eSigning WPT Made Easy

Esigning wpt

hi I'm Sabine Yaakov this presentation is entitled wireless power transfer circuit theory limitations of the classical design now a wireless power transfer system consists of two antennas one of the transmitter connected to the driver here and then there is the receiving antenna connected to receiver which has a rectifier some other electronics in it in it in order to extract the power from the antenna now the transfer of energy from the transmitter to the receiver is based on a inductive or magnetic transfer and that is that the transmitter is generating a magnetic flux which is then penetrating through the receiving antenna and in therefore generating a voltage here which actually transferred the power to the receiver now unfortunately part of this flux is actually not common to the tool and it's sort of closed here around the transmitter so therefore not all the flux is actually moved to the receiver and in order to quantify it we have this coupling coefficient term such that K times the total flux is the common or the flux which is common to both the transmitter every receiver while 1 minus K flux total flux is the part which is not common and the stray flux here around the transmitter itself now the way to analyze the circuit like this a classical way textbook approach would be by mutual inductance we have two inductor these are the antennas and in order to optimize the circuit would see it later we put a capacitors primary secondary and then we have unfortunately the pathetic resistances of the wire of the antennas and interconnection and then obviously we have this mutual inductance which is equal to K times the square root of the product of these inductances now in order to simplify the presentation here and the expression I'm assuming the diesel are equal this is not losing any generality there's just simplifying the discussion now RL actually represent the load now loads are not resistors usually they are active circuit like with a rectifier and a charger or switch mode converter but it is customary to replace the actual and nonlinear load by a equivalent code so called our AC load which represents the power consumption the power consumption of the load so this is why I am going to use this R sub M as representing the load now the method for analyzing circuit like this can be based on actually separating the two coupled pairs and then introducing a dependent dependent voltage sources defined by j omega m i2 which is current secondary in J Omega M i1 are of the primary these inductors are no more coupled this both these voltage sources or dependent voltage sources are actually taking care of the original coupling here so the current in the secondary is the voltage divided by the total resistance okay total resistance here or I should say total impedance here is ISA s which is equal to the sum of these components and therefore the voltage source at the secondary which is J Omega M times i2 and i2 is this value and since we have products of two J's and there is a minus here this becomes resistive this part and then divided by this of s and I 1 so this voltage source actually is this value here so we can call this as the pins the values the units are really ohms because this is M square divided by ohm and this a big sort of they reflected impedance to the primary and the voltage force is this reflected resistance times I won so in fact the voltage source can be as shown here can be replaced by any pittance Z sub R heart which is equal to this value here I've shown here Omega K here have the two inductance s divided by this V sub s and this is the value here this is the total impedance here now obviously you know to get a maximum current in the primary you'd like these are to be as small as possible and this can be done by running the system at residence such that Z sub s in this sub s these two components will cancel each other the impedances if it is at resonance and you'll be left just be with the resistive component so this is actually what is done the components go back for seconds of these two reactive elements else' basis of s and these two are chosen such that they are equal to the excitation frequency of the excitation frequency is adjusted to death and therefore we have now a resistive circuit circuit none of the reactive element is shown anymore we have just the parasitic resistors they reflected this is now a resistance R L Prime into sum of these two the same thing goes for the secondary we have this positive resistance and the actual RAC or they register which represent the load now the power that will be delivered to the load is I 2 pi sub 2 squared times the load okay now I 2 squared is of course the voltage divided by the total impedance and this voltage is a function of the current at the primary so we end up with this expression I 1 squared times this expression and then I have to correct it for the fact that I've actually calculated the power for the total branch and to get just this portion here I have to do multiply it by this ratio let's not worry about this because at high efficiency this is approaching 1 so let's talk about this part here now what we see here is very interesting that we see that this resistor here this reflected register actually represent the power consumption of the system because we see that I squared I 1 square times this one is indeed the power that is delivered to the load so this actually simplifies matter because then we can not just look at the primary in here and optimize the circuit in terms of this like that value in terms of the efficiency and the power transfer so this is what we are going to do now we then have the problem that we have a source we have a prosthetic resistance and we have this reflected resistance so for high efficiency we would like to have this term much larger than this pathetic resistance that is shown here this one we'd like to have it much much larger than this one and the same thing goes for the secondary we like this resistor to be larger than this pathetic resistance now there is not much we can do here because this is what we choose RL that's it so here is where we can concentrate and understand what's going on and maybe optimize the system by looking at the primary so from here we find that Arab Prime has to be much smaller than this value this all this is known this is the operating conditions and this is also the part that ik resistance now how small should it be it really depends on the efficiency that we are looking for and the efficiency is the ratio I mean this is the primary efficiency of the primary it is the dis resistance divided by the resistance plus the paralytic resistance and therefore this we can sort of quantize this this relationship and say that given a certain efficiency we would like this value to be about this value well this is the efficiency / 1 - efficiency time to depart attic resistance or this is now expressed as our prime has to be for this target efficiency