AC Generator AVR

If it works fine then no need to ask questions,be happy with what you have.

Well there are a whole lot of things that it could be that explains why the manufacturer wants the exciter field windings to have no less than 10 Ω resistance. They all hinge on the fact that some part of the system will no longer will act linearly if the resistance is less. I suspect that the magnetics can saturate if the resistance is less than 10Ω. But a few of the other possibilities could be an over heating problem of the windings, the exciter control circuitry might be damaged, the carbon brush contacts will self heat to the point of ignition, the engineers wished to annoy people by not explaining generator design to the customer.

GA from me.

The AVR output circuitry must have an internal resistance of ~10Ω therefore, if the external resistance (here the exciter stator) is less than that it will make the AVR to dissipate more power internally that it is giving out. In this case, 1.25A^2 x 4Ω = 6.25 W given out while dissipating internally 1.25A^2 x 10Ω = 15.63W. For a full 4A, 16x10 = 160W inside against 64W in the exciter.

Not correct.

You are assuming that inside resistance is 10 ohms.

Inside voltage should be 60V or more when you have open circuit for it is deliver 6A into 10 ohms load. Power 60VX60V/10 Ohms = 360W.

Something is going wrong inside as it delivers only 1.2AX1.2AX4 Ohms= 5.76W

What is happening to the power driving capability of the device?

If it is a current source then 6A must be delivered and if it is a voltage source then 60V must appear.

Perhaps it tried to deliver 60V and then power becomes 60VX60V/4 Ohms = 900W.

Perhaps its overcurrent sensing device trips and you get some minimum 1.2A still from a nearlt dead system.

Using device improperly is not engineering and it is called :JUGAD" in India, which means somehow using things to get some results for some time. Things appear working for a while and sometime better than Original.

"Not correct" ??

First of all, please think about it before making such a statement:

1- AVR = automatic Voltage regulator: It regulates the Output Voltage of the main alternator by CONTROLLING (Varying) THE Current in the Field Coils. The current is here the subject. Therefore, The AVR is a Current Source: The Output Voltage of the AVR is Varied in such a way to Increase or reduce the current into the exciter field which has a fixed resistance of 4 Ω here.

2- Therefore, the Voltage output of the AVR is not fixed to 60 V DC. It Cannot be otherwise how do you regulate ...??

3- The Exciter Output Voltage will Vary and produces a varying current into the main field. Which brings you back to the fact that the AVR is basically a current source.

4- Now: Since the exciter in this discussion has a resistance = 4 Ω , and the nameplate says that the full load exciter field current = 4.25 A, the required voltage from the AVR will be 17VDC (4.25 A x 4 Ω), and the nameplate reflect this(!).

The New AVR is rated at being able to supply (SINK) 6 A and needs to be connected to a minimum resistance of 10 Ω. There is no mention of the output voltage from this AVR. This is because you can calculate it to be : 6 A x 10 Ω = 60 VDC . But this does not meen that the voltage is fixed!! It can even be higher if the AVR is driven to sink more than 6A into the 10 Ohms it is rated for.

The Problem is this: The AVR Output stage has an internal resistance as seen from the outside. This resistance determines the internal losses that will be dissipated when operating. The 10 Ω value means that if the Load resistance is less, the internal losses will become higher than the power supplied to the load at whatever level, from no load situation (minimum excitation current) to the full load situation (max excitation current). This ia the main reason why this value is given by the manufacturer. Usually, this is the point where the internal loss = the external load power used. { It could be less if the manufacturer chose to so that the ratio could be less internal loss for a load: 50/50, or 40/60...}

Therefore, Hooking the AVR to a 4 ohm exciter stator will make it work less efficiently since it will spend more energy internally than it is delivering to the load. If at full load, when the exciter field will be taking 4.25A, the power delivered is (4.25)² x 4 Ω = 72.25 W, the internal dissipation will be (4.25)² x 10 Ω =180.625 W.

Now you say "Who told you it is 10 Ω?". The manufacturer said "Not Less than 10 Ω ...". Therefore, whether it is 10 or not is irrelevant at this point for you to argue about it. Suffice it to say that the manufacturer is warning you that the efficiency between dissipated internal power and the delivered power will be less than the designed value. This might lead to a hotter AVR if you reached the 6A: Load will be 6A*6A*4 Ω = 144 W into the exciter and possibly 6A*6A*10 Ω = 360 W being dissipated internally. If not that much, then surely more than the designed value! { suppose internal R = 4 Ohms, then 144W is max at 6A when giving 6A*6A*10 Ω = 360W to load, and eff.=71.4% by design BUT here it will be 50% Only}.

Conclusion: Whatever is the AVR internal R at the Output, by design, the Manufacturer is telling you that if you hooked it to an R < 10 Ω, you loose on efficiency and might exceed the dissipation rating of the AVR. ( No warranty....).

I hope that I was clear enough to dissipate the "Not Correct" Statement...

I think you have done good amount of analysis and you know in details now.

Voltage regulators are often designed with very low internal resistance except that they may trip on over current or over heating.

When warning is for not less than 10 ohms, that clearly means that short circuit protection does not exists or something may get into irreversible change.

For a variable voltage regulator in overload condition and at high preset voltages problems will be worst than what one will see at lower preset voltages. Sometime regulators are meant for high forced currents even at lower voltages down to zero volt as in the case of capacitor changing AVR, then there is full force current at all voltages. However, it all depends on high this device is designed.

I think, warning means real warning, so never one should ignore. AVR without short circuit protection is unimaginable. However there were crow bar like protection that can make some fuse to blow permanently for safety or some PTC or NTC thermistor or resettable or non-resettable fuse to come into action.

Have a look at the circuit manual and perhaps more details may be there to figure out. I am only telling how these circuits are designed and not beyond.