5KW brushless 3PH generator output voltage regulator and load-dump over-voltage protection required

I am tasked with the job to design a regulator circuit for a brushless 5KW

3PH AC burshless generator that gets it's output rectified to produce a 340VDC power source.

This is my first expereince with a generator.

Coming from a background in linear and switchmode power supplies, the first thing I notice is how slow a generator responds to changes in excitation coil current; and how awfully large the output ripple with a rich splash of harmonics relating to the number of poles in the output and exciter windings.

The big issue I am facing is load dump transient response. The generator takes several hundreds of milliseconds to reduce it's output voltage after a

15 amp load is removed from the 340VDC rectified output.

Odd I find because the excitation coil is shut off by the regulator circuit alomst instantly and I can see it decays to nothing in just under 100ms.

I notice the generator output can handle going from no load to full 15 amp load in less than 100ms.

I am told the rotating excitation coil inside has some diodes on it, and that this is a typical construction of a brushless generator.

Why is the load dump response time so slow, and what can be done to speed it up? What type of output over-voltage protection schemes are typically used in a generator based DC power supply? I can't have downstream electronics modules connecting to my output getting blown-up! Yet to clamp or limit or otherwise regulate or filter 5KW of power for almost 1/2 second long transients where the outptu voltage can double, seems like an expensive prospect?

If anyone knows a better news group or this type of question please do suggest. I appreciate this is borderline power electrical and not entirely electronics. However, I'm looking for answers that are more creative than just to to use a large series filter inductor, shunt capacitor bank, and over-voltage cut-out relay. My application is cost and size and weight sensitive.

Suggestions anyone?

snipped-for-privacy@direct.ca

Reply to
positivebalance41m
Loading thread data ...

The problem you have is due to the effects of inductive time constants in the circuit. There is energy stored in the magnetic field, and the machine will continue to generate until the field has decayed.

Look for a figure called the synchronous reactance for the generator - this is the effective Thevenin impedance in series with the internal voltage source. It is usually relatively high, and you will find that if the output current is cut off, the terminal voltage of the generator will rise rapidly - maybe by 100% -until the voltage regulator can cut the field current. You then have to discharge the field in some way to get the voltage down.

With a brushless machine, you have got what amounts to two machines to control. You have access to the field windings of a small AC generator that develops an ac voltage ono the shaft of the machine. This is rectified by rotating diodes, and the current is then fed to a second winding providing the main field. A change in the current through the controlled winding can remove the exciting field, and can be arranged to extract energy from this very rapidly, but you have got no access to the main field, and the current in this (and therefore the terminal voltage) will decay at a rate set by the L/R ratio of the circuit. There is not much you can do to speed up this response.

Bruce.

positivebalance41m wrote:

Reply to
Bruce Durdle

The machine has to dissipate a large amount of stored energy before its output goes down. There are ways of making this happen more rapidly but they are rather rough. Kind of like sticking a stick into the spokes of your bicycle in order to slow it down. Capacitance is the cure for inductance just as a spring is used to cure the thump of inertia.

You said it took a long time to reduce the voltage after removing a 15 Amp load. I think what you meant that the voltage spiked and took a long time to recover to normal. With inductance you essentially have a constant current supply. Voltage rises to whatever is needed to force the prior current through whatever load is left. When you disconnect the last load the spike can be very large indeed. This spike lasts until the residual energy is dissipated. A big capacitor will absorb the spike. The solution in my industry is to have the DC supply charging a bank of batteries. (This is called an uninterruptible power supply (UPS).) In this way you have the generator regulated to keep the batteries on charge while the batteries supply the load through an electronic regulator. No, its not cheap.

Walter.

electronics

Reply to
Walter Driedger

This would be typical for a small generator. Large ones could take 10 seconds or more to decay.

As mentioned by someone else, this only reduces the output of the brushless exciter. The main field winding, internal to the rotor, takes a lot longer to decay.

The brushless exciter can raise the main field voltage to several times its rated value very quickly, but it can't reduce it to less than zero in a normal generator. If you took off the exciter and put on slip rings, driving the main field directly, you could force the voltage negative and get much faster response. Some large exciters have controllable devices on the shaft to get faster response, but this will be too complex for a small machine.

Correct; they rectify the output of the exciter.

The voltage rise on load dump will be about the same magnitude regardless of what you do to control it, although you can make the duration shorter. You could apply an inductive load to the output momentarily on detecting high output voltage -- a reactor and some SCR or triac switches would do the job without dissipating any major amount of power. Come to that, even just the triacs might do the job if the output reactance is high enough (short the output for some part of a cycle).

Alternatively, could you use a controlled rectifier to limit the voltage rise on the DC bus? High voltage SCR's are fairly inexpensive.

The simplest and most common way to solve this problem is to use a generator with lower output impedance (i.e. a larger one, usually).

Hope this is helpful.

Regards, Allen

Reply to
Allen Windhorn

I suggest you buy one of the many excellent choices from Canada or the US. This will be far more cost effective than designing one from scratch.

TZ

electronics

Reply to
Zman

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.