H-Bridge Circuit Board Layout

[GeorgesWelding]

Now that I seem to have solved my IGBT switching issues I was having in my other thread, still testing though, I've started to design the circuit board for the full version of my H-Bridge. I figured a double sided circuit board would be much better to use than the extremely terrible layout I'm using on my prototype now. I did a little bit of reading about H-Bridge layout and I decided that for me the easiest way to design the board would to put all of the high-power traces on one side, and all of the signal traces on the other. I still have a lot to learn about PCB design so if my thinking is wrong, please correct me on it. Also, is my choice in IGBT going to be ok? I was told these would be find for what I was doing, but if there is a better choice please let me know.

Attached is the PDF of the H-Bridge with 2 paralleled IGBTs making up each switch, the free-wheeling diodes, filter caps, 2 storage caps(2-4 more(if needed) will be mounted off the board as well) and the IGBT drive resistors and diodes. The maroon color is the top copper layer, green is the bottom. I designed this using ExpressPCB. I have a copy of Multisim & Ultiliboard 10 but EPCB's simple interface seemed to be much easier for me being new to PCB design. The IGBTs(IRG4PSC71UDPBF/Super-247 Case) will be mounted on the top of the board with their backs facing up and the heat-sinks will be bolted on top of those. The FW diodes will also be mounted to the top with a tiny heat-sink behind each one(Will that be enough? There will be very good fans attached.) All the other components will be mounted from the bottom. C1 & C2 are 1000uF/400VDC storage caps(Again, 2-4 more of these(if needed) will be mounted off-board but still very close). C3 & C5 are 1uF/400VDC filter caps, and C4 & C6 are 2.2uF filter caps. D9-16 are MUR1560 diodes(My IGBTs are Co-Packed with FW diodes internal to their case as well). The small resistors and diodes are the IGBT drive components. The W# holes are where the input power and output power leads will be bolted to the board and the MT# holes are just mounting points. This board will be mounted bottom-up and the control board that will house the driver ICs will be mounted above this board so the driver leads from the control board to the bridge board shouldn't be more than an inch or so each(I know they all should be the exact same length). I can't think of anything else right now. I'm currently transferring the schematic of this bridge board from paper into the computer, I just drew up the board first as I wanted some insight on it. Thanks in advance for looking. Also, I know some of the green traces are hidden, but it should be pretty obvious where they go, if not I can post up a PDF of just the bottom side of the board.

Also, since I used the ExpressPCB software to design this board, they do offer a service to actually make this board(have to order atleast 2) but has anybody used this service before and how is the quality of their boards? Do you think it would be a good place to have these prototypes made? Their software quoted me $169.95 including shipping for 2 boards.
-Brad

View attachment Bridge Board (Dual IGBT).pdf

 

[GeorgesWelding]

I just realized I posted this in completely the wrong section. Mods feel free to move this thread to the appropriate location. Very sorry.

 

[Orson Cart]

Good idea to get rid of the extra diodes, the ones in the igbt's should be fine. Your layout is still quite spread out, but might work if the turn on is slow enough. Given your igbt's are 60A @ 100C case, four should be sufficient, no need to place the other lot for initial testing.
You need to have at least 600nS of dead time with these devices - make sure this is what you have!.
Ideally keep your switching freq below 40kHz.

 

[GeorgesWelding]

Well in that case, back to the drawing board...Literally. =) My controller's standard dead-time isn't enough for these IGBTs at that frequency either. 2% DT nets ~440ns(If I calculated right), while this can be increased, maybe I should keep looking at device datasheets. I haven't ordered anything yet.

And thank you Orson. You have been a great help so far.

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Ok. I found a couple alternative IGBTs. I haven't had any luck finding ones to carry all the current without paralleling 2 of them together that also switch at the speed I desire them to. Here are the alternates that I've found. I think they might work if I double them up. Some have internal diodes(which I prefer) some don't.

Turn-Off and Fall Times are @ 150*C.

IRG4PC50WPbF - 600V, 27A@100*C, Turn-Off: 210nS, Fall Time: 62nS. Datasheet says they can be hard switched to 150KHz. No internal diode. Don't think they make this model with a diode.

