+ Post New Thread
Page 1 of 4 1 2 3 ... LastLast
Results 1 to 20 of 68
  1. #1
    Full Member level 5
    Points: 2,985, Level: 12
    Achievements:
    7 years registered

    Join Date
    Sep 2008
    Posts
    292
    Helped
    0 / 0
    Points
    2,985
    Level
    12

    Relationship between Q factor and the coupling factor of two wireless coils

    Hi All, I am testing the data transfer between two wirelss coils.
    The input sinusoidal signal of the primary coil is from the function generator, whose peak to peak level is 10V and frequency is adjustable between 1MHz to 5MHz.

    Click image for larger version. 

Name:	CF.JPG 
Views:	8 
Size:	51.0 KB 
ID:	136808

    I am using inductive coupling method, to place a cap in parallel with the Tx and Rx coils. This means I can only get resonance at a frequency within the input frequency range. Therefore, Q values at different input frequency shoudl be different.

    When I have the distance between two coils fixed (say 1 cm), and two coils aligned with each other, I found the peak to peak amplitude of the sinusoidal signal in Rx are relatively constant (for example, 4V Vpp at Rx when Tx has 10V Vpp) over the entire f range.

    If we assume the Vpp at Rx can be determined by L_rx, L_tx, and the coupling factor, can I conclude that the coupling factor is a weak function of Q?

    I only know the coupling factor is related to the distance between the coils and the coils' diameter. However, I did not find the equation that relates the coils' coupling factor and the Q. I am not sure if my conclusion based on the exerimental result is correct. Can anyone advise me or refer me to some related papers/books?

    Thank you!

  2. #2
    Advanced Member level 3
    Points: 4,040, Level: 14

    Join Date
    Dec 2015
    Location
    Madrid, Spain
    Posts
    724
    Helped
    179 / 179
    Points
    4,040
    Level
    14

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by bhl3302 View Post

    I am using inductive coupling method, to place a cap in parallel with the Tx and Rx coils. This means I can only get resonance at a frequency within the input frequency range. Therefore, Q values at different input frequency shoudl be different.
    Inductive coupling is when there is no capacitor to tune the circuit.
    If any capacitor is there in order to tune the circuit to 1 frequency, then it is called resonant coupling, not inductive coupling as the Qi standard uses.

    Quote Originally Posted by bhl3302 View Post
    If we assume the Vpp at Rx can be determined by L_rx, L_tx, and the coupling factor, can I conclude that the coupling factor is a weak function of Q?

    I only know the coupling factor is related to the distance between the coils and the coils' diameter. However, I did not find the equation that relates the coils' coupling factor and the Q.
    Depending on how you have defined the Q factor (it can be Q factor of coil, or the Q factor = Q_tx * Q_rx or some Q related to your wireless link) it is dependent on the values of the inductors. The coils geometry influences the "k" factor as you said. Coils geometries influences the inductance. I conclude then, the Q factor is related to "k" somehow.

    "k" is related to coil geometry --> Coil geometry is related to the Inductance --> Inductance is related to Q factor

    ====> "k" is related to Q factor.



  3. #3
    Full Member level 5
    Points: 2,985, Level: 12
    Achievements:
    7 years registered

    Join Date
    Sep 2008
    Posts
    292
    Helped
    0 / 0
    Points
    2,985
    Level
    12

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by CataM View Post
    Inductive coupling is when there is no capacitor to tune the circuit.
    If any capacitor is there in order to tune the circuit to 1 frequency, then it is called resonant coupling, not inductive coupling as the Qi standard uses.


    Depending on how you have defined the Q factor (it can be Q factor of coil, or the Q factor = Q_tx * Q_rx or some Q related to your wireless link) it is dependent on the values of the inductors. The coils geometry influences the "k" factor as you said. Coils geometries influences the inductance. I conclude then, the Q factor is related to "k" somehow.

    "k" is related to coil geometry --> Coil geometry is related to the Inductance --> Inductance is related to Q factor

    ====> "k" is related to Q factor.
    Thank you CataM! I am trying to explain the observation on bench using the theory. But I am not sure what is the right way to do it.

    If my parameters are like this:
    (1)The Tx coil is Wurth 760308101302 and Rx coil is Wurth 760308102212.Both coil are 5.4uH.
    (2) both caps are 10pF to resonant with coils at 3MHz.

    How can I explain that I get similar Vpp at Rx when input frequency is 1MHz, 3MHz and 5MHz on bench?



