Three Phase Rectifiers 1# Half Wave Rectifiers with Resistive Loads

Summary

This lecture delves into three-phase half-wave rectifiers with resistive loads, a fundamental rectifier type for three-phase voltage sources. It explains the balanced nature of three-phase sources and how they differ from single-phase rectifiers. The script covers the circuit diagram, phase voltage equations, and the point where phase voltages cross. It also discusses the output voltage ripple and calculates the average output voltage, highlighting the inherent advantage of three-phase rectifiers in providing a smoother output without additional filtering.

Takeaways

• 🔌 The lecture discusses three-phase half-wave rectifiers with resistive loads, which are a type of rectifier for three-phase voltage sources.
• 🌐 A three-phase voltage source is typically balanced, meaning the three voltage signals are separated by a third of a cycle.
• 💡 The reason to use a three-phase rectifier could be to convert a common three-phase AC source to DC voltage for industrial applications.
• 📈 The three-phase half-wave rectifier has one diode per voltage source, making it the simplest type of three-phase rectifier.
• 🔄 The phase voltages are represented by equations: V_a = V_s sin(ωt), V_b = V_s sin(ωt - 2π/3), and V_c = V_s sin(ωt + 2π/3).
• ⚙️ The voltages cross each other at points that are symmetrical and can be calculated to understand when one phase voltage exceeds another.
• 📊 The output voltage of the rectifier is the greatest of any of the three phase voltages at any point in time, resulting in a fluctuating output.
• 🔽 The minimum output voltage occurs where the phase voltages cross each other, which is 0.5 times the peak input voltage (V_s).
• 📉 The output voltage ripple fluctuates every third of a cycle because it's equal to the greatest of the phase voltages at any point in time.
• 📊 The average output voltage can be calculated by integrating the waveform over a period, resulting in approximately 0.82699 times V_s or (3√3/π) times V_s.
• 🔌 Three-phase rectifiers inherently provide a better output voltage with less ripple compared to single-phase rectifiers, even without additional voltage filtering.

Q & A

What is a three-phase half-wave rectifier?

A three-phase half-wave rectifier is the simplest type of rectifier for a three-phase voltage source. It has one diode per voltage source and is used to convert AC voltage from a three-phase system into DC voltage.

Why might one use a three-phase rectifier?

One might use a three-phase rectifier in industrial applications where three-phase sources are common but a DC voltage is needed. It allows the creation of DC voltage from a three-phase AC source.

What is a balanced three-phase voltage source?

A balanced three-phase voltage source is one where the three-phase voltage signals are a third apart from each other every period. This means that the voltages are equally separated from each other.

How are the phase voltages represented in a three-phase system?

In a three-phase system, the phase voltages are represented as VA = Vs * sin(ωt), VB = Vs * sin(ωt - 2π/3), and VC = Vs * sin(ωt + 2π/3), where Vs is the peak voltage and ω is the angular frequency.

What is the significance of the point where the phase voltages cross?

The point where the phase voltages cross is significant because it indicates where the voltage of one phase equals another, which affects which diode is forward-biased and thus which phase voltage is present at the output.

How is the voltage at the point where the phase voltages cross calculated?

The voltage at the point where the phase voltages cross is calculated by setting the equations for two phase voltages equal to each other and solving for the angular frequency ωt. For example, setting VA equal to VC and solving for ωt gives ωt = π/6.

What is the voltage at which the three phase waveforms cross each other?

The voltage at which the three phase waveforms cross each other is 0.5 Vs, where Vs is the peak of the three-phase input voltages.

How does the output voltage of a three-phase half-wave rectifier behave?

The output voltage of a three-phase half-wave rectifier is equal to the greatest of any of the three phase voltages at any point in time, which results in a fluctuating output voltage that is equal to the peak of the voltage sources when they are greater than the others.

What is the output voltage ripple for a three-phase half-wave rectifier?

The output voltage ripple for a three-phase half-wave rectifier fluctuates every third of a cycle and ranges from the peak of the voltage sources (Vs) to the voltage at which the phase voltages cross each other (0.5 Vs).

How is the average output voltage of a three-phase half-wave rectifier calculated?

