Three Phase Rectifiers 1# Half Wave Rectifiers with Resistive Loads
Summary
TLDRThis 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
- ๐ The lecture discusses three-phase half-wave rectifiers with resistive loads, which are a type of rectifier for three-phase voltage sources.
- ๐ A three-phase voltage source is typically balanced, meaning the three voltage signals are separated by a third of a cycle.
- ๐ก The reason to use a three-phase rectifier could be to convert a common three-phase AC source to DC voltage for industrial applications.
- ๐ The three-phase half-wave rectifier has one diode per voltage source, making it the simplest type of three-phase rectifier.
- ๐ The phase voltages are represented by equations: V_a = V_s sin(ฯt), V_b = V_s sin(ฯt - 2ฯ/3), and V_c = V_s sin(ฯt + 2ฯ/3).
- โ๏ธ The voltages cross each other at points that are symmetrical and can be calculated to understand when one phase voltage exceeds another.
- ๐ 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.
- ๐ฝ The minimum output voltage occurs where the phase voltages cross each other, which is 0.5 times the peak input voltage (V_s).
- ๐ 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.
- ๐ The average output voltage can be calculated by integrating the waveform over a period, resulting in approximately 0.82699 times V_s or (3โ3/ฯ) times V_s.
- ๐ 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(ฯt), VB = Vs * sin(ฯt - 2ฯ/3), and VC = Vs * sin(ฯt + 2ฯ/3), where Vs is the peak voltage and ฯ 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 ฯt. For example, setting VA equal to VC and solving for ฯt gives ฯt = ฯ/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 * โ3 / 2ฯ * 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
๐ 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.
๐ 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. The่ฎฒๅธ 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. The่ฎฒๅธ 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.
๐งฎ Calculating the Average Output Voltage
The่ฎฒๅธ 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. The่ฎฒๅธ 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. The่ฎฒๅธ 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.
๐ 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. The่ฎฒๅธ 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.
Mindmap
Keywords
๐กThree-phase half-wave rectifier
๐กResistive loads
๐กBalanced three-phase voltage source
๐กDiode
๐กRectifier
๐กVoltage source
๐กOutput voltage ripple
๐กPeak voltage
๐กSymmetric waveforms
๐กAverage output voltage
๐กIndustrial application
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
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