Efficient 3φ Converter System Using Diode: A Comprehensive Guide

A 3φ converter system using diodes is a common configuration for converting alternating current (AC) to direct current (DC). There are several types of three-phase diode converter configurations, including half-wave, full-wave, and bridge rectifiers.for understanding the working of three-phase controlled rectifiers, it is better to commence with the study of 3ϕ ac to dc converters using diodes.

Fig(i). Three-phase half-wave rectifier with (a) common cathode arrangement and (b) common anode arrangement.

In Fig(i) are shown the simplest three-phase rectifier circuits with half-wave configuration. The power circuit of Fig(i)(a) consists of three diodes A1, B1, C1 connected to a common load. The other load terminal must be connected to neutral N of the supply for return of load current. As the cathodes of three diodes are connected together, circuit of Fig(i)(a) is also called common cathode circuit for a three- phase half-wave rectifier. The three- phase supply voltage is shown as \(v_a\) (=\(v_{\text{an}}\) voltage between A and N), \(v_b\), \(v_c\), in Fig(ii) (a).

Fig(ii). (a) Line to neutral voltages (b) and (d) diode conduction and (c) and (e) load voltage \(v_0\).

The rectifier element connected to the line at the highest instantaneous voltage can only conduct. In Fig(i)(a), a diode with the highest positive voltage will begin to conduct at the cross-over points of the three-phase supply. It is seen from Fig(ii)(a) that diode A1 will conduct from a \( \omega t = 30° \) to \( \omega t = 150° \) as this diode is the most positive as compared to the other two diodes during this interval. Diode B1 will conduct from of \( \omega t = 150° \) to 270° and diode C1 from \( \omega t = 270° \) to 390°. The conduction of diodes in proper sequence is shown in Fig(ii)(b). When a diode is conducting, the common cathode terminal P rises to the highest positive voltage of that phase and the other two blocking diodes are reverse biased. The voltage across the load \( V_0\) follows the positive supply voltage envelope and has the waveform as shown in Fig(ii)(c). It should be noted that voltage of the neutral point. N is taken as zero and is given by the reference line \( \omega t \). The voltage of point P of Fig(i)(a) is shown by \(v_a\), \(v_b\), \(v_c\) etc. above the reference line in Fig(ii)(c). It is seen that for one cycle of supply voltage, output voltage has three pulses, the circuit of Fig(i)(a) can therefore be called a 3-phase 3-pulse diode rectifier or 3-phase half-wave diode rectifier.

Fig(iii). Evolution of 3-phase six-pulse rectifier. (a) Circuits of Fig(i)(a) and (b) connected in series (b) circuit of (a) rearranged and (c) 3-phase full-wave bridge rectifier obtained from (b).

For the rectifier circuit of Fig(i)(b), called common anode circuit for a 3-phase half-wave rectifier, a diode will conduct only during the most negative part of the supply voltage cycle. This means that a diode will conduct when the neutral is positive with respect to A, B or C. Therefore, for the voltage waveform of Fig(ii)(a), diode \(C_2\) conducts from \( \omega t =90°\)  to 210° as this is the most negative as compared to other two diodes during this interval. Diode A2 conducts from \( \omega t = 210°\) to 330° and diode B2 from \( \omega t = 330° \) to  450°  and so on. Each diode conducts for 120° in both the circuits of Fig(i). The load voltage \(V_0\) derived from Fig(ii)(a), follows the negative supply voltage envelope and has the waveshape as shown in Fig(ii)(e) for the diode configuration of Fig(i)(b). Here again voltage of the neutral point N is fixed at zero by reference line \( \omega t \) in Fig(ii)(e). The voltage of point Q of Fig(i)(b) is shown by \(v_b\), \(v_c\), \(v_a\) etc. below the reference line, in Fig(ii)(e).

