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Sam Ben-Yaakov
Добавлен 20 авг 2008
This is an engineering education channel covering the area of analog and power electronics. The channel is meant to explain basic and advanced topics in an intuitive way with minimum number of equatios. Comments are welcome, and I try to answer all queries. Welcome abroad.
Intuitive answer to riddle and simulation
The ridddle video
ruclips.net/video/Sl7dv780MgY/видео.html
This presentation assumes a basic knowledge in magnetic circuits. A forthcoming video will include an explanation of magnetic circuits theory and applications.
ruclips.net/video/Sl7dv780MgY/видео.html
This presentation assumes a basic knowledge in magnetic circuits. A forthcoming video will include an explanation of magnetic circuits theory and applications.
Просмотров: 1 369
Видео
A riddle for the magnetics-philes but open to all
Просмотров 1,3 тыс.12 часов назад
A riddle for the magnetics-philes but open to all
A comprehensive user-friendly simulation model of ceramic capacitors
Просмотров 1,5 тыс.День назад
Relevant references Zeltser and S. Ben-Yaakov, "On SPICE Simulation of Voltage-Dependent Capacitors," in IEEE Transactions on Power Electronics, vol. 33, no. 5, pp. 3703-3710, May 2018, doi: 10.1109/TPEL.2017.2766025. S. Ben-Yaakov, "Some Observations on Loss and Hysteresis of Ferroelectric-Based Ceramic Capacitors," in IEEE Transactions on Power Electronics, vol. 33, no. 11, pp. 9127-9129, Nov...
Common mode chokes: There is more than meets the eye
Просмотров 4,3 тыс.14 дней назад
#EMI #switchmpde #noise #PWM #coupledinductor
Improved intuitive average model of PCM converters
Просмотров 1,9 тыс.21 день назад
#PCM #PWM #switchmode #Buck # acsimulation #averagemodel
Practical circuits tutorials : Flyback, power
Просмотров 2,6 тыс.28 дней назад
#Topswitch # Flyback #commonmode #TL431 #offline
Transformer saturation and gapped core current transformer
Просмотров 2,9 тыс.Месяц назад
#magnetics #transformer #currenttransformer #CT #magneticsaturation
Gain margin in PWM converters
Просмотров 3,1 тыс.Месяц назад
#switchmodeconverters #pwm #phasemargin #gainmargin #stability
Testing the transient response of PWM converters by LTspice
Просмотров 2 тыс.Месяц назад
Relevant videos Why is the overshoot a function of the phase margin in closed loop feedback systems ruclips.net/video/DYQxMieRA6I/видео.html DC regulation of PWM converters ruclips.net/video/x6O3WArZ-xk/видео.html #phasemargin #PWM #stepresponse #PWMstability
Illustration of sampling delay in PWM convertersSamp delay final
Просмотров 2,2 тыс.Месяц назад
Relevant videos Modeling and control of PWM converters - Tutorial - Part 2 VM control ruclips.net/video/bGj1huz2Log/видео.html Analog average simulation of digital controllers ruclips.net/video/C-4V5MQhSA0/видео.html
Current sharing of parallel connected power supplies by the droop method
Просмотров 3,2 тыс.Месяц назад
Current sharing of parallel connected power supplies by the droop method
Power flow considerations in power filters
Просмотров 2 тыс.2 месяца назад
Relevant videos: BJT filters and capacitance multipliers ruclips.net/video/4h8DGb-s0i4/видео.html DC regulation of PWM converters ruclips.net/video/x6O3WArZ-xk/видео.html
BJT filters and capacitance multipliers
Просмотров 3,3 тыс.2 месяца назад
resources.pcb.cadence.com/blog/2019-designing-a-capacitance-multiplier-as-a-power-supply-filter
DC regulation of PWM converters
Просмотров 2,4 тыс.2 месяца назад
Conclusion High LG improved DC regulation Feedforward can help to improve input DC voltage regulation as well as attenuate output voltage disturbance due to input voltage step
Absolute maximum ratings: II. Power MOSFET
Просмотров 2,4 тыс.2 месяца назад
Absolute maximum ratings: II. Power MOSFET
Absolute maximum ratings: I. Op. Amp.
Просмотров 1,6 тыс.3 месяца назад
Absolute maximum ratings: I. Op. Amp.
