zener clamping - Page 1
Forum section: electronics / projects

author: thefamilyman
author_rank: Regular Contributor
timestamp: February 16, 2016, 12:02:04 am
content: Good evening gentlemenA client has been having problems with over voltage on their DC motor controller circuit and wondering how's the best way to implement zener clamping.a bit of an overview on their basic system setup:high powered DC motor ~35kWMotor controller nominal voltage of ~470V DCpower is supplied by a 60KVA diesel generator running at 330V AC at 60Hz, cable run of about 70m of 25mm2 copper.AC voltage is then fed into a DC rectification box; inside is a 150A 3 phase bridge rectifier with 2* 350V 2200uF capacitors.the problem they have is that when the DC motor is turned off via the motor controller the DC voltage shoots up to over 550VDC which is dangerously close to the absolute maximum voltage of the controller.the AC side of the rectifier does not change by much between pre-load, on load and directly after load. The diesel generator only overshoots for about half a second while the AVR counters the load, this could cause the DC to charge the capacitors to a dangerously high voltage of over 550VDC.Can a high power zener diode(s) can be used to clamp the sudden surge to say a safer voltage of 530VDC while the generator stabilizes?If so, what precautions of design considerations should be taken into account?I hope this makes sense.Many thanksPatrickthe client is wanting to

author: langwadt
author_rank: Super Contributor
timestamp: February 16, 2016, 12:50:38 am
content: Quote from: blueskull on February 16, 2016, 12:19:13 amthat would most certainly explode, you never want to run an IGBT in linear mode

author: thefamilyman
author_rank: Regular Contributor
timestamp: February 16, 2016, 12:53:01 am
content: Quote from: blueskull on February 16, 2016, 12:19:13 amthanks for the reply, i was also thinking of some sort of transistor switching like this too.just a question, the 550VDC would only be the power stored in the caps as they DC motor is off and no current drawn through the motor controller, thus the power needed to be dumped by the zener is not 35kW right?during the motor operation the DC and AC voltage is stable and a safe operating value, only until the motor is instantaneously stopped the DC voltage suddenly shoots up (due to the generator AVR not reacting fast enough and perhaps the inductance in the long cable run) and is stored in the large capacitors.

author: langwadt
author_rank: Super Contributor
timestamp: February 16, 2016, 01:19:06 am
content: Quote from: blueskull on February 16, 2016, 01:11:36 amgate voltage follows the supply so it'll be linear, there is not thing to make it snap on and snap off

author: dom0
author_rank: Super Contributor
timestamp: February 16, 2016, 01:27:55 am
content: Honestly when I read about >xx kW DC motors I already know the whole concept is probably garbage. Doctoring on issues of such a thing is throwing money away ; remedy the root cause and be done.

author: langwadt
author_rank: Super Contributor
timestamp: February 16, 2016, 01:30:40 am
content: Quote from: thefamilyman on February 16, 2016, 12:53:01 amif the generator is generating 35kW and you suddenly stop using that I think the clamp would havehandle 35kW at least initially as the generator ramps down.what about the energy in the motor?

author: uncle_bob
author_rank: Unknown
timestamp: February 16, 2016, 01:39:37 am
content: Quote from: thefamilyman on February 16, 2016, 12:02:04 amHiThe problem you are running into is commonly called a "load dump". A bit of searching on Google under that term may help give you some background on the nature of the problem and common limits concerning what is to be expected. Your situation is from a motor load, but it (if I understand correctly) is a load none the less.If all of your data is correct and I understand correctly, after the load is taken off line, your 4,400 uf of caps have 550 volts on them. You want to get it back to a more normal voltage.Your AC is at 330V, so I'm guessing you full wave rectify that. If so 466 V would be a good guess at the normal no load DC voltage.Net result is that the caps are about 90V above where they should be at no load. A magic circuit that dropped them 90V would take care of the problem. From Q=CV, you get 0.39 coulombs. That gets you 35 Joules of energy to be disposed of with a magic / perfect circuit. If you spent a second discharging things it would be 35W. If it pulled to ground (as it almost certainly must) you have about 200 joules to deal with.Ok, but .. a second?? didn't we already fry the controller? Indeed you need to know how fast the transient that charges the capacitor is. It could (and likely is) a 150A pulse for a lot less than a second. The key issue is that the energy involved is the same if it is fast or slow. The peak watts change. The watts averaged out over a second do not change.So ... an answer ... A resistor that will handle 150A (for a short while) at 460V. A 3 ohm part is about right for the job. You need a switch on the bottom end that turns full on quickly (FET or whatever) If it has a 1V drop, you will get a very brief 150W pulse of heat into it. If it is over in 50us a surprisingly small heatsink may work fine. Things like bond wires and absolute current limits are not quite as easy to get around.Yes that's only one way to do it. It makes a boat load of unsupportable assumptions. Given my track record for math errors today, it would be *very* wise to check my numbers.There are a few issues like getting the switch turned off well before you get down to 460V. To some degree that drives you to a larger resistor and less current (You need the voltage to go up in order to detect the event). Net result is that the stuff above gets a bit easier rather than a bit harder. Switching losses in the FET do count, you may need a bit bigger heatsink ...details do not always work out in your favor.Bob

