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(Magnetic Coupling) Why connect the capacitor in parallel to the inductor?

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hikaru1221
#1
Sep19-10, 01:05 PM
P: 799
I attach below the picture showing the principle of magnetic coupling in witricity (source: Electrons Unplugged by David Schneider, IEEE Spectrum, May 2010). I'm wondering why they connect the capacitor in parallel with the inductor. Why not in series? Any opinion? Thank you
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Carl Pugh
#2
Sep20-10, 11:01 AM
P: 384
The red capacitor is in parallel, and the blue capacitor is in series.

This is an ancient circuit that was used with vacuum tubes.
The red capacitor is in parallel for low impedance. (Vacuum tube plates operated best when connected to a low impedance). Also if capacitor was in series, there would not be a path for the tube's plate current.
The blue capacitor is in series to obtain the highest voltage on following tube control grid.

Would anyone like to bet that more than half the people, who comment on this post, will say that both capacitors are in parallel?
hikaru1221
#3
Sep20-10, 11:12 AM
P: 799
Thank you

Quote Quote by Carl Pugh View Post
The red capacitor is in parallel, and the blue capacitor is in series.
The picture doesn't show the rest of the circuit (I don't know what the rest is). From the picture, there are 2 wires connecting the blue capacitor to something else, so can it still be in series with blue inductor?

This is an ancient circuit that was used with vacuum tubes.
The red capacitor is in parallel for low impedance. (Vacuum tube plates operated best when connected to a low impedance). Also if capacitor was in series, there would not be a path for the tube's plate current.
The blue capacitor is in series to obtain the highest voltage on following tube control grid.
Sorry, I'm new to this field, so can you elaborate on this?
From what I understand, there is a capacitor connecting in parallel to the filament in this picture: http://vi.wikipedia.org/wiki/T%E1%BA...acuum_tube.png
I'm not sure about the other capacitor, and also, how the vacuum tube is related to witricity.

vk6kro
#4
Sep20-10, 09:43 PM
Sci Advisor
P: 4,032
(Magnetic Coupling) Why connect the capacitor in parallel to the inductor?

That circuit shows two parallel tuned circuits (both tuned to the same frequency) that are loosely coupled and lightly loaded to produce selectivity.

Coupling is relatively inefficient since a lot of the input power is lost and does not appear in the output. However for a relatively small range of frequencies, there is a path from input to output.

It was common to add a small capacitor between one end of each coil to improve efficiency.

These circuits were used in valve amplifiers which were fixed tuned to a frequency like 455 KHz, and radio signals were then mixed with an oscillator to produce input to a number of such narrow band amplifier stages. This gave the receiver relatively good selectivity.

Valve amplifiers generally had to have high impedance loads because they had a lot of internal resistance and were unable to supply large currents. They were operated at high voltages because of this.
Mike_In_Plano
#5
Sep20-10, 11:17 PM
P: 560
I don't think these capacitors are getting enough credit. This circuit has intrinsic value and is still around for induction heating, transdermal power transfer, and even for remote charging circuits.

Essentially, the inductor stores energy as magnetic field. This is greatest when it experiences the highest current.
The capacitor stores energy as electrical charge. Likewise, this is greatist when the voltage across the capacitor is maximum.
When you place an inductor and capacitor in parallel, the energy can transfer from one to the other, alternating back and forth. Assumming that nothing is present to consume the energy, a small signal source can build large energies in this circuit known as a "tank."
Why is this important? Well, one good use is that the tank does build up energy only around a certain frequency. Thus, it is frequency selective and can be used in a filter. This is often done with two tanks - as shown in your diagram.
However, that's not the only use. Two tanks are often used to convey power from one region to another. For example one tank can be implanted within a person to supply power to an electronic implant. The other tank can then transfer the power from a regular battery through the implanted circuit.

Without the capacitors, neither side would perform efficeintly. The power transmitting side would require that all the electrical current be derived from an electronic circuit. Instead, most of the current oscillates too and from a capacitor.

On the other side, the pick-up inductor will have a significant amount of self inductance that it does not share with the external coil. The capacitor on this side serves to cancel some of the inductive reactance from the self-inductance. Thus improving the power transfer.
hikaru1221
#6
Sep21-10, 12:46 AM
P: 799
I really appreciate your replies, but I'm absolutely new to this, so can someone please explain why we connect them in parallel but not in series? I don't quite get the answer for this from the previous posts (sorry!), though they are very informative.

