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Registered Member #54436
Joined: Sun Feb 08 2015, 10:33PM
Location: Spain
Posts: 9
Hello IH family: Trying to follow the Campbell project, I built the hybrid drivers for the IGBT’s, (M57962L driver and VLA106-15242 DC-DC converter) but I’ve found a strange behaviour in them: the output signals they delivery to the gates is wider than the input ones. It seems those drivers delivery the signals referenced to the negative cycle of the input pulse… which is always under the theory 50%, so both channels of igbts are shorted for a time, see pictures. I’m scared to feed the bricks under that short-circuit… I’ve tried with different oscillators: the mindchallenger PLL, another one based on a IC3525, other on the pwm494, also a 555 timer, all with and without gate transformers, etc. Problem persists… Even using the 494 as pwm, I’ve seen the narrower is the input, the wider the output. Datasheets don’t clarify anything… I’ve tried to contact to Mr. Campbell, but, unfortunately it seems he is no longer active in IH world… and what is worse, will never finish his wonderful ReactorForge project. So, I come here hopping someone else has worked or somehow involved to those hybrid drivers, for any help if ever found the same problem. Thank you in advance..
Registered Member #54436
Joined: Sun Feb 08 2015, 10:33PM
Location: Spain
Posts: 9
Hello Parasole; Yes I read that datasheet, and above all I stuffed that hybrid drivers following strictly Campbell's BOM and pcb layout... What I find hard to believe is, as I said, the out pulses provided by this drivers are WIDER than the input ones, the signal from an oscillator. So it always will be a shortcircuit time. Maybe it works as a kind of "oposite", taking the negative cicle of the wave to drive the gate signals. In this case, I dont't know how to generate a wave with wider negative pulses... Yours is some different to mine; test it in a breadboard and tell me the result... Thank you.
Registered Member #54647
Joined: Wed Mar 18 2015, 02:52PM
Location:
Posts: 13
Ramonn wrote ...
What I find hard to believe is, as I said, the out pulses provided by this drivers are WIDER than the input ones
hm, looked once again to your signals, did you try to change the duty cycle for your signal? Try for example 30/70 instead of 50%... Then you may adjust your input signal duty cycle to the desired dead time interval, I believe this is the way... From data sheet it looks like the rise time and propagation time could be different within their limits, then no surprise that there might be differences in output signal: “L-H†Propagation Time tPLH VI = 0 to 4V, Tj ± 85°C — 1.0 1.5 µs “L-H†Rise Time tr VI = 0 to 4V, Tj ± 85°C — 0.6 1.0 µs “H-L†Propagation Time tPHL VI = 0 to 4V, Tj ± 85°C — 1.0 1.5 µs “H-L†Rise Time tr VI = 0 to 4V, Tj ± 85°C — 0.4 1.0 µs
Registered Member #54436
Joined: Sun Feb 08 2015, 10:33PM
Location: Spain
Posts: 9
Hello; having another look to the paper, I've noticed that the built-in optocoupler seems really working referenced to negative cycle, see picture... pin 14 is constant 5V supply, and pin 13 is driver signal, so the opto only will be 'on' during negative cycle. I've tried a pwm pulse, it confirms that: decreasing the duty increases the output!! I'm thinking to change polarity; 5V to pin 13 and signal to pin 14. Let's see what happens...
Registered Member #54647
Joined: Wed Mar 18 2015, 02:52PM
Location:
Posts: 13
Ramonn wrote ...
Hello; having another look to the paper, I've noticed that the built-in optocoupler seems really working referenced to negative cycle, see picture... pin 14 is constant 5V supply, and pin 13 is driver signal, so the opto only will be 'on' during negative cycle.
Now I got what do you mean by negative cycle Actually it is not negative, at the input it is logical level and there is high (Vcc) and low level (ground). Than yes, while having feeding positive voltage supply, signal side active level is low, and this is all you need to know, just provide low active with requited duration. Obviously, you may ground pin 13 than feeding high active level to pin 14 will do the job, just make sure you will provide enough current, usually active high is having lower load capability than active low, just due of the way output stage is done on ICs...
Registered Member #54436
Joined: Sun Feb 08 2015, 10:33PM
Location: Spain
Posts: 9
Yes, I said it wrongly!, both my english and electronics knowledge are small, :( meant "low level".... I changed pins as I said, but as I start from the simple oscillator output from the PLL 4046 chip, I have to "duplicate" it, with a transformer, obtaining two signals to feed the two pairs of IGBT gates (180 degrees de-phased), what leads me to the original problem: overlapping open-times for the igbt's. But I've discovered that decreasing de voltage of the pcb, decreases the overlapping, or putting a resistor in one input of the transformer. But doing this the signal is weak and jitters... I'm going to get mad!!. -Have you tested yours?
Registered Member #54647
Joined: Wed Mar 18 2015, 02:52PM
Location:
Posts: 13
Since you use 4046, I would add an comparator for duty cycle adjustment, then one more invertor for having 180 degree bridge signals... Another option would be wiring few logic elements and introduce dead time... Do some google search, there are plenty of examples how to do this...
I don't know when I will test my drivers, still to busy... From other side I would not have this issue since I am intending to use controller and will have full control over my driving signals...
Registered Member #54436
Joined: Sun Feb 08 2015, 10:33PM
Location: Spain
Posts: 9
Ok, thanks for advises, and good luck in your project. If you get succes in that controller, will share it? As I understand, the 4046 PLL never has a good or total apliance for our DIY induction heaters, just getting bigger our boxes of blown transistors
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Induction heater: hybrid drivers strange behaviour
Actually it is not negative, at the input it is logical level and there is high (Vcc) and low level (ground). Than yes, while having feeding positive voltage supply, signal side active level is low, and this is all you need to know, just provide low active with requited duration. Obviously, you may ground pin 13 than feeding high active level to pin 14 will do the job, just make sure you will provide enough current, usually active high is having lower load capability than active low, just due of the way output stage is done on ICs...
