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Brushless Exciter and Field Current

Brushless Exciter and Field Current

Brushless Exciter and Field Current

(OP)
I have a gap in my knowledge about the interaction of brushless exciters and generator field.  I have tried to find some text on line and elsewhere to show me the theory about how field current is generated when an exciter is brushless.  

Now I know that the AVR field current is applied to the exciter stator and, due to the rotation of the exciter rotor,  transformer action results in an AC current being induced on the exciter field. This in turn is rectifed by rotating diodes and sent to the generator rotor.

But.

The actual exciter current AND voltage is much lower than the generator field current and voltage so this extra power comes from the prime mover.

Can somebody explain the physics of how this is happening?

Primarily,

How does increasing the exciter current by say 10% result in  a generator field current increase of 10%?  Lets say the exciter current is 20A and the field current 1500A. If I increase the exciter current the field current must increase also.  This implies a change in load angle of the generator.  

Why does such a small increase in exciter current result in an increase of shaft power?  What is the interaction between the increase of exciter current and field current?

Hope you generator gurus can help.

 

RE: Brushless Exciter and Field Current

Quote:

The actual exciter current AND voltage is much lower than the generator field current and voltage so this extra power comes from the prime mover.

The MAIN field current is equal to the generator field current.  But the field current you are talking about is the PILOT field current.  It is applied to the exciter field.

The output of the brushless exciter never leaves the rotor and is generally not even directly measured.  The brushless exciter is a second smaller generator mounted on the same shaft as the main generator.   

RE: Brushless Exciter and Field Current

In the beginning there were exciters with brushes. The old brushed machines had a small DC generator on the back end of the shaft. The AVR supplied the field current to the DC exciter. This had a commutator and brushes to deliver the main, DC current. This current then was directed to another set of brushes riding on slip rings to power the main field.
A small current in the stationary field of the exciter would develop a greater current and more power to supply the main field. You are correct that the power to generate the excitation current is part of the mechanical load on the prime mover.
Then came the brushless exciter. This is so similar to the original exciter that now old machines are often converted to brushless by modification of the original DC exciter.
The exciter is now a small alternator. A disk called the diode plate is added to the shaft. It has a set of diodes to rectify the output of the AC exciter. This rectified output is fed directly to the main field.
As for the similarities between brushless exciters and brushed exciters, remember that the main difference between the rotor of an alternator with a stationary field and a DC generator is he manner of extracting the energy produced. Brushes and slip rings are used in an alternator and brushes and a commutator are used for a DC machine. If you tap two opposite commutator segments and rotate the armature the taps will deliver AC. Normally slip rings are used to extract the AC when the AC is being developed in the rotating member. In the brushless exciter, the AC is being rectified by the diodes on the revolving diode plate and the resulting DC fed to the rotating main field, no brushes, commutators or slip rings are needed.
On a practical scale, winding shops that specialize in generator work will rewind the rotating armature of the DC exciter for AC and remove the brushes, commutator and slip rings. A diode plate, which may be a aluminum plate, and diodes are installed.
If the original exciter is mounted outboard, the shaft will be drilled to allow the field conductors to pass the bearing. The high inductance of the field winding provides all the filtering that is needed. I understand that three commutator segments may be tapped and rectified and will give adequate performance in many applications but I have not seen this done.
A number of machines that i serviced and repaired were sent out and modified from brushed to brushless. The original AVR was kept in service. I understand that the exciter field was left untouched, but am not certain. These were in the 40 KVA to 60 KVA range.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Brushless Exciter and Field Current

A brushless ac exciter works the same way as the main generator. The only difference is that while the main generator has a rotating field and stationary armature, the brushless exciter has a stationary field and a rotating armature (with a rotating rectifier to convert AC to DC).

In both cases, the excitation power is only a fraction of the individual generator capacity. Ultimately, almost the entire excitation power comes from the prime mover.