  4. #4
    Advanced Member level 3
    Points: 4,040, Level: 14

    Join Date
    Dec 2015
    Location
    Madrid, Spain
    Posts
    724
    Helped
    179 / 179
    Points
    4,040
    Level
    14

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by bhl3302 View Post
    How can I explain that I get similar Vpp at Rx when input frequency is 1MHz, 3MHz and 5MHz on bench?
    Experimentally: Do a Bode plot of the magnitude of your link.
    Theoretically: knowing "k" and all other parameters of your design, find V_rx / Vin and do a Bode representation. If there is no improvement at the resonant frequency, I guess your circuit is too tightly coupled and hence detuned already due to the frequency split phenomena. But as I said, that is just a wild guess.



    •   Alt11th March 2017, 00:57

      advertising

        
       

  5. #5
    Advanced Member level 2
    Points: 4,327, Level: 15

    Join Date
    Aug 2015
    Posts
    667
    Helped
    268 / 268
    Points
    4,327
    Level
    15

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    The Q of the coils and any components connected to them, is to a very large degree independent of the coupling between coils, when an adjacent coil pulls power out of a driven coil - it may look like the Q is reducing - indeed the Q of the overall system is - as you are introducing a load or losses - but the Q of the individual coils is independent of coupling.



  6. #6
    Full Member level 5
    Points: 2,985, Level: 12
    Achievements:
    7 years registered

    Join Date
    Sep 2008
    Posts
    292
    Helped
    0 / 0
    Points
    2,985
    Level
    12

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by Easy peasy View Post
    The Q of the coils and any components connected to them, is to a very large degree independent of the coupling between coils, when an adjacent coil pulls power out of a driven coil - it may look like the Q is reducing - indeed the Q of the overall system is - as you are introducing a load or losses - but the Q of the individual coils is independent of coupling.
    Hi Easy peasy, thank you for your help. Sorry I do not link your explanation to my observations. When we calculate Q, we use the frequency information and the LC components. So since I have an input frequency range, I suppose to get a range of Q. How can I use it to explain my observation that the Vrx at different frequencies are almost have the same amplitude?

    - - - Updated - - -

    Quote Originally Posted by CataM View Post
    Experimentally: Do a Bode plot of the magnitude of your link.
    Theoretically: knowing "k" and all other parameters of your design, find V_rx / Vin and do a Bode representation. If there is no improvement at the resonant frequency, I guess your circuit is too tightly coupled and hence detuned already due to the frequency split phenomena. But as I said, that is just a wild guess.
    Hi CataM, I do have a frequency response analyzer in hand, are you talking about the measurement like this?
    Click image for larger version. 

Name:	bode.png 
Views:	6 
Size:	258.2 KB 
ID:	136811
    Based on your understanding, if I get a flat gain within the target frequency range, can you conclude the frequency split phenomena ? From the time domain, when I have 10 V peak to peak signal at the Tx, it is already attenuated to 4V. And the distance between the coils is 1cm, I am not sure if this can still be treated as "tightly coupled" case.
    I should be able to get the bode tomorrow. Would you be able to review it? Thank you!



    •   Alt11th March 2017, 04:00

      advertising

        
       

  7. #7
    Advanced Member level 5
    Points: 13,005, Level: 27

    Join Date
    May 2015
    Location
    Melbourne, Australia
    Posts
    2,213
    Helped
    748 / 748
    Points
    13,005
    Level
    27

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Easy P. is quite correct.

    What happens at great distance is that coupling is weak and the two tuned circuits can be peaked independently, but energy coupling is poor because of the distance.

    As the two tuned circuits are moved closer the amplitude increases and you reach a condition known as "critical coupling".

    If you move the tuned circuits even closer together a very strange thing happens. You get two new frequency peaks either side of the original resonant point, and a dip in the middle.
    Its called "over coupling" and is actually quite useful for increasing the bandwidth of some kinds of inductively coupled filters.

    http://www.ee.bgu.ac.il/~intrlab/lab...20circuits.pdf

    In actual practice there are so many secondary effects such as the Q of the two circuits and coil geometry, its pretty difficult to predict a result. It really needs to be tested.
    Cheers, Tony.



