Thank you so much professor. Its the best in the internet.
@sambenyaakov3 жыл бұрын
Thanks
@oussenisawadogo3733 жыл бұрын
Thanks a lot for your incredible work . May god bless and protect forever
@sambenyaakov3 жыл бұрын
Thanks for kind note.
@sivakrishna28223 жыл бұрын
It's an excellent video dear Prof. Thank you very much! Ofcourse, very eager to watch more such informative videos from your end.
@sambenyaakov3 жыл бұрын
Many thanks!
@electronicaindia3 жыл бұрын
In the end of the presentation you have shown in halfbridge configuration there is negative voltage peak has seen ,but it can be eliminated using series resistor to save the driver. There is an alternative way to save that using TVS diode to clamp the negative transient, without inserting a resistor in the path....and than k you for your wonderful presentation as always.
@sambenyaakov3 жыл бұрын
Thanks for participation and contributing,. Indeed, the peak can be chopped but the purpose in video was to show the implication's when a resistor is added.
@vladimirm42713 жыл бұрын
Very interesting! Thank you,professor !
@sambenyaakov3 жыл бұрын
Thanks Vladimir. How are you?
@vladimirm42713 жыл бұрын
@@sambenyaakov Aefut mehahagim ))
@amitt1113 жыл бұрын
Thank you professor! Great presentation as usual 😊
@sambenyaakov3 жыл бұрын
Thanks
@raviselvans19633 жыл бұрын
Thank you very much professor.
@sambenyaakov3 жыл бұрын
👍
@melplishka59783 жыл бұрын
Ty. Again very well explained.
@sambenyaakov3 жыл бұрын
Thanks
@bernard.tomasevic Жыл бұрын
7:06 Why is the Vgs = Vt a crucial time for Ig calculation?
@sambenyaakov Жыл бұрын
This is when the switching transition occurs
@bernard.tomasevic Жыл бұрын
@@sambenyaakov is that the point at which Ig is at its peak? Why wouldn’t we calculate like: Ig = Vcc / Rsum?
@sambenyaakov Жыл бұрын
@@bernard.tomasevic at this point the voltage of the gate is Vt, the peak current is when Vgs=0
@electronic79793 жыл бұрын
Helpful video 👍 I liked it
@sambenyaakov3 жыл бұрын
Thanks
@smity22552 жыл бұрын
Thank you professor for the video. I have question. How about putting the resistor only on the source pin? but not in the gate. ie at 19:34 remove the Rm and replace by Rson and Rsin.
@sambenyaakov2 жыл бұрын
There is a need for separate Ron and Roff for each transistor because parallel connection of MOSFETs is not desirable due to Vth spread.
@Antyelektronika3 жыл бұрын
Hi Profesor. In minute 13:43 you said (8-0.5)/1 give us 7A. But what with this 10R resistor?, he will not play in current value? If we will take large amount of current from this two capacitors, the have ESR and he will not decrease amplitude of current? As always thank you for very educational video
@justpaulo3 жыл бұрын
The capacitors start charged at 8V. That is why the 10R resistor is there - to charge them initially. So the initial current (8-0.5)/1 comes from the capacitors and goes down tending to ~(1.18-0.5)/1 = (8-1.18)/10, and in that case the current comes from the 8V supply. ESR could have an impact, but you can get good caps with ESR in the order of 0.1 ohm.
@sambenyaakov3 жыл бұрын
Thanks for input.
@sambenyaakov3 жыл бұрын
Thanks for participation. The pulse is short so the voltage on capacitor barely changes (ESR is small compared to internal resistance of driver). The function on the 10R is to charge/discharge the cap during the rest duration.
@nhanle44033 жыл бұрын
Thank
@sambenyaakov3 жыл бұрын
👍🙏
@HadeedSher3 жыл бұрын
In some books it is written that the gate resistance is used to dampen the oscillations caused by the junction capacitance and the stray inductance. I am curious that how is the presented theory in harmony with this statement. Thank you
@sambenyaakov3 жыл бұрын
Thanks for bringing up a very good point which I have not covered in video. The problem of oscillation is in parallel the issues I have discussed and stems from the fact that the drive lines have a stray inductance. Resistance will dampen the oscillation if the quality factor Q is reduced to about 1 or below.
@hadeedsher42703 жыл бұрын
@@sambenyaakov Thank you for the reply Prof.
