Sebastian Valerio Alejo
Oscar Alberto Gomez Rodriguez
Half Wave Single Phase Rectifiers Workshop
Source and load characteristics for all exercises.
•
AC voltage source: 110 Vrms at 60 Hz
•
Consumption of load 1 (W): 50 W; (resistive load)
•
Consumption of load 2 (W): W ?; (highly inductive, coil only) Consumption of load 3 (W): 50 W;
•
(26mH combined resistive and inductive load connected in series)
Tasks to do.
Characteristics of the rectifier diode.
1) Half-wave rectifier resistive load
First of all, we proceed to obtain the values of the RMS voltage and average voltage
• •• =
• ••• =
•• =
•
•• =
two
110 ∗ √2
•
110 ∗ √2
two
= 49.52 •
= 77.78 •
After this we proceed to extract the value of the resistance of the power that is required of us
P=
• ••• =
•
two
two
two
• • −−−−−− → • =
•• =
4•
4•
Once the resistance is found, the power factor is found
(110√2)
4 (50)
two
= 121
• two
•
S = • ••• • ••• =
2√2•
1
• two
•
•
fp = = 4•
•
two =
1
• •two =
√2
two
≅ 0.7071
√2
2√2•
With the data found, we proceed to calculations l at average current and RMS
• •• =
•• =
110 ∗ √2
••
121•
•• =
• ••• =
= 1.29 •
110 ∗ √2
2•
= 0.64 •
2 (120.99)
Finally we proceed to find the apparent power:
•=
• two
•
(110√2)
=
2√2•
two
2√2•
= 70.71
To derive the Fourier series, we exp r That is the function of the rectified wave like this:
0 ≤ •• ≤ •
• (•) = {(110√2) sin ••,
0,
• ≤ •• ≤ 2•
Where the Fourier sum would be the following:
∞
•
• • ( ••) = 0 + ∑ • ••• (••) + •• • ••• (•••)
•
two
•=1
The period would be expressed like this:
• = 2•
Where the limits of integration would be placed from 0 to π, since from π to 2π the value is 0; • • would be determined
like this:
two •
two
• • = ∫ • (••) cos (•••) ••• =
•0
2• 0
(110√2) •
=
•
•
∫ • • sin (••) cos (•••) •••
110√2 (1 + (−1) •)
∫ sin (••) cos (•••) ••• = -
•
0
• two - 1
and the • • So:
two •
two
• • = ∫ • (••) sin (nwt) ••• =
•0
=
2• 0
(110√2) •
•
•
∫ • • sin (••) sin (•••) •••
110√2
∫ sin (••) sin (•••) ••• =
•
0
From this it can be concluded that:
110√2 (1 + (−1) •)
••={ -
•
• two - 1
,
•≠1
0,
•=1
(
1
•-1
(0) -
1
•+1
(0))
• • = { 110√2
two
0,
•≠1
,
•=1
Where the value at n = 1 was determined thanks to the limit found by Symbolab. Returning with the
Fourier sum, we have:
∞
•
• • ( ••) = 0 + ∑ • ••• (••) + •• • ••• (•••)
•
two
• • ( ••) =
•=1
110√2 110√2
+
•
two
110√2
••• (••) -
•
∞
(1 + (−1) •)
∑
•=2
• two - 1
From the above it can be deduced that the magnitudes of the first 2 harmonics will be:
• = 0 −−−− → • 0 = • •• =
110√2
•
= 49.52
• = 60 −−−− → • 1 = • ••• = 55√2 = 77.78
These first 2 harmonics describe the average voltage and the RMS voltage as indicated in the
equation. Then, from the summation, we proceed to obtain the value of the missing harmonics
(LOSSES):
• = 60• −−−−− → • • = -
110√2
•
10
∑
•=2
(1 + (−1) •)
• two - 1
As it is an even summation, all odd n within the sum will give 0 so the odd numbers are discarded for
the following harmonics, then it would have to:
• = 120 −−−− → • 2 = 33.01 • • = 240 −−−−
→ • 4 = 6.6 • • = 360 −−−− → • 6 = 2.83V • =
480 −−−− → • 8 = 1.57V • = 600 −−−− → • 10
= 1V
All these values were found thanks to the Symbolab mathematical tool. Now we proceed to place
the data produced by the simulation and compare:
It is observed that it actually gave the previously calculated values and that they also coincide with the
calculated voltages ( • •• • • •••) at the beginning of the document.