Muthu
www.edison.co.in

RE: Brushless Exciter and Field Current

Refer to the attached link:

http://kato-eng.com/pmg.PDF

The PMG is on the same shaft as the main generator that is a separate source for AVR. AVR converts that to a small DC that feeds the field of the exciter that produces larger AC source that gets convered to DC main field via rotor mounted rectifier.

  

Rafiq Bulsara
http://www.srengineersct.com

RE: Brushless Exciter and Field Current

Pilot exciter, PMG, brushless exciter.
Two diferent animals.
The AVR needs power. It may get it's power from the main generator output, or a dedicated Permanent Magnet Generator may be used to supply the AVR. The PMG has rotating permanent magnets and the output is taken from a stator winding straight to the AVR. The PMG is often outboard on the end of the shaft but may be installed inboard. A PMG is a factory installed option.
The brushless exciter is part of the machine. It is most often installed inboard so as to leave space outboard for the possible addition of a PMG. Older machines particularly those that have been converted from brushed to brushless may have the exciter outboard in which case a PMG is not possible.
I am not sure what a pilot exciter is, possibly the PMG, or possibly the first stage of a multi stage excitation system on a very large machine. Scotty, are you familiar with the term?

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Brushless Exciter and Field Current

(OP)
Hi Guys thanks for the response but I'm not quite getting the answer I need.  

To DPC - I know I am refering to the pilot field current.  I am calling this the "exciter current" as opposed to the generator field current.

My question is what is the physics behind the fact that if I increase the "Pilot Current" the main field current increases?  This increase in power is provided by the prime mover but how? (don't say by adding more steam please)  What about increasing the pilot field current suddenly requires the generator load angle to change?

So it must go like this:

Increase in pilot current => decrease in load angle=> Increase in mechanical input (to maintain original load setpoint) => increase in load angle (i.e more torque)=> more generator field current.

Quote - Edison
"A brushless ac exciter works the same way as the main generator"

OK but a generator increases it's power output by an increase in mechanical input.  A brushless exciter increases the field current from power elsewhere.  Clearly 15A and say 100V on the exciter can't result in 1500A and 1000V on the generator field without some other power source.

Thanks for your patience guys but it's still not clear to me how an adjustment of current to the exciter results in an increase of generator field current.


 

RE: Brushless Exciter and Field Current

You seemed to have wrong the notion that excitation power should be equal to the main power of the alternator (whether brushless or the main generator). The excitation establishes the rotating flux, which in turn induces the voltage in the armature.  

As the load is applied to the main generator stator, the armature tends to demagnetize due to armature reaction, which would tend to reduce the generator voltage. The AVR senses this and produces more output to the ac exciter field, which turn makes the ac exciter armature to generate more amps to feed to the main generator rotor to counteract the demagnetization due to armature reaction.

Ultimately all this power has to come from the prime mover.

Muthu
www.edison.co.in

RE: Brushless Exciter and Field Current

A generator is a type of magnetic amplifier. before the development of electronic drives DC generators direct coupled to DC motors were used for variable speed drives. The speed of a DC motor could be controlled from zero to 100% by varying the voltage applied to the generator field. A refinement was the amplidyne drive. The amplidyne is a special DC generator with a very high gain. An input signal of a few Watts would produce an output signal of a few thousand Watts. When this output was applied to a generator field it could control a motor of about 30,000 watts.

Remember your basic theory. EMF depends on the strength of the magnetic field and on the effective speed of the conductor through the field.

When a load is placed on an alternator the current increases. The current causes an internal voltage drop. This voltage drop subtracts from the EMF and the terminal voltage drops. When the AVR senses this drop in voltage it increases the voltage to the field of the exciter. The increased voltage causes an increase in current and field strength. This causes an increase in the output voltage and a corresponding increase in the current through the main field and an increase in strength of the main field that will raise the EMF enough to compensate for the internal voltage drop of the main generator.