The inner works of Peak Current Mode (PCM) control
Просмотров 3,5 тыс.3 месяца назад
The inner works of Peak Current Mode (PCM) control
Estimating by time domain simulation the LoopGain transfer function
Просмотров 2,4 тыс.3 месяца назад
Estimating by time domain simulation the LoopGain transfer function
How to get equally spaced data in LTspice?
Просмотров 1,2 тыс.4 месяца назад
How to get equally spaced data in LTspice?
Study of thermoelectric coolers by an LTspice model
Просмотров 1,4 тыс.4 месяца назад
Study of thermoelectric coolers by an LTspice model
A behavioral LTspice model of phase change materials (PCM)
Просмотров 1,1 тыс.4 месяца назад
A behavioral LTspice model of phase change materials (PCM)
Back to basics #2: Practical capacitor charging losses
Просмотров 2,4 тыс.4 месяца назад
Back to basics #2: Practical capacitor charging losses
Back to basics: Practical capacitor charging currents by LTspice demonstration
Просмотров 2,6 тыс.4 месяца назад
Back to basics: Practical capacitor charging currents by LTspice demonstration
I2t. What? Why? When?: The electric fuse case
Просмотров 4 тыс.4 месяца назад
I2t. What? Why? When?: The electric fuse case
Synchronous sampling and anti aliasing filter: a demonstration by LTspice simulation
Просмотров 2,3 тыс.5 месяцев назад
Synchronous sampling and anti aliasing filter: a demonstration by LTspice simulation
Inductor discharge: answer to riddle
Просмотров 1,7 тыс.5 месяцев назад
Inductor discharge: answer to riddle
Correction of thermistor temperature sensors response time: the passive analog solution
Просмотров 1,5 тыс.5 месяцев назад
Correction of thermistor temperature sensors response time: the passive analog solution
Modeling the time response of thermistors (NTC) temperature sensors
Просмотров 1,6 тыс.5 месяцев назад
Modeling the time response of thermistors (NTC) temperature sensors
What about a full bridge rectifier with capacitors across each diode?
I would go by basic MMF balance in an ideal transformer. If assuming high mu_r core, then 3 MMF sources in an ideal transformer will be in parallel. So that N*Is = N*I1 = N*I2 => Is = I1 = I2. Also with energy conservation power input must be total power output, so Ps = P1+P2 or Vs*Is = V1*I1 + V2*I2 => Vs = V1+V2 = I1*R1+I2*R2. Now : 1) Vs = Is*R1+Is*R2 = Is*(R1+R2) = Is*5 => Is = Vs/5 = 10/5 = 2A = I1 = I2 and V1 = 2*4 = 8V and V2 = 2*1 = 2V 2 & 3) if R1 is shorted => V1 = 0 but I1 != 0 but it also holds that Is = I1 = I2 due to MMF balance so Vs = V1+V2 = V1 + 0 = 10V. I1 = V1/4 = 2.5A = Is = I2
Cycle by cycle current protection is possible with average current mode control if you feed the unsmoothed signal (containing triangle wave) to a comparator.
Hi professor. Related to this topic, DC-blocking capacitors in DAB converters.There are people that indicate that DC bias can be solved with a gap in the transformer, others say that DC capacitor is always necessary. I thing that resolution of PWM in microcontrollers is going to always generate some DC and in transients very likely a DC half-cycle applied, and hence saturation. Can you dedicate a video commenting misconceptions on this issue?
👍🙏❤️
Excellent explanation, thank you
Simply the most smart professor on this planet! 🎉❤
I did get the right answers using magnetic circuit analysis as well as by power balance when assuming the 3 currents are identical. But, one thing that bothered me at first was that such analysis "contridics" basic three-winding ideal transformer models including those used in many simulation platforms such as PLECS and PSIM (there isn't any contradiction - see further on for an explantion). For a 3-winding transformer, you get V1=V2=V3 and the currents at eact side are determined by the loads with the source current determined by power balance. Transformer model: V1:V2:V3 = N1:N2:N3 and magnetic circuit model: I1:I2:I3 = N1:N2:N3 So why is there a contradiction? The answer is simple: The configurations of the magnetic circuits assumed for this case (all 3 MMF are in parallel therefore must be equal if N1=N2=N3, or, the total MMF is the average of the three). The assumed configuration for the transformer is all 3 windings are on the same leg -> series connection of 3 MMFs. Parallel MMF connection -> electric loads are in series (post reflections). Serries MMF connection -> electric loads are in parallel (post reflections).