Now for the terms of the current the following is taken into account:
•
•=
•
Taking this into account, we then proceed to modify the following term found previously:
• • ( ••) =
110√2 110√2
+
•
two
••• (••) -
110√2
•
∞
(1 + (−1) •)
∑
• two - 1
•=2
Which gives:
• • ( ••) =
110√2 110√2
+
••
2•
••• (••) -
110√2
••
• • ( ••) =
∑
(1 + (−1) •)
•=2
Replacing the R we have to:
110√2 110√2
+
••• (••) 121•
(121) 2
∞
110√2
121•
∞
∑
•=2
• two - 1
(1 + (−1) •)
• two - 1
From the above it can be deduced that the magnitudes of the first 2 harmonics will be:
• = 0 −−−− → • 0 = • •• =
110√2
121•
= 0.404A
• = 60 −−−− → • 1 = • ••• =
55√2
121
= 642.82 ••
As with voltage, the first 2 harmonics of the current describe the average current and the RMS
current as indicated in the equation. Then, from the summation, we proceed to obtain the value of the
missing harmonics (LOSSES):
• = 60• −−−−− → • • = -
110√2
121•
10
∑
•=2
(1 + (−1) •)
• two - 1
As it is an even summation, all odd n within the sum will give 0 so the odd numbers are discarded for
the following harmonics, then it would have to:
• = 120 −−−− → • 2 = 262.82 •• • = 240 −−−−
→ • 4 = 54.56mA • = 360 −−−− → • 6 = 23.38
mA • = 480 −−−− → • 8 = 12.99 mA • = 600
−−−− → • 10 = 8.26 Tue
All these values were found thanks to the Symbolab mathematical tool. Now we proceed to
leave the simulation data and compare:
It is observed that they effectively agree with the calculated data and with the calculated currents ( • •• •
• •••) at the beginning of the document.
Now moving on to the simulation we have to:
To get the simulated apparent power multiply the simulated RMS current and voltage at the
source in Multisim:
• •••••••• = • ••• ••••••••
Variable
• ••• •••••••• = 637.918•• ∗ 109.848• = 70.074••
Calculated value
Simulated value
• •• in the load
49.52 [V]
49.0651 [V]
• ••• in the load
77.78 [V]
77.1881 [V]
• •• in the circuit
404 [mA]
405,497 [mA]
• ••• in the circuit
642.82 [mA]
637,918 [mA]
P
50 [W]
49,232 [W]
S
70.71 [VA]
70,074 [VA]
fp
0.7071
0.91675
It is then observed that the calculations are in accordance with the measurements made except for
the power factor, this may be due to the fact that the missing power factor is consumed by the
harmonics (energy loss) which also occupy a power factor in the circuit. .