Load angle and speed is a separate topic. It is dependent on the governor action of the prime mover.
When the load on a generator set is increased, the governor will act to maintain the speed of the set. Losses and field energy are a small part of the overall load that the governor sees when the external load is increased.
That is, if the external load is increased by 90 kW, the load on the prime mover will be increased by 90 kW plus the losses. The increase in load on the prime mover for a 90 kW increase in electrical load will be closer to 100 kW (assuming 90% efficiency).
 

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Brushless Exciter and Field Current

(OP)
Thanks guys,

But lets try it one more time.  Warcross and Edison, thanks for your patience but you both are explaning WHAT happens.  This I know.  I've been working in power plants around the world for 20+ years.  What I don't know (for sure) is HOW it happens

Quote:

"When the AVR senses this drop in voltage it increases the voltage to the field of the exciter. The increased voltage causes an increase in current and field strength. This causes an increase in the output voltage and a corresponding increase in the current through the main field and an increase in strength of the main field that will raise the EMF enough to compensate for the internal voltage drop of the main generator"

This I understand no problem, but how does increasing the exciter (pilot) current by say 10% result in the main field current increasing by 10%.  

Clearly the power fed to the exciter is much less than the power on the field.  When the exciter current is increased The rotor is still turning at 3600RPM but the field current is increased at a multiplier of the exciter current.  The question is how?  It comes from the prime mover ok, but how does the increase in field current happen? If I have 15A exciter current and 100V exciter voltage HOW is this transformed to 1500A and 1000V on the field?

Edison.

I don't, as you state,  have the notion that the exciter power should equal the main field power.  I have given examples of exciter current 15A and field current 1500A.  I know by experience what happens,  just not how.

So I am asking for your patience and please to not get into what happens.  I know increase in exciter current=increase in field current.  If somebody can explain how or point me to a link or text I would appreciate it.

RE: Brushless Exciter and Field Current

The pilot exciter is still an AC generator,( in case of the Kato alternaor posted ) The AC is then converted to DC via rectifiers. Are you asking how AC generator works? Or DC, if it were to be DC generator. It could be a DC generator in theory, but rectification is much simpler.

PMG is a separate AC generator that provides source for the field of the exciter.

In a PMG machine with brush less exciter there are three generators. The PMG, the pilot exciter and the main.

Without the PMG you still need to find a source for the exciter's field, mounted on stator.



Rafiq Bulsara
http://www.srengineersct.com

RE: Brushless Exciter and Field Current

How does the main generator produce much more current than the field?  The mechanical energy of the prime mover is converted to electrical energy by driving the magnetic field of the rotor across the armature.  The exciter is the same, except the prime mover drives (turns) the armature through the stationary magnetic field.  In both cases, an increase in power applied strengthens the magnetic field, which requires more power to drive it at the same speed, while tending to increase the voltage.  Since the impedance of the main feild is the same, the current goes up.

RE: Brushless Exciter and Field Current

The pilot field produces a set of fixed magnetic poles on the stator.

The prime mover turns the rotor which moves a set of winding past these magnetic poles. The speed the windings move cause the windings to produce a voltage. The windings have a certain impedance. Add a fixed load and you get an output current.

Now, increase the pilot field current 10%. The magnetic field will increase 10%. The voltage the windings produces will increase 10%. The output current will increase 10%.

