AC to DC rectification
- Thread starter R o y a l
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ptonsparky
Tom
- Occupation
- EC - retired
Peak voltage is 1.41 * RMS.
1.41*600= 846
WAG is your SCR is picking up somerwhere along the waveform to get you the 750.
Been to long since school and have CRS anyway. The guys that know should show up some time today.
1.41*600= 846
WAG is your SCR is picking up somerwhere along the waveform to get you the 750.
Been to long since school and have CRS anyway. The guys that know should show up some time today.
Hi everybody ,
Does any one knows the calculation of The 3 Phase Silicon-Controlled Rectifier .
I saw one rectify 600 VAC to 750 VDC .
so,
750/600=(1.25 --->is this factor .) if yes from where we get this factor mathematically.
Regards.
The value in my textbook shows Vdc=1.35*Vrms. So for an RMS AC voltage of 600V I would expect to see a DC rectified voltage of 810V. It is possible that the SCR's are only triggering on part of the waveform to give this reduced voltage of 750V.
Hi everybody ,
Does any one knows the calculation of The 3 Phase Silicon-Controlled Rectifier .
I saw one rectify 600 VAC to 750 VDC .
so,
750/600=(1.25 --->is this factor .) if yes from where we get this factor mathematically.
Regards.
Depends on the exact circuit. If its just an SCR circuit with an output filter, it would depend on when the SCR's fire, and the output filter details.
More likely, you are looking at a switching mode power supply. Then, by controlling the firing of the SCR's along with some inductor ratios, you can get any output voltage you want. Step-up or step-down.
Steve
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
101124-1335 EST
Royal:
You can get any output voltage up to a maximum defined by the supply voltage with a simple phase shift controlled rectifier. So there is no single answer to your question.
Rather than throw some numbers at you I will provide a couple references. What you really need is a basic understanding of how various diode rectifier circuits work. Knowing this, understanding how a load interacts, and what change an SCR makes vs a diode, then with this basic understanding you can reason thru new circuits that you may encounter.
Pages 312 and 313 in section on Rectifiers and Filters in "Reference Data for Radio Engineers", fourth edition, 1960, International Telephone and Telegraph Corp.
Chapter 1. Polyphase Rectifiers, "Theory and Application of Industrial Electronics", Cage and Bashe, First Edition, 1951, McGraw-Hill.
There should be many other books that will provide comparable information.
.
Royal:
You can get any output voltage up to a maximum defined by the supply voltage with a simple phase shift controlled rectifier. So there is no single answer to your question.
Rather than throw some numbers at you I will provide a couple references. What you really need is a basic understanding of how various diode rectifier circuits work. Knowing this, understanding how a load interacts, and what change an SCR makes vs a diode, then with this basic understanding you can reason thru new circuits that you may encounter.
Pages 312 and 313 in section on Rectifiers and Filters in "Reference Data for Radio Engineers", fourth edition, 1960, International Telephone and Telegraph Corp.
Chapter 1. Polyphase Rectifiers, "Theory and Application of Industrial Electronics", Cage and Bashe, First Edition, 1951, McGraw-Hill.
There should be many other books that will provide comparable information.
.
Jack Linville
Member
- Location
- Indianapolis
Rectifier Information & Theory
Rectifier Information & Theory
Although some may be skeptical, Wikipedia is really a good source of information on almost any subject.
http://en.wikipedia.org/wiki/Rectifier
Rectifier Information & Theory
Although some may be skeptical, Wikipedia is really a good source of information on almost any subject.
http://en.wikipedia.org/wiki/Rectifier
jumper
Senior Member
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- 3 Hr 2 Min from Winged Horses
I beg to differ. Anybody can post on that POS site.
Use it as a reference in an academic or professional paper/presentation and you will get nailed.
skeshesh
Senior Member
- Location
- Los Angeles, Ca
It depends on if what you're looking at is really a rectifier or a power converter. It could get pretty complicated depending on the device. "fundamentals of power electronics" is a book I used in school a few years back and it's pretty excellent in explaining the theory. You better sharpen a bunch of pencils and refresh yourself on transistors though if you really want to dive into it since the math can get pretty messy.
dbuckley
Senior Member
- Location
- Canterbury, New Zealand
If you have a rectifier (assume full bridge) fed with a sine wave followed by a filter capacitor of adequate size you'll get smooth DC out, at around 1.414 the RMS AC voltage. Essentially the cap charges to the peak voltage of the AC half-cycle.
If the cap is insufficiently large, then the cap will be discharged by the load just past the peak of the AC half-cycle. So you'll get some lesser voltage out, which is a sort of (warning: bad terminology) "average" voltage, depending on lots of things, both about the real voltage present, and the way the measuing instument works.
On the discharge, the capacitor and the load form a CR network, and you can calculate that, so you can work out for any time past the half-cycle peak what the actual voltage across the cap will be.