At 7:04 I've got confused. Why is V1+V2=10V? In an ideal transformer with 1:1:1 ratio I would say that V1=V2=10V. Now this ferrite does not make an ideal transformer but is it really that far from it to completely change the spirit of the equation? I would expect a small voltage drop when the winding gets loaded but both should still be close to 10V.
My understanding is that based on the magnetic circuit, the three leg MMFs are equal, MMF=nI so currents are equal (I think) If input power is 10I, output power cannot be 2*10I, the power must be divided. Only way to do that is divide voltage, because V1*I + V2*I = 10*I
This is not a transformer configuration! See my explanation above...
Your solution is quite valuable to me. I want to replicate your result but waveforms of Vplus, ID, IL1, IL2 do not match yours. Could you send me your LT spice sim files. I would really appreciate it. Thank a lot
Great Video and explanation. Thank you for taking time.
Thanks for the explanation sir. At 6:53 , If the right side winding is opened instead of shorting, then as the center, left and right are in series, no current shall flow in center and left winding. Is it correct sir? In short, if R1 is opened instead of shorted, what are the answers to the above 3 questions.
Let me take a stab before Prof gets to it.. based on the presented logic it initially looks like you are correct. 2. R1=inf I=0 because I=10/(inf+R2) =0A 3. V2 = 0×4 = 0V Like you I don't really like this answer. How could a wire breaking on one side prevent any power input entirely. I think something else happens in this case. Prof neglected reluctance in normal circuits, which was the underlying assumption that I1=I2=I. I think it cannot be neglected in this case. MMF B2=0AT The right arm will flux saturate, producing some voltage V1 depending on the ferrite. Some current <2.5A will flow to I2. I believe you would have to simulate. (Fingers crossed I've redeemed myself here Prof!)
I'm not sure my sign conventions for the fluxes are the right way round but here's my take anyway.. Let: V0 = 10V Then: (0) V0 = -NAdB0/dt (1) V0 = -N(A/2)dB0/dt -N(A/2)dB0/dt Assume direction of fluxes of each section are such that from (1) we get: (2) V2 = N(A/2)dB1/dt +N(A/2)dB0/dt (3) V1 = -N(A/2)dB1/dt +N(A/2)dB0/dt where there is a common flux - Φ0 flowing between each half of the centre section and each of the two side sections - Φ1 flowing between the two side sections (2)-(3) => (4) V2-V1 = -NAdB1/dt (2)+(3) & (0) => (5) V2+V1 = V0 (4)+(5) => (6) V0 = 2V2 +NAdB1/dt (6)+(0) => V0 = V2 +N(A/2)dB1/dt -N(A/2)dB0/dt & (3) => V0 = V2 -V1 & (5) => V2+V1 = V2 -V1 => V1 = 0 & (5) => V2 = V0 = 10V. But my flux sign convention can't be correct as it introduces an asymmetry in a symmetrical circuit, right?
Hold on to my answer video
I suppose that at sufficiently low frequencies the limiting factor becomes the dielectric breakdown voltage. Can we speak of a typical frequency when that happens ? And if so, what is its value/range ?
V and I are two independent limits , At low frequency V is the limit. At high I.
1) V1 = 2 V and V2 = 8 V The voltage of each secondary winding is related to the impedance reflected in the primary winding. 2) I1= 2.5 A The current of secondary 1 is equalised to the current delivered by secondary 2. 3) V2 = 10 V Since secondary 1 is short-circuited, the impedance seen from the primary is the load of secondary 2. Due to the 1:1 ratio, the voltage is the same as that of the primary.
👍👏Thanks for participating.
Only from reading the first couple of comments i learned a lot totally new insights. Flux splitting and the summing of outputvoltages and so on. The drawing did me think of 3fase transformers from wich i never understand how the different fases can use the same core. I look so much out for the answervideolesson, in fact as always
Thanks for participating. Hold on to my answer video.
1) 10V 2) 2.5A 3) 10V The reluctance imposed by R1 is n^2.s/R1, where s=j.2.pi.f The exact overall formula using non zero leg reluctances is tedious but straightforward.
Thanks for participating. Sorry , 1 is incorrect, Hold on to my answer video.