IRG4PF50WDPbF - 900V, 28A@100*C, Turn-Off: 270nS, Fall Time: 190nS. Datasheet says they can be hard switched to 100KHz. Has internal diode.

IRG4PC50UDPbF - 600V, 27A@100*C, Turn-Off: 240nS, Fall Time: 130nS. Datasheet says they can be hard switched to 40KHz. Has internal diode.

Now. The IRG4PC50WPbF would be ideal as far as the speed is concerned, but since it lacks the internal diode I would have to include them on my board which would spread out my layout considerably(basically the board I posted above) since I would have to run 2 of these paralleled together to get the amperage I need. The third one on the list just seems too slow in my opinion, so the second one on the list, the IRG4PF50WDPbF, seems like it might be the one I will have to use, although I will have to increase my dead-time slightly. Since it does have the internal diode I won't have to include them on my board which should allow me to tighten up the layout considerably. I'm just not sure. What do you think Orson? Also, do you have any recommendations for an IGBT that will handle the current I need and still be fast enough to run at 40-45KHz?

 

[Orson Cart]

Hello there, surely the TL494 has a deadtime adjust pin that allows you to get all the deadtime you need? You can then run your original choice of igbt, except the switching losses will start to dominate above 40kHz.

How much primary current do you wish to switch? 10A? (2300VA from a 10A, 230V wall socket?) The IRG4PSC71UD should just handle 10A at 45kHz, 325V bus,

 

[GeorgesWelding]

Well the final full power version of this machine will probably draw around 50A Max. on the primary side from a 230V wall outlet(USA). Most other inverter welders I've researched that output 250A have an inrush current of 45A and an operating current of 35-40A when outputting the full power. I'm looking to output 275-300A so I will probably be drawing 40-50A on the primary side at full power, depending on how efficient I can get this to operate.

The IRG4PSC71UDPbF datasheet states that it shouldn't be used for hard-switching above 40KHz where as the IRG4PF50WDPbF says it's good to 100KHz. I will still need to increase the dead-time for either of those IGBTs, but if I used the IRG4PF50WD I wouldn't be running it above it's recommended limit. Also, I have a question, in the datasheets there is a graph that has Load Current vs. Frequency, and on this graph, as the frequency increases the Load Current decreases significantly. Now, does this mean that the current carrying capability of said IGBT will be way under its rated current since I will be running them that fast?

Let me post up the drawing of the new board that would use 2 parallel IRG4PF50WDPbF and I'll also post the board that uses single IRG4PSC71UDPbF for you to look at the layouts of each. I tried to make them as compact as possible, but I'm no expert on board design, esp when it comes to high-power applications. Also, I doubt that these prototype boards from ExpressPCB won't handle the amount of current that I'm going to need to pass through the bridge board. Their website says that their traces are 0.0017" thick. Still though due to their price and being able to order just 2 at a time, I think they will make good prototypes. Then after I get my design perfected I can order a fully custom board with thick enough traces to handle the possible 50A of bridge current. Well, back to work on my board drawings.

 

[GeorgesWelding]

Here's the newest versions of my bridge boards. One is for a single IGBT the other is for 2 IGBTs paralleled together.

I switched the traces to the opposite sides from the originals as the first boards I posted were completely wrong. Power traces are now on the bottom of the board and the driver traces are now on the top. The IGBTs are mounted on the top of the board with their backs facing up and their leads will be bent over flat against the power traces on the bottom side and soldered solid. The two storage caps and 4 filter caps will also be mounted on the top(and glued to the board) and soldered on the bottom. The driver components will be mounted through the bottom and soldered to their traces on the top side. I'm tinkering with case designs in Solidworks yet, but I still think I will be mounting the power board upside down(IGBT Heatsinks facing downwards) with the controller board mounted a couple inches above it so that the driver leads between the control board and power board will all be very short.

View attachment Bridge Board (Dual IGBT & No Diodes).pdf
View attachment Bridge Board (No Diodes).pdf

 

[Orson Cart]

Also, I have a question, in the datasheets there is a graph that has Load Current vs. Frequency, and on this graph, as the frequency increases the Load Current decreases significantly. Now, does this mean that the current carrying capability of said IGBT will be way under its rated current since I will be running them that fast?