    •   Alt11th March 2017, 07:17

      advertising

        
       

  8. #8
    Super Moderator
    Points: 231,463, Level: 100
    Awards:
    1st Helpful Member

    Join Date
    Jan 2008
    Location
    Bochum, Germany
    Posts
    39,962
    Helped
    12204 / 12204
    Points
    231,463
    Level
    100

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    I guess "testing the data transfer between two wireless coils" involves operating it without a receiver coil load. That's different from the intended operation of these coils for wireless power transfer, where a coil Q of about 100 is almost irrelevant because the real load dominates all circuit losses.

    Secondly you are operating the coils in a higher frequency range where ferrite losses are probably dominant.

    Finally "coupling factor" in your post seems to refer to tx/rx voltage ratio of a resonant circuit, which is not the same as "k" in usual coupled inductor models. Your definition of Q is neither quite clear. It's usually described as frequency dependent parameter of a single inductor, Q = jωL/ESR. How do you determine it?



  9. #9
    Advanced Member level 3
    Points: 4,040, Level: 14

    Join Date
    Dec 2015
    Location
    Madrid, Spain
    Posts
    724
    Helped
    179 / 179
    Points
    4,040
    Level
    14

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by bhl3302 View Post
    (2) both caps are 10pF to resonant with coils at 3MHz.
    Why 10 pF and not 521.2 pF ?



  10. #10
    Full Member level 5
    Points: 2,985, Level: 12
    Achievements:
    7 years registered

    Join Date
    Sep 2008
    Posts
    292
    Helped
    0 / 0
    Points
    2,985
    Level
    12

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by CataM View Post
    Why 10 pF and not 521.2 pF ?
    Hi CataM, I found my mistake in the initial calculation for the cap. I chosen the wrong cap which results in a resonance at 20MHz. For some reason I cannot change the cap value now, do you think it makes sense that I do not have too much variation for the waveforms between 1MHz to 5MHz?

    - - - Updated - - -

    Quote Originally Posted by FvM View Post
    I guess "testing the data transfer between two wireless coils" involves operating it without a receiver coil load. That's different from the intended operation of these coils for wireless power transfer, where a coil Q of about 100 is almost irrelevant because the real load dominates all circuit losses.

    Secondly you are operating the coils in a higher frequency range where ferrite losses are probably dominant.

    Finally "coupling factor" in your post seems to refer to tx/rx voltage ratio of a resonant circuit, which is not the same as "k" in usual coupled inductor models. Your definition of Q is neither quite clear. It's usually described as frequency dependent parameter of a single inductor, Q = jωL/ESR. How do you determine it?
    Hi FvM, you are right. I am defining tx/rx voltage ratio as the "coupling factor" instead of the one for the coupled inductor. The coils I am using can operate in several MHz level. (The Tx coil is Wurth 760308101302 and Rx coil is Wurth 760308102212.Both coil are 5.4uH.)
    In the real application, there is another DC block cap connected to the output of the receiver, and then feed as the input of a logic inverter. If we assume the Cgs of the input MOSFET of the inverter is still in pF level, Cgs will connect in series with this 10nF DC blocking cap, then stay in parallel with that 10nF cap C_SEC. I think the overall load of the secondary data coil is still several pF level. Am I correct?
    For a RLC in series circuit, I think Q is still calculated as jωL/(ESR+all parasitic). CataM just found out a mistake in capacitance calculation of my case, which means my RLC should peak at 20MHz. If we use this Q definition, its level between 1MHz to 5MHz should be pretty constant. Assuming the distance between my coils is kept at 1 cm, does the relatively low but constant Q level explain the constant peak to peak rx voltage I observed on bench?
    Thank you!

    - - - Updated - - -

    Quote Originally Posted by Warpspeed View Post
    Easy P. is quite correct.

    What happens at great distance is that coupling is weak and the two tuned circuits can be peaked independently, but energy coupling is poor because of the distance.

    As the two tuned circuits are moved closer the amplitude increases and you reach a condition known as "critical coupling".

    If you move the tuned circuits even closer together a very strange thing happens. You get two new frequency peaks either side of the original resonant point, and a dip in the middle.
    Its called "over coupling" and is actually quite useful for increasing the bandwidth of some kinds of inductively coupled filters.

    http://www.ee.bgu.ac.il/~intrlab/lab...20circuits.pdf

    In actual practice there are so many secondary effects such as the Q of the two circuits and coil geometry, its pretty difficult to predict a result. It really needs to be tested.
    Hi Warpspeed, should I expect to get the bode plot in showing all three conditions if I play with the distance between the coils?