Finally, the waveforms of the circuit are shown:
2) Inductive load (26mH) half wave rectifier
In this case, all the current goes through the diode and reaches the coil, in such a way that the
equation that represents the voltage is:
• • = • • without ••
However, despite the fact that all the current can pass through the diode, it no longer has a negative
cycle, so it has an offset in the signal and the equation for its maximum current value is:
2 ((110) (√2))
2• • =
••=
2•••
= 31.74 •
2• (60••) (26••)
Now, the function of the current is given by:
•
• • = • (1 - cos (••))
two
Because in the voltage in the load the wave is purely sine it can be said that:
• •• = 0 • ••• = • = 110•
•
√2
However, in the current despite also being a sine wave, it has an offset so if it is going to have
average current and the conditions to find the RMS current will also change:
1•
• •• ( ••) = ∫ • (••) ••• =
•0
=
2• 0
2• 31.74
1
2• 0
∫
two
1•
• ••• ( ••) = √ ∫ (• (••)) ••• = √
•0
=√
2• • •
1
1
2• 0
∫
(1 - cos (••)) •••
two
(1 - cos (••)) ••• = 15.87 •
1
two
2• 0
2• • •
∫
(1 - cos (••)) two •••
two
2• 31.74
∫
two
(1 - cos (••)) two ••• = 4.88 •
For reactive and apparent power you have to:
• = • ••• ∗ • ••• ≈ 536.8••
•
• = • ∗ ••• () ≈ 536.8 [•••]
•• = 0
two
Since the voltage is a pure sine wave, it is expected that when applying Fourier, a single harmonic will be
seen at the frequency of 60Hz with the amplitude of the RMS voltage.
It is observed that indeed said analysis is close to what has already been described before.
Harmonics do not disappear completely because the diode acts as a very small resistor.
In the case of current, everything changes:
∞
•
• • ( ••) = 0 + ∑ • ••• (••) + •• • ••• (•••)
•
two
Where:
two •
two
• 0 = ∫ • (••) ••• =
•0
•••
•=1
(1 - cos (••)) ••• =
∫
2• −• two
two •
=
• • = lim (
•→1
31.74 •
•
two
•••
(1 - cos (••)) cos (•••) •••
∫
2• −• two
31.74 2•
∫ (1 - cos (••)) cos (•••) ••• = ••••••••••
2•
0
∫ (1 - cos (••)) cos (•••) •••) = - 15.87
−•
∫ (1 - cos (••)) ••• 2•
−•
= 31.74 •
• • = ∫ • (••) cos (•••) ••• =
•0
31.74 •
two •
two
• • = ∫ • (••) sin (nwt) ••• =
•0
=
2• 0
2• • •
∫
(1 - cos (••)) sin ••• •••
two
31.74 2•
∫ (1 - cos (••)) sin ••• ••• = 0
2•
0
Replacing these values in the Fourier summation we have:
∞
•
• • ( ••) = 0 + ∑ • ••• (••) + •• • ••• (•••)
•
two
• • ( ••) =
31.74
two
•=1
- 15.87••• (•••) = 15.87 - 15.87••• (•••)
It is observed that in fact the harmonics in the current have the same magnitude although they
differ in sign.
Regarding the simulation, you have to:
Variable
Calculated value
Simulated value
• •• in the load
0
12.556mV
• ••• in the load
110V
110V
• •• in the circuit
15.87A
15,324 A
4.88A
4.88 A
0
- 1,184 mW
• ••• at
circuit
P
S
536.8VAR
564VAR
fp
536.8VA
520VA
• •• in the load
0
0.025
Finally, we proceed to make the corresponding graphs:
This voltage peak is due to the diode, since being not ideal, it has a very small resistance that causes
part of the voltage to be stored in the coil, so when it returns to its positive cycle it is released forming a
peak in the voltage. . Also this would explain some of the harmonics already seen in the Fourier
spectrum earlier.
It is then observed that indeed all the current passes through the diode, but with an offset whose
maximum agrees with the calculated one.