The faster you run them at a given current, in a hard switched topology, the higher the switching losses, therefore the less dissipation available for ON state current carrying losses - hence the lower current rating at faster switching speeds - total device dissipation and therefore die temp is fixed at a maximum for a given heatsinking arrangement.

The 4 igbt board looks good, on the power side, to draw 45A from a 220/230 power point into a bridge rectifier/cap arrangement, the power factor is around 0.6, giving 230V x 50A rms say = 11500VA drawn from the mains, x 0.6 = 6900 watts available to the converter, /375A on the o/p = 18.4V at the o/p at full power (higher at no load before striking).
If you try to draw more power than 6900W, then you will draw more primary current from the AC supply and sooner or later trip the breaker or blow the supply fuse.

 

[GeorgesWelding]

Output will be approximately 275A not 375A. But still, given the 6900W available to the inverter stage of the welder and a 275A output, that would leave 25V at full output which would be perfectly fine for TIG welding. I'm starting to get confused. 8-O

As for the lower current rating at higher frequencies. Won't that mean I will need many more IGBTs to carry the amount of current I will need? Still confused...

I do know that this guy's circuit: **broken link removed** only has a 16A bridge on the input and he's pulling a MAX of 150A at the output with a 20:6(3.333) turns ratio on the transformer. He is also only running 2 IRG4PC40W IGBTs at 42KHz. I honestly don't know what he is actually pulling from the mains, but his bridge didn't explode in the 150A test video. I do have a 100A/600V bridge for my mains rectifier as I figure over-kill never hurt anybody.

Sorry for the lack of knowledge. I'm starting to understand why even the cheap-o Chinese made inverter TIG machines are $1700 and the top of the line american ones are $10 grand!

 

[Anna Conda]

As for the lower current rating at higher frequencies. Won't that mean I will need many more IGBTs to carry the amount of current I will need? Still confused...

Yes, if you really want all that current at 45kHz, then you need more devices in parallel (//) however making sure they share their on state current is another thing.

You can buy high current mosfets at 500/600V that would do the job (less turn off losses) but there are quite a bit more expensive than the igbt's.

100A input bridge is a good idea (w h/sink).

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http://danyk.wz.cz/svar_en.html

That web page is really interesting, I'm not sure how his circuit survives given the caps are miles from the power devices, at least his current limit works.!!!

Also his o/p ammeter may not be measuring the true average current, as it appears to be an AC one off a switchboard, which may or may not read true average in this situation.

 

[GeorgesWelding]

I guess the inverter welder schematic I posted above was making 150A short-circuit current, the working/welding current was only 100A. I still feel as though my goal of 275A output is reachable. I'm back to doing more research. I could always drop the frequency to 25KHz(still above audible range) and just use a bigger switching transformer to eliminate some of the problems of the current carrying capability of the bridge. At the slower frequency I could still use the IRG4PSC71UDPbF within it's rated switching range, and at 25KHz that IGBT is capable of carrying ~22-25A according to the graph in the datasheet. I think that is the way I'm going to go. My original design called for 25KHz anyways, I didn't intend to go with the higher frequency until 3/4 way though my design but now I think it might be easier to go back to my original 25KHz frequency.

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His output is DC current so there should be a constant current draw at the output. If you look at his wiring closely, his welding leads go through that ammeter in series. While it may not be making the full 100A he claims, he is burning a 3/32" welding rod(although he has the electrode negative polarity while it should be positive) which requires about 70-80A to burn away that quickly on the wrong polarity. You are correct though, his circuit layout is really absurd to say the least.

 

[Orson Cart]

Just reading the above, you should really have a current limit on the primary side drive - which you can then set lower for testing, else if you short the o/p while welding you will blow the works up.
A current transformer (25kHz) on the transformer primary is one of the best ways, full wave rectified then a burden resistor, going back to the TL494 so that if instantaneous current exceeds "x" the devices turn off for that half cycle, it was on the video posted on that site above, showing the duty cycle dropping to <10% when the welder was "stuck" between strikes. This will save you buying a lot of igbt's and o/p diodes.