has to be about this value all of these are known is their running frequency mutual inductance prosthetic resistance and the target efficiency so let's have a look now at some green numbers here let's assume that we are running the system at 150 kilo Hertz these two inductances are 20 micro Henry positive resistances are 100 million point 1 ohm and let's do the calculation for a coupling coefficient of 0.5 now first thing I'm calculating Omega M which turns out this is the 2 point 1 point 5 10 to the 5th this is the one frequency and Kate it comes up to be 10 ohm square it's 100 ohm square so we want to keep this relationship and taking a 90 percent the target then we find that this should be around ten times the resistance the positive resistance and I find that they are a prime that if the total resistance here should be around 100 this is for K point five four point eight it comes up to be 250 it's a different number now what about the power that we can transfer now power is as we have said approximately being divided square divided by this resistance well if you like to have it more accurate you have to multiply by the efficiency actually taking into account the losses here in our case it turns out to be mean over one times the efficiency so say one vote will sort of will be a little bit less than one part for one vote excitation from here we find that the power is proportional to our n because the larger there are end the smaller would be this value and the larger will be P out so we can to put her adhere to the nominator and see that the larger the power the larger is our LD larger the power but what about losses now we know that the smaller is this value the larger would be the losses or we were like this value to be large and therefore the loss is a sort of 1 over this value and again the losses seem to be proportional to our rap so here's the problem we are locked here if we increase our L we increase the power but the loss is increased so the efficiency will go down and vice versa so there is a fine area here's an area that can work with fairly high power and fairly high efficiency and if you move from it you are in a problem then there is a question of the quality factor the quality factor of this circuit with the inductor and capacitor is the this series resonant network will be the Omega L over R I'm neglecting the pathetic resistance and since this is omega L this is also mega n lighter I end up with this and in this particular case we are now looking at this numerical situation we find that the Q is around 10 this is fairly high this means that small deviation and the excitation frequency or the tuning will move us from the optimum point so this is another issue that one has to take into account so let's have a look now at some of the simulation of this circuit for the particular parameters that I've taken as an example so here we have the inductances the this capacitor were chosen to match it to a 150 kilo Hertz this is the running frequency here and here is our is the parameter the losses says what has me do here in 100 milliohms here so for K point 5 the value we've chosen we know that the sort of optimum value is 100 ohm we see the indeed the efficiency is as we expected or we a target that it's about 0.9 and the power as we have actually predicted Li is approaching one ground so this is really very consistent with what we have said now what happens if we are moving from this optimal point sorry the case that I've run it for 100 oh this is the nominal value and then 1k so let's look at here first the 1k brings up a much higher power this is two and a half what has compared to one run but see the efficiency very bad it's point five so you see that increasingly with the store is indeed increasing the power as we expected but also increasing them also so therefore the efficiency went down quite a bit now I've also added a resistor of 10 ohms now the efficiency is very good very excellent or not 100% but the power that we can get is very low here it is and I expanded scale here so this is for 10 on the efficiency this is the 100 ohm which is 2.9 this is almost 100% but the power we can get is very very low as compared to only few milli watts 100 milliwatts perhaps as compared to what we actually lower them that as compared to what we have here so we see that indeed the system is very very sensitive to the value of the resistor given the operating point the frequency the inductances and of course most important is the coupling coefficient this is taking into account the coupling coefficient and now what happens if the coupling coefficient is changing so I'm keeping now the resistor to be 100 ohms and changing the coupling coefficient 0.2 0.4 in point 8 now the nominal value for which we have chosen 100 ohm was 0.5 okay so what we see here is again a variation 4.4 which is close to 0.5 we see pretty good efficiency pretty good efficient this is the point 8c and the power is not so bad it's also about one what is the point for now what happens with the point eight this is actually very good coupling much better but again since it is not optimized we have indeed a a higher efficiency to 0.82 high efficiency almost one artisan but power level is very low and for the point two we have a lower efficiency but higher power so we have much higher power but lower efficiency so again we see that once you move from this optimal point you get into trouble so what can be done one way to go if people have shown in the number of papers is to do impedance matching this could be done by a passive network such that the impedance that the system would see would be changed to the optimal one or by a active like switch mode converter which can also reflect a resistance which is different from the actual resistance so what are the conclusions here we see that there is a very large load dependence here and one have to bear in mind that no more loads practice are not resistors they are say constant power loads and then the extra resistance is the power divided by hi square for the hour battery charging and in this case they resistance is something like the voltage of the battery divided by the current times some constant so we don't deal with resistors as loads we deal with active circuitry which has a variable reflected load so there might be a need in order to optimize the system for an impedance matching between the actual load you have and the optimum load for the system especially if indeed you are under changes of the coupling coefficient that is the distance so the distance is really a major player here and the larger the distance the more complicated becomes the situation and you'd expect the lower power transfer and lower efficiency and then there is the issue of the high Q which means that a deviation of frequency and components might actually move you from the optimal point so this brings me to the end of this presentation I hope you found it interesting and perhaps it will be useful to you in the future