and associated 𝑢𝑇s for the following dual full-bridge thyristor converters. Take a Dd0y1 three-phase transformer to feed two bridges with bridge-2 connected to d-winding and bridge-1 to y-winding. Each output of the transformer is having equal rms line-voltage 𝑉𝐿 . Net internal inductance on d-winding is 𝐿𝑠2 and that on y-winding is 𝐿𝑠1. Delay angles for bridge-2 and bridge-1 are 𝛼2 and 𝛼1, respectively.

(i) Dual series-bridge converter with constant load current 𝐼𝑜 with concurrent control. (ii) Dual series-bridge converter with constant load current 𝐼𝑜 with sequential sub-mode 1 control. (iii) Dual series-bridge converter with constant load current 𝐼𝑜 with sequential sub-mode 2 control (inverter mode operation). (iv) Dual parallel-bridge converter with constant load current 𝐼𝑜 with concurrent control and equal sharing of current. (v) Dual inverse-parallel-bridge converter with constant load current 𝐼𝑜 with bridge-2 current 𝐼𝑜2 and bridge-1 current 𝐼𝑜1.

You may use appropriate standard expressions for the relevant cases without internal inductances and modify them to account the effects of 𝐿𝑠2 and 𝐿𝑠2, giving reasons.

The Dual full bridge Converters with suitable examples and sketches are discuss here advanced. Dual full bridge Converters is needed to read.