 

[GeorgesWelding]

In his circuit he uses a current transformer on the primary side of the switching transformer to regulate the output current. The way it is set up his controller runs the minimum duty-cycle that it takes to make the amount of amperage set by the current transformer's adjustment POT. Regulating the primary side current loosely regulates the output current. It should be steady enough though for stick welding. I'm going for much tighter current control do to the way TIG welding works.

For my current control I intend to run a DC current shunt(0.00015Ω) placed after the rectifier/inductor combo that follows the switching transformer. Measuring the voltage across the shunt will give me an accurate measurement of the current flowing through it. The voltage drop across the shunt is 75mV when there is 500A passing through it. I will only be running a max of 250-300A so @ 300A the voltage from the shunt will be only 45mV. This small signal will be amplified through an op-amp(Gain variable with a 25t trimmer) and will become a 0-5V signal that I will then pass to the POS terminal of ErrAmp1 of the TL494. On the NEG terminal of ErrAmp1 will be the reference voltage for comparison that will be set by the current control knob on the panel or varied with the foot-pedal amperage control so that the amperage can be adjusted while welding is taking place. The TL494 will respond by raising the duty-cycle of the output until the voltage from the current shunt op-amp is equal to the reference voltage set by the current control knob circuitry. Now if a short is placed across the output terminals of the welder(stuck the electrode to the work) then the current should spike severely, also spiking the voltage across the current shunt. This will in-turn cut back the duty-cycle of the TL494 to whatever the minimum required duty-cycle is to maintain whatever the output current is set at.

Now as a precaution I will do as you said and place a current transformer(I have a few, or I'll wind one) on the leads going to the switching transformer and measure the primary side current directly as well. I can then do as you said and full-wave rectify the CT's output and condition it to a usable signal(0-5V is what I like) and pass it to the TL494's POS terminal of ErrAmp2. (What is a burden resistor?) Then I can set a reference voltage on the NEG terminal of ErrAmp2 that way if the POS terminal exceeds the NEG terminal's reference voltage then the duty-cycle will be cut back. The way the ErrAmps work is that which ever amplifier is demanding the LEAST duty-cycle is the one that controls the TL494's output. So this should work fine. Under normal operation the current control amplifier(ErrAmp1) will be demanding less duty-cycle than the primary-side current limiter so the current control will work normally. Under a condition where the amperage at the primary side of the switching transformer spikes above the preset limit then ErrAmp2 will demand a lower DC than the current control amp(ErrAmp1) and the DC will be cut back to limit the primary side current. Only thing is I will have to make sure the primary side limiter is set high enough that I can achieve the full output amperage set by ErrAmp1 without triggering the maximum primary current set by ErrAmp2. This shouldn't be a hard thing to accomplish with some 25t trimmers to give me precise control over the adjustments.

Now since there is absolutely no voltage regulation here is what I'm doing to keep the duty-cycle of the TL494 from running full-blast when the output is activated but before an arc is struck(Output is on but no load across it, therefore no current draw) I have an Open-Circuit Voltage(No-Load Voltage) limiter hooked to the Dead-Time Controller pin of the TL494. I am sensing the voltage at the output of the switching rectifier and passing it through a voltage divider(Made up of 1MΩ and 56KΩ resistors) into a TL431A programmable shunt regulator. When the voltage through the divider which is applied to the TL431A exceeds 2.5V, the TL431A begins to conduct which then forward-biases a 2N3906 transistor which pulls the DTC pin of the TL494 up to the 5V reference voltage effectively disabling the TL494's output transistors. I tried this circuit on my 15A test rig with huge success. Without this circuit attached the duty-cycle of the TL494 runs full blast trying to maintain the output current. Since there is no load across the output this is impossible to do so it just runs at it's max of 48% and the OCV is approx. 100V. With this circuit attached when the output voltage of the switching rectifier exceeds ~50V the limiting circuit is triggered which adds dead-time, effectively reducing the duty-cycle, and limiting the output voltage. As soon as you attempt to strike an arc with the welding torch, the sudden load across the output drops the voltage below 50V, disabling the OCV limiter, and the current control circuits hooked to the ErrAmps takes over regulating the output current. Since the arc voltage of TIG welding is between 18-25V for an inverter machine, limiting the OCV to 50V doesn't effect how the machine welds. The OCV only serves two purposes, to set the DC to a minimum amount(It takes very little DC to maintain 50V at the output with no load) and since the OCV is limited to ~50V the machine is slightly safer.