The average output voltage is calculated by taking the integral of the output voltage waveform over one period and dividing by the period. The result is approximately 0.82699 Vs or 3 * √3 / 2π * Vs.

Why does the three-phase rectifier provide a better output voltage without filtering compared to a single-phase rectifier?

The three-phase rectifier inherently provides a better output voltage without filtering because it uses a three-phase voltage source, which results in a smoother output voltage waveform with less ripple compared to a single-phase rectifier.

Outlines

00:00

🔌 Introduction to Three-Phase Half-Wave Rectifiers

The lecture introduces three-phase half-wave rectifiers with resistive loads. It contrasts these with single-phase rectifiers and explains that three-phase rectifiers are used when a DC voltage is needed from a three-phase AC source, which is common in industrial applications. The concept of a balanced three-phase voltage source is discussed, where the voltages are equally spaced by one-third of a cycle. The lecture then delves into the circuit diagram and input voltage equations for the three-phase system, emphasizing the phase shifts and the points where the phase voltages cross each other.

05:02

📈 Analyzing Voltage Crossings and Output Voltage

This section discusses the calculation of the point where the phase voltages cross each other, which is crucial for understanding the behavior of the rectifier. The讲师 calculates the voltage at the crossing point to be 0.5 times the peak voltage (Vs). The output voltage of the rectifier is described as being equal to the greatest of the three phase voltages at any given time. The讲师 explains how the diodes in the circuit conduct based on which phase voltage is the highest, resulting in an output voltage that fluctuates every third of a cycle, with a ripple that varies between the peak voltage sources and the crossing point voltage.

10:02

🧮 Calculating the Average Output Voltage

The讲师 proceeds to calculate the average output voltage of the rectifier. This is done by integrating the output voltage waveform over one complete cycle and dividing by the period. The讲师 breaks down the integral into several parts, corresponding to the different segments of the cycle where either the a, b, or c phase voltage is the greatest. The讲师 then sums up these integrals to find the average output voltage, which is approximately 0.82699 times the peak voltage (Vs). This calculation shows that the three-phase rectifier provides a smoother output voltage compared to a single-phase rectifier without the need for additional filtering components.

15:05

🌉 Transition to Full-Wave Three-Phase Rectifiers

The final paragraph summarizes the benefits of three-phase rectifiers, noting that they inherently produce a better output voltage waveform compared to single-phase rectifiers, even without voltage filtering. The讲师 then transitions the discussion to the next topic, which is the full-wave three-phase rectifier with resistive loads, hinting at further exploration of rectifier types and their characteristics.

Keywords

💡Three-phase half-wave rectifier

A three-phase half-wave rectifier is a type of electrical circuit that converts AC (alternating current) from a three-phase system into DC (direct current). It is the simplest form of rectifier for a three-phase voltage source. In the context of the video, this rectifier is used in industrial applications where three-phase sources are common and a DC voltage is needed. The script explains that each phase of the three-phase system has one diode, and the rectifier operates by allowing current to flow through the diode when its respective phase voltage is the greatest.

💡Resistive loads

Resistive loads are loads in an electrical circuit that consume power and oppose the flow of current, similar to how a resistor does. They are characterized by their resistance, which is the ratio of voltage across the load to the current through it. In the video, resistive loads are mentioned as the type of load for which the three-phase half-wave rectifier is being discussed, implying that the output DC voltage will be used by such loads.

💡Balanced three-phase voltage source

A balanced three-phase voltage source is one where the three phase voltages are equal in magnitude but are displaced by one-third of a cycle relative to each other. This balance is crucial for the proper operation of three-phase systems. The video script describes how the three-phase voltage sources are separated by two pi over three, which is a characteristic of a balanced system.

💡Diode

A diode is a two-terminal electronic component that allows current to flow in one direction while blocking it in the opposite direction. In the context of the video, diodes are used in the rectifier to allow current to flow only when the phase voltage is higher than the other two phases, effectively 'rectifying' the AC input into pulsating DC.

💡Rectifier

A rectifier is a device that converts alternating current (AC) to direct current (DC). The video focuses on three-phase rectifiers, which are used to convert three-phase AC to DC. The script explains that the three-phase half-wave rectifier is the simplest type of three-phase rectifier, with one diode per voltage source.

