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Answer to riddle: Why a fast diode in a gate driver bootstrap circuit?

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Sam Ben-Yaakov

Sam Ben-Yaakov

2 жыл бұрын

Relevant video:
Gate driver bootstrap riddle: Why a fast diode?
• Gate driver bootstrap ...
Relevant publication:
N. Krihely and S. Ben-Yaakov, “Simulation bits: adding the reverse recovery feature to a generic diode,” IEEE Power Electronics Society Newsletter, 25, 2, 26-30, 2011.
www.ee.bgu.ac....

Пікірлер: 71
@dixsusu
@dixsusu 2 жыл бұрын
All 18 minutes pass by like it was 5 minute talk presentation . Thank you Sam .
@sambenyaakov
@sambenyaakov 2 жыл бұрын
🙂Thanks
@_-Skeptic-_
@_-Skeptic-_ 2 жыл бұрын
one of the most under rated channels. I'm writing this comment in the hope to make your channel reach more people, cause youtube only suggests clown channels. I'm really grateful for all the information you are providing us with.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Thank you for kind note and support. I hope it will have an effect.
@mancio92M
@mancio92M 2 жыл бұрын
one of the best channels of explanation, complete, precise and extremely detailed.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Thanks😊
@griswoldmark4258
@griswoldmark4258 2 жыл бұрын
One of the best technical series on power electronics available. Dr Ben Yaakov is just exellent!!
@sambenyaakov
@sambenyaakov 2 жыл бұрын
🙏🙂
@mohalothman99
@mohalothman99 2 жыл бұрын
Thanks for explaining! It is really precious
@sambenyaakov
@sambenyaakov 2 жыл бұрын
You were quick😊Thank you
@taki_maciek4799
@taki_maciek4799 2 жыл бұрын
Thank you! You've answered a question that I won't even figure out to ask. Very interesting material :)
@sambenyaakov
@sambenyaakov 2 жыл бұрын
👍😊
@sayantansinha292
@sayantansinha292 2 жыл бұрын
Thank you very much Prof. for this video, without this I never could know about this hurdle hidden under an apparently simple looking riddle.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
👍
@theoldbigmoose
@theoldbigmoose Жыл бұрын
At 70 I still learn much from your presentations Professor!
@sambenyaakov
@sambenyaakov Жыл бұрын
🙏😊
@tseckwr3783
@tseckwr3783 2 жыл бұрын
Professor Sam. Thanks for another fine video.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Thanks
@David_11111
@David_11111 2 жыл бұрын
yay ... perfect timing for what I am doing today :) thankyou
@sambenyaakov
@sambenyaakov 2 жыл бұрын
👍😊
@nyyotam4057
@nyyotam4057 2 жыл бұрын
Thanks professor. Now I understand that you did not intend to get as far here, as the class lecture. Only to discuss the reverse current in itself. Definitely a good lesson in humility for me 🙂. Thanks for the detailed explanation.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
👍
@neduodumbu1969
@neduodumbu1969 2 жыл бұрын
Thank you Prof. Watching your explanations makes me wish to physically attend your class. I hope I will get the opportunity someday.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
😊I hope so. You never know.
@savassahin6371
@savassahin6371 2 жыл бұрын
Thank you very much Professor, sharing absolute clear knowledge and teaching, i learned a lot from your videos.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Thanks for kind note.
@2meters2
@2meters2 Жыл бұрын
Sam, thank you so much ! This was brilliant ! I am working on a prototype power converter with a 400 V high rail, using a bootstrap circuit for the upper transistor. For the past two days I've been trying to figure out why there is a large spike (only 20 nsec long) during on-switching of the upper transistor, which even affected my 15 V driver supply (which got a little jolt). I've been thinking about the reverse-recovery for the diode (I used an old 1N4007 that I had laying around), but did not think that made sense since the diode should not be conducting at that time... Now I know my problem, and your explanation was (as always) impeccable. Thank you very much !
@sambenyaakov
@sambenyaakov Жыл бұрын
😊Thanks for sharing
@xDR1TeK
@xDR1TeK 2 жыл бұрын
We are humbled by your wisdom professor.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
A bit exaggerated. Thank youI was really touched.
@xDR1TeK
@xDR1TeK 2 жыл бұрын
@@sambenyaakov in no way I could have imagined that the LS body diode will influence the bootstrap diode recovery.
@RaananL
@RaananL 2 жыл бұрын
As usual, very informative and well explained !