author: langwadt
author_rank: Super Contributor
timestamp: February 16, 2016, 01:40:23 am
content: Quote from: blueskull on February 16, 2016, 01:27:45 amthere is feedback, when the IGBT turns on the voltage drops, lowering the gate driveso it'll settle at some linear spot

author: thefamilyman
author_rank: Regular Contributor
timestamp: February 16, 2016, 02:07:28 am
content: thanks everyone for the replies.I'm not an electronic engineer, so the concepts i understand just not the finer details so that's why i'm asking you guys for expert help.I'll get the voltage and current curves from the client this afternoon and post them to show a bit more information.I came across an IGBT from IR that's raded for DC switching (marked as a relay replacement) AUIRGDC0250, would this help?

author: langwadt
author_rank: Super Contributor
timestamp: February 16, 2016, 02:19:58 am
content: Quote from: blueskull on February 16, 2016, 02:01:49 amall the that zener will do is keep you from blowing up the gate assuming it has the right voltage

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 16, 2016, 02:32:08 am
content: Surely this is more fundamental: a control problem? Why does the motor have to shut off suddenly? Why would you even want it to? (This situation being a prime example of why you wouldn't!)Among passive solutions, it might be reasonable to inductively ballast the generator, by using a saturable reactor into an LC resonant tank. With the right choice of values, it can be made to resonate just below line frequency, at nominal voltage; and therefore will draw little current. Then, when voltage swells, the inductance drops, pushing resonance up to line frequency, sucking extra amps from the generator. (Unless the resonant circuit has a heavy load, and even heavier construction, the current draw will be largely reactive. I don't know the behavior of diesel generators, to be able to tell if this would help or not. Or if the line frequency is even stable enough for this to work.)Or if such a network is used as a pass element, then you have a ferroresonant transformer, which tends to regulate the output voltage (albeit proportional to line frequency). These are extremely bulky and hot, for this kind of power level... but just saying, there may be a passive solution.As for active solutions, I agree, you don't want to use an IGBT here, not with such a simple circuit anyway. And it wouldn't work to use just one, with hysteresis (so it's only on or off), because you're just trading one load step for another: it can't just stay on, it has to turn off eventually! It may end up working anyway (hysteretic control will result in PWM operation), or it may make things worse (pulsating load upsetting the generator, regulator or governor?).If the controller will deliver full power at a lower voltage, I'd suggest simply turning down the whole thing. Adjust the generator regulator, or add a transformer.Tim

author: thefamilyman
author_rank: Regular Contributor
timestamp: February 16, 2016, 02:43:12 am
content: Quote from: T3sl4co1l on February 16, 2016, 02:32:08 amthe control of the motor is not up to me, the client has finely tuned the characteristics of the pump for their application.the application of using a generator is for their test site.Unfortunately they've put the pressure on us to develop something with in the next few days, these motor controllers are custom designed, built they cant afford to blow them up.we're already changing the AVR in the generator from an analog to a fast reacting digital AVR to hopefully react faster but they still want something that can clamp down any high transient voltages.I know it might be asking a bit much for you guys to help but we've all have had this experience when the client puts the pressure on.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: February 16, 2016, 02:51:43 am
content: Hi group,Where did the motor controller come from? and how was it meant to be used?Typically the dc voltage will come rectifying 3-phase.Common voltages are 400V and 480V line-to-line.If you rectify these with a capacitor input filter you get 1.34 times the line-to-line voltage, under load. The capacitor voltage will peak charge to 1.41 x Vac (line-to-line) under no load.For a 400V system, the no load rectified voltage is 400 x 1.41 = 564V d.c. (nominal). Add 10% for line variation and you are at 620V d.c.For the 480V systems the corresponding numbers are 676V and 743V.IGBTs normally come in 600V and 1200V devices, there are other voltages.I would be using at least 800V worth of electrolytics in these applications, either 2x 400V or 4x 200V parts with ballast resistors.The voltage rating of the motor drive doesn't seem rightRegards,Jay_Diddy_B