@wordsoccer7473 жыл бұрын
Thank you Professor! Just curious why half the energy is lost during turn on?
@sambenyaakov3 жыл бұрын
See kzbin.info/www/bejne/ep22nqF7rseliKc There are many more videos on the subject in my KZbin channel
@guitardenver13 жыл бұрын
At 3:43 you say: (Paraphrasing here) "Before Q1 is turned on, Q2 was turned on. Current was flowing through Q2. Then, Q2 is turned off, and we start a dead time. This will cause Q2s parasitic diode to forward bias." If, Q2 was conducting, that means the switching node was positive (+HV). So when Q2 is turned off, the switching node will go negative to keep the current going in the same direction. That means Q1s parasitic diode with forward bias, not Q2s. We then proceed to turn on Q1 once the dead time is over. Could you help me out reasoning how Q2s diode get forward biased after Q2 is turned off? Wouldn't it be Q1s diode that forward biases?
@sambenyaakov3 жыл бұрын
"the switching node will go negative to keep the current going in the same direction." Not so. The current is maintained same direction by D2. To self commutate the mid point voltage you have to charge the parasitic capcitor with a current of OPPOSITE direction. Thanks for intertest. capacitors
@therealspixycat4 ай бұрын
Should you consider the 10ohm gate resistor when determine the Ton and Toff current??
@sambenyaakov4 ай бұрын
Please indicate to which slide number or time in video you are referring to.
@therealspixycat4 ай бұрын
@sambenyaakov in slide 15 and 16. The 10 ohm resistor to the threshold battery replacement. As far as I understand you suggest that an additional gate resistor is not required because the on and off current are within reasonable limits? I try to understand how to calculate the gate resistor values. Thanks!
@eduardinification3 жыл бұрын
Thank you professor. Would be nice a video on configurable current mode MOSFET drivers, pros and cons, etc... I have used them sucessfully to tune the rise and fall time to make the system EMC compliant.
@sambenyaakov3 жыл бұрын
Thanks for suggestion. Will consider.
@VladislavMocluza12 күн бұрын
One thing which I can not understand: why is it so important to discharge gate faster than charge it? I understand that it improves power dissipation of the MOSFET. But why we can not do the same with a charge? What is the difference between charging and discharging that we are doing different time to them?
@sambenyaakov12 күн бұрын
This has to do with the direction of current and diode reverse recovery.
@VladislavMocluza5 күн бұрын
@@sambenyaakov Thank you for the answer. Do you have a more detailed video about this topic on your channel?
@sambenyaakov2 күн бұрын
@ kzbin.info/www/bejne/aKC8ioyMqpl0oc0
@mohammadhassanzade68933 жыл бұрын
Thank you for your great work i think RG of mosfet(that write value in datasheet of each mosfet) must use for external resistor switch on and off true ?
@sambenyaakov3 жыл бұрын
But in many datasheets you are given range of Rg , so which one to choose?
@mohammadhassanzade68933 жыл бұрын
@@sambenyaakov I think we must add RG in data sheet to external resistor that find with equation so If RG is 2ohm and we find Ron resistor 12, better use 10ohm in circuit and if for roff if find with equation 4ohm, good choise for roff external is 2ohm Ron(from equation)=Ron(external)+RG Roff(from equation)=Roff(external)+RG
@mohammadhassanzade68933 жыл бұрын
@@sambenyaakov for range of RG I think better choose is take maximum resistor that write in the data sheet and add value to external resistor to achieve equation resistor
@sambenyaakov3 жыл бұрын
I now understand that by RG you meant the internal gate résistance. Strictly speaking you are right but in practice, the other inaccuracies make the precise calculation meaningless.
@mohammadhassanzade68933 жыл бұрын
@@sambenyaakov thank you for good reply i got it :)
@nachiketadeshmukh84443 жыл бұрын
Hello, Prof. Are you suggesting that the threshold voltage of the FET and Miller plateau voltage of the FET are equal? In my limited understanding, the miller plateau voltage is greater than or equal to the threshold voltage.
@sambenyaakov3 жыл бұрын
Thanks for participation and input. The plateau is at voltage that needed to sustain the current during transition. Since the gm curve is rather fast rising the difference from Vt is small.
@eduardinification3 жыл бұрын
BTW professor, are you sharing the presentations in PDF format somewere? thanks a lot!
@sambenyaakov3 жыл бұрын
No, sorry.