The three-phase half-wave rectifier circuits of Fig(i)(a) and (b) can be connected in series as shown in Fig(iii)(a). In this series connected circuit, load current can exist even without neutral N. For example, when diode Al is conducting, the return path for the current is through diode B2 from of \(\omega t = 30°\) to 90° and through diode \(C_2\) from \( \omega t = 90°\) to 150°, see Fig(ii)(b) and (d). Supply point A connected to the anode of diode Al is the same as that connected to the cathode of diode A2. The neutral wire can thus be eliminated and the circuit of Fig(iii) (a) can be redrawn as shown in Fig(iii)(b). This circuit can further be rearranged to that shown in Fig(iii)(c). The only difference between Fig(iii)(a) and Fig(iii)(b) and (c) is that load voltage is equal to line to neutral voltage in Fig(iii)(a) and it is line to line voltage in Fig(iii)(b) and (c). The circuit configuration shown in Fig(iii)(c) is called 3-phase full-wave bridge rectifier or 3-phase six-pulse bridge rectifier. Note that diodes A1, B1, C1 of the bridge will conduct when supply voltage is the most positive where as diodes A2, B2, C2 will conduct when supply voltage is the most negative. Diodes A1, B1, C1 may therefore be called a positive diode group and A2, B2, C2 a negative diode group. The voltage across load will always be direct emf with the polarity of P positive and that of Q negative as shown.

Line to neutral, or phase, voltages of Fig(ii)(a) are again drawn in Fig(iv)(c) as \(v_a\),\(v_b\),\(v_c\) Fig(ii)(b) and (d) are combined and redrawn as shown in Fig(iv)(b), which show that for \( \omega t = 0°\) of  to 30°, diodes C1, B2 conduct together; for \( \omega t = 30°\) to 90°, diodes A1, B2 conduct together and so on. Each diode conducts for 120°. At the instant marked 1, diode B2 is already on and the conduction of diode C1 stops and that of Al begins. The magnitude of load voltage \(v_1\) at instant 1 is therefore given by

\[ V_1 = V_{mp} \sin(30^\circ) + V_{mp} \sin(90^\circ) \]

Solving this:

\[ V_1 = V_{mp} \times \frac{\sqrt{3}}{2} + V_{mp} \times 1 \]

\[ V_1 = \frac{\sqrt{3} + 2}{2} \times V_{mp} \]

So, \( V_1 = 1.5 \times V_{mp} \).

At the instant marked 2, the load voltage has a magnitude of

\[ V_2 = V_{mp} \sin(60^\circ) + V_{mp} \sin(60^\circ) \]

Solving this:

\[ V_2 = V_{mp} \times \frac{\sqrt{3}}{2} + V_{mp} \times \frac{\sqrt{3}}{2} \]

\[ V_2 = \sqrt{3} \times V_{mp} \]

So, \( V_2 = \sqrt{3} \times V_{mp} \).

At the instant marked 3,\(V_3 = 1.5 \times V_{mp}\)

Here \(V_{mp}\) is the maximum value of phase (or line to neutral) voltage.

The voltage of the load terminals P and Q of Fig(iii)(c) is shown in Fig(iv)(a) , This figure also reveals that at the instants marked 2, 4, 6, 8, 10 etc, the load voltage has a magnitude of √3\(V_mp\) and at instants the marked 1, 3, 5, 7, 9, 11 etc., the magnitude of load voltage is \(1.5 V_mp\).The load voltage, or the rectified output voltage, \(V_0\) can therefore be plotted as shown in Fig(iv)(c). Each diode conducts for 120°. It should be remembered that in Fig(iv)(c),\(v_ab,v_ac,v_ba\) etc. ,are line voltages whereas in Fig(iv) a), \(v_a,v_b,v_c\) are phase voltages. The dual subscript ab in \(v_ab\) may be taken to denote that as per the first subscript 'a', diode connected to phase A from the positive group, ie. Al conducts and as per the second subscript 'b' diode connected to phase B from the negative group, i.e. B2 conducts. For example, for voltage \(v_{cb}\) diode C1 from positive group and diode B2 from negative group conduct.

Fig(iv). (a) 3-phase input voltage (b) Conduction sequence of positive and negative group of diodes (c) Output voltage waveform of 3-phase six pulse diode bridge

In Fig(iv)(c), voltage of terminal Q is shown at zero potential by straight reference line \(\omega t\), whereas the voltage of terminal P is shown by line values \(V_cb, Vac ,V_ac\) etc. In fact, if waveform of voltage of terminal Q in Fig(iv)(a) is made a straight line, Fig(iv)(c) is obtained.

Fig(iv)(c) shows that there are 6 pulses for one cycle of supply voltage. Thus, this bridge can be called a 3-phase six-pulse diode rectifier. The object of this article has been to examine the evolution of a 3-phase bridge rectifier from a 3-phase half-wave rectifier.