A pure signal can be observed without loss of energy
3) Combined load half wave rectifier
So, first of all, the function that this rectifier describes is:
• • sen ••,
• •• ( ••) = {
0,
0 <•• ≤ •
• ≥ •• ≤ 2•
Where:
••
• = ••• −1 () = ••• −1 (
•
2• (60••) (26••)
) = 80.83•10 −3
120.99
• = • + • = • + 80.83•10 −3 = 3.2224 • = √ (••) 2 + • 2 = √ ((2• ∗ 60••) ∗ 26••) + (120.99) 2 = 121.39
two
With these values we proceed to extract the average voltage:
••• (••),
• •• ( ••) = {
0 ≤ •• ≤ •
• ≤ •• ≤ 2•
0,
•
(110) 3.2224
∫
2•
0
•
• • =• • ∫ ••• (••) ••• = • •• =
2• 0
••
• ••• =
=
two
110 ∗ √2
two
= 77.78 •
In the case of current, its function is as follows:
• • ••• (••) • •• + −•••
• • ••• (•• - •),
• (••) = {•
0 ≤ •• ≤ •
•
0,
−20.99••
9.8
• (••) = {1.28 ••• (••) •
••• (••) ••• = 49.44
• ≤ •• ≤ 2•
+ 1.28 ••• (•• - 0.0808),
0,
0 ≤ •• ≤ 3.2296
3.2296 ≤ •• ≤ 2•
With Beta expressed in radians; now the average current is as follows:
• •• =
• •• =
1
2• 0
1
2• 0
•
∫ • (••) •••
3.2224
∫
1.28 ••• (••) •
And the RMS current is:
• ••• = √
1
2• 0
•
∫ • two( ••) •••
−20.99••
9.8
+ 1.28 ••• (•• - 0.0808) ••• = 443.30mA
3.2224
1
• ••• = √
2• 0
∫
two
−20.99••
9.8
(1.28 ••• (••) •
+ 1.28 ••• (•• - 0.0808)) •••
= 683.10mA
•• =
•
50 •
=
• ••• • •••
(
110
√2
= 0.941
•) (683.10 ••)
To derive the Fourier series, the rectified wave function is expressed as:
(110√2) sin ••,
• (•) = {
0 ≤ •• ≤ •
• ≤ •• ≤ 2•
0,
Where the Fourier sum would be the following:
∞
•
• • ( ••) = 0 + ∑ • ••• (••) + •• • ••• (•••)
•
two
•=1
The period would be expressed like this:
• = 2•
Where the limits of integration would be placed from 0 to π, since from π to 2π the value is 0; • • would be determined
like this:
two •
two
• 0 = ∫ • (••) •••) =
•0
2• 0
two •
two
• • = ∫ • (••) cos (•••) ••• =
• 0
=
=
(110√2) •
•
2• 0
•
∫ ••• (••) ••• = 98.87
•
∫ • sin
• (••) cos (•••) •••
∫ sin (••) cos (•••) •••
0
55√2 (−•••• (• - ••) - cos (• - ••) + •••• (• + ••) + 2 - •••• (• + ••))
• (• + 1) (- • + 1)
and the • • So:
two •
two
• • = ∫ • (••) sin (nwt) ••• =
•0
=
=
2• 0
(110√2) •
•
•
∫ • sin
• (••) sin (•••) •••
∫ sin (••) sin (•••) •••
0
55√2 (sin (• - ••) (• + 1) - sin (• + ••) (−• + 1))
From this it can be concluded that:
• (• + 1) (• - 1)
••
55√2 (−•••• (• - ••) - cos (• - ••) + •••• (• + ••) + 2 - •••• (• + ••))
={
,
• (• + 1) (- • + 1)
•••••••,
55√2 (sin (• - ••) (• + 1) - sin (• + ••) (−• + 1))
••={
•=1
•≠1
,
• (• + 1) (• - 1)
•≠1
•••••••,
•=1
Where the value at n = 1 was determined thanks to the limit found by Symbolab. Calculating the
different harmonics, the first 2 harmonics will have the following value:
• = 0 −−−− → • 0 = • •• =
110√2
•
= 49.52
• = 60 −−−− → • 1 = • ••• = 55√2 = 77.78
As happened with the purely resistive circuit, its first 2 harmonics are close to the amplitude of the
average and RMS voltages respectively. We proceed to extract the value of the missing harmonics
(LOSSES):
••=
••=
For this part things change; unlike the purely resistive circuit that
only harmonics appeared in • •, and in • • did not appear, here harmonics appear both in • • like in • •, In
addition, harmonics also appear in the odd numbers of the
• •, which did not happen in the purely resistive circuit, with all this said it translates into greater losses.
