Yes

Induction cooker is a delicate design. By changing the gap between the cookware and the coil, the igbt would burn, let alone replacing the coil.

@@SiliconSoup THe problem I encountered so far is that it is too smart and any notices difference and gives no cookware error.. :/

If the magnetic vector potential varies with time the effective E field cannot be conservative. Only if the vector potential is time invariant is there an exclusive scalar conservative potential difference between two points. The voltmeters in Lewin's experiment measure the Ohmic voltage drop along the particular segment of the closed loop path being probed. Voltage and potential difference aren't the same things when the magnetic vector potential varies with time. It's possible to have a (somewhat unusual) situation in a circuit where there is no potential difference between two disparate points on the circuit, even though continuous current flows between those two points and Joule heating occurs in the intervening conducting path. The International Electrotechnical Commission (IEC) defines voltage and potential difference differently. These definitions can be found at the Electropedia website under the heading IEC 60050, subject area 121 Electromagnetism. Some may argue there is a hidden scalar PD between the two points in Lewin's experiment and it is this PD that obeys KVL. You may be aware that physicists and electrical engineers sometimes have different definitions of KVL and this can lead to confusion.

I'm sorry maybe I didn't explain myself very well but for "newly defined field" I meant the sum of E plus the time derivative of magnetic vector potential A which is always conservative. I took your point about the difference in the definitions given by IEC of voltage and electric potential and I think that in this terms we can agree that LKV for potential differences always holds and is endeed used to deals with real problems.

OK. You are referring to the scalar gradient component of the effective field at any location. Yes, this is conservative and the closed line integral of this field component considered in isolation is zero. This is one workaround for saying KVL holds in the Lewin circuit. Trouble is this doesn't account for the total E field at every location around the Lewin loop. As I indicated, there are special situations in which this scalar field component is everywhere zero even though continuous current flows in a circuit path where energy is dissipated. So you can have zero potential difference everywhere and yet there is a voltage difference between any two points. Hence the requirement for the IEC definitions. So the sum of (IEC) voltages arising from the total E field encountered around the Lewin circuit is non-zero.

You can think on it as a workaround but I find it important in circuit theory becouse allows schematization of the different contributions of the fields which usually coexist into separate ideal component, for example the contribution of the E field in the wire due to Ohm's law in an equivalent resistence and all the contribution of the dA/dt along the path in an equivalent inductance. This component has their proper costitutive relations among voltage and current and for more complex circuit you can implement standardized analysis method based respectively on LCK and LKV.

@@pietrodaprile1490 If you have some time, it's enlightening to replace Lewin's loop with a homogeneous resistive circular wire loop of uniform conductivity and cross-sectional area. If the homogenous circular loop is concentric with the primary (assumed confined) solenoid magnetic field flux, then the analysis is quite straightforward, but the results can be surprising if you've not done the exercise. The analysis can be further simplified by assuming the confined primary flux is increasing at a constant rate. Hence the rate of change of the vector potential external to the primary solenoid boundary is a constant value along any circular path at a fixed radial distance from the solenoid longitudinal axis. Whilst ever rate of change of the vector potential remains constant, the induced loop current is also constant, meaning there is no self-inductive effect along the loop path. This was the methodology adopted by Robert Romer in his now well-known paper which predated Dr Lewin's demonstration by many years.

Who is not right, Dr. Lewin or electroboom?

@@SiliconSoup The two results of measurement are different (0.9V and -0.1V) but the real voltage is only one. The method of measurement alters the results. Both of them are wrong or nobody. They speak about different things. We have some circuits in a plane. One voltmeter is located in left side and the second one in the right side. They measure different values, right. Try to measure with a voltmeter located in a perpendicular direction of that plane. In this case, the voltmeter will indicate the right value of voltage D-A, that is 0.4 V. KVL always works, I do not see any problem. Dr. Lewin is a little bit misterios, I guess.

​@@Pseudonimul_meuvoltage is multivalued when the electric field is not conservative. So both 0.1 and 0.9 are correct. One is correct for the path going through the smaller resistor, the other is correct for the path going through the bigger resistor.

@@copernicofelinis The voltage AD is only one. Obviously.

