How to drive IGBTs (sinewave dimming) - Page 1
Forum section: electronics / projects

author: stefan_o
author_rank: Regular Contributor
timestamp: April 14, 2017, 12:31:55 am
content: Hi,I like to create a sinewave dimmer (230V, 10A, no phase cutting, just PWM) using IGBTs. I never worked with IGBTs before but there seem to have much higher switching losses compared to FETs and I'm not sure how to calculate the drive componets.I found this circuit of a sinewave dimmer:https://www.uploadarchief.net/files/download/sinewave-dimmer_2010.pdfI did not find any more information on that circuit (is it just an idea or has someone actually build it or is it reverse engineered?), but I used it as a starting point.It is designed for just 5A, but I would like to have 10A. It seems that IGBTs are more efficient than FETs, I think theInfineon IKP15N65H5is well suited (for hard switching applications, low losses, designed for 15A, "best in class"). Instead of the obsolete Analog ADUM1230 in the original schematic I found theTexas Instruments UCC20520: Very similar device, but one major advantage: Just one input and programmable deadtime, so now deadtime handling in µC (a software crash won't release magic smoke).I attached my current circuit. What I would like to know:-How to calculate the IGBT drive resistors (R2, R3, R9-R14)?-What current do I need in the voltage bridge with the zener diode (what value to choose for R4/R6)?-The IGBT datasheet states 0.17mJ switching energy: 0.17mJ * 40kHz * 4 = 27.2 Watts => Does the transformer needs that much power? If not, how to calculate transformer?If you have other ideas or if you see problems in the circuit, please tell me.Best regards,Stefan

author: Ian.M
author_rank: Super Contributor
timestamp: April 14, 2017, 01:25:46 am
content: Way over-complicated. The topology you probably want is an IGBT shorting the + and - terminals of a bridge rectifier. Stick a RC snubber across the IGBT. Connect the AC terminals of the bridge in series with the load. Either drive the IGBT gate via a pulse transformer so you don't have to provide a floating low voltage supply for the gate driver, or use an optocoupler and a gate driver IC, referenced to the bridge - terminal powered by an isolated DC-DC converter. PWM it however you like - 555 or MCU or whatever. It only needs a single PWM line with no worries about shootthrough or deadtime.Silicon Chip published the above arrangement as a 10A motor speed controller back in Feb 2014.http://www.siliconchip.com.au/Issue/2014/February/230V-10A+Speed+Controller+For+Universal+Motors%2C+Pt.1Unfortunately the circuit you want is behind a paywall.

author: stefan_o
author_rank: Regular Contributor
timestamp: April 14, 2017, 04:42:52 pm
content: Thanks for your fast answer. I have seen circuits like that as well, similar to this one:With the difference to put the load on the other side of the rectifier and switch IGBT with a small transformer (if you just change the load to the other side you won't be able to power 100% anymore).My thoughts about it (could be wrong):-Higher losses, because two diodes and IGBT instead of one diode/IGBT in the power path-I need 5 high current semiconductor instead of 4 (ok, I could use diodes with one common connection)-Is is possible to power any load (transformers, electric transformers), no matter how voltage/current relate in a period?-More filtering to avoid noise back to supply.Of course no chance of shoot through, but that could also be achieved by removing half of the original circuit I posted?Still the question remains, how do I calculate the gate drive resistors and the power needed for the drive circuit?Best regardsStefan

author: Ian.M
author_rank: Super Contributor
timestamp: April 14, 2017, 05:16:44 pm
content: That's a completely different circuit topology - it puts DC through the load (actually full wave rectified 50Hz mains chopped by the PWM, but who's nitpicking?). You'll have something like 310V peak on C1, so it only can supply about 3mA of gate current so turnon will be rather slow. R1 is dissipating 0.9W and with >200V across it needs to be a high voltage resistor.The IGBT wrapped in a bridge can drive any load, provided the repetitive pulse current is limited to what the diodes and IGBT can handle, and back-EMF from inductive loads is clamped or otherwise damped below the peak inverse voltage rating of the diodes, and the voltage rating of the IGBT. Of course, the clamping/damping may introduce high losses . . . .Your circuit has the extra pair of IGBTs shorting the load to allow free-wheeling so may well have lower losses for high currebt inductive loads. The alternative is to tune the load with a shunt capacitor to near resonance at the mains frequency which improves the power factor and gives the current in the load inductance simewhere to go at IGBT switchoff.Silicon Chip's version was a 10A universal motor controller, and you don't get much nastier loads than that. Unfortunately I don't recall how they controlled the back-EMF and I'm not sure which EPE isuue it was republished in to take another look at it.To calculate the drive circuit power, you have to study the IGBT datasheet S.O.A curves to see how long you can afford to spend switching, then calculate the amount of gate charge you have to slam in-to/out-of the gate to do that.

