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wind interface box: a circuit to divert excess current when a certain voltage is exceeded

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IamSmooth

Wed Nov 19 2008, 10:08PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

I need to design a protective circuit for my inverter. It can't take more than 600v or it is toasted. I will set a cutoff at 500v. when 500v is reached the circuit should divert excess current to a dump load. When the voltage falls below, say 450v, the load is removed and all the current flows through to the inverter. I was going to use some zeners attached to the base of some transistors as a reference. When the voltage is reached current will start to flow and the transistor will turn on and conduct. This would drop the voltage below the zener cutoff and I'll see oscillations which is not good.

Can anyone help with suggestions on how I could approach this? I guess I could use comparators. My goal is to divert excess power away from the inverter; I would prefer not to have a relay that switches from inverter to dump load.

Dr. Slack

Thu Nov 20 2008, 08:16AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

What you're describing is a shunt regulator, or are you? Do you actually want the 500v shunt, 450v off hysteresis thing, or is your generator ahigh enough output impedance so that it could be loaded down safely by a linear shunt regulator. Assuming the latter then ...

You could design a stable one or an unstable one, but the chances are that something thrown together from zeners going to the base of a transistor (a so-called amplified zener) will be stable. When it starts to turn on, it turns on gradually, and the input voltage drops just enough to stablise the current and the zener voltage in balance. It's a good policy to have a base-emitter shunt resistor to draw some zener current even when not regulating, so the turn-on is better defined.

If you are not familar with system stability, then a good rule of thumb is to have a feedback circuit dominated by just one time constant (the so-called dominant pole approach). If you have two or more similar time constants, then you have an excellent chance of building an oscillator. Generally, a zener-transistor shunt is fast enough that other time constants will be the lowest frequency effect, your generator output impedance into the load capacitors for example. If the transistors are too slow and you do get oscillatoions, then they can be speeded up and the gain reduced (at the expense of regulation) by using a series emitter feedback resistor.

Don't whatever you do try to "tame" things by putting a bit of shunt C on the base of the transistor, that will increase delay, put another time constant in the circuit, and make oscillations much more likely. You can tame at this point if you put a *very* large amount of C here, and slow it down so much that this time constant becomes the dominant pole. However, then you have a system which is very slow to react.

A FET would work quite well in this duty, and perhaps is easier to get in the voltage and current that you need than a bipolar, and easier to parallel to get the dissipation.

If the stability concepts are new to you, then a good way of getting oscillations is to put extra stuff like comparators in the circuit without knowing precisely what they are doing. It's more phase shift and delay round the circuit, which is gnerally bad, and more opportunity to get things wrong.

BTW, a nice thing to use for a dump load is a light bulb, or several in series to get the voltage. You have a good idea of what the safe dissipation will be, and you can see when it's dumping and by how much. OTOH, a constant value load resistor will cause less peak dissipation in your transistor with its straight load line.

If the 450v/500v thing is that there is no shunting below 450v, and the voltage shall never exceed 500v at the generaot's maximum output current, then it's very straightforward to design a nice stable finite gm shunt to acheive that.

OTOOH, would a series regulator be better? Less heat to get shot of in gale conditions. Have a 480v string of zeners with a biassing resistor from the input, and a FET or bipolar as a source/emitter follower. As long as it has enough Vds/ce to withstand the generator maximum output, it would feed the inverter a moderately constant voltage, losing only a few volts Vgs/be at low generator output. And abosultely guarranteed stability. Or is the increased generator loading at high output part of its overspeed protection?

IamSmooth

Thu Nov 20 2008, 12:35PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Neil, thank you for your answer. I am leaning towards the zener shunt with some power bipolars. The generator will usually not exceed 300v and will hopefully put out 800-1000w. It is capable in higher winds to put out 1200-1500w, although it would be a strong wind. It is possible, although unlikely in this scenario, for the voltage to exceed 500v with a load from the grid. The big concern is if I lose grid power and the inverter goes offline. Then, the grid load is gone, the turbine can overspeed and the open voltage can easily exceed 500-700v. Once the grid comes online the voltage would destroy the inverter.