@@Pseudonimul_meu saying that there is only one voltage when the field is not conservative is like saying that the amount of gas required to go from LA to NY is the same no matter which route is followed. It should be clear that if you go via Chicago you need a certain amount of gas, but if you go via Anchorage, or Buenos Aires, you need a different amount of gas. Obviously.

6:00 Current = integral of density vector * distribution. Preform this at each at any point times that on the opposite side of the equation multiple by resistance of the wire you’d get V and - V across any two opposite points of the wire. Assuming the magnitude of charge density is the same.

8:20 you are not testing in an enclosed loop.

8:35 I believe the convolution of volt meter 1 and volt meter 2 would be the exact same graph as the one they were compared to, effectively reproducing the entire signal.

I used Azoteq IC to make the video. For single touch as an example, Azoteq offers IQS227. The theory I presented in the video works for all brands and implementations, whether it is the charge transfer or relaxation oscillation methods.

@@SiliconSoup Thank you very much, very nice of you

Do you believe there there is a voltage distributed along the coil if the coil is made of perfect conductor? To me, ohms law applies and there is no voltage built-up along the coil, but yet there is a voltage across the coil terminals.

Thanks for sharing. Kearney is knowledgeable and rigorous. I read his explanation carefully and use it as reference to check my understanding of the topic. Copernico Felinis is another person I like to compare notes with.

The great American Electrical Engineer Dr. Joseph Slepian was (among his other roles) the Associate Director at Westinghouse Research Laboratories (1938 to 1956). He had an extraordinary knowledge of electrodynamics - such as in transformer design. He was awarded the Edison Medal in 1947 “for his practical and theoretical contributions to power systems through circuit analysis, arc control and circuit interruption” I recall a written dialogue between him and another British expert in electromagnetism, Professor E. G. Cullwick in the May 1950 issue of the AIEE’s Journal. They were debating the role of the resultant electric field along the unloaded secondary winding of an energised transformer. Both agreed that there was a negligible resultant electric field along the transformer unloaded seconding winding. In essence the question under consideration was that if there was negligible electric field along the winding how was it that an electromotive force could be measured across the winding terminals. Slepian wrote: “Now, for the electromotive force of an open-circuited complete transformer winding, I certainly will not integrate the electric field along the winding from one end to the other, and call it the electromotive force. As Professor Cullwick says, I'll only get zero. I define electromotive force only for closed paths of integration. However, if I close my path of integration through the coil by a path outside the case, where the field has a potential, then the integral obtained is independent of where I take my closing path in the external region. Thus, I give meaning to the engineer's electromotive force of an open winding when that winding can be closed by a path external to the machine, by defining it as the integral of the electric force around the path so closed. It will of course be the same as the potential difference between the coil terminals, this potential difference being defined over paths lying external.” He continues: “For a part of a coil which cannot be closed by such an external path (for example a half-turn), I give no interpretation of what an engineer might mean by the induced electromotive force. Actually, such an electromotive force never appears in any final design or calculation of the engineer.” So, neither Dr Slepian nor Professor Cullwick suggested that a transformer winding became anything other than a low resistance conductor along which there was negligible resultant electric field and thus negligible voltage gradient. This remains true even when the secondary is on load. The induced secondary emf appears almost in its entirety across the external terminals and nowhere else. It is for this reason that we may, for analytical purposes, model a transformer’s terminal conditions as a localised source of emf. We may then remain blissfully unaware of the electrical field conditions along the (often lengthy) secondary winding proper.

@@trevorkearney3088 I’m a bit unsure if you’re agreeing with me or disagreeing. I’m also not sure if you can assert that neither of them imply that the voltage across the secondary becomes anything but negligible, so I’m fairly sure I’ve misunderstood your point.