As in single-phase conveters, the average output voltage in a 3-phase diode rectifier can be obtained by considering the output voltage wave over one periodic cycle.

For a 3-phase diode rectifier of Fig(i), the periodicity is 120° or 2π/3 radians as per Fig(ii)(c). Here the output voltage comprises of phase voltages \(v_a,v_b,v_c\) and its average value \(V_0\) is given by

 \(V_{o} = \frac{1}{\text{periodic time}} \times \int_{a_1}^{a_2} V_a \, d(\omega t)\) =

\[ \frac{1}{\frac{2\pi}{3}} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} V_{mp} \sin(\omega t) \, d(\omega t) \]

In the above expression, \(v_a\) is zero at \(\omega t = 0\); therefore \( V_{mp} \sin(\omega t) \) is written for \(v_a\). Further, \(v_a\) appears from \(\omega t = 30°\)to 150° in the output voltage waveform, these are therefore the limits of integration

\[\frac{3}{2\pi} V_{mp}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \sin(\omega t) \, d(\omega t) \]

Using the trigonometric identity \(\int \sin(x) \, dx = -\cos(x) + C\), we have:

\[\frac{3}{2\pi} V_{mp} \left[- \cos(\omega t)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\]

\[\frac{3}{2\pi} V_{mp} \left[- \cos\left(\frac{5\pi}{6}\right) + \cos\left(\frac{\pi}{6}\right)\right]\]

 \[\frac{3}{2\pi} V_{mp}\left[\cos\left(\frac{\pi}{6}\right) - \cos\left(\frac{5\pi}{6}\right)\right]\] 

we can first simplify the cosine terms:

\[\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\]

Now, let's substitute these values into the expression:

\[\frac{3}{2\pi} V_{mp}\left[\frac{\sqrt{3}}{2} - \left(-\frac{\sqrt{3}}{2}\right)\right]\]

Simplify the expression inside the brackets:

\[\frac{3}{2\pi} V_{mp}\left(\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\right)\]

\[\frac{3}{2\pi} V_{mp} \cdot \sqrt{3}\]

Now, let's calculate the result:

\[\frac{3}{2\pi} \cdot V_{mp} \cdot \sqrt{3} = \frac{3V_{mp} \cdot \sqrt{3}}{2\pi}\]

To find \(v_{hp}\), we need to know the value of \(V_{mp}\) and the frequency of the waveform. Assuming \(V_{mp}\) represents the peak voltage and \(\omega\) represents the angular frequency, we can calculate \(v_{hp}\) using the formula:

\[v_{hp} = \frac{V_{mp}}{\sqrt{2}}\]

Let's substitute the given value into the equation:

\[v_{hp} = \frac{3V_{mp} \cdot \sqrt{3}}{2\pi \cdot \sqrt{2}}\]

\[v_{hp} = \frac{3V_{mp} \cdot \sqrt{6}}{2\pi}\]

So, \(v_{hp} = \frac{3V_{mp} \cdot \sqrt{6}}{2\pi}\).

 \( V_{mp} \) is the maximum value of phase voltage \(V_{ph} \).

For a 3-phase six-pulse diode rectifier, the output voltage of Fig(iv)(c) consists of line voltages \(v_{ab},v_{ac},v_{bc},v_{ba}\) etc., and its average value \(V_0\) is given by 

\[ V_{o} = \frac{1}{\text{periodic time}} \times \int_{a_1}^{a_2} V_{ab} \, d(\omega t) \] 

The value of \(v_{ab}\) at \(\omega t = 0°\)  is  \( V_{m} \sin(30^\circ) \) =  and periodicity is 60° or π/3 radians.

 Given the expression:

\[ V_o = \frac{1}{\text{periodic time}} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} v_{ml} \sin(\omega t + 30^\circ) \, d(\omega t) \]

with \(\text{periodic time} = \frac{\pi}{3}\), let's solve the integral with respect to \(d(\omega t)\).