@sambenyaakov
@sambenyaakov 2 жыл бұрын
👍🙏
@zaikindenis1775
@zaikindenis1775 2 жыл бұрын
Thank you for the video! I recently found that LTspice had always possibility to simulate reverse recovery using parameter vp. But documentation and reference about it appeared only in version Ltspice17 and not in Ltspice4.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Vp is for shaping the second half of the reverse recovery. I yet to see a SPICE model of a commercial a diode that simulate accurately the reveres recovery.
@mand6029
@mand6029 Жыл бұрын
a great video thanks Prof
@sambenyaakov
@sambenyaakov Жыл бұрын
Very welcome. Thanks.
@seanm8030
@seanm8030 2 жыл бұрын
Good explanation.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Thanks.
@sudheerkumar5966
@sudheerkumar5966 2 жыл бұрын
Very good 👍
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Thanks
@eduardinification
@eduardinification 2 жыл бұрын
Thank you professor. I understand that this is true provided that the dead time period is short enough to prevent the bootstrap capacitor from being totally charged. Right?
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Indeed .
@esijal
@esijal 2 жыл бұрын
Very nice professor 🙏, I believe for GaN Fets the effect of extra charging in the dead time period would be much worse as the reverse voltage drop could be in the range of (Vth+Vgs_off)>Vd_MOSFET... (For the case of outgoing load current)
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Good point. Thanks. But...bootstrap supply is not recommended for GaN drivers
@hamidk4772
@hamidk4772 2 жыл бұрын
Outstanding Job x 2.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
ThanksX2 😊Hamid
@electromaster9836
@electromaster9836 Жыл бұрын
Perfect presentation (as always). Could you confirm that with LLC or CLLC, as the current is reversed before closing high side, reverse diode recovery is not involved? In other words, no need to have a fast bootstrap diode with resonnant converters.
@sambenyaakov
@sambenyaakov Жыл бұрын
When the current is lagging there is no dead time after turn off so the extra charging does not exist so no need for fast diode.
@jackywang1717
@jackywang1717 2 жыл бұрын
Thank you very much! My question is: does schottky have better performance than fast diodes?
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Yes (up to 100V), but you need to choose a low leakage device.
@jackywang1717
@jackywang1717 2 жыл бұрын
@@sambenyaakov thanks you!.got it
@Chris_Grossman
@Chris_Grossman Жыл бұрын
A Schottky diode does not have any minority carrier charge storage, it is a majority carrier device. The TT term in the PDS5100 model makes no physical sense.
@sambenyaakov
@sambenyaakov Жыл бұрын
As know all diodes share same templates which are tuned to each diode
@pigrew
@pigrew 2 жыл бұрын
This makes me wonder about the tradeoffs of Schottky vs fast PN diodes in this application. Schottky should have higher leakage but I don't think this would be a significant disadvantage?
@tonysonglalala
@tonysonglalala 2 жыл бұрын
Should be ok if you have enough bootstrap cap.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Schottky is better (up to about 100V), choosing a low leakage device.
@Trading_Mindfully
@Trading_Mindfully Жыл бұрын
Hi Prof.... Its a great video... I have a practical issue in my project.... I'm using bootstarp gate driver for pmsm motor inverter. Issue is as and when I turn ON inverter voltage (270vdc) my low side gate signals are distorting. Could you please share possible reason
@sambenyaakov
@sambenyaakov Жыл бұрын
Is the input to driver Ok? You have sufficient deadtime?
@tamaseduard5145
@tamaseduard5145 2 жыл бұрын
👍💖🙏
@sambenyaakov
@sambenyaakov 2 жыл бұрын
🙏😊👍
@pratheeshvidya9385
@pratheeshvidya9385 Жыл бұрын
Sir, what should be the voltage rating required for this diode? Do we have to consider the drain voltage?
@sambenyaakov
@sambenyaakov Жыл бұрын
It should be the bus voltage (Vb+Vcc)-Vcc
@pratheeshvidya9385
@pratheeshvidya9385 Жыл бұрын
@@sambenyaakov Thank you sir
@JacquesMartini
@JacquesMartini 2 жыл бұрын