If the cap is insufficiently large, then the cap will be discharged by the load just past the peak of the AC half-cycle. So you'll get some lesser voltage out, which is a sort of (warning: bad terminology) "average" voltage, depending on lots of things, both about the real voltage present, and the way the measuing instument works.
On the discharge, the capacitor and the load form a CR network, and you can calculate that, so you can work out for any time past the half-cycle peak what the actual voltage across the cap will be.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
101201-2035 EST
Do the British generally call them CR instead of RC? So it would be the CR time constant?
.
Do the British generally call them CR instead of RC? So it would be the CR time constant?
.
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Interesting note:
Some years ago, a friend and I were experimenting with audio amplifier bridging. We made a high-ampacity dual-polarity power supply using a pair of transformers with the secondaries in series as a grounded center-tapped source, and full-wave rectification with two high-value caps on each polarity's bus.
The transformers each came with a capacitor for connection to a tertiary winding, but we didn't connect them at first. We noticed that the supply voltages sagged readily when using the amps, so we decided to connect the caps and measure again. The DC voltages then remained at the peak, even when loaded.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
101201-2336 EST
Larry:
Where those essentially ferroresonant regulating transformers? Like Sola.
For a while Stancor was making such a transformer for regulating DC output relative to input AC variations. These did not change the output impedance and so were not regulating the DC output voltage relative to load variations on the output.
.
Larry:
Where those essentially ferroresonant regulating transformers? Like Sola.
For a while Stancor was making such a transformer for regulating DC output relative to input AC variations. These did not change the output impedance and so were not regulating the DC output voltage relative to load variations on the output.
.
ohmhead
Senior Member
- Location
- ORLANDO FLA
Hi everybody ,
Does any one knows the calculation of The 3 Phase Silicon-Controlled Rectifier .
I saw one rectify 600 VAC to 750 VDC .
so,
750/600=(1.25 --->is this factor .) if yes from where we get this factor mathematically.
Regards.
I have built lots of ac to dc power supplies when you install just the diodes on the secondary of a transformer you will get voltage of that voltage of that transformer not pure dc but dc .
Your next step is a cap or caps this is to smooth out the dc ripple of the rectifier this actually gives you a higher voltage then the transformer secondary you started with say it was 24 volts ac sec when it comes thur your rectifier with the capacitors it will be 26 or 28 volts a little higher then before the caps . the capacitors will keep voltage up not down no formulas this is from actually working and building power supplies hands on .
So what you see is normal
Last edited:
Jack Linville
Member
- Location
- Indianapolis
Wikipedia
Wikipedia
Jumper,
Did you find anything in error in the Wikipedia article I linked? Or did you even look? Often an article written by a credible source will have a large set of links to other "academically acceptable" references.
Just because something isn't "approved" doesn't mean it is in error.
It seems anybody can post on this site too. with some moderation, but I wouldn't refer to it as a POS.
Wikipedia
Jumper,
Did you find anything in error in the Wikipedia article I linked? Or did you even look? Often an article written by a credible source will have a large set of links to other "academically acceptable" references.
Just because something isn't "approved" doesn't mean it is in error.
It seems anybody can post on this site too. with some moderation, but I wouldn't refer to it as a POS.
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
I admit I did not look at the reference, nor am I going to.
I will apologize if in any way if I insulted you somehow, I was bashing wikipedia not you. Your profile is impressive and I did not mean to imply anything towards you.
I am too old, stubborn and set in my ways. I want an encyclopedia and not a wikipedia. I personally know a PhD who writes for Encyclopedia Britannica, so I am partial to more authoritive sites.
The internet has burned me too many times, so I am skeptical.
Well, you both get to be right ;-)
For both single and three phase the peak at no load will be sqrt(2) or 1.41. For three phase the voltage drops to 1.35 times the AC voltage once the rectifier is loaded.
This depends on what type of bridge you are using, three pulse, six pulse etc...
For example a six pulse rectifier will have a dc voltage of 1.35 times the AC rms voltage as you mentioned, however if you use only a 3 pulse rectifier on a three phase circuit then the dc ouput voltage will only be 0.675 times the AC rms voltage.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
101203-1117 EST
To calculate the various values that have been presented above you need to define the question and then analyze the resulting wave form.
Some of the above comments relate to the peak voltage of the sine wave source. These all imply a capacitor input filter that as a first approximation, if the capacitance is large enough, captures the peak voltage of the waveform. In this case for any of the normal rectifier circuits the DC output voltage equals the peak voltage of the sine wave or Vrms/0.707 or Vrms*1.414 . So from a 120 V supply for any of 1/2 wave, full wave, 3 phase, and 6 phase the result is 120*sq-root of 2 = 169.7 V.