RE: Brushless Exciter and Field Current

Let's not get into the PMG. A PMG is an option that improves performance but is not needed for basic generation.  
An AVR needs a source of power. This AVR supply power is commonly supplied from the main output of the generator, possibly through a transformer. The addition of a PMG and an AVR designed for PMG supply improves motor starting and short circuit performance. If either the PMG or the PMG AVR fails, the set may be operated with a conventional AVR until parts arrive.
When the AVR senses a drop in voltage it increases the output voltage to the exciter field. The increased voltage results in increased current. This causes an increase in the output voltage of the exciter. This increased voltage causes an increase in current in the main field. The increased magnetic field causes an increase in the EMF of the main generator.
A basic generator is a conductor moving in a magnetic field. When the strength of the field is increased the EMF is increased. At each step the EMF is increased by increasing the strength of the magnetic field. The actual relation ship between the field current and the strength of the field is a design consideration. For instance the current required to produce a given field may be reduced at the design stage by using a winding with a lower I2R, reducing the air gap, using a better grade of iron. The strength of field required for a given EMF may be reduced by using a larger radius or increasing the speed. (1800 RPM vs 3600 RPM).
In the normal operating range, the current that a generator produces is dependent on the impedance of the load.
EMF is divided between the load and the internal losses in the generator. As the load increases the internal voltage drop increases (I2R, I2Z). This subtracts from the EMF and drops the load voltage or the terminal voltage. The field strength is increased to increase the EMF so that the terminal voltage is maintained.
In an automobile, the terminal voltage of the charging device whether a generator or an alternator has always been controlled by varying the field current/strength. The speed variations add an extra dimension but this also is compensated for by varying the field strength. (The three brush generator that Stienmetz designed for Henry Ford is an exception but has been gone for over half a century so let's not go there)
In a generator the exciter is either a generator or an alternator. The exciter is cascaded with the main generator. The control loop monitors the voltage of the main generator and controls it by controlling the current applied to the field of the exciter by the AVR.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Brushless Exciter and Field Current

GTstart - I suggest the book "Performance and Design of AC machines" by MG Say to understand how the ac machine works. The theory cannot be explained in a blog post, in a nutshell, as it were.

Muthu
www.edison.co.in

RE: Brushless Exciter and Field Current

GTStartup,

Hmmmm. Well I'll have a stab at it myself here – I think I know what you are getting at: how can a small increase in the value of the Field current lead to an large increase in the output current (load dependant) when, with all things being equal, you cannot get something for nothing, and this additional 'power' has to come from somewhere – right? I believe you have been a little confused in your posts when you talk about 1500A Field current though – I believe you are referring to Stator current, as the field currents are relatively small.

To get all this terminology correct let's break things down (some repeat of what has been said before):
A PMG (or maybe a PT with Field Flashing) supplies AC to a Voltage Regulator. This AVR outputs DC to the stationary Exciter Field. This excites the rotating Exciter Armature which generates 3 phase AC through the Rotating Rectifiers, to the Main Generator Field (all via the Rotor). This Generator Field then excites the Generator Armature (Stator windings) and provides all the juice.

Taking into account the values of current (1500A) that you were talking about, I have based the following on the fact you are referring to the Generator Armature current (machine output). If you are referring simply to the Exciter field and current, then you can apply the same principals but on a smaller scale.

Well.... the formula for the output voltage of the generator (EMF) is V = -NdΦ/dt
Without wanting to bring Einstein maths into it, you can see that the EMF if proportional to the Flux Density (Φ), and inversely proportional to the rate of change of speed (d/dt). The N in the formula is the number of turns in the machine and the minus sign before it is because the EMF always opposes what developed it – but this is not important for this particular discussion.

As you increase the current from the Exciter Armature (by increasing the Exciter Field current from the AVR), the strength of the Generator Rotor Field increases following the basic laws of electro-magnetism. This increase in Flux density also increases the generated voltage according the formula above.

Now, OK you might say, but it still doesn't fully account for this 'massive' increase in output power, but remember that the output current is dependant on the load, so the machine will only generate this extra power if the load is demanding it, and so, we will go from there.

The load has increased and is demanding more power, this causes the Generator speed to decrease due to Armature Reaction (demagnetization). The Engine load controller (eg Woodward 2301A on Diesel Gens, LSM module of Solar's etc) senses this decrease in speed and signals the Governor to open the fuel valve and bring the engine back up to speed, which it will do. The energy in this additional fuel that is used is basically proportional to the electrical load that has increased (please everyone – this is a very simplistic analogy I know, but hopefully suitable for this purpose) and this is where I think you are aiming at. You are not getting electrical power for 'free', but rather using more fuel, and more excitation, to generate this power.