author: thefamilyman
author_rank: Regular Contributor
timestamp: February 16, 2016, 03:01:11 am
content: Quote from: Jay_Diddy_B on February 16, 2016, 02:51:43 amHi Jay_Diddy_BIt's all custom made, including the motor.the motor controller is designed for around 470VDC, with an upper limit of 580VDC before damage.the generators are custom wound and controlled to supply 3 phase AC to the 3 phase rectifier.i didn't design the rectifier circuit but here's a pic of it

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 16, 2016, 03:08:08 am
content: Well, if this is just for testing... why don't you rent a lab supply? A few kilobucks/month should be more than enough to cover things, and it'll get you through development.Tim

author: thefamilyman
author_rank: Regular Contributor
timestamp: February 16, 2016, 03:11:38 am
content: Quote from: T3sl4co1l on February 16, 2016, 03:08:08 amno onsite mains power hence the generators, and 470VDC @ 80A lab supply would be a bit hard to come by.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 16, 2016, 03:17:49 am
content: Well if you didn't ask, it's no one else's fault...I've heard of industrial scale DC lab supplies before, though I don't know of a source offhand. I'm sure they're out there! Such a supplier may also have a fully portable option (i.e., including the generator, or at least offering support for use with your existing generator).Tim

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: February 16, 2016, 03:19:52 am
content: Hi,If you want to try a quick fix, try increasing the number of capacitors in the rectifier circuit. 2x 2200 uF 350V (in series) is only 1100uF. I would try adding another 3 sets, for a 4 in parallel, 2 in series arrangement. This will give you 4400uF.Regards,Jay_Diddy_B

author: bills
author_rank: Frequent Contributor
timestamp: February 16, 2016, 03:19:57 am
content: I used to work for a newspaper and the printing press had 5-150 hp dc v/s motors. we had a module made bycrydom that would shunt the back emf from the motors as they decelerated.Even those would release the magic smoke , but the drives would still be ok.

author: uncle_bob
author_rank: Unknown
timestamp: February 16, 2016, 05:57:40 pm
content: Quote from: Jay_Diddy_B on February 16, 2016, 03:19:52 amHiI missed the "2200's in series" part in my math earlier.If you plug 1/4 the capacitance I used (I assumed parallel) then all the energies and power's scale by 1/4. The current is still the same. That gets a snub circuit down to a pretty reasonable size.That said, for a test environment and never to be used again .... put in a bigger capacitor stack.Bob

author: thefamilyman
author_rank: Regular Contributor
timestamp: February 16, 2016, 07:59:10 pm
content: morning.last night during some tests i adjusted the stability of the generator's AVR and governor which has helped a lot with the overshoot but still has the overshoot.looking into the rectifier box more, the caps are actually big 22,000uF 350V in series and a 100uF 800V Polypropylene cap across output.as promised i've got the voltage and current curves for one of the tests.the blue line is the current on the motor controller, max draw is 79Ared line is DC voltage measured at the output of the rectifier box. Nominal voltage of 480VDC, minimum voltage of 419VDC and max of 527VDC.each division is 0.5sec.as you can see on the red line at the end of the motor shutdown the voltage peaks to over 500VDC, the client wants to clamp that to around 500-520VDC to protect the motor controller.

author: SeanB
author_rank: Super Contributor
timestamp: February 16, 2016, 08:10:20 pm
content: Have a look for inverter brake chopper resistors and the associated control gear. They are designed to do just what you are doing, dump the extra energy that a motor inverter dumps on the supply capacitors stopping a motor very quickly. Normally used on inverters where you do not have a few so you can share the DC bus, or where you are sharing but the energy returned is more than the other drives can use. They are available in assorted sizes from industrial inverter suppliers.