@@sambenyaakov1) V1=Vs*R1/(R1+R2)=2, V2=Vs*R2/(R1+R2)=8
@@sambenyaakovV1=Vs*R1/(R1+R2)=2V, V2=Vs*R2/(R1+R2)=8V, Is=2A
If R1 & R2 are infinite (I1=I2=0), then the answer is easy: the magnetic flux (Φ = B.A) generated by the centre winding splits equally between the other two core legs, and from Faradays law (V = NdΦ/dt) each of these windings sees half the voltage, ie: V1 = V2 = 5Vac. However, this flux distribution is not maintained under the loaded condition, since I1 and I2 are not zero. In the reluctance diagram, the total MMF generated by centre leg sees two other MMFs in series. Since the nmbr of turns are the same, the currents must be the same, and the two voltages have to add to 10V. So we solve two simultaneous equations: (1) V1 + V2 = 10. (2) I1 = I2 (3) V1 / R1 = V2 / R2, where R1=1Ω, R2=4Ω. Writing V2 in terms of V1 from (1), and subbing into (3) we get: V1 / 1Ω = (10 - V1) / 4Ω --> V1= (2.5 - V1 /4Ω) --> 1.25x V1 = 2.5 , So V1 = 2, and V2 = 8. And I1 = I2 = 2A. Power: P1 = V1.I1 = 2V x 2A = 4W. P2 = V2.I2 = 8V x 2A = 16W. Total power = 20W, so 2A in the centre winding (ignoring magnetising current). If R1=0, then this situation **defines** the winding voltage to be zero , so V1=0, this means zero magnetic flux, so all the flux (and hence all of the primary voltage) is moved over to V2, so: V2=10V, and I2 = 10V / 4Ω = 2.5A. Power = 10V x 2.5A = 25W. Same for R2=0, V2 is forced to 0, so: V1=10V, and I1 = 10V / 1Ω = 10A. Power = 10V x 10A = 100W. Another way to look at it: let's leave R1=1Ω, and let R2 vary from infinite to zero. For R2 open ckt (infinite), then all the voltage appears at V2, & V1=0. Total load power is zero. As R2 reduces, then V2 voltage starts to reduce, and V1 starts to increase, and load power increases. At R2=R1=1Ω, both the voltage and the currents will be equal, the voltages will sum to 10V, so 5V each, and 5A each, so 25W per winding, 50W total. As R2 reduces toward a 0, V2 reduces while V1 increases, until at R2=0 we get V2=0, V1=10V, I1=10A, and Power = 100W. So power increases quite rapidly as R2 reduces from 1Ω to 0Ω. We can think of this as a way of controlling how magnetic flux is split up from one path to another path.
Thanks for participating. Sorry incorrect, Hold on to my answer video.
I do not know, but here is my 2 cents (PS: I see I'm completely wrong by other's answers... Therefore I look forward for you video) -- If the core was ideal ( *no saturation* ): 1) V1 = V2 = 10/2 Vac (b/c the flux of the center portion splits 50/50 onto the other legs) 2) I1 → ∞ 3) V2 would still be 10/2 Vac -- With real ferrite core (i.e. *with saturation* ): 1) V1 = V2 = 10/2 Vac (if R1 and R2 are high enough for the core not to saturate) 2) As R1 → 0 the current in the "primary" increases. Eventually the core saturates and I think that then I1 will reach a maximum I1 = Isat and it will remain that way. V1 will drop to 0 as R1 → 0. 3) When the core saturates the magnetic flux will be Φsat and will stop increasing. V2 therefore will clamp also to V2_sat. Bonus: when the core saturates I_primary → ∞ (i.e. will become a short and bad things will happen)
Thanks for participating. Sorry incorrect, Hold on to my answer video.
1. Since the ratios are 1:1:1, V1 & V2 are same as the driving source or 10V. 2. Assuming an ideal transformer and windings, i.e no other resistances than the values of R1 & R2, and that all the core limbs are equal x-section shorting out R1 will completely short out the flux and draw unlimited current. 3. Because of a short on R1 (all flux diverted to R1) and therefore same on the driving source there can't possibly be a voltage across R2, so V2 = 0V. I'm now ready for the bad news as I've never considered this scenario before.