The conversion primarily being fuel converted to electric.

Hope this helps. If not post back and we can have another crack at it :)

RE: Brushless Exciter and Field Current

(OP)
Thanks once again. I am piecing it together.

Lets clarify some things.

I'm not talking about how a generator works per se.  I undertand the relationship between load angle, torque and power.  

Corodist.

Sorry to mislead but I was actually talking about field current not stator current. 1500A is more typical of large static machines but it was just an example.  but I think you see what the crux of my question is.  

Let me make a stab at the mechanism and you can tell me where I am wrong.

1. AVR applies a small current  (lets say 20A at 100V) to the exciter.

2. This power is applied to the field using the exciter and diode wheel.

3.Now how does this small power input become larger?  It must come from the interaction of this initial current, the magnetic flux it generates and the torque of the generator. Simlar to how current is induced in the rotor of an induction motor but intead of porportional to slip it is proportional to load angle.
 


 

RE: Brushless Exciter and Field Current

I think you may have cause and effect mixed up a little.
Generator output: Terminal voltage, not EMF.
Your formula is for EMF. s thyeload increases two things happen, one electrical and one mechanical. Forget armature reaction,let's keep it simple. You have more load on the set and need more fuel to maintain your speed. The governor takes care of this. Purely mechanical.
The second effect is internal voltage drop. The greater the current the greater will be the internal voltage drop. The internal voltage drop is supplied by the EMF so it subtracts from the EMF ans results in a lower terminal voltage. The voltage regulator sees the drop in terminal voltage and increases the field strength of the exciter. This results in a higher EMF generated by the exciter. The higher EMF results in greater current flowing in the main field and increased magnetic flux which in turn raises the EMF to compensate for the internal voltage drop and maintain the terminal voltage.

This is starting to remind me of a fellow asking how a generator got its magnetism when it started. After a few hours of telling him things he already knew it turned out that the answer he was looking for was:
"When the set is first built or when it loses its residual magnetism you re-establish the residual magnetism by connecting a car or truck battery to the exciter field for a few seconds."
I seems that we are giving the right answers to the wrong questions.
BTW I neglected armature reaction to keep it simple, but if armature reaction reduces the EMF, the terminal voltage will drop proportionally. The AVR will see this and will increase the excitation to maintain the terminal voltage.
The voltage regulator is monitoring and controlling the terminal voltage, not the EMF.
 

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Brushless Exciter and Field Current

(OP)
Thanks Warcross,

You are probably right. The problem is not the answers but the questions. It's still not clear how 20A and 100V to the exciter can result in 10 times those values on the field.  It seems to defy laws of conservation of energy. (However I know this "exta" power comes from the prime mover I know how it happends on the stator, just not the field.)

 Your statement

"The voltage regulator sees the drop in terminal voltage and increases the field strength of the exciter. This results in a higher EMF generated by the exciter. "

Is fair enough, but the exciter power is not equal to the field power and this difference in power comes from the turbine and somehow is converted to electical power on the field.  So there is my dilemna

Nevertheless I think the failure to resolve this is mine and I need to do some studying.  Perhaps by finding the answer I will know how I should have asked the question.



 
Thanks for all your help

RE: Brushless Exciter and Field Current

Ahh let's start over.
Forget the complications of an exciter cascaded onto an alternator, lets just look a simple permanent magnet alternator.
Magnets are spinning in such a way as to induce an EMF in a winding. Once a load is connected, the current in the windings creates a magnetic field that opposes the field of the magnets. The heavier the load the greater the current and the greater the field opposing the rotation of the magnets. Further, this opposing field serves to buck or reduce the effective field of the permanent magnets.
A reduced field results in reduced EMF.
Solution, use an electromagnetic field. The power that the field uses is not related to the power output of the generator. We may increase the strength of the magnetic field to compensate for armature reaction and internal voltage drop.
Stronger magnetic field equals greater EMF. More field current equals a stronger magnetic field. (Did I say that backwards?)