author: dom0
author_rank: Super Contributor
timestamp: February 16, 2016, 09:18:13 pm
content: Quote from: T3sl4co1l on February 16, 2016, 03:17:49 amUp to 200 kW cont. DC output or so:http://www.heinzinger.de/en/produkte/hochleistung.php

author: thefamilyman
author_rank: Regular Contributor
timestamp: February 16, 2016, 09:25:08 pm
content: Quote from: dom0 on February 16, 2016, 09:18:13 pmtime, availability and cost prohibitive.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: February 16, 2016, 09:39:08 pm
content: hi,I think the best way to address this is to ramp the pump controller down, a speed sufficiently slow that the AVR on the generator can track the change.With the abrupt stop, the total time is around 20 seconds. If you take the last second and stretch it 2 seconds, ramping the current to zero, you will have a total time of 21 seconds, but you would probably get the overshoot under control. You would start the tapering process when the transfer is 95% complete.You may not need to do this with the LiPo battery pack. This is a work around because the generator is not a battery.Any other methods would take too long to engineer.I was thinking about a Buck regulator or a disconnect switch, but this would have to handle 80A for 20 seconds, not trivial.Regards,Jay_Diddy_B

author: uncle_bob
author_rank: Unknown
timestamp: February 16, 2016, 11:38:01 pm
content: HiOk, so we are over voltage for a bit less than a half a second?If the clamp is at 520 V and it needs to dump 79 amps, that would be 40KW for 1/2 second.A 50W zener is doing pretty well. Anything under 5V is not worth playing with. 10A is about all a stack will be good for. 520V per stack is 100 diodes plus a 2 ohm 200W resistor. Roughly 800 diodes would do it. Even at 1K pcs they still will cost you over $10 each. Simple to do, easy to wire up. Drop them into an insulated heatsink and away you go. Essentially guaranteed to work, grab the company visa card and have them air ship you $15,000 worth of diodes.Bob

author: langwadt
author_rank: Super Contributor
timestamp: February 16, 2016, 11:52:23 pm
content: Quote from: thefamilyman on February 16, 2016, 07:59:10 pmhead over to digikey with a creditcard and buy more capacitors, it'll help absorb the spike in voltage and should be similar towhat you would see with a battery

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 17, 2016, 01:28:24 am
content: Quote from: langwadt on February 16, 2016, 11:52:23 pmUm, for a wholesecond?80A for 1 second is 80 coulombs. At delta ~50V, that's 1.6F!Tim

author: uncle_bob
author_rank: Unknown
timestamp: February 17, 2016, 01:36:01 am
content: Quote from: T3sl4co1l on February 17, 2016, 01:28:24 amHiYes and that is a fallacy in my numbers for clamping as well. What we observe on the plot is a decaying burst from the inductance and the capacitors discharging from the resistor(s) across them (and maybe other things like the controller).The real current spike duration is (at this point) unknown. It is a pretty good bet that if I get a bump of X with caps = Y then with caps = 2*Y, I'll get a bump of X/2.Bob

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 17, 2016, 01:41:37 am
content: Quote from: blueskull on February 17, 2016, 01:38:23 amYeah, no kidding! Lead acid will take it, and is cheap, but gets beat to shit in the process... This is absolutely the domain of supercaps, but you need a metric shitload of them to handle it. So you still have the same problem with cost and availability, as with the power supply (which I'm still not sure was fully considered and not just dismissed out of hand..?!). A technical term.Tim

author: langwadt
author_rank: Super Contributor
timestamp: February 17, 2016, 01:56:54 am
content: Quote from: T3sl4co1l on February 17, 2016, 01:28:24 amit getting late so I might screw up but:80C * 50V = 400J0.5*11000e-6*500V^2 = 1375J0.5*11000e-6*550V^2 = 1663J

author: thefamilyman
author_rank: Regular Contributor
timestamp: February 17, 2016, 02:05:21 am
content: just an idea:would a high energy Varistor work?http://www.littelfuse.com/products/varistors/industrial-high-energy/da.aspx