  11. #11
    Advanced Member level 5
    Points: 13,005, Level: 27

    Join Date
    May 2015
    Location
    Melbourne, Australia
    Posts
    2,213
    Helped
    748 / 748
    Points
    13,005
    Level
    27

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Not sure about a Bode plot, but the amplitude versus frequency effect of over coupling is certainly very readily visible on a spectrum analyser.

    The main thing is that with very loose coupling right up to critical, both tuned circuits resonate quite independently of each other.

    With tighter coupling you are then dealing with a "tightly coupled system" with a very high degree of interaction. You loose the independence, and each resonant circuit heavily loads the other.

    Phase through a coupled radio system is not really relevant where there is only one signal path.
    Phase only becomes important when there is something else to compare it to.
    Cheers, Tony.



  12. #12
    Full Member level 5
    Points: 2,985, Level: 12
    Achievements:
    7 years registered

    Join Date
    Sep 2008
    Posts
    292
    Helped
    0 / 0
    Points
    2,985
    Level
    12

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by Warpspeed View Post
    Not sure about a Bode plot, but the amplitude versus frequency effect of over coupling is certainly very readily visible on a spectrum analyser.

    The main thing is that with very loose coupling right up to critical, both tuned circuits resonate quite independently of each other.

    With tighter coupling you are then dealing with a "tightly coupled system" with a very high degree of interaction. You loose the independence, and each resonant circuit heavily loads the other.

    Phase through a coupled radio system is not really relevant where there is only one signal path.
    Phase only becomes important when there is something else to compare it to.
    Hi Warpspeed, thank you so much for your help. I just got the bode plot, it may tell us something.
    I am testing Vout/Vin while the setup is the same as #6 in this thread. All the other circuits are operating.
    I did not see the frequency split phenomenon, but I did see the frequency of the peaking is moved towards lower frequency level as I reduced the distance between the coils. I did some similar time domain measurement and confirm the movement of Q does influent the amplitude of the received sine signal at the secondary coil.
    This is my results when the distance between the coils are 1cm, 0.5cm and 0.1cm. When I meaured the time domain waveform previously, I used 1cm as the distance.
    Click image for larger version. 

Name:	bode_measure.png 
Views:	9 
Size:	100.2 KB 
ID:	136830
    My questions regarding this measurement is
    (1) For 1cm case, if we relate this peaking in f domain to the peak to peak amplitude measured in time domain, that may explain why I can get relatively constant peak to peak in 1MHz and 3MHz, but I supposed to get higher Vpeak to peak at 5MHz, which actually not the case I see. Do you have any idea what can explain the discrepancy between the prediction of this bode plot? I used maximum V_AC in the bode test (1V) to get this bode result, and I am using a 10 V V peak to peak (5V Vpeak) in time domain to get ~4V Vpeak to peak at the frequency of 1MHz, 3MHz and 5MHz.
    (2) what could be the explaination that the reduction of the distance between the coils cause the movement of the peaking towards the lower frequency level?
    Thank you!

    - - - Updated - - -

    Quote Originally Posted by CataM View Post
    Why 10 pF and not 521.2 pF ?
    Hi CataM, I uploaded the bode plots in three different distance measurement in a previous reply to Warpspeed. Thank you for pointing out the cap selection issue. However, in using 10pF cap, I can get a resonant at around 6MHz when the distance is around 1cm. And this peaking is moved towards the lower frequency level when the distance is reduced. Do you have any idea how it comes out like this? Thank you!



  13. #13
    Super Moderator
    Points: 231,463, Level: 100
    Awards:
    1st Helpful Member

    Join Date
    Jan 2008
    Location
    Bochum, Germany
    Posts
    39,962
    Helped
    12204 / 12204
    Points
    231,463
    Level
    100

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Resonance frequency shifts because the inductance increases with reduced coil distance.

    I'm not sure if the in circuit Q observed in your measurements has much to do with coil losses. I guess it's mostly determined by the generator impedance, not yet mentioned in the thread.



  14. #14
    Full Member level 5
    Points: 2,985, Level: 12
    Achievements:
    7 years registered

    Join Date
    Sep 2008
    Posts
    292
    Helped
    0 / 0
    Points
    2,985
    Level
    12

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by FvM View Post
    Resonance frequency shifts because the inductance increases with reduced coil distance.