author: stefan_o
author_rank: Regular Contributor
timestamp: April 14, 2017, 08:03:26 pm
content: QuoteI noticed that. That's what I meant with "load on the other side of the rectifier"I read the free pages and I noticed that there are two controls speed and feedback. Unfortunately I could not even find a part of the circuit for free, no large picture of the PCB either. My idea is to have a dimmer for any load, so I cannot specifically tune it to one special load.I foundthis document for Deadtime calculation, but nothing about how to calculate the gate resistors. I read somewhere (cannot find source anymore), that you need a minimum resistor to be able to drive the IGBT in a safe way and of course the higher the resistor, the slower the switching. Also that you add an diode with second resistor to be able to discharge faster. The driver I found (UCC20520) can handle up to 4A / 6A, so even 10 Ohm would be enough.For total energy: Datasheet says 38nC charge:38nC * 40kHz = 1.52mAIs that correct? How do they get to the 0.17mJ switching losses also in the datasheet?0.17mJ * 40kHz = 6.8 Watts1.52mA * 18V < 0.03 WattsEven If I need factor 2 for charge and de-charge, still factor 100 less? What is correct here? If the 1.52mA are correct, even with losses at the resistor and the current for the driver, a 100mA transformer should be more than enough.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 14, 2017, 08:45:21 pm
content: Switching loss is collector power dissipation, not gate drive. You need that much heatsinking.Your circuit is missing a "DC" Link capacitor (which will of course be AC in this case), EMI filtering, and some kind of current protection (IGBTs will poof in ten microseconds if you aren't watching collector current).With these additions, you should be good to go.The nice thing about this kind of circuit: it's not just a dimmer, it's a switching VARIAC. It's still limited to whatever current the transistors can handle, so for the selected parts, it doesn't have any current gain. But it might be handy to pick more robust parts, so that more current is available at lower voltages.You can also use active current limit as a feature: inrush limiting. (Given that stubborn loads may not start up at all, without inrush. Another good reason to use oversized transistors?)Since you've got semiconductors across the line, don't neglect transient protection, either. IGBTs aren't rated for avalanche, but even if you were using MOSFETs rated for avalanche, they'd not even begin to withstand a proper surge...Note that MOV clamping voltage will necessitate higher voltage devices. That is, for 240VAC operation, you need 340V peak operating voltage, so a "360V" MOV would be reasonable. This will have a peak clamping voltage around 800V, so that 1000 or 1200V devices are desirable. This makes IGBTs all the better choice, since they have lower voltage drop than MOSFETs.Finally, when -- not if -- the transistors or MOV fail, you need a fuse on the input. Possibly two fuses, one for each line, if you can't guarantee which lines are hot and neutral (which is used in some configurations).Tim

author: Ian.M
author_rank: Super Contributor
timestamp: April 14, 2017, 08:55:17 pm
content: I mis-remembered the schematic. The Silicon Chip project, as republished in the March 2015 issue of EPE, feeds the load with pulsed DC in a similar manner to your reply #2 circuit. I apologise for misleading you, and now cannot remember where I'd seen the IGBT or MOSFET wrapped in a bridge as an AC dimmer.Obviously I cant post a scan of the Silicon Chip circuit here due to copyright, but I can describe the circuit to you:The mains is bridge rectified (35A) for the DC bus, which is allowed to swing with the full-wave rectified waveform, as it only has a 100nF cap across its output. Everything is referred to the bridge - terminal as local 0V. Another capacitively coupled bridge (with the caps acting as droppers from L and N) provides a shunt regulated 15V supply to power the control circuit and an IR2125 gate driver with current lmiting. The IR2125 drives the gate of a STGW40N120KD IGBT via a 4R7 resistor. There is a 0.01R resistor, emitter to 0V, providing current sense feedback to the IR2125, and also via a sampling and signal conditioning circuit to the controlling MCU. There is a series RC snubber 33R 3W, 10nF collector to 0V. The load is wired bridge + to IGBT collector with a SL32 10015 NTC thermistor for inrush limiting and a STTH3012W free-wheeling diode.Whether a single IGBT wrapped in a bridge (or two, one as a series pass device and the other driven antiphase to provide a free-wheeling path across the load) is more viable than your original circuit depends heavily on the current you are designing for and the availability of cheap, reasonably fast low Vf individual diodes, half bridges or bridges. If you can use dual diodes and skip the free-wheeling circuit, you'll only need three power devices. OTOH if you are using modules or worse 'hockey-puck' IGBTs and diodes, your four IGBT topology will be cheaper and probably lower loss, though the body diode Vf drop, may not actually be less than the 2*Vf drop if a Schottky bridge is used.Edit: IR2125 part no. corrected

