Wireless power Transfer (WPT): Circuit theory limitations of the classical design
Summary
TLDRLa présentation de Sabine Yaakov sur le transfert de puissance sans fil met en lumière les limitations théoriques du design classique. Elle explique que le transfert de l'énergie repose sur une interaction magnétique entre deux antennes, le transmitteur et le récepteur, où un coefficient de couplage détermine l'efficacité de la transmission. L'analyse traditionnelle repose sur la mutualité d'induction et l'optimisation du circuit avec des condensateurs primaires et secondaires. L'efficacité est influencée par la résistance des fils et des interconnexions, et la puissance fournie au chargeur dépend de la division du courant secondaire par la résistance totale. Pour une haute efficacité, il est souhaitable d'avoir une grande résistance de charge par rapport à la résistance apparente. L'augmentation de l'inductance peut augmenter la puissance mais aussi les pertes, réduisant l'efficacité. Le facteur de qualité du circuit est élevé, ce qui signifie une sensibilité aux déviations de fréquence. Des simulations montrent la dépendance de la charge et l'importance du coefficient de couplage. L'impédance adaptative, passive ou active, peut être utilisée pour optimiser le système. La conclusion souligne l'importance de l'adaptation du système en fonction de la charge et du coefficient de couplage, en particulier avec les variations de distance.
Takeaways
- 📡 La transmission sans fil d'énergie est basée sur la transfert inductif ou magnétique, où le générateur de flux magnétique est connecté à l'antenne d'émetteur et pénètre l'antenne du récepteur pour transférer la puissance.
- 🔗 Le coefficient de couplage (K) est utilisé pour quantifier la quantité de flux magnétique partagée entre l'émetteur et le récepteur.
- ⚙️ L'analyse classique du circuit utilise la mutualité d'induction et optimise le circuit avec des condensateurs primaires et secondaires pour atteindre la résonance et minimiser les pertes.
- 🔩 La résistance apparente (R_L') reflète la charge réelle, et pour une haute efficacité, il est souhaitable que cette résistance soit bien plus grande que la résistance parasitaire.
- 🔋 La puissance fournie à la charge dépend de la division du courant secondaire par la résistance totale, et est donc fonction de la courbe de l'efficacité.
- 🔢 La sélection de la valeur de la résistance pour maximiser l'efficacité dépend de la fréquence opérationnelle, de l'induction mutuelle et de l'efficacité cible.
- ⚖️ Un compromis doit être trouvé entre l'augmentation de la puissance et la minimisation des pertes, car une augmentation de l'indutance (L) augmente à la fois la puissance et les pertes.
- 🔍 Le facteur de qualité (Q) de la circuit est élevé, ce qui signifie que des déviations de fréquence ou de composants peuvent déplacer le système loin du point optimal.
- 📉 L'efficacité et la puissance transférée sont très sensibles à la valeur de la résistance et au coefficient de couplage, qui est influencé par la distance entre l'émetteur et le récepteur.
- 🔧 L'assortiment de l'impédance peut être une solution pour améliorer l'efficacité du système, que ce soit par un réseau passif ou un convertisseur actif.
- 🔗 La charge réelle n'est pas une résistance mais un circuit actif avec une résistance variable, ce qui nécessite une optimisation du système pour correspondre à la charge optimale, surtout en cas de variations du coefficient de couplage.
Q & A
Quelle est la base du système de transfert de puissance sans fil décrit dans la présentation?
-Le système de transfert de puissance sans fil est basé sur la transfert inductif ou magnétique, où l'antenne de l'émetteur génère un flux magnétique qui pénètre l'antenne du récepteur pour transférer la puissance.
Comment le coefficient de couplage est-il utilisé pour quantifier le flux magnétique dans le système de transfert de puissance sans fil?
-Le coefficient de couplage, noté K, est utilisé pour quantifier le flux magnétique commun aux deux antennes. K fois le flux total représente le flux commun, tandis que (1 - K) fois le flux total représente le flux dérivé autour de l'émetteur.