All I know is that the frequency of the source (not shown on the left of the picture) matches the natural frequency of the LC branch (red L + red C or blue L + blue C), so that resonance happens, which first maximizes the power transferred from the source to the left LC branch, then maximizes the power transferred from the left LC branch to the right LC branch. But I cannot see the difference between connecting in parallel and connecting in series.
vk6kro
#7
Sep21-10, 01:25 AM
Sci Advisor
P: 4,032
You can use either, but for best selectivity, the tuned circuit has to have very little resistance in series with it (around the loop formed by the coil and capacitor) and preferably infinite resistance across (in parallel with) the coil or the capacitor.

So, it depends on how you are driving the circuit and what load you are putting on it.

In this case, it was for a valve (vacuum tube) driving a valve.

So, it was being driven by a fairly high impedance, so you wouldn't want that in series with the coil and capacitor.
And, it had a high impedance input of a valve as its load, so again you wouldn't want that in series with your coil and capacitor.

So, it worked best with two parallel tuned circuits.
hikaru1221
#8
Sep21-10, 02:33 AM
P: 799
I see. Thank you vk6kro

I have another question: How do people take into account the mutual inductance of the 2 circuits (or 2 inductors) in selectivity?

This question came when I did some simple calculation about this problem. I simplified the situation to do some estimation, and my calculation can be found in the attached file. In case 2 (series case), I can show that the current through the left circuit is equal to zero! Normally in RLC circuit, when resonance occurs, the current amplitude only depends on R (I=U/R). However in this case, due to mutual relationship, it drops to zero theoretically.
(by the way, this is the file I sent to my professor regarding the same issue, so you may find this unrelated to the question above, but still related to the topic)
Attached Files
File Type: pdf Untitle.pdf (177.6 KB, 17 views)
Blenton
#9
Sep21-10, 11:10 AM
P: 193
Im also interested with this same thing. I'm reading a book "Inductive powering: basic theory and application to biomedical systems - koenraad".

It says that high efficiency series resonant links can only be achieved for low Rload values as high values dampen the secondary tank resonance.

Rload = wLs / Q

Since high link efficiencies require high values of Q, the value of Rload must be small which seems to be why parallel resonant secondaries are used for high load values.


Whats also interesting is that it mentions that only the secondary should be at resonance, non resonant primaries are more efficient at higher values of X = k^2*Q1*Q2 of the coil quality factors. But I believe this is for well coupled coils, primary and secondary resonance are more efficient at lower couplings.

I'm still going through it, going to take a while to understand it well.
Mike_In_Plano
#10
Sep21-10, 10:25 PM
P: 560
Wow, someone wrote a book about inductive power transfer in biomed. We had to do it from scratch back when ;')

Hikaru pointed out something important. At steady state, the series resistor, R, doesn't make a difference. However, there are a few requirements for this to be true:

1. The system must be at resonance - deviation will result in losses
2. The system cannot supply power to a load
3. The assumption is only valid during steady state operation - step changes in source excitation will result in temporary losses as the tank voltage adjusts to the source voltage.

Regarding the biomedical power transfer, I used to design these for a little company starting with A, ending in S and having only one additional letter ;)
The difficulty is sending power through people is that you have very little mutual inductance with reasonable diameter coils and seperations on the order of an inch or more. You start with a great many amp turns on the primary to ensure that you get sufficeint magnetization within the secondary. Then, the secondary has significant series inductance that acts as a series impedance with the incoming signal. Thus, you end up with parallel tuning on each side.
On the primary side, the turns of the primary inductor resonate with a parallel low loss capacitor such that high currents are maintained in the winding. Without the resonance, such high currents must travel through the driving circuitry, and the loss is high.
On the secondary side, the reactive impedance of the self inductance can be "tuned out" with a corresponding amount of capacitance reactance. In reality, the capacitance is never ideal because the coil to coil coupling varies from patient to patient and run to run.
Typically, the load capacitance is chosen for a somewhat worst case. If the coils are seperated a slight bit beyond this, the power transfer drops off rapidly. If the coils are brought closer together, the capacitance value is again wrong and the capability to transfer power is less than it could have been simply because the capacitance loading.


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