    I'm not sure if the in circuit Q observed in your measurements has much to do with coil losses. I guess it's mostly determined by the generator impedance, not yet mentioned in the thread.
    Thank you FvM, would you explain a little bit more regarding the inductance increase as distance reduces?

    I am using a function generator with a BNC cable (and 50 Ohm attenuators) to connect to the TX coil. Would you teach me how can I evaluate or estimate the loss? According to the bode, is the loss acceptable?

    Thank you!



  15. #15
    Advanced Member level 3
    Points: 4,040, Level: 14

    Join Date
    Dec 2015
    Location
    Madrid, Spain
    Posts
    724
    Helped
    179 / 179
    Points
    4,040
    Level
    14

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by bhl3302 View Post
    do you think it makes sense that I do not have too much variation for the waveforms between 1MHz to 5MHz?
    Yes. Actually, if tuned perfectly i.e. very little parasitics, on that range of frequencies shows a flat band.
    As the coupling factor is increased (magnetic coupling factor, not Rx/Tx), as FvM said, the peak shifts. Maybe parasitics play an important role on your circuit which shows a peak at 6 MHz.
    Quote Originally Posted by FvM View Post
    Resonance frequency shifts because the inductance increases with reduced coil distance.
    Do you mean Mutual inductance, right ?

    - - - Updated - - -

    Now that you know that you placed the wrong cap, could you do as well the bode plot around the 21 MHz freq and see what shows there?



  16. #16
    Full Member level 5
    Points: 2,985, Level: 12
    Achievements:
    7 years registered

    Join Date
    Sep 2008
    Posts
    292
    Helped
    0 / 0
    Points
    2,985
    Level
    12

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by CataM View Post
    Yes. Actually, if tuned perfectly i.e. very little parasitics, on that range of frequencies shows a flat band.
    As the coupling factor is increased (magnetic coupling factor, not Rx/Tx), as FvM said, the peak shifts. Maybe parasitics play an important role on your circuit which shows a peak at 6 MHz.

    Do you mean Mutual inductance, right ?

    - - - Updated - - -

    Now that you know that you placed the wrong cap, could you do as well the bode plot around the 21 MHz freq and see what shows there?
    Hi CataM, my frequency response analyzer can only provide results up to 15MHz, so I am not able to see the plot at 21MHz. However, your question did remind me, is it possible that frequency splits already happened when I have the distance 1cm? This means all I saw in the bode plot are the lower frequency component of the splitted frequency, while the upper frequency component is higher than 21MHz? Thank you!



  17. #17
    Advanced Member level 3
    Points: 4,040, Level: 14

    Join Date
    Dec 2015
    Location
    Madrid, Spain
    Posts
    724
    Helped
    179 / 179
    Points
    4,040
    Level
    14

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by bhl3302 View Post
    is it possible that frequency splits already happened when I have the distance 1cm? This means all I saw in the bode plot are the lower frequency component of the splitted frequency, while the upper frequency component is higher than 21MHz? Thank you!
    No because the right hand side of the bode plot decreases very fast. Moreover, in the parallel-parallel resonant topology is more difficult to see or to meet the condition for the 2 peaks, in contrast of the series-series topology.
    I think somehow you have reached the ~130 pF with parasitics... otherwise, I do not know why the resonance is at ~6 MHz. In that circuit, at first sight you already have 10 pF + 6.4 pF (parasitic of the coil), 7.5 pF for Tx and Rx respectively.

    To make sure that the true resonance of the circuit is around 6 MHz, do a Bode plot at a very large distance (i.e. very loosely coupled where the true resonance is not influenced) e.g. 7 cm or so. Then, the true resonance of the circuit is at that frequency. Make sure this time you are able to see up to 15 MHz because it will shift from upper frequency to lower.
    Last edited by CataM; 12th March 2017 at 20:32.



  18. #18
    Full Member level 5
    Points: 2,985, Level: 12
    Achievements:
    7 years registered

    Join Date
    Sep 2008
    Posts
    292
    Helped
    0 / 0
    Points
    2,985
    Level
    12

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by CataM View Post
    No because the right hand side of the bode plot decreases very fast. Moreover, in the parallel-parallel resonant topology is more difficult to see or to meet the condition for the 2 peaks, in contrast of the series-series topology.
    I think somehow you have reached the ~130 pF with parasitics... otherwise, I do not know why the resonance is at ~6 MHz. In that circuit, at first sight you already have 10 pF + 6.4 pF (parasitic of the coil), 7.5 pF for Tx and Rx respectively.

