5KW brushless generator needs load dump overvoltage protection

Brushless generator? You're probably hooped, since you can't force the field winding voltage negative in order to force the field current to decay faster. I would put a buck converter (or an off the shelf SMPS) downstream and live with the overvoltage.

When clamped, the voltage won't double, but the decay period might be longer.

2500joules - that looks like a 6inch long by 1inch diameter resistor ~ 27 ohms 50W. You might have got by with a smaller part constructed out of SiC (once refered to as carborundum), but they seem to be available only on special order from the present manufacturers.

Control voltage with a 400V sensing circuit and a 600V mosfet switch with hysterisis (on at 425V, off at 385V). Maybe tweak 'on' to respond to dv/dt. You'd have to examine switch losses, if it spent much time in a linear mode at turn-off. The turn-off time would have to be geared to the generator's real response time to non-double voltage output - takes some testing.

Kanthal Globar offers ceramic parts:

No relevant emi effects for a single surge event.

Fairly simple configuration - should not affect MTBF, as it doesn't normally function, but still one more power switch stressed to 60% at

380V, when off.

RL

Terry, the power resistor and MOSFET switch (with hysteresis) that R.Legg suggests sounds uncomplicated and easy to implement. But you may find several FETs wired in parallel are required to handle 15A. For example the classic IRF840, IR's largest 500V FET in the old days, is rated at only 5.1A when operating hot at 100C junction temp, which is the temperature the junction would be after running a while at 5A.

Here's the calculation for that: Ron = 1.75 * 0.85-ohm max = 1.5 ohms (fig 4 in datasheet), so P = I^2 R = 38.7W for 5.1A. With R_theta_JC = 1.5C/W (w/ insulator) we calculate a 58C increase over the FET's case temperature. Say the heat sink contact-point rises to 70C over a 22C ambient, then Tj rises to its 150C limit. This would be a worst-case equilibrium result for an indefinite-duration situation.

If the heat sink can be counted on to stay cooler for a short time, say a 38C rise, then you're allowed a 90C junction-to-case gradient, which works out to 60W and 6.3 amps max. So sorry, at least three parallel IRF840 FETs (four would be better) would be required for a safe 500V holdoff and 15A long-duration switching.

There are modern high-voltage FETs that can handle more current (but they're often hard to get). For example, IRF's IRFPS40N60K is a 600V FET that can handle 15A continuously with a safety margin, if properly mounted. Allied Electronics has 39 in stock, selling for $12.60 each. By contrast the IRF840 is widely available at low prices. Allied has them at only $1.33 each. Digi-Key's website says they have 38543 !! in stock at $2.05 each. Newark has the IRF840A, an advanced version of the '840 requiring less gate charge, for $1.56 each.

Thanks, - Win

whill_at_picovolt-dot-com

There seem to be innumerable switching techniques that can be used. One way would be to trigger a buck regulator on OVP, and the regulator switch is normally bypassed by a polyfuse which is forced open by an OVP crowbar. Some variation on this anyway- : Please view in a fixed-width font such as Courier.

(+)o----------+--polyfuse---+ HYSTERETIC VOLT | | CONTROL | | | +----+ +--+---+ | 375---------------- |OVP | switch | | | | / \ / \ | | -------> | IGBT | | | / \ / \ |CTL | | | | | / \ / \ +----+ +--+---+ | 340------------------- | | +-|>|---+-----| Vout | | | | === | | | | | | | (-)o-+-------+-------------+------->

Acckk!! A 5kw regulator.

That's a more simple 5kW regulator than most. But the IGBT will get hot! And it'll stay hot as long as the gnerator is in use, whereas the switched-resistor and FET will be cool.

BTW, thanks for the insight; an IGBT would be much more suitable than a power MOSFET to switch the resistor, because a single low-cost part can do the job nicely. For example, an IRG4PC30F is a 600V IGBT that likely dissipates under 25 watts at 15A, and costs $3.08 from Newark.

Thanks, - Win

whill_at_picovolt-dot-com

Terry,

I'm wondering where energy is coming from to send output up to 2X. Could solution be simple as increasing your LC filter capacitor across output of generator to snub energy stored in winding reactance or filter inductance? If you double C and 1/2 L, what happens?

regards, Bob

It appears the generator output sags under its 15A max load, and soars when the load is removed. Can the increase be due simply to an inductance flyback?

Terry, does the engine speed slow automatically after the load is removed, to bring the output voltage back down to spec.?

Thanks, - Win

whill_at_picovolt-dot-com

excitation

Is this a design where the field you are controlling is the field of a pilot exciter? You may have found that without some way to control the field directly on the rotating portion the decay is slow. This is why some large generators have SCR controls on the shaft - particularly on water turbine generators where the time constant can be longer. Not really practical for a generator of this size.

Shouldn't be problem in a rectified output application, but yeah, you should be able to see the number of poles, and if you look close you may even be able to see the ripple from the stator slots.

generator

recovery

typically

electronics

You may want to look at using a SCR rectifier bridge instead of a diode bridge to produce the 340V DC. For normal operation the SCRs can be gated full on. Under a load dump the SCRs can be turned off. I would think that 1200V SCRs that could hold off the voltage rise are much cheaper than any sort of clamp circuit. You could use a 1/2 controlled rectifier module (3 SCRs for the positive side, and 3 diodes on the negative side) for the rectification to hold the price down. For an example, look at

L/C filter should be able to limit the dV/dt rate to a value slow enough that the voltage rise won't be a problem for the 1 cycle response time of the SCR bridge. You could use 3 zero crossing SCR optocouplers to turn on the SCRs if the voltage is OK, and turn off the opto's if the voltage climbs too high; I would think that a simple hysteresis control would work to limit the voltage rise until the regulator could regain control.

One thing to worry about is the startup of the regulator. Many regulators use the residual magnetism of the exciter to provide the power to start up the regulator and build up the output voltage. If you intend to use this method, you will need to make the regulator operate at low output voltages, often less than 5% of normal. Also be aware that if the generator isn't used for awhile, it may not pick up at; requiring "flashing" the field. (Briefly applying an external voltage to the exciter field). The other option is to power the regulator off of the engine starting circuit.

In article , positivebalance41m wrote: [snip]

That L/C filter might also cause trouble, because at No-Load it is a peak detector, and there is no means of reducing the voltage on the C.

Have you experimented with other load reductions? ie, Going from 100% Load to X% Load (in closed-loop or switching down the primary exciting current appropriately), where X is down around 10 or 20%. Is the transient Over-V more acceptable for these steps?

Doing this experiment could point you towards the value of the transient dummy load that is required to keep the voltage across the C within spec.

Note that, as soon as you put a C on the output, you also have to do a fast discharger of that C anyway, for safety reasons.

Have sent email confirm several solutions easy

Vdc from a 3-phase bridge rectifier is roughly Vac(L-L)*1.35. So 340Vdc requires a nominal 252Vac under normal operation.

A single half-cycle that charges that C to 650Vpk would be equivalent to 0.707*650, or about 460Vac L-L. [see note]

Silly Question Time.... Terry, exactly what is the No-Load terminal voltage of that alternator under the equivalent of Full-Load excitation? Is it really as high as 460Vac?

Note: Sinewave sums are only approximations when applied to small alternators, because the crest-factor changes with loading. The waveshape is often 'peaky' at No-Load, with the tips flattening on load, especially rectifier loads.