💡Voltage source

A voltage source is an electrical component that provides a specified voltage to a circuit. In a three-phase system, there are three separate voltage sources, each 120 degrees out of phase with the others. The video script discusses how these voltage sources are used in the three-phase half-wave rectifier to generate DC voltage.

💡Output voltage ripple

Output voltage ripple refers to the fluctuation in the output voltage of a rectifier, which is a result of the rectification process. The video script explains that the ripple in a three-phase half-wave rectifier is less severe than in a single-phase rectifier, as the output voltage is always the greatest of the three phase voltages at any point in time.

💡Peak voltage

Peak voltage is the maximum voltage in an AC waveform. The script mentions that the voltage at which the three phase voltages cross each other is 0.5 times the peak voltage (vs). This is an important value because it represents the minimum voltage level at which the transition between diodes occurs.

💡Symmetric waveforms

Symmetric waveforms are those where the positive and negative halves of the wave are mirror images of each other. The video script uses the term to describe the crossing points of the phase voltages, indicating that the waveforms are symmetrical around the midpoint of the cycle.

💡Average output voltage

The average output voltage is calculated by integrating the output voltage over one complete cycle and then dividing by the period of the cycle. The video script provides a detailed calculation of the average output voltage for the three-phase half-wave rectifier, showing that it is approximately 0.82699 times the peak voltage (vs).

💡Industrial application

Industrial applications refer to the use of electrical systems in manufacturing and other large-scale operations. The video script mentions industrial applications as a context where three-phase rectifiers are commonly used, highlighting their practical relevance in providing DC voltage from a three-phase AC source.

Highlights

Introduction to three-phase half-wave rectifiers with resistive loads.

Comparison of three-phase rectifiers to single-phase rectifiers.

Explanation of the simplicity of the three-phase half-wave rectifier.

Industrial applications for three-phase rectifiers where DC voltage is needed.

Assumption of balanced three-phase voltage sources.

Description of the phase separation in a three-phase system.

Circuit diagram and input voltages for the rectifier.

Equations for the three phase voltages.

Graphical representation of where the three phase voltages cross.

Calculation of the voltage at the crossing point of phase voltages.

Explanation of the output voltage being the greatest of the three phase voltages.

Operation of diodes in the rectifier circuit.

Description of the output voltage ripple.

Calculation of the average output voltage.

Integral calculation for the average output voltage over a period.

Result of the average output voltage being approximately 0.82699 times the peak voltage.

Comparison of the output voltage ripple of a three-phase rectifier to a single-phase rectifier.

Advantage of three-phase rectifiers in reducing voltage ripple without filtering.

Transition to the discussion of full-wave three-phase rectifiers with resistive loads.

Transcripts

in this lecture we're going to be

talking about three-phase half-wave

rectifiers with resistive loads

so so far we've talked about

single-phase rectifiers throughout the

course but now we're going to talk about

the three-phase equivalent of those

rectifiers

so similar to the single-phase half-wave

rectifier that we looked at

the half-wave three-phase rectifier is

the simplest type of rectifier for a

three-phase

voltage source and also similar to the

single-phase

rectifier the three-phase halfway

rectifier has one

diode per voltage source and it's also

the simplest type of three-phase

rectifier there is

and so the reason why you may want to

use a three-phase rectifier is that

let's say for example that you have an

industrial application

where three-phase sources are common but

you need a dc voltage

so you can use a three-phase rectifier

to create that dc voltage

now when we talk about three-phase

voltage source we typically assume that

they're what's called

balanced meaning that the three-phase

voltage signals are a third apart from

each other

every period so let's say for example

that you have a period that's 2 pi

in length then the voltage sources are

going to be separated by 2 pi

over 3 and 4 pi over 3 meaning that

in one period of 2 pi you're going to

have

the three phase voltage sources equally

separated from each other so let's go

ahead and draw the circuit

and the input voltages for this

rectifier

so you can see here that the three-phase

voltage sources are separated from each

other by two pi

over three equally so they start every

one third of a cycle so for our case

one cycle is two pi so the b phase

voltage starts at two pi over three and

the c

phase voltage starts at four pi over 3.