I have fashioned a dump load using 16 100ohm/100w ceramic resistors. I don't have the final resistance, but it will be somewhere between 75 - 125ohms/1600watts.

Please correct me if I'm wrong from this point: I have some 100v/1watt zener diodes. I was going to make a string of 500v with a 1M/0.25W resistor for current protection. The end of the zener was going to go to the base of a parallel array of power transistors which would connect the CE through my dump load. I am unclear as to how much resistance to add to the BE as you suggested. Maybe you can clear that point for me.

Thanks again.

Dr. Slack

Thu Nov 20 2008, 03:06PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

1600w at 500v is 3A. If you use high voltage bipolars, then what is their beta? It may only be 50 or lower, and then you'd need 3/50 = 60mA base current flowing, which would dissipate 0.06*500 = 30 watts in your zeners. I think this is why I was leaning towards FETs, you don't need current to turn them on. You could use one of your high voltage transistors in a Darlington connection, to improve the beta. Or you could use a source follower FET to drive the bases.

What's this 1M current protection resistor? If in series with your zeners then you won't get sufficient base current without a follower to boost the current into the bases.

What causes the potentially high output voltage? Is it lack of resistive load on a high output impedance generator, or is it too much speed on a low output impedance generator? If the latter, then your job is simply to limit the max speed of the generator. It would be cooloer for your power electronics to switch your resistor loads on and off, with some kind of hysteresis. If the loads did switch second by second and the generator speed hunted, if max voltage was constrained then is that a problem? With a linear shunt, then under certain mild overspeed conditions you could get 25% of 1600watts to dissipate from your bipolars.

IamSmooth

Thu Nov 20 2008, 06:43PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

The excessive voltage is the loss of a load, allowing the rotor to spin without any back emf impedance. I have an emergency disc brake, but this is another topic.

I like your idea of using FETs. After reading your comments I was neglecting to consider the Beta, although I probably was going to use a darlington setup. The 1meg resistor probably won't allow enough leakage current. The FETs sound good, but I have a concern. I don't work with them much, but don't they just TURN ON? I liked the bipolar idea because they would gradually turn on, allowing me to bleed some of the current to the dump load and still allow the inverter to work if the grid was active. Would the FET just open and divert all power away? If it did, then the voltage would drop below 500v and I would oscillate between it opening and closing, which I don't want. If so, maybe the darlington setup is better?. If the FET would work, any suggestions on one that I could get from Mouser, or should I just pick one that can handle the voltage and current?

Dr. Slack

Fri Nov 21 2008, 08:24AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

No, FETs turn on just as smoothly as bipolars, in fact they're the output device of choice for high quality audio amps because they are more tame and easier to drive, and the "problem" of FETs in switching circuits is the difficulty of slamming them through the linear amplifying region as quickly as possible. On any FET data sheet, the gm gives the change in drain current per change in gate voltage. If you already have the bipolars, then the easiest and cheapest way to make a linear current shunt would be to use one FET to darlington your bipolars as below. If you haven't got them yet, then the circuit would work just as well with Nch FETs in the power position, with small changes in resistor values, and the option of deleting the first FET.

Here's my suggestion of what could work. To be quite honest, I've just dashed this down in a few minutes, so it's likely that some of the extra components I've put in for flexibility are not really necessary, or some bits could be more elegant, but it will do the job. It really can be a very crude design because by definition a current shunt is 100% inefficient, and your application has huge voltages lying around which dwarf tempcos and variable Vgs.