It is true that the losses seem few, but this
it is because the angle • (80.83•10 −3 ••• ≈ 4 ••••••) it is small so the losses are somewhat negligible.
However, where that angle
• If it were larger, it would cause greater amplitudes in the harmonics and therefore greater losses, that
is why a flying diode is used for this type of circuit, since that simple diode will reduce losses.
With everything said above, we proceed to make the Fourier comparison:
It is observed that indeed these values are quite close to the
harmonics calculated in • • because the values of the • • they are small and therefore negligible.
Now for the terms of the current the following is taken into account:
•
•=
•
Taking this into account, we then proceed to modify the following term found before r iormente:
• • =•
55√2 (−•••• (• - ••) - cos (• - ••) + •••• (• + ••) + 2 - •••• (• + ••))
Which gives:
• (• + 1) (- • + 1)
• • •=
55√2 (−•••• (• - ••) - cos (• - ••) + •••• (• + ••) + 2 - •••• (• + ••))
•• (• + 1) (- • + 1)
Replacing the R we have to:
• • •=
55√2 (−•••• (• - ••) - cos (• - ••) + •••• (• + ••) + 2 - •••• (• + ••))
121• (• + 1) (- • + 1)
the same is done for him • • and you •have to:
• • •=
55√2 (sin (• - ••) (• + 1) - sin (• + ••) (−• + 1))
• (• + 1) (• - 1)
And as with the voltage, we proceed to remove the harmonics from the current, it is not a mystery that the first 2
harmonics are going to give the current, they are going to give the rms voltage and current consecutively:
• = 0 −−−− → • 0 = • •• =
• = 60 −−−− → • 1 = • ••• =
110√2
121•
55√2
= 0.404A
= 642.82 ••
121
Now the missing harmonics of the current will be calculated (LOSSES)
••=
••=
As it happened with the voltage, in the current it is the same analysis that is that a
Unlike the purely resistive circuit that only harmonics appeared in • •, and in • • did not appear, here
harmonics appear both in • • like in • •, What's more
harmonics also appear in the odd numbers of the • •, which did not happen in the purely resistive circuit,
with all this said it translates into greater losses.
It is true that the losses seem few, but this is due because the angle •
( 80.83•10 −3 ••• ≈ 4 ••••••) it is small so the losses are somewhat negligible. However, where that
angle • If it were larger, it would cause greater amplitudes in the harmonics and therefore greater
losses, that is why a flying diode is used for this type of circuit, since that simple diode will reduce
losses.
With everything said above, we proceed to make the Fourier comparison:
As can be seen as with voltage, current also has values similar to • • Y • • it is negligible for having
small magnitudes
Finally, the simulation is shown with its respective comparisons.
The voltage lag is effectively observed by the •
Variable
Calculated value
Simulated value
• •• in the load
49.44V
49.05V
• ••• in the load
77.79V
77.167V
• •• in the circuit
404mA
404.846mA
640.8mA
635.907mA
P
50W
48.86W
fp
0.941
49.21VA
• ••• at
circuit
4) Combined load half wave rectifier flying diode
• •• =
• •• =
•• =
110√2
•
•
110√2
•• =
••
= 49.52 •
• (120.99)
= 409.26 ••
The Fourier sum is as follows:
• • ( ••) =
110√2 110√2
+
•
two
••• (••) -
110√2
•
∞
∑
•=2
2• •
(• two - 1) •
From this it follows that the magnitude of the 2 first or s harmonics is:
• = 0 −−−− → • 0 = • •• =
• = 60 −−−− → • 1 = • ••• =
110√2
= 49.52 V
•
110√2
= 55√2 •
two
The following harmonics are calculated (LOSSES):
Now we proceed to compare with the simulation and the following is obtained:
It is then observed that those harmonics that appeared at point 4 of the combined half-wave
rectification load were effectively erased as the analysis had been done at that point.
