@8:45, how is 1/I(B2) plot the impedance, not the ESR?

Dear professor, It's nice to see this thermal modeling approach! This approach is exciting!Have you published any literature on this method? Can you give some references for this method? I want to quote this method😀

Nicely explained. I use your videos to relearn my lost knowledge. Thanks for taking the time to make them. Best wishes from Norway.

I think you accidentally drawn a reversed second winding direction in the first schematic of the video.

Another excellent tutorial on the inner working of transformers.

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Prof, could you take a look at a question someone asked on the riddle, about open circuiting one leg of the parallel MMF situation? This video cleared up a lot, thank you again, but that situation remains unintuitive to me. Based on another comment about parallel MMFs reflecting to input in series, one open circuit must stop current in all branches. I now think it must relate to 9:20 here about the pure inductor case, that only Lm would have current and none flows in the loads.

Number 2?🤭

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Now translate to the temperature of the transistor junction.

Sir, if the right side is shorted and left is opened, then V2 will be 10 V. How is this possible? Please explain.

Sir, what does the meaning of your statement in the end of this video, "With the B=1mT, this is really core that can sustain a much high power?" Does that mean that the lower/smaller the B is the higher the voltage we can feed to it?

Sir please average model of peak current mode control active clamp forward converter.. please

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?

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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.

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.

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?

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 ?

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.

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

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.

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.

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)

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.

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).

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.

V1=V/(1+R2/R1)=2V, V2=V/(1+R1/R2)=8V, 2.5A, 10V. Thank you for the video!

1. Amp turns must balance,half Mmf on each leg so 10v on both for conservation of energy. 2. As I 2. 3.10v .

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.

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.

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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.

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.

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?

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

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

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?

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.

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.

professor,what's V(EXC) of B4 in your LTspice simulation in your presentation？