Apologies if I've caused confusion. In your original post you stated that “... a conductor exposed to a changing magnetic field is no longer a conductor and has to be treated as a transformer instead.” Does a conductor in the vicinity of a time varying magnetic field embody a localised source of emf? - what another KZbinr called a “hidden battery”. Every other non-electromagnetic source of emf that I can think of has a localised region of discontinuity or a physical transition zone. Therein a non-conservative transfer of energy occurs which establishes a potential difference at the source terminals. This includes phenomena such as an electrochemical effect (battery), thermoelectric (Seebeck) effect (thermocouple , Peltier device), photovoltaic effect (solar cell) and electromechanical effect (piezoelectric device, Van de Graaff machine). No such localised region of discontinuity occurs in an energised transformer winding or AC generator winding and yet a potential difference arises at the device terminals. How do we account for the emf that arises in a closed circuit loop linking a time-varying magnetic field? The most common explanation found in advanced texts on electromagnetic theory is that an induced electric field appears together with the time varying magnetic field. The closed line integral of this electric field equates to the emf. It's often stated that the time varying magnetic field causes the induced electric field. But such causality is questionable. An induced electric field can appear at a point in space where there is a confined time varying magnetic field in the surrounding region, but there is negligible magnetic field at that point and other similar points. Consider (by way of example) an energised primary winding formed over a highly permeable toroidal magnetic core. A single turn secondary loop passing through through the core window manifests an induced emf across its end points. Actually what is observed on a voltmeter is a charge based scalar potential difference - as Silicon Soup points out. I can take a small magnetic field probe and insert it into the toroidal core window near the aforementioned secondary wire. I might wonder at the apparent lack of any magnetic flux adjacent to the wire and yet there must be a substantial magnetic field within the permeable core. I've actually verified this experimentally. So if there is negligible magnetic field adjacent to the wire as it transits the core window, what is causing the emf observed across the wire end points - other than the total electric field encountered everywhere along the wire path AND including the E field in the space between the wire ends? Richard Feynman’s Lecture 22 on AC Circuits includes a discussion of the source of the terminal voltage of an inductor. Feynman notes there is negligible electric field strength along the wire path forming the inductor. Since voltage is defined as the path integral of the electric field strength, there cannot be a significant voltage gradient along the wire path. Of course, Feynman also notes that, so long as we consider only the voltages from conservative fields at component terminals around a closed circuit, then KVL holds. The problem with the single loop case in question is we have no option but to traverse the loop conductor, where the meaning of KVL is ambiguous owing to the non-conservative induced electric field.

@@trevorkearney3088 I’m afraid I still don’t understand your argument. Voltage is, as you said, defined as the line integral of the electric field across a component. I don’t understand what could be incorrect about just modelling the entire loop as consisting of the resistors in question and any number of secondary coils. It doesn’t really matter in my opinion, that the voltage gradient across a coil is very low as a coil can certainly express a voltage difference between it’s terminals. That is all that’s necessary for KVL in circuit analysis. Also, the voltage you measure across the single wire through the toroidal window is caused by the magnetic vector potential which is not enclosed by the toroid and is not measured by the field probe (hall effect, i’m assuming) as the total b-field outside the coil is still zero. My point is that with the single circular loop in Lewin’s experiment, with two resistors connected opposite one another and by two half-circle conductors, KVL will hold due to the two conductor portions forming coils with an induced emf, and their total voltage exactly matching that of the resistors. This has to be the case, how could it possibly be anything different? What Dr. Lewin’s conclusion hinges upon is the difference in the measurements across the probes involved. But that difference is also explained by the probe wires acting as secondary coils; there are multiple people who have confirmed this experimentally by slowly moving the coil wires to not intersect the b-field in a way that generates emf, and the entire difference in measurement disappears and KVL holds. So if the discrepancy is entirely caused by the probe wire placement, isn’t it far more reasonable to assume everything is due to faulty modeling of the system? Instead of concluding that the system breaks a very robust generalization that is derived from Maxwell’s equations?

I would agree that KVL holds around the loop if we consider only the scalar potential differences as we traverse the loop. The experiments you mentioned using (for example) fromjesse's "Lewin Clock" are intended for that purpose. As far as I understand the matter there is no standing voltage across the wire segments in the Lewin experiment - Ohm's law must be satisfied if we consider the resultant E field along the wire segments. The line integral of the resultant electric field encountered in a complete traversal of the loop will not be zero. The line integral of the electrostatic field only taken around the loop will be zero.

Hahaha, you are right.

I am not sure about the maximum temperature the coil can tolerate, but as soon as the thermistor there senses 175C, we shut off the induction.

