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measuring the resonance of a circuit

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dr.power

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Hi there,

If I have got a black box with a lot of RLC components inside (all RLC components are in series and parallel together), Then how do I measure the resonance frequency and of course impedance of 2 output leads of it by an scope?
I myself have a function generator and 2 multimeters.

Thanks.
 
Depending on the circuit configuration there may be one or more series resonances and/or one or more parallel resonances. Apply a frequency sweep to the input and look for any dips or peaks in the magnitude of the output.

Alec
 
Depending on the circuit configuration there may be one or more series resonances and/or one or more parallel resonances. Apply a frequency sweep to the input and look for any dips or peaks in the magnitude of the output.

Alec

Thanks sir, But actually I think there would be a NET resonance for the whole circuit, right?
Where is input?!
 
Thanks sir, But actually I think there would be a NET resonance for the whole circuit, right?
Where is input?!
There is not necessarily one "NET" resonance. Depending upon the circuit, it could have more than one resonant point.

If you circuit has only one pair of terminals then you use that to determine the resonance(s). Just apply a frequency sweep from a generator with a series resistor and measure the circuit voltage for dips and peaks as suggested. You many have to try different values of series resistance to get a good indication of the resonance(s).
 
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Assuming the capacitance input to scope is much less then capacitance of RLC, feed an AC generator through about a 33k resistor to top of LRC stack and look for output peak on scope.
 
Thanks guys for all your inputs, you are really helpful.

Well I thought there is a net equivalent containing a circuit having a net equivalent frequency (actually I thought a complicated RLC circuit will be simplified as a series or parallel RLC circuit having just one resonance frequency). I will try looking around to find an almost simple circuit having more than just a an RLC and put it here to investigate it.
 
Assuming the capacitance input to scope is much less then capacitance of RLC, feed an AC generator through about a 33k resistor to top of LRC stack and look for output peak on scope.

Thanks, thats right but I am not sure about the resistor effect on the RLC circuit specially in the impedance or phase of it?
 
Ok here is what I have found, It is very simple but I am not sure if it is a good circuit having more than one resonance freq? If so What are they?
 

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The can be a series resonant frequency between Ls and Cs, and a parallel resonant frequency between Ls and Cd.
 
Thanks, thats right but I am not sure about the resistor effect on the RLC circuit specially in the impedance or phase of it?

It is not likely that you will have an inductor with a Q that will give you an Rp greater then about 20k ohms. You can go higher or use a series cap that is < 1/100 of the parallel resonating cap. If you want to measure tank circuit Q then the loading must be 3 to 7 times the effective Rp of the tank (mostly inductor Q) so you don't degrade Q measurement too much. (measure -3 db bandwidth points to get Q). Usually the capacitor can be considered very high Q so circuit Q is dominated by coil Q.

If it is a 'black box' and you don't know L or C values then you can figure their values by doing a delta F measurement when a known smaller parallel external cap it added to circuit. This is what is done when measuring RF crystal parameters to derive Co, Cm, Lm, and Rs of crystal. You need a phase meter across external series reference resistor to nail the exact Fs frequency.
 
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The can be a series resonant frequency between Ls and Cs, and a parallel resonant frequency between Ls and Cd.

Thanks, The above pic shows the structure of an ultrasonic piezo sensor, Saying the resonance frequency of the ultrasonic sensor is A, Then there should be oJust one NET resonance frequency for a such model (for instance a 40KHz ultrasonic sensor has a peak resonance at 40KHz which I call that frequency the net frequency of the sensor). right?
 
Another question which arises is. Why does an ultrasonic piezo sensor says to be an capaciotor or behaves as a capaciotor? The model shows that It is not just a capacitor, unless we say that in resonance frequency the Ls and Cs cancel each others and then we just have Cd and Rs, right?
 
You can look for resonances by applying sine voltages of various frequencies to your "black box", which is apparently a piezoelectric sensor, and looking for peaks or dips in the current drawn from the signal source.

It is much more convenient to use a special instrument intended to measure impedance at various frequencies. A good quality instrument of this sort can plot the impedance versus frequency and display it in graphical form.

Just to show you what can be done, I've attached 3 images. The first shows the impedance of a parallel connection of a 470 µH inductor and a .033 µF capacitor, with the frequency varied from 10 kHz to 500 kHz. You can see that resonance occurs at about 40 kHz and the maximum impedance is about 7 kΩ.

The second image shows the same capacitor and inductor connected in series, and the impedance for the series combination plotted over the same frequency and impedance range. The minimum impedance is almost as low as 1 Ω, which is the equivalent series resistance of the inductor at 40 kHz.

The third image shows the same series connection of a 470 µH inductor and .033 µF capacitor, with an additional capacitor of .5 µH connected in parallel with the series combination. This is the circuit you show in post #8, with Cd = .5 µF, Cs = .033 µF, Ls = 470 µH and Rs = 1Ω. This is the sort of impedance curve you would expect to see from a typical resonator as used in modern electronics, be it a quartz crystal or a piezo sensor, etc. You can see that there is both a parallel resonance and a series resonance, very close together in frequency.

In the next post I will show the actual impedance of a piezo sensor.
 

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Another question which arises is. Why does an ultrasonic piezo sensor says to be an capaciotor or behaves as a capaciotor? The model shows that It is not just a capacitor, unless we say that in resonance frequency the Ls and Cs cancel each others and then we just have Cd and Rs, right?

A piezo sensor behaves as a fairly large capacitor with several resonant circuits more or less well coupled to the terminals.

Look at the circuit you attached to post #8; the equivalent value of Cd is large enough that its impedance is the dominant characteristic at all frequencies. This is because of the way the sensor is constructed. It is probably a disc (or some similar geometric shape) of piezoelectric ceramic material with conductive electrodes deposited on each side. That's exactly how ceramic disc capacitors are constructed. The difference is that the material in the piezo sensor is permanently polarized (see: https://en.wikipedia.org/wiki/Electret), whereas the material in a ceramic disc capacitor is not permanently polarized. The piezo sensor is built just like a capacitor, so we shouldn't be surprised if its impedance looks mainly capacitive.

Now, the piezo sensor can vibrate mechanically and those vibrations can cause (or result from) a voltage at the terminals. But, since a disc (or other shape) can vibrate in many different ways (see: https://en.wikipedia.org/wiki/Normal_mode), we would expect to see many resonances in the impedance curve of the piezo transducer. The overall impedance of the piezo sensor trends downward as the frequency increases, which is characteristic of a capacitor.

The attached image shows the impedance vs. frequency of a Matsushita 40 kHz piezo sensor.

Typically, one series resonance/parallel resonance pair is dominant, and is the one we want to use. The other resonances are called "spurious" resonances.

When you ask about the "net" resonance, perhaps you are referring to the lowest frequency normal mode, the dominant resonance in a disc.

If the sensor (resonator) is a high Q device, the series and parallel resonances will be very close together, and detecting them separately will require the use of a signal generator that can be set to a particular frequency with a resolution of a few Hz, and doesn't drift when you take your hand off the tuning knob.
 

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