@tamaseduard51453 жыл бұрын
🙏🙏🙏❤️🙏🙏🙏
@sambenyaakov3 жыл бұрын
👍😊
@lowrdson40003 жыл бұрын
Incredible video! I have one question. When calculating the gate current, will it be more accurate to consider the Miller Plateau voltage instead of the threshold voltage?
@sambenyaakov3 жыл бұрын
When gm is high, as typical of power transistors, the threshold and Miller plateau voltages are about the same.
@khadimusyaffa31273 жыл бұрын
Thank you professor, this is the best explanation about gate resistors in the internet. However, i have a question sir. How to determine the turn-on time to avoid the reverse-recovery? is there any range for that?
@sambenyaakov3 жыл бұрын
Reverse recovery is a function of the forward current and dI/d Hence, for a gives forward diode current the way to reduces di/dt. One way is by slowing down the turn on of the complementary transistor.
@khadimusyaffa31273 жыл бұрын
@@sambenyaakov Is there any explanation about how much "slowing down" is sir?
@clifforddicarlo91789 ай бұрын
How practical is it to find a 7.2 Ohm resistor, Ben-Yaakov?
@sambenyaakov9 ай бұрын
I can only assume that you did not watch the video carefully. Why would you have a problem with a 7.2 Ohm resistance of the Rds(on) of the gate driver as given by the manufacturer and copied in the presentation, Clifford?
@kamparaju13 ай бұрын
Hi professor, For new SiC modules do we need to maintain Rgon higher than Rgoff.
@sambenyaakov3 ай бұрын
Absolutely
@huanzhou47682 жыл бұрын
Hi Professor. I have one question during test in the labo when using Mosfet SIC NVHL160N120SC1 for the full bridge ; For the commad Vgs (-5V & 20V), during turn on, there is Peak Voltage found on the Vgs , which can be higher than 25V; So there is risk to destroy the Mosfet due to limited max 25V from the datasheet; I do the simulations and this peak voltage on the Vgs coming from the stray indutace of the pins of TO247 ( NVHL160N120SC1); My solution is to increase Rgon value to slow the switching on ; do you think is a good way? thanks
@sambenyaakov2 жыл бұрын
I do not think that the reason is inductance in package, probabely the traces. Yes, adding R reduces Q and hence damps oscillation - at the expense of a slower drive and higher switching losses.
@marcinszajner29243 жыл бұрын
Great video, this help new people in energy electronics safe a lot of time. I have one question about gate charge circuit, many time I saw resistor between gate and source (most popular 10kohm). I thing this is useless element if we use dedicated driver for mosfet, but nice to know your opinion about this element.
@sambenyaakov3 жыл бұрын
Hi Marcin, thanks for participating. The function of the parallel resistor is to provide a low impedance, sort of ESD protection and bleeding of leakage current of gate. Indeed, when there is a driver it is supposed to provide the low impedance path. But...if the auxiliary power supply is off? the impedance might be high and EMI might turn on or even damage the gate. And more severely if one applies the high voltage to power transistor (with no auxiliary power to driver) ,the spike via the "Miller" cap might be a problem.
@marcinszajner29243 жыл бұрын
Hi Sam, Thanks for respond. That nice to know, Thanks!
@SefaOralz10 ай бұрын
Sir, during verification there is 10R resistor after 8V source. Why you didnt take 10R resistor into the consideration and divide it only by 1ohm ??
@985051772298505908183 жыл бұрын
Why this diode turns on momentarily when we turn on lower fet ? When upper fet is on this diode is reverse biased when lower fet turns on anode of diode is pulled down to ground so it’s still in reverse biased mode why it shows current peaking ?
@sambenyaakov3 жыл бұрын
Reverse current is during the dead time
@985051772298505908183 жыл бұрын
@@sambenyaakov dead time is when both fets are off state .. so diode becomes forwards biased momentarily ?
@Harishbhardwaj0003 жыл бұрын
Sir inverter me Mospet k gate par kitne ohm ki regitance lage gi
@sambenyaakov3 жыл бұрын
?
@adampawowski13692 жыл бұрын
Hello professor, I have a question about this moment shown in the video kzbin.info/www/bejne/d6KUeqpsd96Ld9k. I understand that delta time refers to the voltage rise time of the VGS and not the fall time VDS. In the case where we assume the fall time for the voltage VDS, we will use the charge accumulated in the plateau region in the equation. Am I right?