Iam afraid I have to disagree. 1.) The startup problem exist only in theory and simulations, since in a real circuit, the driver voltage is applied a LOT of time BEFORE the circuit starts switching. When the low side driver is ON, the bootstrap capacitor has all the time in the world to charge. If both drivers are in tristate and so both MOSFETs are off, the problem will be there. But this is a rather questionable design. 2.) The polarity of the load current or type of load (resistive or inductive) does NOT matter! So does the voltage of MID. When the low side is ON, the voltage of MID will be somewhere around 1-2V plus (negative output current) or 1-2V minus (positive output current, diode conducting for IGBT, maybe reverse mode of MOSFET) . But this does NOT directly define, whether reverse recovery in the bootstrap diode occurs or not. It only amplifies the effect in some cases. 3.) You own measurement shows the real cause of reverse recovers. A very short time between the LOW side going off and the high side going on, in your screenshot ~400ns. When the carging process is not finished before the high side goes on, reverse recovery will occur in the bootstrap diode. This translates to very high duty cycles close to 100% at very high PWM frequencies. 4.) What puzzles me is, that in your measurement the charging process of the bootstrap capacitor starts when the LOW side goes off. This is odd. Usually, this process starts as soon as the low side goes on, since it needs the low side mosfet conducting or at least the MID voltage driven low, which can also happen through an inductive load. I guess your setup uses MOSFETs, and when the low side goes off, the MOSFETs leaves reverse conduction mode (low R_DS_ON) and the body diode has to carray all the inductive output current, which cases the MID voltage to incease with negative polarity! This will start an additional charging process for the bootstrap capacitor.
@sambenyaakov
@sambenyaakov 2 жыл бұрын
Thanks for input. 1. Indeed the startup issue is more of a theoretical case and was introduced just as a background. In any event, as stated in the video this possible process is of no sinificance.. As for the other parts of your comments, it seems that you are thinking about an IGBT which is not a bidirectional device and hence the diode of lower device is conducting during the LS duration and there is no voltage step in the mid point. As explained, the reverse recovery is due to this step in a MOSFET. This has nothing to do with the duty cycle. The simulation and experimental were with 50% DC. As for current direction, funny, despite my explanation, the presented simulation, slide 19, (which is supported by measurements) you argue that the direction does not matter? Come on. Show us a real proof that you are right. As for the experimental measurements they support one to one the explanation and simulation. May I suggest that you watch again the video with an open mind and the understanding it talks about MOSFETs .
@JacquesMartini
@JacquesMartini 2 жыл бұрын
@@sambenyaakov OK, I watched it again an I get your point. And yes, unconsciously I was a little bit too much focused on IGBTs, since I use them in my design. 😵Just a small additional note. You say, that the MID voltage goes negative, when the lower MOSFET turns off. It was already negative, when the MOSFET was conduction (inverse mode), but will go even more negative, when only the diode is conducting. But still, the reverse recovery will also take place for very high duty cycles, since the recharge process will be still ongoing then the MID voltage goes high, independend of output current direction. So after all, we both are right. 😀
@sambenyaakov
@sambenyaakov 2 жыл бұрын
@@JacquesMartini Indeed, if the charging is not completed within the low side duration there will also be a reverse recovery process. But... the series resistor to the floating capacitor must be sized such that the charging is 'completed' for the expected maximum (HS) duty cycle. Otherwise, the capacitor will not be charged to the full voltage and upper transistor will not get proper drive which will increase losses (and this is for max DC which normally corresponds to max current}. I think we have gotten it straighten out now😊😊
@RakeshYadav-lu9oc
@RakeshYadav-lu9oc 5 ай бұрын
I have one request, please also upload the simulation file l
@sambenyaakov
@sambenyaakov 5 ай бұрын
Send me your email and I will try to look it up. No promise
@freedomisfood6966
@freedomisfood6966 6 ай бұрын
Hi master many persons that have excercise in electronics activate in youtube.must her edourment no any result.please as a sample rather than theory go to action an define good base driver
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