Completely wrong above. Just did some actual measurements - did not have access to a suitable ferrite core but had a pair of steel laminated UNICOREs from AEM Cores on which I wound up 3 coils as in the video, 23 turns each. also the signal source I have was incapable of driving such low resistances shown in the video so I scaled the values to 1k for R1 and 4k for R2 keeping the same ratio of loads. in addition had to keep the drive signal to the middle coil at 1V rms to prevent overloading of the drive circuit when R1 shorted. The input frequency is 10kHz. 1. Measured V1 at 433.2 mV rms and V1 at 417.5 mV rms. 2. R1 shorted: I1 measured at 17.4 mA rms. 3. R1 shorted: V2 measured at 787mV rms. As these are real values with real lossy magnetics etc, I'd be curious to see how they compare to the theoretical values using the resistors and input I used. PS sorry for the James T. Kirk method of solving problems.😊 Cheers Mark G
Sorry incorrect. Hold on to my answers video.
@@sambenyaakov Thanks for the response, look forward to the answers video. Just out of curiosity, other than a possibility of errors in measurements is there a problem of scaling by using different values in my attempt? EDIT: Further investigation revealed some issues with my ac voltmeter, I also recalibrated my Rigol scope. Using only the scope to do all the measurements I found that if the two secondary windings are either unloaded or loaded with equal resistances, the outputs end up being equal but only half the value of the applied voltage. this makes sense as the total applied flux is split in two equal values, one left and the other right of center limb - this is only true if the L & R magnetic path areas are identical - my cores seem to have a small difference between the two halves. Unbalancing the two loads unbalances the voltages in the direction where the lower resistor develops a lower voltage and the opposite for the higher value resistor. The sum of the two voltages add up to the value on the primary winding. If either of the outside windings is shorted out the opposite winding ends up with almost full primary voltage across it, an ideal transformer without any losses I suspect would get a full primary voltage on the winding with resistive load whose current would be V2 / R2. so in case of R2 = 4R and Vmax across it = 10V, this would end up at 2.5A Not sure what the current I1 would end up being, don't have a clamp meter to do the measurement, replacing R1 with a much lower resistor and measuring the voltage drop across it suggests that the current increases. Will be quite interesting to see the math behind it, hopefully it's not beyond my level of understanding. Thanks for a nice little head scratcher leading up to a weekend. Cheers Mark G
@@markg1051 Thanks for your thoughts. Have a nice weekend
@@sambenyaakov you too.
If the coupling is ideal (infinite Ur core), then because the magnetic circuit is of a 3 MMFs connected in parallel, the two resistors will reftect to the voltage source side as a single series resistor of 4 + 1 = 5Ohms, therefore the source current as well as the two resistors' currents would be 10/5 = 2A (1:1:1 reflection). Thus, V1 would be 2*4 = 8V, and V2 would be 2*1 = 2V. When R1 is shortet (=0Ohms) then the total reflected resistance would be 4+0 = 4 Ohms. and the 3 currents would be 10/4 = 2.5A, thus. V2 would be 2.5*4 = 10V, which makes the magnetic circuit behave as a two-winding Xformer with a 1:1 ratio. The 3rd shorted winding will cancel out (opposite) it's leg's flux (shorted winding leg flux will be zero, thus circuit will become a simple ideal transformer).
Good answer! thanks for participating.
Assuming unity coupling factor (otherwise it is incorrect) : 1. The load on V1 is 4 times greater, so 4*V1 = V2, as well as V = V1 + V2, yielding V1 = 2Vac, V2 = 8Vac. 2. If R1 = 0, V1 = 0, so V2 = 10Vac, therefore I2 = 2.5A. We must therefore have I1 = 2.5A. 3. V2 = 10Vac.
Good answer! Thanks for participating.
V1=V/(1+R2/R1)=2V, V2=V/(1+R1/R2)=8V, 2.5A, 10V. Thank you for the video!
Because, V1+V2=V from flux split, and i+i2=0, i+i1=0, i1=i2, V1/R1=V2/R2 assuming mu=infinity, from mmf theorem.
Correct and elegant. Thanks for participating.
@@zaikindenis1775could ypu explain why does voltage divide this way (R1/R2) instead of R1/(R1+R2) like parallel currents through resistors would?
@@electrowizard2000 Actually, I have written it confusing. 1/(1+R1/R2) is the same as R2/(R1+R2).
1. Amp turns must balance,half Mmf on each leg so 10v on both for conservation of energy. 2. As I 2. 3.10v .
Sorry incorrct. Hold on to my answer video. Thanks for participating .