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: February 17, 2016, 02:17:17 am
content: Quote from: thefamilyman on February 17, 2016, 02:05:21 amShort answer is: NoReason: There is too much of gap between the voltage that they start to conduct at and the voltage they will clamp at. If you choose one that is too low in voltage it will conduct during normal operation. Even with relative low current it will overheat and die a miserable death.If the clamping voltage is too high it will not save your motor driver.From the datasheet:If you chose one for a working voltage of 560V, It will clamp at 1130V.Assuming that you could find one that clamped hard enough, to absorb 600V at 80A for 1 second = 48 Kilo Joules. The varistor is rated for 0.6 kJ. The magic smoke will be released.Have you considered decelerating the pump instead of shutting it off abruptly. In my opinion this is the best solution. The AVR on the generator will be able to keep up with the reduction in current and all will be happyJay_Diddy_B

author: thefamilyman
author_rank: Regular Contributor
timestamp: February 17, 2016, 02:33:19 am
content: hmmmi would love the motor to decelerate but for the application it has to stop abruptly.

author: thefamilyman
author_rank: Regular Contributor
timestamp: February 17, 2016, 02:42:49 am
content: maybe i'm just throwing out stupid ideas now....how about high powered TVS diodes (to make up the break down voltage) into a high power resistor?http://www.littelfuse.com/~/media/electronics/datasheets/tvs_diodes/littelfuse_tvs_diode_ak6_datasheet.pdf.pdf

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 17, 2016, 06:22:26 am
content: Quote from: langwadt on February 17, 2016, 01:56:54 am400J would be from 0 to 50V. It's considerably harder to go from 500 to 550, because you have to pile on top of all the charge that went before!So the delivered energy is more like (80 to 0A) * (500 to 550V) * (0 to 1 sec) = 21kJ. Actual might be more than double, depending on just how you end up dealing with it (for instance, letting it up very slowly, over several seconds, to avoid any overshoot at all).The biggest problem with seconds-long transients is,they aren't transient. You've got real heat spreading and dissipation issues, in almost all kinds of devices. There are high energy bulk resistors that could handle the energy directly, but otherwise, you basically need something that can handle the wattage continuously (e.g., a string of stainless wire, long enough and thick enough).For reference, 21kJ is enough to vaporize 10 grams of water, or melt 63 grams of ice, or raise 50 grams of water from 0 to 100C. Or raise the temperature of 1 kg of aluminum by 23C.Tim

author: Seekonk
author_rank: Super Contributor
timestamp: February 17, 2016, 06:32:47 am
content: Another something to throw out there. This is a PWM shunt used for windmills. Use whatever FET or IGBT you want and a load resistor. Ready made if you don't want to build one. Modify the input and power with an external source. This is a really novel circuit and not what you expect at first glance.http://www.ghurd.info/

author: uncle_bob
author_rank: Unknown
timestamp: February 17, 2016, 12:38:16 pm
content: Quote from: Seekonk on February 17, 2016, 06:32:47 amHiI think that the real issue is the cost of the giant bank of semiconductors used as a heat dissipation element. You need a *lot* of parts and that costs.Bob

author: Ian.M
author_rank: Super Contributor
timestamp: February 17, 2016, 01:41:18 pm
content: ... which is why you *DON'T* dump to a semiconductor load. I've taken some measurements off the plot in post #24 and there is 0.17 seconds from when the load current cuts off to when the voltage exceeds 480V, and 0.24 seconds to 500V. There is therefore plenty of time for a control system to react. PWMing a high current IGBT module to dump up to 150A from the cap bank into a massive 3.2 ohm load resistor would do the job nicely. Trigger it on a rapid drop in load current, with PWM duty cycle chosen to maintain the total current initially then taper it to zero over a period of several seconds to allow the AVR system time to react.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: February 17, 2016, 01:42:14 pm
content: Hi,I am going to throw this idea out there for a review by the group.How about:The circuit is simplified. A typical circuit will contain RC damping circuit, MOV etc.The idea is to replace the 3-phase rectifier with a rectifier consisting of diodes and SCRs. Sometimes this is called a semi-controlled rectifier.This arrangement requires a single gate drive circuit, reference to the positive output. The control circuit would contain a reference a comparator and an oscillator to drive the pulse transformer. The SCR would be gated on if the output voltage is below the maximum voltage.The advantages of this circuit:1) Very little extra power loss. Just SCRs versus Diodes2) No great dissipation problems3) Relatively simple control scheme4) SCR/Diode modules are available in high voltages / high currentJay_Diddy_B

author: ConKbot
author_rank: Super Contributor
timestamp: February 17, 2016, 01:50:27 pm
content: This is not a new or particularly difficult requirement. Buy a brake resistor controller(or multiple) and brake resistor(s)http://www.automationdirect.com/adc/Shopping/Catalog/Drives/AC_Drive_%28VFD%29_Spare_Parts_-a-_Accessories/GS_Braking_Options_%28Braking_Units_-a-_Resistors%29/Braking_Options_%28All_GS_Drives%29/GS-4DBUas an example. Make sure the controller can adjust to the range you need.