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Update: Orson, you said that my small(4 switch) bridge board had an OK layout? What could I do to improve the layout of the 8 switch board just in case I have to parallel the switches together. As I said the IRG4PSC71UDPbF IGBT will carry ~22A at 25KHz but I don't think that is going to be enough to hit my goal of 275A of output. Also, I am a big fan of overkill when I build things because I'd rather be able to handle more current than I need than not be able to handle enough.

Thank you guys for all of your help so far. I really hope that after a few months I will have a full on TIG welding machine to show you guys!

 

[Anna Conda]

Interesting, I presume the 494 is on the mains side, driving the igbt's? and your o/p current sensing is on the o/p (isolated side), similarly for your o/p volts sense?

Are you going to tie the 0v o/p to the rectified mains 0V? You would only do this once.

In order to have the stable regulated current control on the o/p that you seek, your error amp will need to integrate the difference (error) signal over quite a few switching cycles, i.e. it will take a little while to respond to a short, while it will take a lesser time for the igbt's to go west - unless your fusing is pretty good.

This is why all successful designs of this type use cycle by cycle current mode to control the igbt's, that, and it gives balanced drive to the transformer, so the transformer does not saturate (see UC3856 data sheets and app notes) and cause an igbt failure.

If your TL494 implementation is quick enough (e.g. responds in 200-300nS after the primary current exceeds 20A say) then you might just get away with it and give time for the secondary side current limit to take over, but it may not be very stable over that time.

BTW "burden resistor" is that resistor with which you burden or load the o/p of a CT, so for pri side I sensing, using a 200:1 CT, 20A on the 1 turn thru, becomes 100mA on the sec side, if you burden this with 1ohm you get a 100mV signal, usually you would FWR (full wave rectify) the CT first with high speed diodes, e.g. 20-45V 1A schottkies, and then place the resistor. Say the diodes are 0.45V each + the 100mV = 1V needed on the CT sec for 1/2x25kHz, 20uS, so the CT needs to be rated for at least 20VuS (ideally the flux in the CT should be <50mT peak).
I assume you know this but do not leave the CT open circuit on the sec (200T) side, burden or short if not using.

Good luck and we all follow with interest.

 

[GeorgesWelding]

Hrm...Well I could always ditch the shunt and go with a current transformer on that side, but what about the V-Sense?

Here's the thing though, I can take the black wires(0V) from a PC power supply and touch them to the case(earth ground) and nothing happens. I'm basically building a very high current PC power supply, I'm using a PC PSU controller even. If I use two Isolated 15V supplies, one to drive the logic, and one to drive the HV side of the IR2110, each will have different grounding points, why can't I tie the negative lead of the rectifier that is after the switching transformer to the earth ground?

The negative lead from the rectifier that is after the power transformer in the welder I have in the garage is tied to the earth ground, it's NOT tied to the 0V side of the mains rectifier though.

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BTW, thanks for the info about the "burden resistor," that was VERY helpful explanation. Thank you!

 

[Orson Cart]

I'm not sure you understand that you can't run a signal from the isolated o/p to the 0V of the rectified mains, as this would connect the mains to your output. (I think Anna was saying you would only do this once as the shock you would receive would teach you not to do it again).
You need a zener / optocoupler circuit to get the effect across the isolation barrier for the Vo limit, same for the o/p current limit, an opto is needed to transmit the o/p of the error amp back to the TL494. There is nothing wrong with connecting your o/p to earth, this is a separate thing.

The primary CT is a good idea...!

Re layout for 8 igbt's, the key is to have the devices as close together as possible still allowing for good heatsinking, good and large de-coupling nearby, and to minimise the loop area of the drive to the transformer.

 

[GeorgesWelding]

I understand that I can't connect an output signal to the 0v from the mains rectifier. I have 3 wires from the mains though, +325, 0v, and an earth connection. The TL494 is powered from an isolated supply(transformer) though with its 0v at earth ground. If connecting the o/p to earth is fine then the o/p and the TL494's supply both have the same 0v point(earth) then I don't see the issue. I had no intentions of connecting anything to the 0v of the mains rectifier except for the inverter bridge. Now if I tie the logic 15v supply to earth then I will have to run a separate isolated 15v supply to feed the HV side of the IR2110 because that supply will need its 0v tied to the 0v of the mains rectifier/0v of the bridge to trigger the igbts.