First, let's simplify the expression within the integral:

\[ \sin(\omega t + 30^\circ) \]

Using the angle addition formula for sine, we have:

\[ \sin(\omega t + 30^\circ) = \sin(\omega t) \cos(30^\circ) + \cos(\omega t) \sin(30^\circ) \]

\[ = \frac{\sqrt{3}}{2} \sin(\omega t) + \frac{1}{2} \cos(\omega t) \]

Now, we can rewrite the integral:

\[ V_o = \frac{1}{\frac{\pi}{3}} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} v_{ml} \left(\frac{\sqrt{3}}{2} \sin(\omega t) + \frac{1}{2} \cos(\omega t)\right) \, d(\omega t) \]

\[ = \frac{3}{\pi} \left[ \frac{\sqrt{3}}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} v_{ml} \sin(\omega t) \, d(\omega t) + \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} v_{ml} \cos(\omega t) \, d(\omega t) \right] \]

We can now solve each integral separately.

1. For \( \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} v_{ml} \sin(\omega t) \, d(\omega t) \):

\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} v_{ml} \sin(\omega t) \, d(\omega t) = - v_{ml} \cos(\omega t) \Bigg|_{\frac{\pi}{6}}^{\frac{\pi}{2}} \]

\[ = -V_{ml}\left( \cos\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{6}\right) \right) \]

\[ = -v_{ml}\left( 0 - \frac{\sqrt{3}}{2} \right) \]

\[ = \frac{v_{ml} \sqrt{3}}{2} \]

2. For \( \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} v_{ml} \cos(\omega t) \, d(\omega t) \):

\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} v_{ml} \cos(\omega t) \, d(\omega t) = v_{ml} \sin(\omega t) \Bigg|_{\frac{\pi}{6}}^{\frac{\pi}{2}} \]

\[ = v_{ml} \left( \sin\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{6}\right) \right) \]

\[ = v_{ml} \left( 1 - \frac{1}{2} \right) \]

\[ = \frac{v_{ml}}{2} \]

Now, let's substitute these results back into the expression for \(V_o\):

\[ V_o = \frac{3}{\pi} \left[ \frac{\sqrt{3}}{2} \cdot \frac{v_{ml} \sqrt{3}}{2} + \frac{1}{2} \cdot \frac{v_{ml}}{2} \right] \]

\[ = \frac{3}{\pi} \left[ \frac{3v_{ml}}{4} + \frac{v_{ml}}{4} \right] \]

\[ = \frac{3}{\pi} \cdot \frac{4v_{ml}}{4} \]

\[ = \frac{3v_{ml}}{\pi } \]

So, the solution for \(V_o\) is \(\frac{3v_{ml}}{\pi} \).

Given:

\[ v_{ml} = \sqrt{2} v_{l} \]

We want to prove:

\[ \frac{3v_{ml}}{\pi} = \frac{3\sqrt{2} v_{l}}{\pi} \]

First, we substitute the expression for \( v_{ml} \):

\[ \frac{3\sqrt{2} v_{l}}{\pi} = \frac{3\sqrt{2} (\frac{v_{ml}}{\sqrt{2}})}{\pi} \]

\[ = \frac{3v_{ml}}{\pi} \]

Therefore, \( \frac{3v_{ml}}{\pi} = \frac{3\sqrt{2} v_{l}}{\pi} \) .

\(v_{ml}\)is the maximum value of line or line to line voltage \(v_l\).

Advantage

• Higher efficiency compared to half-wave rectification.
• Produces a relatively smooth DC output voltage.
• Suitable for various industrial applications requiring three-phase power conversion.

Disadvantage

• Requires six diodes, which increases cost and complexity compared to single-phase rectifiers.
• Higher ripple compared to more advanced rectification techniques like pulse-width modulation (PWM).

Three Phase Converters Using  MCQ Objective Questions

Q1:The capability of a power supply to maintain constant voltage or current at the output despite change in supply's load is known as _______.

1. Input Regulation
2. Line Regulation
3. Load Regulation
4. Source Regulation

Answer (Detailed Solution Below)

Option 3 : Load Regulation

Concept:

Load Regulation:

• Load regulation refers to the ability of a power supply to maintain a constant output
• voltage or current despite changes in the load connected to it.
• In other words, it is the measure of how well a power supply can maintain its specified
• output voltage or current within a certain tolerance range as the load changes.
• Load regulation is an important characteristic of a power supply, as it determines the
• stability and reliability of the power supply's performance in various operating conditions.
• The capability of a power supply to maintain constant voltage or current at the output despite changes in the supply's load is known as "regulation."
• A power supply with good load regulation will provide a stable output voltage or
• current, regardless of changes in the load, while a power supply with poor load
• regulation may exhibit variations in output voltage or current with changes in the load,
• which can affect the performance of the connected devices or systems.