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 14, 2017, 10:08:14 pm
content: Quote from: blueskull on April 14, 2017, 08:58:02 pmIt certainly will, but not for long, with 15A rated transistors.Tim

author: stefan_o
author_rank: Regular Contributor
timestamp: April 15, 2017, 12:15:44 am
content: Thanks for your great input!QuoteFuses of course, they are not shown in the circuit, because they will not be PCB mounted. Any hint on how I should design EMI filtering?QuoteThat was my idea:Output=Input*PWMdutyOf course there are some additional losses, but mostly load independentQuoteThank you very much! So this circuit is not really what I want, but they also have the free wheeling in it, which is more complex in case of AC.QuoteIt seems that a FET like theInfineon IPP65R110CFDis better suited, with 0.11 Ohm there are just 11 Watts loss at 10A, which is less than the IGBT, not even talking about switching loss...It is rated 20A at 100°C continuous and 99A pulsed, that should be enough even for devices with major inrush current.So by replacing the IGBTs by these FETs I can solve problems with output filtering, inrush current etc.. Do I still need some kind of protection or is this FET overrated enough?

author: Ian.M
author_rank: Super Contributor
timestamp: April 15, 2017, 12:37:56 am
content: AAGH. I cant seem to type part numbers correctly tonight. The gate driver in the Silicon Chip circuit is Infineon'sIR2125which does cycle by cycle current limiting by shortening the gate drive pulse if the current threshold is exceeded.Due to your active free-wheeling shunt topology, if you are doing cycle by cycle current limiting, you need to do so in the control circuit, not in the MOSFET driver as you need to generate the complementary waveform with deadtime inserted for the shunt. If you include a small amount of extra inductance in the load circuit to limit dI/dt, your control circuit should have time to react even if the output is suddenly hard-shorted.

author: stefan_o
author_rank: Regular Contributor
timestamp: April 15, 2017, 02:23:24 am
content: QuoteQuoteIf the voltage drop increases, can I use a varistor to clamp it? There are ones with 10V clamping voltage (like this one). 10V at 0.11Ohm are about 91A, that should be fine? A super rapid fuse needs max. 10ms to break at 10x rated current, varistors only need to survive that time.Do I still need some active circuit? What I like about the Texas Instruments UCC20520 is, that I only need one PWM input and the deadtime/second output is handled by it, so no worries in the µC code. But no desaturation protection.QuoteCan you point me there, whenever I google Wolfspeed and Infineon I only get financial stuff (Infineon wanted to buy Wolfspeed, but weren't allowed, US national security concerns...)

author: Ian.M
author_rank: Super Contributor
timestamp: April 15, 2017, 02:46:01 am
content: No. you cant clamp it - if you did it would short the MOSFET whenever the Vgs went over your 11V varistor breakdown voltage. Instant smoke!However you can monitor the Vgs drop with the MOSFET on - its a bit tricky as you need to sample it excluding the initial turnon transient and of course turnoff, and use that to shut off the PWM pulse early. IMHO its not worth it - its simpler to sense the current for cycle by cycle limiting so you can totally avoid desaturation rather than shutting off if it becomes excessive.

author: stefan_o
author_rank: Regular Contributor
timestamp: April 15, 2017, 03:16:51 am
content: QuoteYes, maybe it's time for me to go to bed now, forgot that the idea of a FET is, that it can also be switched off...QuoteWhat's the best way to do it? A shunt resistor, some op amp and an optocoppler connected to the disable pin of the UCC20520 (will disable within 20ns according to datasheet)?QuoteIt should run safe & stable, industrial grade is a bit overkill. Both devices seem to be a lot more complex and need more additional components. But I would prefer to have all the "dangerous" stuff hard-wired, so whatever my software does, it cannot blow up the circuit.