and so let's write the equations for

these three phase voltages

so we can say that va is going to be

equal to

vs sine omega t

so this is the same as we have looked in

the single phase

example so remember that the input

voltage for that was also vs sine omega

t

but then vb

is going to be equal to vs

sine omega t minus

two pi over three

so this represents the shift of the b

phase voltage with respect to the a

phase voltage

and then v c is going to be equal to vs

sine omega t

plus 2 pi over 3.

so you may be wondering well why not

minus 4 pi

over 3 since it starts at it starts

going positive at 4 pi over 3.

well plus 2 pi over 3 and minus 4 pi

over 3 is the same thing remember that

sine waves repeat

so if we were to say we could also say

here

minus 4 pi over 3 but saying plus 2 pi

over 3 is equivalent of that

and so i'm going to define it as plus 2

pi over 3.

and so one important thing here is to

define where the

three phase voltages cross so

essentially this point

right here where for example

the a phase voltage so remember that

this is the a phase voltage va

this is v b and this vc

so we want to see where they cross each

other so for example

if we look at this point right here

this would be the point where va crosses

vc and you can kind of see that

graphically in this waveform you kind of

know that that's going to be pi

over 6 in other words half of pi over 3.

but let's calculate it

so in order to do that we can say well

va is equal to vc at that point

so let's set va equal

to vc and solve for omega t

so this will be equal to vs

sine omega t is going to be equal to

vs sine

omega t plus 2 pi over 3.