If instead, the question is what is the average DC voltage from the rectifier, then you need to look at the waveform and calculate the average value from this. The average is determined by the area under the curve divided by the averaging period. Working on a full cycle period the averaging time is 2*Pi.
The area under a sine wave can be mathematically determined with integral calculus. The integral of the sine is = -cos . To calculate the area under the curve the integral has to be evaluated at the beginning and ending points of interest.
The integral is evaluated by determining the value at the upper limit of the range and subtracting the value at the lower limit of the integration range.
For a half wave rectifier the area under the curve is -cos 180 - (-cos 0) = -(-1)-(-1) = 2 . This is the evaluation of the half sine wave for the period 0 to 180 deg. Pi radians = 180 deg. Note the integral over the second half of the waveform is zero because the voltage is zero. Since the area is 2 and the averaging period is 2*Pi the average value is 1/Pi = 0.318309886 . The average value relative to the peak voltage of the sine wave is 0.318309886*Vpeak. To get the value relative to the RMS voltage you divide by (1/2)*sq-root of 2. The result is 0.450158158*Vrms .
For the full wave rectifier it is obvious that the value is double that of the half wave because the area doubles. The results are 0.636619772*Vpeak and 0.900316316*Vrms. Thus, at 120 Vrms the average DC output voltage is 120*0.9 = 108 V DCaverage. If you connect an inductive load, such as a clutch, supplied from a bridge rectifier from 120 V single phase supply, then the DC voltage across the clutch coil would be 108 V minus the voltage drops in the diode rectifier. So about 106 to 107 V.
Now to the 3 phase 3 diode rectifier. This is one circuit of what may be described as an "n-phase half wave rectifier". Draw the composite rectified waveform. Conduction from any one diode is for 120 deg. Thus, the first diode conducts over the range 90+/-60 or from 30 to 150 deg. In radians this is from Pi/6 to 5*Pi/6. The area under this curve is -(-0.866)-(-0.866) = sq-root of 3 = 1.732050808, and the averaging time is 4*Pi/6 = 2.094395102. Thus, the average voltage is 0.826993343*Vpeak, and 1.169545202*Vrms.
From the information I provided you can try to figure out the 6 phase 6 diode half wave rectifier. This waveform is the same as a 3 phase full wave rectifier.
Also I make calculation and logic errors so see if there are any mistakes.
.
To calculate the various values that have been presented above you need to define the question and then analyze the resulting wave form.
Some of the above comments relate to the peak voltage of the sine wave source. These all imply a capacitor input filter that as a first approximation, if the capacitance is large enough, captures the peak voltage of the waveform. In this case for any of the normal rectifier circuits the DC output voltage equals the peak voltage of the sine wave or Vrms/0.707 or Vrms*1.414 . So from a 120 V supply for any of 1/2 wave, full wave, 3 phase, and 6 phase the result is 120*sq-root of 2 = 169.7 V.
If instead, the question is what is the average DC voltage from the rectifier, then you need to look at the waveform and calculate the average value from this. The average is determined by the area under the curve divided by the averaging period. Working on a full cycle period the averaging time is 2*Pi.
The area under a sine wave can be mathematically determined with integral calculus. The integral of the sine is = -cos . To calculate the area under the curve the integral has to be evaluated at the beginning and ending points of interest.
The integral is evaluated by determining the value at the upper limit of the range and subtracting the value at the lower limit of the integration range.
For a half wave rectifier the area under the curve is -cos 180 - (-cos 0) = -(-1)-(-1) = 2 . This is the evaluation of the half sine wave for the period 0 to 180 deg. Pi radians = 180 deg. Note the integral over the second half of the waveform is zero because the voltage is zero. Since the area is 2 and the averaging period is 2*Pi the average value is 1/Pi = 0.318309886 . The average value relative to the peak voltage of the sine wave is 0.318309886*Vpeak. To get the value relative to the RMS voltage you divide by (1/2)*sq-root of 2. The result is 0.450158158*Vrms .
For the full wave rectifier it is obvious that the value is double that of the half wave because the area doubles. The results are 0.636619772*Vpeak and 0.900316316*Vrms. Thus, at 120 Vrms the average DC output voltage is 120*0.9 = 108 V DCaverage. If you connect an inductive load, such as a clutch, supplied from a bridge rectifier from 120 V single phase supply, then the DC voltage across the clutch coil would be 108 V minus the voltage drops in the diode rectifier. So about 106 to 107 V.
Now to the 3 phase 3 diode rectifier. This is one circuit of what may be described as an "n-phase half wave rectifier". Draw the composite rectified waveform. Conduction from any one diode is for 120 deg. Thus, the first diode conducts over the range 90+/-60 or from 30 to 150 deg. In radians this is from Pi/6 to 5*Pi/6. The area under this curve is -(-0.866)-(-0.866) = sq-root of 3 = 1.732050808, and the averaging time is 4*Pi/6 = 2.094395102. Thus, the average voltage is 0.826993343*Vpeak, and 1.169545202*Vrms.