author: langwadt
author_rank: Super Contributor
timestamp: February 17, 2016, 02:22:19 pm
content: Quote from: T3sl4co1l on February 17, 2016, 06:22:26 amthere got to be something missing then because the voltage only rises to ~550V, if there was really 21kJ to dump and the only place to do it is the capacitors thevoltage would rise much more

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: February 17, 2016, 02:50:39 pm
content: Hi,It is kind of tricky to figure out how much the voltage will rise by.In a typical three phase generator there are two main frequencies:1) The mechanical frequency that you are applying to the generator.2) The electrical frequency of the output.The difference between the two is theslipfrequency.Theslipfrequency depends on the size of the machine and the loading.The larger the machine the smaller the slip frequency. The lighter the load the smaller the frequency.The output voltage is governed by Faraday's Law of Induction.V= N d Phi /dtPhi is the Flux density x Area.If you have a generator with an output frequency 60Hz and 3Hz slip frequency. The mechanical frequency is 63Hz.Now if you off-load the generator. The electrical frequency will increase to 63Hz and the voltage will increase by 63/60, since the rate of change of flux has increased.This is in addition to any voltage increases from the rectifier changing from averaging to peak detecting and any increases due to the reduction in current and the output impedance of the generator.The voltage rise is bounded.-------------------------------The Brake controller, proposed by ConKbot, is really an IGBT switch and a large load resistor. The load resistor is capable of dissipating about 10 kW continuously. There is some pro-rating due to the low due cycle. The idea is to replace the load current drawn by the pump motor with a resistive load until AVR can slow down the generator.This should work in this application. It requires a load bank capable of dissipating around 10 kW continuously, 50 kW for a second or two.Regards,Jay_Diddy_B

author: thefamilyman
author_rank: Regular Contributor
timestamp: February 17, 2016, 08:45:34 pm
content: Morning allAfter a lot of research and consulting with some power design companies it seems a simple and reliable option is a Selenium voltage suppressor.(example)

author: uncle_bob
author_rank: Unknown
timestamp: February 17, 2016, 10:59:17 pm
content: Quote from: thefamilyman on February 17, 2016, 08:45:34 pmHiIf that beast ever goes over power .... clear the room *quickly*. The smoke from a burning Selenium pile is *nasty*.Bob

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: February 18, 2016, 12:35:30 am
content: Hi,Reading this catalog:http://www.dalitech.com/Resources/HVCA_CKE_Catalog_2007.pdfpage 77 onwards.If the Selenium voltage suppressors are similar to these, they have a working voltage to clamping ratio of 1 : 2.66This makes them unsuitable for this application that requires 'precision' clamping just above the working voltage.The solution proposed by ConVbot in the earlier message is capable of this. This a simplified schematic of the Brake resistor controller:The OP-AMP circuit can be set with the precise threshold that you need. The multi-kilowatt resistor array takes care of the power.Regards,Jay_Diddy_B

author: uncle_bob
author_rank: Unknown
timestamp: February 18, 2016, 01:39:57 am
content: HiSelenium is pretty much bound by physics. A 450V part clamps at 1200 V max (at rated current). That's on page 76. After looking at all the other products, you would *think* there would be a bunch of graphs for them ... not so much. Figure 113 on page 48 shows them having an "alpha" of 8. The graph implies that a 1A to 10A change gives them an 80V swing and thus their alpha = 8 number. By implication (no other data) running up to 100A would be a bit over a 200V increase. That is all on their example part at 300V.Boosting it all up to 450V multiplies everything by 1.5. Now the first bump is 120V not 80. The second bump for the 100A now is a 300V rise. If you idle at 10A, you are putting 400+ W into the device steady state. The rise from there is still 180V ...Will these gizmos help? Maybe. Will they do the job by themselves? .... it sure does not look like it.Bob

author: Ian.M
author_rank: Super Contributor
timestamp: February 18, 2016, 03:19:34 am
content: @Jay_Diddy_B: That simple circuit wont work. The problem with the over-voltage surge is that it is a load dump transient from the generator (the charted current didn't go negative so it cant be from regenerative brakeing of the pump). Your circuit will of course shunt that surge, but it doesn't taper the load on the generator, so when the voltage drops below the threshold again and it cuts out, it will trigger yet another load dump surge that it will have to shunt.It is ESSENTIAL that any surge control system tapers the load (e.g. by PWM, or by switching out the resistor array one series string at a time) over a period of several seconds to avoid causing another load dump.