author: NiHaoMike
author_rank: Super Contributor
timestamp: April 15, 2017, 03:21:09 am
content: A friend of mine had that idea of a "solid state variac" many years ago, and while it's easy in theory, the devil is in the details. Namely how to deal with the transitions, which requires either a very difficult to achieve near zero dead time or clever gating of the signals so there is always a freewheeling diode in play. Things get even messier when trying to account for reactive loads.Instead, use the 4 transistors in a H bridge and add a capacitor and bridge rectifier. Or cut it down to two transistors and add two capacitors and two diodes. And that can control both the output voltage and frequency, greatly increasing versatility.

author: stefan_o
author_rank: Regular Contributor
timestamp: April 18, 2017, 01:36:49 am
content: Quote from: Ian.M on April 15, 2017, 02:46:01 amI made an addition to my circuit, it contains twoTI INA200current sense amp/comparators for both directions and an optocoppler that is connected to the disable pin of the UCC on the other side. I only measure at the side when the load is powered, not freewheeling, how should current on free-wheeling be any higher? (except you connect a power source).Using the optocoppler I can also directly signal overcurrent to the µC. I should add an optocoppler in other direction to reset the comparator using the µC. I attached the circuit, is it a good idea?Quote from: NiHaoMike on April 15, 2017, 03:21:09 amIsn't it possible to clamp output voltage at for example 400V to protect the FETs against voltage spikes? What other transitions do you speak of?

author: stefan_o
author_rank: Regular Contributor
timestamp: April 18, 2017, 02:11:56 am
content: Quote from: blueskull on April 18, 2017, 01:44:26 amI rather blow up some FETs than myself... But using saltwater as resistor with moveable electrode was still a common method for dimming in theaters about half a century ago. But I highly doubt that any installation is still in use.

author: Zero999
author_rank: Super Contributor
timestamp: April 18, 2017, 07:54:59 am
content: Quote from: stefan_o on April 18, 2017, 02:11:56 amThere's still one where I work, in a motor speed control application. Quite often there's no point in upgrading infrastructure, when it works and is reliable, so obsolete technologies often remain in place for many years.

author: Zero999
author_rank: Super Contributor
timestamp: April 18, 2017, 03:42:36 pm
content: Quote from: blueskull on April 18, 2017, 08:01:47 amI've no idea of the power consumption. I doubt it's an AC motor, probably DC. Yes, replacing the motor with a modern one (if there is a suitable unit which would be a mechanical fit) and drive would save power and money but it's nothing compared to the amount of power the whole site uses.

author: stefan_o
author_rank: Regular Contributor
timestamp: April 19, 2017, 09:06:26 pm
content: QuoteI looked at a selection guide and it says CFD2 is for soft-switching, but this is a hard-switching PWM application. For this kind of application Infineon recommend the P6 or C7 series. The C7 series is way more expensive than the P6, but has lower losses. Is there a special reason that you recommend the CFD2 series?

author: stefan_o
author_rank: Regular Contributor
timestamp: April 19, 2017, 11:22:47 pm
content: Quote from: blueskull on April 19, 2017, 09:23:38 pmI compared the datasheets of P6 and C7, it seems the only benefit of the C7 over P6 is the lower Rds(on), the capacity of the C7 is slightly higher. For currents with maximum 10 A continuous this is not so important (5W vs. 10W). Any special reason why I should choose C7 over P6, the C7 costs more than twice as the P6.I added a reset optocoupler to the circuit and designed PCB (see attachments), is this an appropriate design? For input EMI filtering I will use a standard filter and add two fuses. The two holes in the middle are for a heatsink.Edit: I did not Kelvin connect the amp to the shunt resistor, because I do not want to precisely measure the current, but just set a threshold for shut-down at ~20A.

author: stefan_o
author_rank: Regular Contributor
timestamp: April 20, 2017, 12:19:14 am
content: Quote from: blueskull on April 19, 2017, 11:38:57 pmGate charge is pretty clear, but what exactly is Qout?Quote from: blueskull on April 19, 2017, 11:38:57 pmYes, like a resistor, there is a resistance and a maximum allowed power based on heat dissipation, on that you can calculate Imax. I think 10 Watt heat dissipation can be done without aggressive thermal design, for 20A (40 Watts), the situation is different.