To obtain the harmonics of the current, the fourier summation is divided over the voltage and the following
is obtained:
110√2 110√2
+
••
2•
• • ( ••) =
110√2
••• (••) -
••
Replacing the R we have to:
• • ( ••) =
110√2 110√2
+
121•
242
••• (••) -
110√2
121•
∞
∑
•=2
2• •
(• two - 1) •
∞
∑
•=2
2• •
(• two - 1) •
The first 2 harmonics are then found and it has to be:
• = 0 −−−− → • 0 = • •• =
• = 60 −−−− → • 1 = • ••• =
We proceed to find the missing harmonics lost:
110√2
= 409.23mA
121•
110√2
= 642.82••
242
Now we proceed to compare with the simulated data:
It is observed that it actually gave the calculated values; what the flying diode does is bring the angle
closer • to zero which reduces the energy lost, this analysis has already been done in point 4.
The simulation as the waveforms are as follows:
5) Full wave rectifier resistive load
2 (110√2)
2• • =
• •• =
•
• ••• = 110•
• •••••• =
• ••• =
• two
•=•=
2•
•
• two
• −−−−
→•
2•
• ••• =
110
•
242
(110√2)
2 (242)
= 99.03
=
= 454.52 ••
two
= 50 ••
• two
•
2 ∗ • ••••••
=
(110√2)
2 ∗ 50
two
= 242 Ω
•• = =
•
fifty
=1
•
fifty
To derive the Fourier series, the fu n tion of the rectified wave like this:
• (•) = (110√2) sin ••
Where the Fourier sum would be the following:
∞
•
• • ( ••) = 0 + ∑ • ••• (••) + •• • ••• (•••)
•
two
•=1
The period would be expressed like this:
• = 2•
Where the limits of integration would be placed from 0 to π, since from π to 2π the value is 0; • • would be determined
like this:
two •
two •
• • = ∫ • (••) cos (•••) ••• = ∫ • • sin (••) cos (•••) •••
•0
•0
=
2 (110√2) •
220√2 (1 + (−1) •)
∫ sin (••) cos (•••) ••• = - •
•
0
• two - 1
and the • • So:
two •
two
• • = ∫ • (••) sin (nwt) ••• =
• 0
=
2• 0
(110√2) •
•
•
∫ • sin
• (••) sin (•••) •••
110√2
∫ sin (••) sin (•••) ••• =
•
0
(
1
•-1
(0) -
1
•+1
(0))
From this it can be concluded that:
110√2 (1 + (−1) •)
••={ -
•
• • = { 220√2
two
,
•≠1
0,
•=1
• two - 1
0,
•≠1
,
•=1
Where the value at n = 1 was determined thanks to the limit found by Symbolab. Returning with the
Fourier sum, we have:
∞
•
• • ( ••) = 0 + ∑ • ••• (••) + •• • ••• (•••)
•
two
• • ( ••) =
•=1
220√2 220√2
+
•
two
••• (••) -
220√2
•
∞
∑
•=2
(1 + (−1) •)
• two - 1
From the above it can be deduced that the magnitudes of the first 2 harmonics will be:
• = 0 −−−− → • 0 = • •• =
220√2
•
• = 60 −−−− → • 1 = • ••• = 110√2 = 155.156•
= 99.04
These first 2 harmonics describe the average voltage and the RMS voltage as indicated in the
equation. Then, from the summation, we proceed to obtain the value of the missing harmonics
(LOSSES):
• = 60• −−−−− → • • = -
110√2
•
10
∑
•=2
(1 + (−1) •)
• two - 1
As it is an even summation, all odd n within the sum will give 0 so the odd numbers are discarded for
the following harmonics, then it would have to:
• = 120 −−−− → • 2 = 66.02 • • = 240 −−−−
→ • 4 = 12.12 • • = 360 −−−− → • 6 = 5.66V •
= 480 −−−− → • 8 = 3.14V • = 600 −−−− →
• 10 = 2V
All these values were found thanks to the Symbolab mathematical tool. Now we proceed to place
the data produced by the simulation and compare:
As can be seen, they coincide with the calculated data; Here the analysis that is done is that the full
wave signal emits twice as much energy as the half wave signal, but for the same concept also the
harmonics of the lost energy is also twice, so there are more losses than in the half wave.
