Question

Derive the expressions of the average and rms values of the load voltage and current in three-phase uncontrolled bridge rectifiers with resistive load. Draw the waveforms of supply voltage (line-line voltages), transformer phase current (only one phase), load voltage, load current and diode current.

Answer 1

Full-wave Three-phase Rectification

The full-wave three-phase uncontrolled bridge rectifier circuit uses six diodes, two per phase in a similar fashion to the single-phase bridge rectifier. A 3-phase full-wave rectifier is obtained by using two half-wave rectifier circuits. The advantage here is that the circuit produces a lower ripple output than the previous half-wave 3-phase rectifier as it has a frequency of six times the input AC waveform.

Also, the full-wave rectifier can be fed from a balanced 3–phase 3-wire delta connected supply as no fourth neutral (N) wire is required. Consider the full-wave 3-phase rectifier circuit below.

Full-wave Three-phase Rectification

As before, assuming a phase rotation of Red-Yellow-Blue (V_A – V_B – V_C) and the red phase (V_A) starts at 0^o. Each phase connects between a pair of diodes as shown. One diode of the conducting pair powers the positive (+) side of load, while the other diode powers the negative (-) side of load.

Diodes D₁ D₃ D₂ and D₄ form a bridge rectifier network between phases A and B, similarly diodes D₃ D₅ D₄ and D₆ between phases B and C and D₅ D₁ D₆ and D₂ between phases C and A.

Thus diodes D₁ D₃ and D₅ feed the positive rail and depending on which one has a more positive voltage at its anode terminal conducts. Likewise, diodes D₂ D₄ and D₆ feed the negative rail and whichever diode has a more negative voltage at its cathode terminal conducts.

Then we can see that for three-phase rectification, the diodes conduct in matching pairs giving a conduction pattern for the load current of: D_1-2 D_1-6 D_3-6 D_3-6 D_3-4 D_5-4 D_5-2 and D_1-2 as shown.

Full-wave Three-phase Rectifier Conduction Waveform

In 3-phase power rectifiers, conduction always occurs in the most positive diode and the corresponding most negative diode. Thus as the three phases rotate across the rectifier terminals, conduction is passed from diode to diode. Then each diode conducts for 120^o (one-third) in each supply cycle but as it takes two diodes to conduct in pairs, each pair of diodes will conduct for only 60^o (one-sixth) of a cycle at any one time as shown above.

Therefore we can correctly say that for a 3-phase rectifier being fed by “3” transformer secondaries, each phase will be separated by 360^o/3 thus requiring 2*3 diodes. Note also that unlike the previous half-wave rectifier, there is no common connection between the rectifiers input and output terminals. Therefore it can be fed by a star connected or a delta connected transformer supply.

So the average DC value of the output voltage waveform from a 3-phase full-wave rectifier is given as:

Where: V_S is equal to (V_L(PEAK) ÷ √3) and where V_L(PEAK) is the maximum line-to-line voltage (V_L*1.414).

3-phase Rectification Example No2

A 3-phase full-wave bridge rectifier is required to fed a 150Ω resistive load from a 3-phase 127 volt, 60Hz delta connected supply. Ignoring the voltage drops across the diodes, calculate: 1. the DC output voltage of the rectifier and 2. the load current.

1. the DC output voltage:

The RMS (Root Mean Squared) line voltage is 127 volts. Therefore the line-to-line peak voltage (V_L-L(PEAK)) will be:

As the supply is 3-phase, the phase to neutral voltage (V_P-N) of any phase will be:

Note that this is basically the same as saying:

Thus the average DC output voltage from the 3-phase full-wave rectifier is given as:

Again, we can reduce the maths a bit by correctly saying that for a given line-to-line RMS voltage value, in our example 127 volts, the average DC output voltage is:

2. the rectifiers load current.

The output from the rectifier is feeding a 150Ω resistive load. Then using Ohms law the load current will be:

Uncontrolled 3-phase rectification uses diodes to provide an average output voltage of a fixed value relative to the value of the input AC voltages. But to vary the output voltage of the rectifier we need to replace the uncontrolled diodes, either some or all of them, with thyristors to create what are called half-controlled or fully-controlled bridge rectifiers.

Thyristors are three terminal semiconductor devices and when a suitable trigger pulse is applied to the the thyristors gate terminal when its Anode–to-Cathode terminal voltage is positive, the device will conduct and pass a load current. So by delaying the timing of the trigger pulse, (firing angle) we can delay the instant in time at which the thyristor would naturally switch “ON” if it were a normal diode and the moment it starts to conduct when the trigger pulse is applied.

Thus with a controlled 3-phase rectification which uses thyristors instead of diodes, we can control the value of the average DC output voltage by controlling the firing angle of the thyristor pairs and so the rectified output voltage becomes a function of the firing angle, α.

Therefore the only difference to the formula used above for the average output voltage of a 3-phase bridge rectifier is in the cosine angle, cos(α) of the firing or triggering pulse. So if the firing angle is zero, (cos(0) = 1), the controlled rectifier performs similar to the previous 3-phase uncontrolled diode rectifier with the average output voltages being the same.