Q2:In a 3-ϕ half wave diode, rectifier, if per phase input voltage is 200 V then the average output voltage is

1. 233.91 V
2. 116.95 V
3. 202.56 V
4. 101.28 V

Answer (Detailed Solution Below)

Option 1 : 233.91 V

To calculate the average output voltage of a 3-phase half-wave diode rectifier, we can use the following formula:

\[ V_{\text{out (avg)}} = \frac{3}{2\pi} \cdot V_{\text{rms (input)}} \]

Given that the per-phase input voltage \( V_{\text{rms (input)}} \) is 200 V, we can substitute this value into the formula:

\[ V_{\text{out (avg)}} = \frac{3}{2\pi} \cdot 200 \]

\[ V_{\text{out (avg)}} = \frac{3}{2\pi} \cdot 200 \approx 239.13 \, \text{V} \]

So, the average output voltage is approximately 239.13 V, not 233.91 V.

Q3:A 3-phase full bridge diode rectifier is fed from delta-star transformer. The voltage fed to the rectifier has a maximum value of V from line to neutral so that the voltage of phase ‘a’ is V= \(V_m\)sin ωt. The phase sequence is a-b-c. At the instant the phase ‘a’ voltage is passing through zero, what will be the rectifier output voltage?

1. Zero
2. \(V_m\)
3. √2\(𝑉_m\)
4. √3\(𝑉_m\)

Answer (Detailed Solution Below)

Option 4 : √3\(𝑉_m)

Q4:A three-phase uncontrolled diode rectifier supplies a constant load current of 10 A and its supply voltage is 400 V line to line. The laod power or DC power is_______

1. 5402 W
2. 5773 W
3. 7998 W
4. 8436 W

Answer (Detailed Solution Below)

Option 1 : 5402 W

\(V_s\)= 400v

(I_0\)=10A

\(V_0\) = (3*400*√2)÷π

\(V_0 I_0\)=5402w

Q5:How many diodes are used in a three-phase converter, single-phase or double-phase?

Ans:-In a three-phase converter, whether single-phase or double-phase, the number of diodes used depends on the type of rectification employed. 

1. Single-Phase Rectification:

•  In single-phase rectification, each phase of the AC input requires a minimum of two diodes for half-wave rectification or four diodes for full-wave rectification.
• For a three-phase system, if single-phase rectification is used, the total number of diodes required would be 2 or 4 times the number of phases, depending on whether half-wave or full-wave rectification is implemented.

   - Formula:

•   For half-wave rectification: \( \text{Total diodes} = 2 \times \text{Number of phases} \)
•  For full-wave rectification: \( \text{Total diodes} = 4 \times \text{Number of phases} \)

2. Double-Phase Rectification:

•   In double-phase rectification, the AC input is split into two phases, and each phase requires a minimum of two diodes for half-wave rectification or four diodes for full-wave rectification.
•   For a three-phase system, if double-phase rectification is used, the total number of diodes required would be twice the number of phases, regardless of whether half-wave or full-wave rectification is implemented.

   - Formula:

•  For both half-wave and full-wave rectification: \( \text{Total diodes} = 2 \times \text{Number of phases} \)

In summary:

• For single-phase rectification: 2 or 4 times the number of phases, depending on half-wave or full-wave rectification.
• For double-phase rectification: Twice the number of phases, regardless of rectification type.

For example, in a three-phase system:

• Single-phase rectification would require either 6 diodes (for half-wave) or 12 diodes (for full-wave).
• Double-phase rectification would require 6 diodes, regardless of the type of rectification.

Q6:Derive expressions for the average output voltage for a 3-phase three-pulse diode rectifier and a three-phase six-pulse diode rectifier using cosine function for the input supply voltage.

Q7:For a 3-phase half-wave diode rectifier, show the time variation of input voltage and the voltage across one diode for one complete cycle. Find the maximum value of reverse voltage across the diode and its duration for an input supply voltage of 230 V, 50 Hz from a delta-star transformer.

TPCS=Three Phase Converter System

Note:-Comment the answer to question number 6 and 7.