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Thank you for the tips about the layout too!

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If what I said above isn't possible, then is there such a thing as an optocoupler or some type of isolator that has variable voltage that I can use to pass the V-Sense signal? I have no problem using a CT on the output for isolation just like on the input, I'm just unsure of how I would transfer the V-Sense signal since the voltage can be many different values.
Well, actually I just thought of something, I could make a simple circuit using a standard high-speed opto-coupler, where the optocoupler is only triggered once the output goes above 50V which would then trigger the OCV limiter, then once the voltage drops below the 50V threshold the optocoupler turns off until the voltage raises once again. That would work I would imagine and would completely isolate the output from the input via the optocoupler and the current control would be isolated by the current transformer. Thank you guys for helping me think of these things. You guys are excellent.

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Now, since I will be using a current transformer now on both the primary and secondary side of my switching transformer to regulate the output current and limit the maximum primary side current. I want to make sure I understand the usage of a current transformer properly.

As an example for the primary side current limiter. If I used a CT with a turns ratio of 200:1, as you said above it would give an output of 100mA when the current flowing through the primary reached 20A. I will rectify the output of the CT with a full-wave schottky 1A bridge as you suggested. Now if I burden the output of the rectifier with a 50R resistor, this should give 5V across the resistor when the primary current reaches 20A and the CT's secondary current reaches 100mA. Is this correct? What about the voltage drop across the rectifier diodes? If each diode drops 0.45V across it, and the current would be passing through 2 diodes at a time. Would this give a 0.90V drop off of the voltage across the resistor making 4.1V or how does the voltage drop across the diodes effect the output voltage across the burden resistor?

 

[Orson Cart]

Still some issues possibly, how does the 494 drive the primary side igbt's? with a 2110? if so the 494 is connected to the mains, and you cannot connect its 0V line to earth (mains earth). The high side supply to the 2110 has to be completely isolated from anything else, also as it goes up and down 325V at 25kHz you will want to put a common mode choke on the psu wires running to it, (10 turns bifilar on a toroid might just do it also), else you risk radiating into your control (and peoples TV sets).

[Just to avoid doubt, the supply to the TL494 should be completely isolated too, so that you can connect its 0v to the rectified mains 0v without mishap.]

A standard opto and a zener (47V) and some current limiting resistors should suffice to get your OVP circuit working and controlling the TL494, in a quick but linear fashion

The primary side current limit is a given with the CT, however if you want 5V signal + 2 x 0.45 schottky drop, the CT will have to be large enough to support 120V.uS (25kHz), i.e. 6V for 20uS.

Your o/p current limit with shunt is OK, just have to have an error amp driving an opto back to the 494 (to preserve mains to o/p isolation), the 494 op amp for this can be set to unity gain.

Using a 1000:1 CT (25kHz) on the Tx sec gives you 275mA on the CT sec, into 1 ohm this would be 275mV of signal that could be amplified by a good opamp to give 5V full scale - if you want to go that way. With a properly isolated CT and assoc circuitry you can bring this signal direct to the TL494 - except if you want a pot to set the I limit, the pot would be at mains potential!

Good idea to sketch the full circuit and show the isolation barriers on the sketch, e.g. mains Tx, opto's to ensure your finished article is safe.

 

[GeorgesWelding]

I'm sketching up the basic circuit layout right now for you guys to examine.

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When you talk about a common-mode choke on the mains wiring coming into the welder, you're talking about basically a filter to keep the high-speed switching noise from radiating back into the breaker box and interfering with other electronics in the house correct? I have an off-the-shelf mains filter that is rated for 20A that I am going to use on the input for the HF/HV arc starter, but I can easily build a simple one for use on the input for the welder's circuitry. That other inverter welder schematic I posted(the one with the absurd layout) has a noise filter on it's mains wiring. I think he's using a bifilar wound toroid with a 470n/250VAC cap on both sides of it for his noise filter. Also the 20A filter I have has the schematic with component values in the manual for it so I could copy that filter's design and just wind a new toriod capable of handling the higher current.