Quels sont les éléments clés pour optimiser le circuit de transfert de puissance sans fil?
-Pour optimiser le circuit, on utilise la réactance mutuelle et on place des condensateurs primaires et secondaires. On minimise également la résistance des fils et des interconnexions, et on ajuste la fréquence d'excitation pour atteindre la résonance.
Comment la résistance apparente reflétée influence-t-elle la consommation de puissance du système?
-La résistance apparente reflétée représente la consommation de puissance du système. Plus cette résistance est grande par rapport à la résistance parasitaire, plus le système est efficace en termes de transfert de puissance.
Quelle est la relation entre la fréquence d'excitation, la résonance et la puissance maximale dans le circuit?
-La fréquence d'excitation est ajustée pour que les deux composants réactifs annulent mutuellement, ce qui conduit à un circuit résistif. Plus la fréquence est proche de la fréquence de résonance, plus la puissance maximale qui peut être transférée est grande.
Comment la qualité factor (Q) du circuit affecte-t-elle la performance du système de transfert de puissance sans fil?
-Un Q élevé signifie que le circuit est très sensible aux déviations de fréquence ou de composants, ce qui peut entraîner un déplacement du point optimal. Un Q trop élevé peut donc poser des défis pour le suivi de la fréquence et l'ajustement des composants.
Quels facteurs affectent l'efficacité du transfert de puissance dans le système décrit?
-L'efficacité est affectée par la valeur de la résistance, la fréquence d'excitation, les inductances, la distance entre les antennes (coefficient de couplage), et la qualité factor du circuit.
Comment la distance entre les antennes impacte-t-elle le transfert de puissance et l'efficacité du système?
-Une distance plus grande entre les antennes diminue le coefficient de couplage, ce qui à son tour diminue le transfert de puissance et l'efficacité globale du système.
Quels sont les défis associés à l'utilisation de charges variables comme les chargeurs de batteries dans le système de transfert de puissance sans fil?
-Les charges variables, comme les chargeurs de batteries, représentent des circuits actifs avec une résistance reflétée variable, ce qui nécessite un ajustement de l'impédance pour optimiser le système, en particulier face aux changements du coefficient de couplage.
Quelle est une méthode pour améliorer l'efficacité du système de transfert de puissance sans fil face aux changements de charge et de distance?
-L'impédance adaptative, que ce soit par un réseau passif ou un convertisseur de mode actif, peut être utilisée pour s'adapter à la charge réelle et optimiser la performance du système.
Comment la présentation aborde-t-elle la question de la sensibilité du système à la valeur de la résistance et à la fréquence d'excitation?
-La présentation illustre la sensibilité du système à la valeur de la résistance et à la fréquence d'excitation en utilisant des simulations avec des paramètres spécifiques, montrant comment les changements dans ces valeurs affectent la puissance et l'efficacité.
Outlines
🔌 Transfert de puissance sans fil et théorie des circuits
Le paragraphe 1 présente le transfert de puissance sans fil, une technologie qui utilise deux antennes pour transférer de l'énergie de l'émetteur au récepteur par induction magnétique. Le concept de coefficient de couplage est introduit pour quantifier la partie du flux magnétique qui est partagée entre les deux antennes. L'analyse classique de ce type de circuit repose sur la mutualité de l'induction et l'utilisation de condensateurs pour optimiser la performance. Le but est de simplifier la discussion en supposant des inductances égales et de décrire comment l'impédance peut être modifiée pour améliorer le transfert de puissance.
🔧 Optimisation du circuit de transfert de puissance sans fil
Dans le paragraphe 2, l'auteur explique comment optimiser le circuit de transfert de puissance sans fil en utilisant la résonance pour minimiser la perte d'énergie et maximiser le courant primaire. Il est mentionné que pour une efficacité élevée, il est nécessaire d'augmenter la résistance réfléchie par rapport à la résistance parasitaire. L'efficacité du système dépend de la relation entre ces résistances et la charge du récepteur, qui est souvent représentée par une résistance équivalente AC. L'auteur illustre également comment le transfert de puissance dépend de la tension et de l'impédance totale du circuit.