    To make sure that the true resonance of the circuit is around 6 MHz, do a Bode plot at a very large distance (i.e. very loosely coupled where the true resonance is not influenced) e.g. 7 cm or so. Then, the true resonance of the circuit is at that frequency. Make sure this time you are able to see up to 15 MHz because it will shift from upper frequency to lower.
    Hi CataM, thank you so much for your help! Sorry I still have some questions regarding your answers, would you teach me again?
    (1) I model the parasitic resistance (DCR and trace impedance of the pcb) to be in series with the L, and calculate the peaking.
    Click image for larger version. 

Name:	resonance.JPG 
Views:	9 
Size:	106.6 KB 
ID:	136866
    In this plot, the left hand side and right hand side of the peak are pretty symmetric. I did not see the "flat band" on the left hand side, nor "the right hand side of the bode plot decreases very fast". Would you tell me what is wrong in my calculation or derivation?
    (2) regarding the equivalent capacitance of the Rx side, there is another DC block cap connected to the output of the receiver, and then feed as the input of a logic inverter. This is the picture that showed in an earlier post in this treadClick image for larger version. 

Name:	CF.JPG 
Views:	5 
Size:	61.1 KB 
ID:	136867. I am using this inverter (http://www.ti.com/lit/ds/symlink/sn74lvc1g97.pdf).
    My guess is, if we assume the Cgs of the input MOSFET of the inverter is still in pF level, Cgs will connect in series with this 10nF DC blocking cap, then stay in parallel with that 10nF cap C_SEC. So the overall load of the secondary data coil is still several pF level, like what you have pointed out, around 130pF. Is this analysis correct?
    Thank you!

    - - - Updated - - -

    Quote Originally Posted by bhl3302 View Post
    Hi CataM, thank you so much for your help! Sorry I still have some questions regarding your answers, would you teach me again?
    (1) I model the parasitic resistance (DCR and trace impedance of the pcb) to be in series with the L, and calculate the peaking.
    Click image for larger version. 

Name:	resonance.JPG 
Views:	9 
Size:	106.6 KB 
ID:	136866
    In this plot, the left hand side and right hand side of the peak are pretty symmetric. I did not see the "flat band" on the left hand side, nor "the right hand side of the bode plot decreases very fast". Would you tell me what is wrong in my calculation or derivation?
    (2) regarding the equivalent capacitance of the Rx side, there is another DC block cap connected to the output of the receiver, and then feed as the input of a logic inverter. This is the picture that showed in an earlier post in this treadClick image for larger version. 

Name:	CF.JPG 
Views:	5 
Size:	61.1 KB 
ID:	136867. I am using this inverter (http://www.ti.com/lit/ds/symlink/sn74lvc1g97.pdf).
    My guess is, if we assume the Cgs of the input MOSFET of the inverter is still in pF level, Cgs will connect in series with this 10nF DC blocking cap, then stay in parallel with that 10nF cap C_SEC. So the overall load of the secondary data coil is still several pF level, like what you have pointed out, around 130pF. Is this analysis correct?
    Thank you!
    Hi CataM, sorry for my mistake in the 1st question of the last post. I used the absolute amplitude instead of the dB to figure it out. So it is clear for the question 1. Would you advise me for the 2nd question?



  19. #19
    Super Moderator
    Points: 231,463, Level: 100
    Awards:
    1st Helpful Member

    Join Date
    Jan 2008
    Location
    Bochum, Germany
    Posts
    39,962
    Helped
    12204 / 12204
    Points
    231,463
    Level
    100

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Do you mean Mutual inductance, right ?
    Not only mutual inductance. Also the coil main inductance. If you look at the core geometry, it's obvious why.


    1 members found this post helpful.

  20. #20
    Full Member level 5
    Points: 2,985, Level: 12
    Achievements:
    7 years registered

    Join Date
    Sep 2008
    Posts
    292
    Helped
    0 / 0
    Points
    2,985
    Level
    12

    Re: Relationship between Q factor and the coupling factor of two wireless coils

    Quote Originally Posted by FvM View Post
    Not only mutual inductance. Also the coil main inductance. If you look at the core geometry, it's obvious why.
    Hi FvM, would you provide some reference (the name of the paper/textbook) that I can use to know more about this phenomenon? Thank you!



    •   Alt13th March 2017, 00:33

      advertising

        
       

--[[ ]]--