Also, as for the current control and OCV limiter, I've figured out a way to completely isolate the output side of the switching transformer from the controller board. I'm going to use current transformers for both the primary-side current limiting and also for the output side current control. As for the OCV Limiter, I designed a VERY simple circuit that just triggers a PS2561 optocoupler(Is this one fast enough do you think?) which then triggers the Dead-Time Control pin of the TL494. Everything simulated fine in Multisim so I think it will all work. I'm working on the drawings right now so I will post up the schematics soon.

For the CTs I did some calculations and this is what I came up with, so let me know if I did this properly.

For the primary side current limiting I can use a 200:1 CT(This gives 250mA@50A), rectified, then driven into a 2Ω 2W resistor. That should give me a 0-500mV(0-50A) signal which I can then run through a simple op-amp with a gain of 10 to give me the 0-5V signal I desire.

For the output side current controller, I can use a 600:1 CT(This gives 500mA@300A), rectified then driven into a 1Ω 2W resistor. That should also give me a 0-500mV(0-300A) signal which I can do the same as above and amplify it to the 0-5V signal that I desire.

Since the CTs are both isolated from everything, once their signals are amplified by their respective op-amps, I should be able to run that 0-5V signal directly to the TL494 and handle them that way. To set the primary side current limit and to be able to control the output current, I won't be adjusting the signals from the CTs. Those signals will always be 0-5V signals. What I will be adjusting is the reference(compare to) voltage on the opposite pin of their respective ErrAmps in the TL494. For the primary current limit I can set it's compare voltage with a 25t trimmer for very precise control and for the output current control I can set its compare voltage with a POT to set the MAX current, a foot-pedal to adjust the amperage WHILE welding between 0 and MAX(set by the knob) and then I have also designed a simple current pulsing circuit which will pulse the compare voltage between 2 set points giving a very handy current pulsing feature to the welder.

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I got a question. In this schematic that I posted (**broken link removed**) what is the purpose of the 6R8/2W resistor in series with the 4n7/1KV cap across the secondary side of the switching transformer? I'm pretty sure that the 4K7/3W resister across the output terminals is to keep a slight load on the output to keep the voltage from spiking extremely high. Is this correct? The welder I have in my garage has a 250R/100W/WW resistor in parallel with a 25R/50W/WW resistor that is in series with a 30uF/165VAC cap. The cap is pretty large about 6" tall, 2.5" wide, 1" thick and these 3 devices are placed across the output terminals of the welder. They are used as a dummy load from what I was to understand to keep the OCV voltage down to 100V. If you unhook them the OCV jumps to ~150V. This is a transformer based welder though, not an inverter.

Anyways. Here is the layout schematic for my current and voltage sensing isolation circuits. I think they should keep the output side of the switching transformer completely isolated from the logic devices and still provide me with plenty of control over the current and OCV voltage.

View attachment Isolation Circuits.pdf

 

[Orson Cart]

When you talk about a common-mode choke on the mains wiring coming into the welder, you're talking about basically a filter to keep the high-speed switching noise from radiating back into the breaker box and interfering with other electronics in the house correct?

Nope, I was talking about the psu wires that go to the high side drive of the 2110. If you are using a separate supply to power the high side of the 2110.

For the primary current limit I can set it's compare voltage with a 25t trimmer for very precise control and for the output current control I can set its compare voltage with a POT to set the MAX current

So this pot is essentially connected to the mains?

The RC combinations across the o/p of the transformer are called snubbers, they limit the voltage spiking you get at higher powers due to the imperfect action of the o/p diodes, and are often quite necessary to stop the diodes failing short. Often snubbers are placed directly across the diodes.
On 50/60Hz transformers a large C and R might be there to form a minimum load.

The core size (Ae) of your CT's need checking to see if the volt-second product is high enough for your app. dB/dt=V/(N.Ae) dt=20uS, B<50mT, N= 200T or 600T, V = total average CT o/p volts over the 20uS, Ae = core area of CT (m^2).

You might want to take your Vo limit sense from the actual output, and have a cap after the 100Ω to 0v, to smooth things out a little, possibly also a cap across the opto LED for same reason, also a standard opto should be fast enough.