I've put emitter resistors in, nominally a few ohms, to force the currents to share. I am assuming you will need to parallel bipolars for dissipation. Once you know the maximum current you want to sink, then that's the voltage you need across those resistoros, plus a VBE plus a VGS, it's not too critical. Pick D4 a bit bigger than that, but <20v to protect the FET, and pick R10 to sink 1mA or so at that voltage, so that the zeners need to be conducting something higher than leakage before it starts shunting, which sharpens up the knee at the start of conduction. Make R4 the same size, it protects the string from overcurrent in the event of overvoltage. It reduces regulation, but you don't really need 0.1% regulation do you? R12 is pretty nominal, it just shunts leakage away from the bases, 10k perhaps, good practice, not really essential. The FET, which is used as a follower, doesn't need much current (1 amp) or dissipation (10/20 watts maybe), but must have Vds >> 500v. It is darlington connected rather than follower connected, to reduce its disspiation. <edit> Forget the transistor type number, I can't be bothered to delete it </edit>

Because you're making the resistive load from several parallel strings of resistors, you could force sharing by putting one bipolar per string in this case. This way, you only have two wires to your load, and it demonstrates a way of forcing sharing between an arbitrary number of devices. Bear in mind, this form of sharing is only good where you can afford to lose the power in those emitter resisitors. If you want efficiency, then FETs (which share nice), or current balancing transofrmers, or active base balancing might be preferrable.

IamSmooth

Fri Nov 21 2008, 02:57PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

Neil, thanks. I've read it over and most of it makes sense. I have bipolars, but I think I will just go with FETs. I have found several on Mouser with 600-900v Drain-Source voltages and currents in excess of 10A. I believe on has a wattage rating of 400W so I can use four of them and attach them to some nice, large aluminum heat sinks I have lying around.

Just a few points if you still don't mind:

Let's say I design my dump load to be 50ohms/2000w. The generator will most likely max out at 6amps, but I can say 8 to be safe. Does this affect my design, or will this circuit, given that the FETS can handle the power, handle any load value?

Second, if I just use the FETs I believe your last comment said I could eliminate Q1 and replace Q2, Q3, ..., Qn with the FETs, right? Do I still need resistors R1 and R2?

Finally, as I am still reading up on FETs, why do you need D4 and R10? You started by talking about the maximum current I want to sink. In my case it is 8 amps and I want to use a 50ohm dump load giving a voltage of 400v. Using the BJTs did you say I need to pick R1 and R2 to have close to 400v and maximum sinkage (50ohms, roughly)? This confuses me because you said I have to keep the value < 20v.

Dr. Slack

Sat Nov 22 2008, 08:36AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Hi Jonathan,

The benefits of a posted debate are that other people have the chance to learn too, and that you have a measure of protection from my potential stupidity when others can read the thread and say "Whoa, that's dangerous/stupid/whatever".

The basic circuit is good for any conceivable load, just adjust some of the component values and the number of paralleled devices.

The point of some of these extra components is a bit subtle, and not all of them are absolutely necessary for the circuit to function, but they just improve the robustness / likelyhood of not giving trouble.

Apologies, the part numbers in this diagram may be different to those in the first one. I am referring to this diagram.

R1 etc make the active devices share current. Although FETs share better than bipolars with their positive tempco which avoids hogging, they still have varaible Vgsth, which measn that if R1 etc = zero ohms, then one may well be sinking several amps before the others have joined in. The problem with this is that your heatsink may be able to dump a large total power, but only x00 watts per device. To use some actual figures, 40S is a typical FET gm (40 Amps per volt). With only 0.1v difference in Vgsth, one device could be taking 4A more than another, which at 200v (the other 200v on your resistive load) is 800W all in one device.

With R1,2 etc = 5ohms each, this reduces the gm to 0.2, and the 0.1v delta between devices unbalances them by the square root of bugger all. The emitter feedback reduces and controls gm, so that each device is a voltage controlled current sink. It works with bipolars too, such that what was a current input device, when equiped with an emitter resistor, is turned into a voltage controlled current source.