The more loaded a coil is, the more it will deflect the magnetic fields away from itself by creating an opposing field. 1 - 8V+2V. both coils will see half the magnetic flux of the middle one when open circuit. When loaded the more loaded one deflects the field away, so less voltage is induced. 2 - (10V/4Ohm)/2=1.75A. I1=Isupp/2, as it has to run just enough current to match the open circuit magnetic field magnitude. 3 - 10V. The magnetic path towards coil1(the right one) is effectively blocked, leaving out a 2 coil transformer.
2 is off. Thanks.
I would have assumed the flux divides evenly until considering short circuit cases, which tells me flux does not divide evenly. Lower resistance should get less flux. 1. V1=7.2 V2=2.8 (Thus P1=13W, P2=5.2W. Hmm.) 2. Any flux down this leg is resisted by the superconducting shorted turn. Seems undefined? The current has to be nonzero to counteract the input flux. Power is zero. I don’t know! 3. 10V, its a normal transformer on the left half since the right half is effectively disconnected.
Sorry incorrect. Thanks for response.
👍🙏❤️
👍🙏🤞
Since the windings are on opposite sides of the middle core, I *think* it's easier to imagine this as two seperate transformers. . Assuming stable power supply and no saturation, Initial voltage drop across each resistor Voltage drop across R1 should be 10Vac, with I1 10A & I2 2.5A. This already feels like I'm wrong. Shorting either resistor would immediately saturate the ferrite core on that side. But it would also saturate the primary core, dropping the impedance to the DC resistance (0 if ideal). Without a voltage drop on the primary you also can't have a secondary voltage, so.... Well, 0 makes me sure I'm even more wrong. By my calculations, the Professor has just invented the worse current-sense transformer ever conceived. I wouldn't normally post such an obviously wrong answer, but since there's only one other comment so far and I already told that guy that he was totally wrong, it only seems fair to admit that I don't have a clue either.
Drop the assumption that core will saturate. It will not.
@@sambenyaakov But that was the only idea I could come up with! 😁 At this point I either need to find a big ferrite core in my parts bin or wait for the answer video!
@@SkippiiKai Or, if you are familiar with, build a magnetic circuit diagram with reluctances etc. from which the problem and solution can be better envisioned. OR, wait to my video.
@@sambenyaakovIs it possible to see it as two normal transformers with instead of the middle leg the two primaries in series? Is that really an equivalent circuit?
This is a real headscratcher (especially the last two questions). Here are my answers, happy to hear of the problem behind my reasonings, as I am really not sure: 1) V1 and V2 will be 5 Vac with a 180-degree phase shift. This is because the flux splits up from the middle excitation winding, giving half the excitation to each winding. 2) The currents should match in a sub-loop (right side) so allow for a closed magnetic loop path. However, the flux swing will be minimal in this path due to the loss of resistance. So 2.5 A (ac) is based on the exciting winding's current. 3) Because voltage can't be present on V1 (as this requires the resistance), the flux swing has to appear at the left sub-loop. Therefore, it's now a fully coupled circuit, i.e., V2 = 10 Vac.
I think the same way as you. So I agree that V1 = V2 = 5V due to flux splitting and we would have a V2 = 10V if we shorted R1.
1 wrong. 2, 3 OK.
@@SergiuCosminViorel Think in term of reluctances.
@@sambenyaakov reluctances not good. give another terms. reluctances rejected, are meaningless. give me something real
@@SergiuCosminViorel Hold on to m answer video
Thank you, Prof. Ben-Yaakov, very intuitive explanation for this constant on time control for buck converter. Do you know who is the first person to propose this constant on time or constant off time control for buck converter?
No. Will be happy to find out.
@@sambenyaakov Thank you for your reply, Prof. Ben-Yaakov. I tried to find the first reference but not be able to make it. It seems that everyone took it for granted before 1990. But I found one patent from Infineon (US7521913B2) that they use output voltage to adjust the pulse width and frequency during load transient. The technique in this patent is very similar as the COT control during load step up. Will this claim still be valid?
@@jffighting 🤔
At 41:42 , you have calculated switching losses by multiplying switching energy at 200A with switching frequency, although the actual Swithing loss will be an average of the loss during once cycle assuming sinusoidal current. So switching losses would be less than what you have calculated and approx 0.63x10000x0.011 W
This is not entirely correct . At low speed , low elctrical frequency the inverter spends a long time around the phase current peak.