🔢 Exemples numériques et facteurs influençant la performance
Le paragraphe 3 se concentre sur les calculs numériques pour déterminer la taille des composants nécessaires à un certain niveau d'efficacité et de transfert de puissance. L'auteur explore les implications de différentes valeurs de résistance et de coefficient de couplage sur la puissance et l'efficacité du système. Il est également discuté comment l'augmentation de l'inductance peut augmenter la puissance mais aussi les pertes, ce qui peut réduire l'efficacité. La qualité factor (Q) du circuit est également examinée pour comprendre la sensibilité du système à la fréquence d'excitation.
📊 Simulation des performances du circuit et ajustements
Dans le paragraphe 4, l'auteur présente des simulations qui montrent comment les performances du circuit varient avec les paramètres tels que la résistance et le coefficient de couplage. Il est illustré comment les changements de ces valeurs affectent l'efficacité et le niveau de puissance, et comment l'optimisation doit être réalisée pour chaque point. L'auteur mentionne également l'importance de l'impédance adaptée pour s'assurer que le système fonctionne à son meilleur potentiel, que ce soit par un réseau passif ou un convertisseur actif.
🔚 Conclusions et implications pour l'avenir
Le paragraphe 5 conclut la présentation en soulignant la dépendance de la charge et la nécessité d'une optimisation de l'impédance pour les charges pratiques, qui sont souvent des circuits actifs avec une résistance variable. L'auteur met en évidence l'impact de la distance sur le transfert de puissance et l'efficacité, ainsi que les défis posés par le haut facteur de qualité du circuit. Il encourage à considérer ces éléments pour améliorer les systèmes de transfert de puissance sans fil à l'avenir.
Mindmap
Keywords
💡Transfert de puissance sans fil
💡Induction mutuelle
💡Coefficient de couplage
💡Réactance
💡Charge utile
💡Fréquence de résonance
💡Facteur de qualité (Q)
💡Charge constante de puissance
💡Impédance
💡Pertes
💡Distance
Highlights
Wireless power transfer systems consist of two antennas: a transmitter and a receiver, with energy transfer based on inductive or magnetic coupling.
The transmitter generates a magnetic flux that penetrates the receiving antenna, creating a voltage that transfers power.
The coupling coefficient (K) quantifies the common flux between the transmitter and receiver, distinguishing it from the stray flux.
Optimizing the circuit involves adding capacitors to the primary and secondary coils and considering the parasitic resistances of the antennas.
Loads are typically active circuits with rectifiers and converters, often replaced by an equivalent resistive load for analysis.
The circuit can be analyzed by separating the coupled coils and introducing dependent voltage sources, effectively decoupling the inductors.
For maximum current in the primary, the system should operate at resonance to minimize impedance and maximize power transfer.
Efficiency is determined by the ratio of the resistive load to the total resistance, including parasitic resistances.
The power delivered to the load is proportional to the square of the primary current and inversely proportional to the reflected resistance.
Increasing the inductance (L) can increase power but also losses, potentially reducing efficiency.
The quality factor (Q) of the circuit is high, indicating sensitivity to frequency deviations and the need for precise tuning.
Simulations demonstrate the system's efficiency and power output are highly sensitive to the load resistance and operating conditions.
Impedance matching, both passive and active, can be used to optimize the system for different load conditions and coupling coefficients.
The distance between the transmitter and receiver significantly impacts power transfer and efficiency, with greater distances reducing performance.
Active circuitry with variable reflected loads, such as battery chargers, requires system optimization for impedance matching.
The presentation concludes with the importance of considering load dependence, coupling coefficient changes, and quality factor in wireless power transfer design.
Transcripts
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
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