From the information I provided you can try to figure out the 6 phase 6 diode half wave rectifier. This waveform is the same as a 3 phase full wave rectifier.
Also I make calculation and logic errors so see if there are any mistakes.
.
Gentelmen,
Thank you for your answers.
I appreciate every one .
This system that I'm talking about is used in Drilling rigs which use Traction motors in its operation .
and I noticed that most of this rigs (using DC Power) have the same range from 0 to 750 VDC output.when the Power supply is 600 VAC .
I tried all the calculation but I could not get this 750 vdc???
The text below is taken from the man from manufacturer manual:
3 DC Control Unit
3.1 DC Control System
The DC control unit controls SCR components, rectifies 3 phase AC power supply
to provide continuously the variable DC power to the DC motors as figure 3-1
shows. The rectifying unit is a 3 phase full control rectifier bridge formed from 6
SCR components, and the bridge is insulated from AC electric mains by circuit
breakers. The output of the bridge is delivered to the DC motors through the
assignment contactor.
3.1.1 Thyristor Rectifier Bridge
Through the circuit breaker, the 3 phase AC power coming from the AC bus
provides the power to SCR bridge. See figures 021-000. Every phase of AC power
connects two thyristors, namely, phase A connects with A+ and A-SCR, phase B
connects with B+ and B-SCR, phase C connects with C+ and C-SCR. A+, B+ and
C+SCR supply power to DC+bus and A-, B- and C-SCR supply power to DC-bus.
Under the control of the firing pulse added to the gate and cathode of the thyristor,
the DC voltage coming from the rectifier bridge is variable within the range of 0 to
750V.
Thank you for your answers.
I appreciate every one .
This system that I'm talking about is used in Drilling rigs which use Traction motors in its operation .
and I noticed that most of this rigs (using DC Power) have the same range from 0 to 750 VDC output.when the Power supply is 600 VAC .
I tried all the calculation but I could not get this 750 vdc???
The text below is taken from the man from manufacturer manual:
3 DC Control Unit
3.1 DC Control System
The DC control unit controls SCR components, rectifies 3 phase AC power supply
to provide continuously the variable DC power to the DC motors as figure 3-1
shows. The rectifying unit is a 3 phase full control rectifier bridge formed from 6
SCR components, and the bridge is insulated from AC electric mains by circuit
breakers. The output of the bridge is delivered to the DC motors through the
assignment contactor.
3.1.1 Thyristor Rectifier Bridge
Through the circuit breaker, the 3 phase AC power coming from the AC bus
provides the power to SCR bridge. See figures 021-000. Every phase of AC power
connects two thyristors, namely, phase A connects with A+ and A-SCR, phase B
connects with B+ and B-SCR, phase C connects with C+ and C-SCR. A+, B+ and
C+SCR supply power to DC+bus and A-, B- and C-SCR supply power to DC-bus.
Under the control of the firing pulse added to the gate and cathode of the thyristor,
the DC voltage coming from the rectifier bridge is variable within the range of 0 to
750V.
- Location
- San Francisco Bay Area, CA, USA
- Occupation
- Electrical Engineer
Not sure where the 1.35 factor comes from, maybe they are thinking of the peak VDC after a forward voltage drop across the diodes?
But for practical purposes of things we deal with every day, such as VFD and DC drive bus voltages that have smoothing capacitors, the factor is Vrms x 1.41, or Vrms/0.707.
As to why your traction drives only list 750VDC, that actually is not just the rectification constant involved because they also have to deal with tolerance for line voltage fluctuations. So if they design the motor to provide peak output at 750VDC on a 600VDC system, the AC rms voltage can drop to as low as 530V and the traction drive will remain at peak performance. That allows them a -10% voltage tolerance, plus a little fudge room. If the AC rms voltage is at 600V or more, the rectifier just phases back to keep the output at 750VDC.
If on the other hand they designed the traction drive to run at the rectification voltage of 846VDC and the line voltage dropped, the drive peak power would be reduced as well.
But for practical purposes of things we deal with every day, such as VFD and DC drive bus voltages that have smoothing capacitors, the factor is Vrms x 1.41, or Vrms/0.707.
As to why your traction drives only list 750VDC, that actually is not just the rectification constant involved because they also have to deal with tolerance for line voltage fluctuations. So if they design the motor to provide peak output at 750VDC on a 600VDC system, the AC rms voltage can drop to as low as 530V and the traction drive will remain at peak performance. That allows them a -10% voltage tolerance, plus a little fudge room. If the AC rms voltage is at 600V or more, the rectifier just phases back to keep the output at 750VDC.
If on the other hand they designed the traction drive to run at the rectification voltage of 846VDC and the line voltage dropped, the drive peak power would be reduced as well.
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