and so of course the vs cancel out

and if we solve for omega t and this

obviously is going to have several

solutions right because va and vc are

going to cross in multiple points

so they cross over here they cross again

over here but we know that that crossing

is symmetrical in other words

if we take a look at the first point

where they cross

then we know that they're going to cross

again at let's say this second point

right here is going to be

pi plus whatever that point was so if we

calculate this point right here

let's say omega t 1 then we will know

that this point right here would be

pi plus omega t one

so we don't need to calculate every

point we just gotta calculate one and

then we know it's going to be

symmetrical after that

so if we solve for omega t in this

equation at the bottom we would get that

omega t

is equal to pi over six

and again you can kind of graphically

see that from this waveform at the top

because it looks like it's half of pi

over three which is pi over six

and so the reason why i wanted to

calculate that is because that's going

to tell us

what the voltage is when they cross each

other

so now knowing what omega t1 is let me

call this one omega t1

then we can plug it into either equation

of va or vc

and see what the voltage is at that

point so let's use va so we're going to

say that

we're trying to calculate va which is v

as

sine omega t but omega t is pi over 6

so this is going to be equal to 0.5

vs in other words the voltage at which

these three waveforms cross each other

is 0.5 es

so remember that pvs is the peak of the

three-phase input voltages so this point

right here is going to be 0.5

vs and again you can kind of see that

graphically on this waveform because

that looks like it's about half of vs

now as far as the output voltage what

ends up happening is that the output

voltage is going to be equal to the

greatest

of any of the three phase voltages at

any point in time so what i mean by that

is let's say for example during this

portion right here

from here to about here

so from pi over 6 until va crosses vb

the a phase voltage is greater than the

b and c phase voltages

so what that means is that diode d1 is

for bias so the voltage right here

is going to be equal to va because d1 is

on

and so if that's the case then the

voltage

again the voltage on the cathodes of

diodes d2 and d3

are equal to va remember that the

cathode is

this pointer here so cathodes

and anodes on this side so the voltage

at the cathodes

of d2 and d3 is equal to va

and the voltage and the anode of d2 is

equal to vb

and the voltage at the anode of d3 is

equal to vc

and so because vv and vc are lower than

va

then that means that diodes d2 and d3

are reverse bias and they're off

so what happens is that for this portion

of the waveform

diode d1 is on

and diodes d2 and d3

are off now after that when vb becomes

greater than va

so from here to about here

then the same thing happens but for

diode d2 so

diode d2 becomes four bias so the

voltage at the output is equal to vb

which means that the voltage at the

cathodes of diodes d1 and d3

is equal to vb and the voltage at the

anode of d1 is va and the voltage at the

anode of d3

is vc but because those two voltages are

lower than vb

then diodes d1 and d3 become reverse

bias and they're off

so for this part right here d2 is on

and d1 and d3

are off

and then the same happens after that so

d3

would be on d1 and d2

would be off so the diodes can't take

turn when the voltage

for that phase is greater than the other

two phase voltages

then that diode the corresponding diode

for that source

comes on and the output voltage is equal

to that and then when the next phase

becomes greater then the next static

turns on and so on and so forth so what

ends up happening at the output it's

that it's going to look like this

it's going to be equal to whichever

voltage is greater so if i draw the

output voltage here in white

then it's going to look like this

and it's going to repeat after that so

the ripple of the output voltage

kind of fluctuates every third of a

cycle because it's equal to the creator

of the phase voltages at any point in

time and it fluctuates between the peak

of the

voltage sources so vs and then the

minimum is going to be

where the voltages cross each other

because that's when the transition

happens from one diode to the next

so that's going to be equal to 0.5 vs so

the output voltage ripple for v

out would be from here to here

which is 0.5 vs

now let's calculate one more thing let's

calculate what the average of the output

voltage would be

so it would be probably somewhere over

here

would have drawn this dash white line so

let me go ahead and erase a couple

things to make room

so remember that the average of the

output voltage just like in the

single phase examples it's going to be

one over the period

times the integral over the period

of the waveform so v out

d omega t and this would be omega t so

we calculate that for this waveform we

would get the v

out is gonna be equal to

one over the period so the period is two

pi

and we're gonna have to calculate

several integrals

so the first integral is going to be for

this part right here

which again is equal to vc

so from 0 to omega t1 the output voltage

is going to be equal to

vc because that's greater than va and vb

so we're going to calculate the integral

from 0 to omega t1 which is pi

over 6 of

vc which is equal to vs

sine omega t plus

2 pi over 3

d omega t plus

then after that va is greater than vb

and vc so the integral from omega t1 so

pi

over 6 to where they cross

again so this point right here

is going to be equal to and remember

that i said that

this is going to be symmetrical so if we

know that this point right here

um this point right here is 2 pi over 3

then where va and vb cross is going to

be equal to 2 pi over 3

plus pi over 6 and so that's going to be

equal to 5

pi over 6

of va which is vs sine

omega t d omega t

plus then after that vb becomes greater

than va and vc so the output voltage is

going to be equal to vb

so from 5 pi over 6

until this point right here which is

going to be equal to 4 pi

over 3 plus pi over 6 which is equal to

9

pi over 6 of v s

sine omega t minus

2 pi over 3 d omega t

and then the last portion is going to be

vc again

so right here at the end so it's going

to be

from where we stopped so 9 pi over 6

until the end of the cycle which is

going to be 2 pi

of vc so vs sine

omega t plus 2 pi over 3

d omega t so notice that we use

vc for about a third at the beginning

and then two thirds at the end

so if you calculate all this this would

be equal to

so we can take vs at the beginning

so we'll factor out vs so we'll say that

is equal to vs over 2 pi

times the first integral is going to be

0.366

o2 then the two

next integrals are going to be actually

the same and that makes sense because

the width of the portion that we're

using for va and vb

so this and this

are equal to each other so those two

integrals are going to be 1.73

1.73205

plus again 1.73205

plus the last integral which would be

for vc so this point right here

and that's equal to 1.36603

and so all of this would be equal to

zero or approximately equal to 0.82699

vs and this is also going to be equal to

in more exact terms

3 times square root of 3 over

2 pi of vs

so it's going to be the average of the

output voltage for a three-phase halfway

rectifier

and i'm actually not going to calculate

the output current because we know that

for resistive load the output current is

just going to look like the output

voltage

but scaled by the load resistance

so you can see here that for a

three-phase rectifier

without having to add any capacitors for

voltage filtering

the output voltage actually looks better

than for a single phase rectifier

and so that's the nice thing about three

phase rectifiers is that inherently

without even having to

do any filtering for voltage ripple we

get a better

output voltage just by the fact that

we're using a three-phase voltage source

so next we're going to take a look at

the full wave three-phase rectifier with

the resistive load