Dear prof Sam, i always admire your videos but the riddle was a bit misleading because you gave no additinal information about the inductors, i was scratching my head assuming them to be ideal inductors. If you could just tell that they were real inductors with some equivalent load resistor R5, it becomes very simple. The question is if their is no R5, it is obviously going yo explode
Your doubts are about the answer?
Are the doubts about the answer?
Hi sir could please make video on average model of peak current mode Active clamp forward converter
Where is the third part of the explanation?
Well, seeing that there was not much interest in the subject, I gave u😊 Sorry.
Before i can make the effort to follow it, i must do what i must, meaning that if i can bring something new, unheard, unknown to the scientifis world, then it is exactly that, that i must do. And i have something to say about this. Capacitance, as i think of it, can be defined at least in two ways, the way all know, that is the official science, so to speak, and one more way, though it myself, and this way is as valid candidate as the established way, it shows science is still in many parts of it, a matter of approach, not an ultimate and unique truth. The first way to consider capacitance, the already established way, is of a given capacitance, i call it fully formulated capacitance, that defines the ability of... something to store electrical charge, and that storage is gradually used. The second way to consider capacitance, and science could have chosen this instead, is of capacitance not given, that it is continuously formulated, meaning the ability of storing electrical charge is gradually granted, and the storing is not made gradually, but instantly and fully, to the available capacitance.
"meaning the ability of storing electrical charge is gradually granted, and the storing is not made gradually, but instantly and fully, to the available capacitance." This is not defining capacitance but the charging of capacitance, which depends on the charging source not the capacitor.
Thank u for show that cruves, i have a shot now for why this is not working well with high speed dc load. In my wiev this type common mode chokes are do better job before the diode bridge at AC side.
I am showing a DC (battery) source
professor,what's V(EXC) of B4 in your LTspice simulation in your presentation?
Please indicate time or slide number.
@@sambenyaakov LTspice PCM model at 9:41,thank you!professor.
@@luzhouyang5414 it produces REF which is like the output of the voltage error amplifier. Here it is used as a signal to test tracing. E.g. ruclips.net/video/b67u3Y3l26o/видео.html
Your video lecture is really helpful for me. Thank you. - from an power electronics engineer in south korea.
Happy to hear that, Thanks for note.
Sir coming to controlling of llc how does the error Amplifier or compensator ouput gives frequency that we need ,the same compensators are also used for getting duty,how is this possible please do answer or make vedio,please sir.
The output of error amp drives an VCO. Not a duty cycle modulator.
That's fine sir but how the same compensatory circuit is used for getting different control variables like duty, switching frequency .
@@swathi9086 Why same?You need to design the controller per topology based on the particular transfer function of the system.
What is the name of the standard? Where can it be found? Thank you for the video.
There are many types and classes. The most famous CISPR25
Prof. Yaakov is one of the few professors who is consistently working and publishing papers in pure power electronics. Hats off to you Sir. Many have drifted to other domains.
Thanks for the kind note.
I'm curious if you've tried the new LTspice 24? Apparently its major new feature is the "FRA" which extracts bode plots from transient simulations. Haven't tried it myself, but thought it might be of interest to you.
Thanks. First time I hear about it. I guess it is a rebuttal to Qspice
Thank you Professor...can you please make a video on how to simulate switched capacitor filter in LTSpice..
Good subject. Will try. Look up for our papers on average modeling of switched caps converters.
Sure..thank you..
Make a VCO using these capacitors
Good idea. WE have uses it to control resonant converters.
The #1 Professor of Power Electronics! Amazing as always!
Wow😊. Thanks
👍🙏❤️
Thanks
very great,professor,you explain so clearly,i have learned a lot from you,think you!
Thanks for the comment, I appreciate it,.
impressive! and beyond my level. why they don't make simple capacitors, simple behaviour? i never delved deep in this... years ago i tried a voltage multiplier, on maybe 1kHz, maybe 20kHz, i do not remember well, but it was not 100kHz, and it fumegated. tried with electrolitics, and the variations on the last capacitor were in 100V range. i understood there may be some problems, but i did not imagine how big. and at that time, it was no experience gained. as i said, not much thought into it. thanks for the presentation!
Thanks for sharing. Things are less mysterious once you understand the fundamentals
Muchas Gracias Querido Sam Ben !!!! ❤❤❤👍👍👍👍
Thanks
nice explanations. On 5:00 the RDS ON is assumed only dependent on temperature. this should be OK as long as the transistor is operated in saturation.
Indeed. Thanks for input.