author: Connoiseur
author_rank: Regular Contributor
timestamp: February 18, 2016, 03:32:07 am
content: Quote from: thefamilyman on February 17, 2016, 02:33:19 amThis seems like a misoperation of the generator for sure.Is there an active PFC in the rectifier? If the AC input does not register a proportional surge, then the inductor in PFC might be the culprit. Let's hear what others think about it.Quote from: Jay_Diddy_B on February 17, 2016, 02:50:39 pmFriend, what you are talking about applies to induction generators. I don't think they are using it here. Induction generator is seldom used with constant speed (achievable regardless of load) prime movers like diesel engines and steam turbines. It is used where the speed of prime mover varies a lot e.g. wind turbines.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: February 18, 2016, 03:47:08 am
content: Quote from: Ian.M on February 18, 2016, 03:19:34 amYou raise some good points.I believe that the surge comes from off-loading the generator, not regenerative braking the pump, for the reason that you state, there is no negative current.I would be a good idea to taper the load current by PWMing the IGBT. Effectively you would size the load bank to match the load drawn by the pump. Since the pump can not be decelerated, you would ad the resistive load when the pump stops then taper the dummy load current at a rate that the AVR can keep up.I took my inspiration from the braking resistor controller, proposed by ConKbot. The standard application for the braking resistor is to dissipated the energy from regenerative braking. This pump application is not a typical application.Great discussion !!Regards,Jay_Diddy_B

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: February 18, 2016, 04:10:35 am
content: Quote from: Connoiseur on February 18, 2016, 03:32:07 amThere is no PFC. The three phase rectifier has been described in the thread as a 3-phase bridge with 2 x 22,000 uF 350V in series (11,000 uF) capacitor.Friend, what you are talking about applies to induction generators. I don't think they are using it here. Induction generator is seldom used with constant speed (achievable regardless of load) prime movers like diesel engines and steam turbines. It is used where the speed of prime mover varies a lot e.g. wind turbinesIn the first message the power source was described as a 60kVA diesel generator. This is being used with an Automatic Voltage Regulator to make a giant bench power supply. There is a possibility that the generator is really an alternator. The issue is the same, when the generator or alternator is off-loaded the voltage will increase.Regards,Jay_Diddy_B

author: Connoiseur
author_rank: Regular Contributor
timestamp: February 18, 2016, 04:30:58 am
content: Quote from: Jay_Diddy_B on February 18, 2016, 04:10:35 amoops missed that.Quote from: Jay_Diddy_B on February 18, 2016, 04:10:35 amIt is true that the voltage would increase if the load is removed in any type of generator, but synchronous generators also have an AVR built into the excitation system and is pretty fast compared to external voltage regulators, be it servo or relay type.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 18, 2016, 07:54:22 am
content: Two points:- To my knowledge, no passive solution works here. All have curves that are not steep enough. Not Zeners, MOVs, diodes, including selenium...- This is nothing special for a properly designed brake chopper, which is an active device, controlling the dissipation power using feedback from the input voltage, so that it doesn't cause another load dump. Just buy one, or learn how they are designed and design one. As you saw, the changes in voltage are slow, and you are not making an exact regulator, just preventing things from blowing up, so designing the feedback loop is not that difficult.

author: Ian.M
author_rank: Super Contributor
timestamp: February 18, 2016, 08:58:43 am
content: The only passive device that can absorb these surges would be a series bank of 35 12V Lead acid batteries, with a CCA rating>=200A, floated at the DC bus voltage, direct from the bus. As has been previously mentioned, battery life wont be great, but on the other hand, it will vastly decrease the initial sag under load on pump startup, and will provide a supply that is far far closer in characteristics to the LiPO bank that the client is eventually planning to use.However we don't know if they are running different bus voltages for testing (by adjusting the AVR). If so, most simple solutions are out. A current controlled brake chopper wouldn't need readjusting for different bus voltages. It would be simply activated by a step drop in load current, and be PWMed to divert the immediately previous current draw, tapering to zero at a rate the AVR can compensate for.