Let's say that you have 5 parallel FETs with 5ohms each to ground. The total gm is now 1. If you want 8A max, then you want 8v across the source resistors, so with 4v Vgsthmax, that's 12v across R5 and D2. Setting D2 to 12v actually limits your maximum current sink to around 8A. Not so much of an issue in this design with your resistive load, but if used without a load, intending to dump all the power in the semiconductors, it will control the maximum FET dissipation. Put D2 >12v <20v.

The point of R5 is to introduce a clean turn-on threshold. Although a zener supports a constant(ish) voltage at several mA, it will still have quite a high leakage current at a voltage well below its nominal zener voltage, it has a "soggy" knee. Without R5, a few uA leakage through the zeners from 200v, or even tracking across a damp board, would start to turn your shunts on and load the generator way too early. Selecting R5 to require a reasonable current to overcome Vgsth solves this. If you have 1watt 100v zeners, then allocating 300mW to heating at max current is very conservative. At the maximum 12v on D5, that's 3mA, so choose D5 = 3.9k. Now at threshold of a few volts for Vgsth, R5 will need to be supplied with 1mA to generate the 4v drop needed to get the FETs starting to turn on.

The voltage-drop zeners D1 in series with D2 is a very low impedance, and will destroy thenselves in a moment under over-voltage conditions, so R7 is to limit the fault curent that can flow. Obviously it reduces regulation. To get 8A shunt current you need 8v change on the rail without R7. If R7 = 3.9k as well you now need 16v delta to get 8A shunt. But in this application that's tolerable.

IamSmooth

Sat Nov 22 2008, 08:36PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

I'm starting to get most of this. Just a few things left to clear up:

Dr. Slack wrote ...

With R1,2 etc = 5ohms each, this reduces the gm to 0.2, and the 0.1v delta between devices unbalances them by the square root of bugger all. The emitter feedback reduces and controls gm, so that each device is a voltage controlled current sink. It works with bipolars too, such that what was a current input device, when equiped with an emitter resistor, is turned into a voltage controlled current source.

So, without a source resistor the transconductance is 40, but with 5ohms it is 0.1. Is this calculated or did you get this from a table?

I am going to use some real parts for this question. My load will be 25ohms and conduct a maximum of 8A. On digikey's site I found a 600v/460W MOSFET. Here is the page

It lists Vgsth min 3.0v and max as 5v. Figure 7 on page 3 of 5 shows the graphs of I vs Vgs. Between 5.5v and 6v the current is 8A. Is this the value I use for the circuit design: ~5.7v? If so, then I need to pick D2 = 5.7v + 8V = 13.7v, right?

I understand the use of the zeners, but I don't know how some leakage current through D2 affects a voltage-controlled MOSFET? I see how using R5 @ 3.9K sets the value to 12v (3.9K * 3ma), but I am still unclear how it helps.

For those interested in the project his is for, go to

to see my photo diary.

Dr. Slack

Sat Nov 22 2008, 10:46PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

So, without a source resistor the transconductance is 40, but with 5ohms it is 0.1. Is this calculated or did you get this from a table?

The particular FET you have noted has a gm of typical 26, min 16, however when you have added a source resistor, it's approximately 1/R (for gm >> 1/R), so 5ohms gives 0.2S. With 5 FETs and resistors paralleled, you'd get a total of 1S.

Between 5.5v and 6v the current is 8A. Is this the value I use for the circuit design: ~5.7v? If so, then I need to pick D2 = 5.7v + 8V = 13.7v, right?

Given that Vgsth varies between 3v and 5v for your device, there's little point in going for 1/10th volt accuracy. Yes, as long as VzD2 exceeds the worst case (high Vgsth and low gm) voltage you need to get your maximum current, then all's peachy.

I understand the use of the zeners, but I don't know how some leakage current through D2 affects a voltage-controlled MOSFET?

No, it's leakage through D1, with hundreds of volts across it, that needs to be controlled by R5, or it will turn the FETs on preamturely.

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