Now for the terms of the current the following is taken into account:
•=
•
•
Taking this into account, we then proceed to modify the following term found previously:
• • ( ••) =
110√2 110√2
+
•
two
••• (••) -
110√2
•
∞
(1 + (−1) •)
∑
• two - 1
•=2
Which gives:
• • ( ••) =
110√2 110√2
+
••
2•
••• (••) -
110√2
••
• • ( ••) =
(1 + (−1) •)
∑
• two - 1
•=2
Replacing the R we have to:
110√2 110√2
+
••• (••) 121•
(121) 2
∞
110√2
121•
∞
(1 + (−1) •)
∑
•=2
• two - 1
From the above it can be deduced that the magnitudes of the first 2 harmonics will be:
• = 0 −−−− → • 0 = • •• =
• = 60 −−−− → • 1 = • ••• =
110√2
121•
55√2
121
= 0.404A
= 642.82 ••
As with voltage, the first 2 harmonics of the current describe the average current and the RMS
current as indicated in the equation. Then, from the summation, we proceed to obtain the value of the
missing harmonics (LOSSES):
• = 60• −−−−− → • • = -
110√2
121•
10
∑
•=2
(1 + (−1) •)
• two - 1
As it is an even summation, all odd n within the sum will give 0 so the odd numbers are discarded for
the following harmonics, then it would have to:
• = 120 −−−− → • 2 = 262.82 •• • = 240 −−−−
→ • 4 = 54.56mA • = 360 −−−− → • 6 = 23.38
mA • = 480 −−−− → • 8 = 12.99 mA • = 600
−−−− → • 10 = 8.26 Tue
All these values were found thanks to the Symbolab mathematical tool. Now we proceed to
leave the simulation data and compare:
It can be observed then that the components of the current do not change with respect to the half
wave.
Finally, the simulation and the respective graphs are shown:
6) Combined load full wave rectifier
••=
••=
•• =
2 (110√2)
•
•• =
•
2 (110√2)
••
••=••(
1
242•
•-1•+1
Harmonics V: • 2,4,6,8,10 = 66.02•, 13.2•, 5.66•, 3.14•, 2•
= 99.03 •
= 409.23 ••
1
)
• (••) = 99.03 + 66.02••• (2•• + •) + 13.2••• (4•• + •) + 5.66••• (6•• + •)
+ 3.14••• (8•• + •) + 2••• (10•• + •)
••=
••
=
√• 2 + ( 2• ∗ • ∗ • ∗ •) two
••
√242 2 + ( 2• ∗ 60 ∗ • ∗ 0.026) two
Harmonics i = • 2,4,6,8,10 = 262.55••, 52.39••, 22.4••, 12.41••, 7.88•• • (••) = 409.23•10 −3 + 262.55•10 −3 •••
(2•• + •) + 52.39•10 −3 ••• (4•• + •)
+ 22.4•10 −3 ••• (6•• + •) + 12.41•10 −3 ••• (8•• + •)
+ 7.88•10 −3 ••• (10•• + •)
• ••• = √• • +
• two +
•4 + •6
two
two
+
two
= √409.23• 2 +
• 8 + • 10
two
two
262.55• two
two
= 451.29••
two • = 451.29•• two( 242) = 49.29•
• = • •••
• = 110 ∗ 451.29•• = 49.6419 ••
+
52.39• two
two
+
22.4• two
two
+
12.41• two
two
+
7.88• two
two
•• =
49.29
49.6419
= 0.992838
7) Conclusions
a) Full wave rectified waves compared to medium waves generate twice as much energy but also lose
twice as much energy
b) Combined half-wave rectification loads without a flywheel diode have much more energy losses
due to phase shift.