 

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Brushless Exciter and Field Current

In the most basic sense, the DC field current input just creates a fixed magnetic field. That is all it does. If you applied the field current to a stopped machine you would create the magnet field but you would not get any output on the AC leads.

The output power comes from the prime mover moving the rotor windings past the fixed magnetic field or from the prime mover moving the magnetic field past the fixed stator windings. The alternator can be build with the AC coils moving or the DC field moving. In the basic sense, a coil of wire is passed through a magnetic field which induces a voltage in the coil.

If you try to resist this voltage by connecting a load to the coil, then the coil will resist passing through the field or be harder to push through the magnetic field. This resistance to passing through the magnetic field is how the prime mover gets loaded. The prime mover creates or outputs enough power to force the coil to travel through the magnetic field and produce the voltage even when there is a load attached.

So, to address how the output current can increase much more than the field current again.
10% more field current = 10% stronger magnetic field. Now, move the coild past the 10% stronger field at the same speed and you get 10% more output voltage. Connect this 10% higher output voltage to a fixed load and you get 10% more output current. It's not always the same percentage all the way though the system in the real world but let's stick to simple theory for now.

It is the prime mover forcing the coils through the magnetic field that creates the output power so the output power comes from the prime mover. Same exact theory applies to the rotating exciter as to the main alternator. A large machine just has 2 stages, the pilot field current increasing 10% increases the rotating exciter output 10% which forces 10% more current to the main altnerator field which then causes the main alternator to output 10% more voltage which then forces 10% more power to flow into the load.

If you still don't understand this then you really do need a textbook as already suggested. This thread really can't go much further.

Guys, quit trying to discuss how regulators work to maintain the voltage. The basic question is - how does the alternator use the small pilot field current to produce a larger output power?
 

RE: Brushless Exciter and Field Current

I just went back and read your posts again. I think you are missing the fact that the "exciter" is not a transformer. It is a generator and works exactly like the main unit but just on a much smaller scale. It is also built opposite of the main generator - the stator has the field windings and the rotor has the AC windings.

It uses the pilot field and a little prime mover power to generate a larger output which is recitifed and fed to the main field. The main generator then uses this field and the rest of the prime mover power to generate an even larger output which is fed to the AC power grid.

RE: Brushless Exciter and Field Current

(OP)
Ok as you say, lets start over.

Let me ask just this simple question. It might help to focus my point.

How is exciter output power more than its input power?

Ignore any influence of the AVR due to change in load, voltage etc assume it's at a steady state

For definition of terms

Exciter input = VA DC from AVR (Assume it's constant in this case)

Exciter output = VA DC on the field after the diodes
 

RE: Brushless Exciter and Field Current

(OP)
Ahah,

I posted before reading Lionels post.

Now I get it.  Thanks for pointing out where I was going wrong.  The exciter is a generator, driven by the prime mover since it's on the same shaft.  The power from the AVR is the "field current" for the exciter, the output of the exciter is the stator output to a fixed (more or less) resistance of the main field.

I am sure that it's been stated more than one above, I just didn't get it.  So Warcross et al, ignore my last posts.  The penny has dropped.

Thanks!!

RE: Brushless Exciter and Field Current

Just as an FYI, there are synchronous motors built that basically have a rotating 3-phase transformer - well the primary is fixed and the secondary is rotating. The output is fed to rotating rectifiers which feed the motor field. And these don't generate more output power than the input power.

They are built so that the field will remain constant even if the speed of the motor changes so the motor can be used with a VFD and still allow brushless operation. If the field current is constant then the ouput torque of the motor can be constant at all speeds.

The exciter as discussed above would lower the field current as the motor slowed down so it is not nearly as suitable for that application - on a variable torque load it might work.

I guess the same system could be used on a generator and it would be nothing more than straight transformer action but I can think of no advantage to using it in that application.
 

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