The half-bridge topology uses 2 IGBT. It is form commercial induction cooker which can deliver higher power (3.5kW)

@@SiliconSoup I have posted a circuit diagram for a Hatco IRNG-PC1-14 on r/inductioncooking, in case you'd like to take a look. It came with the unit. It's a $1000 single burner unit with 100 power levels spanning 1400w, a minimum constant power of 191w, and single degree temperature adjustments; and 2 IGBT's.

Further yet, the Mirage Pro claims to use 4 IGBT's on a 1400 or 1800 unit. The Hatco units are all 2 IGBT's until 2.4kw, when they hit 4.

I'm interested to read it. Can you give a pointer?

@@SiliconSoup sure & here you are :) www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=becomingborealis.com/wp-content/uploads/2018/02/dollardEm-v3-1.pdf&ved=2ahUKEwiBw4SIpIiGAxW5JzQIHXB4DdMQFnoECAUQAQ&usg=AOvVaw0jEvR4tv54KaFT4sHdeUit

Thank you for this good question.The magnetic field above the glass top attenuates quickly because of free-space permeability. The flux going into the cookware is weak and it cannot induce much current. If the cookware is made of magnetic material with high relative permeability, the flux gets multiplied and there induces more current for heating. In other words, the magnetic cookware enhances the coupling and increases the mutual inductance.

@@SiliconSoup. Hello, thank you for the video. You said the wrong cookware could damage the IGBT by messing with the oscillation too much. For those of us who just want to know how to use and take care of our induction stove and cookware, what do you mean by wrong cookware? Are you talking about the cookware material, it's shape, or both? What is the type of cookware we should and should not use on induction? Is it ok to use a heavy Cast Iron skillet on it? I have two other questions. I hear cheap induction hobs that only work by going 100% power "all the time" and keep switching on and off can damage/warp even Carbon Steel skillets. I have one and I'm afraid of benting mine. How do I know if I have a good induction cooktop or one of these cheap ones? I know I should use a cookware of a size smaller than the coil I'm using, but other than that, what can I do to make sure my cookware won't warp? Also, I like using a stovetop moka "espresso" maker on induction, but I want to use a smaller one and the coil won't detect it. I've tried placing some coins or other "magnetic" steel inside it, but it wasn't enough to make the detection work. How does this work? Does placing several metal objects on top of the coil messes with the magnetic field? Does it have to be just one "big enough" metal piece for it to work? Is there any other way I can "hack" to coil detection in order to get it to detect and head the small stainless steel moka that won't be detected by itself? Magnets do work on the small moka's bottom, but it's just too small. Thank you for your time.

@@Jumbernaut For the coil to detect the cookware, the cookware must be magnetic and resistive, and alloy of iron and chromium is an example. Moreover, you must place the alloy/load just on top of the glass top, not any further away. If it is a bit further, the circuit cannot detect it. The magnetic steel should come as a big chuck (rather than several objects), so that current can flow easily inside, to absorb energy. The transfer of energy to the load damps the oscillation, and the MCU detect to damping to determine whether there is a load. If the material of moka maker can't be detected, then the induction can't heat up the same material either, so there is no point trying to fool the coil to turn on the induction. But I guess if you can submerge a big alloy in water inside the maker, the the cooker can detect the alloy, it will turn on the induction, heat up the alloy and transfer the heat to the water.

@@SiliconSoup. Thank you for your response. Do you think those heavy Cast Iron Skillets, like 95% iron, are good for induction, or to they have too much "magnetic, resistance"? I want my induction cooktop to last as long as possible.

If the DC is 325V, then yes, you can skip the AC altogether. But if the DC is 12V, then the coil inductance and resistance must be much smaller to achieve the same current by the lower switching voltage. Many component values must change.

At 8:00, the waveform shoots up above 1000V. The rating of the IGBT is above 1kV too.

At 5:00 is the schematic. There is a LC filter between the rectified voltage and the coil.

The switching of IGBT invert the DC to AC. When the IGBT is on, current flows through the coil to the ground. When the IGBT is off, the current continue to flow and charge up the capacitor. Some point later the charged capacitor will discharge and the current reverse direction. So there is an alternating current in the coil.

@@SiliconSoup lemmi get to learn about IGBT switching and its inversion ability maybe i will understand better