Discussing primary resistance of a double-tuned ciruit

[brainbaby]

Hi friends,
My question...
At non resonance frequencies the resistance of primary is minimum or maximum?

My analysis...
There are two tuned circuit tuned to same frequency . In primary circuit at low frequency XL is low and large amount of current will flow at the inductor L1, now due to this same current is generated in the secondary circuit because both the circuit are tuned to same resonant frequency. The impedance coupled from the secondary will be of same magnitude but opposite in sign and will cancel the inductive reactance of the primary circuit as discussed per theory. Now since XL is canceled in the primary so no inductive reactance will present which technically means that there will be no driving force which will cause large current to flow in the primary circuit as earlier. This can be thought as an "occurrence of large impedance which cause little current flowing in the inductive branch of primary circuit." So at non resonant frequency the presence of a tuned circuit in the vicinity of other tuned circuit fixed at same resonant frequency will cause high primary impedance to persist which force me to assume that at non resonant frequencies the primary resistance is high..

but according to theory its wrong. The text states that at non resonant frequency the primary resistance is low which causes large current to flow causing occurrence of two humps in the output curve.

Theory

 

[tech99]

I think the book is talking about the impedance in series with the primary (low at resonance) and you are talking about the impedance in parallel with the primary (high at resonance).

 

[brainbaby]

No no that I understand your point that impedance is low at series resonance and high at parallel resonance. But I am referring non resonant cases i.e at high and low frequencies when the dominant impedance is either inductive or capacitive, also individually I am not talking about single resonant circuits rather I am more interested in exploring the effect generated by the presence of another tuned circuit in proximity of a tuned circuit which are tuned to same resonant frequencies..

 

[Baluncore]

Band-pass filters absorb the energy of the unwanted frequency components in their real resistance, or they reflect the energy from their input, back the way it came. Reflection occurs when there is an impedance mismatch.

When discussing the input impedance, what is the load impedance on the output ?
When you ask about the rejection-band resistance, do you not actually mean the input impedance ?

Are the books discussing the same circuit ?
Is it series or parallel resonance ?
I do not think there can be a useful discussion without the circuit diagrams.
We need Fig 35(..cd..) from one book, and Fig 23.32(abc..) from the other.
What are the titles, authors and editions of the books you extracted the partial texts from ?

 

[brainbaby]

Let me elaborate more …

As both the tuned circuit is tuned to same resonant frequency.
At low frequency the secondary inductive reactance XL’ is coupled into the primary as capacitive reactance Xc. This Xc cancels the inductive reactance of the primary XL which is already present there.
This cancelation of impedances brings point A and B to the same potential which prevents the current flowing in the branch 1. No current will flow through C1 as at low frequencies Xc1 is very high. So technically speaking no current is flowing in the primary which means that at non resonant frequencies the primary resistance is high…

What are the titles, authors and editions of the books you extracted the partial texts from ?

The above is my generalization from the text referred from book Radio engineering by Fedrick Terman ..chapter 3 Properties of Resonant circuits..
and Electric circuits by David A bell.

We need Fig 35(..cd..) from one book, and Fig 23.32(abc..) from the other.

Fig 35 c,d

Fig 23.32

 

[Baluncore]

So technically speaking no current is flowing in the primary which means that at non resonant frequencies the primary resistance is high…

It seems that you might mean the input impedance is high. The annotated circuit diagram you give at the top of post #5 has no resistance components and no output impedance specification. That means it has an infinite Q, with a huge circulating energy at resonance that cannot be lost. You continue to confuse resistance R, with impedance Z = R + jX.

The circuit will have low impedance at low frequency since parallel Lp is X_L = 2 π f L.
The circuit will have high impedance at high frequency since Lp Cp is resonant and lossless.
The circuit will have low impedance at high frequency since parallel Cp is X_C = 1 / ( 2 π f C ).

You claim that the two resonant circuits are tuned to the same frequency, which is not the case with the variable M coupled stagger tuned circuits being discussed in the different texts.

 

[brainbaby]

It feels I need to do some more research on this topic...will be back with more inputs

 

[brainbaby]

The circuit will have low impedance at low frequency since parallel Lp is XL = 2 π f L.
The circuit will have high impedance at high frequency since Lp Cp is resonant and lossless.
The circuit will have low impedance at high frequency since parallel Cp is XC = 1 / ( 2 π f C ).

Hi,
The mistake which I made was that i considered the circuit to be a parallel resonant circuit whereas the text is about series resonant circuit...as tech99 said it previously..

At series resonance the reactive impedance cancels each other hence resulting in maximum current, while in parralel circuit the current flowing in two branches the circuit cancels each other, so technically saying there is no current flowing in the circuit and can be thought as a high impedance occurring in the circuit.

Now at frequencies other than resonance a reactive coupled impedance is reflected to the primary and cancels the reactive impedance of the primary which results in minimum impedance in primary causing large current to flow.
While at resonance the impedance that is coupled to primary is purely resistive in nature. So at resonance the primary resistance is maximum and there is a dip in the double hump curve.
But now my question is that at non resonant frequencies why does resistance is not coupled or why do they don't consider it to be coupled by the secondary into the primary since resistance is present everytime as it is the property of the material .??
or
It can be said like...why at non resonant frequencies resistance don't get coupled to the primary circuit as at resonant frequencies only pure resistance get coupled from the secondary and increases the primary resistance...

Text..

 

[brainbaby]

I think the book is talking about the impedance in series with the primary (low at resonance) and you are talking about the impedance in parallel with the primary (high at resonance).

thanks buddy

 

[Baluncore]

Hi,
The mistake which I made was that i considered the circuit to be a parallel resonant circuit whereas the text is about series resonant circuit...

Figs 23.31 & 23-33 in post #5 show an inductively coupled parallel resonant circuit. Maybe the in-circuit voltage source is confusing you. It represents a voltage coupled into the inductor, without that source there could be no simulation.

At series resonance the reactive impedance cancels each other hence resulting in maximum current, while in parralel circuit the current flowing in two branches the circuit cancels each other, so technically saying there is no current flowing in the circuit and can be thought as a high impedance occurring in the circuit.

Parallel resonance occurs when the total reactance is zero. At that frequency the circulating current is high, the parallel voltage is high, but there is little drive needed to maintain oscillation. So the high LC parallel voltage divided by the small input current, makes the input look like a high impedance. You must differentiate between LC circulating energy and the input or output energy.

Where two parallel resonant circuits are coupled, they are tuned to different frequencies. In between those frequencies they have reactances changing in an opposite sense that cancel each other giving a wider pass-band. Above or below the pass-band, the reactances sum to offer a greater mismatch and so reflect a greater proportion of out-of-band energy.

But now my question is that at non resonant frequencies why does resistance is not coupled or why do they don't consider it to be coupled by the secondary into the primary since resistance is present everytime as it is the property of the material .??

The input impedance of the stagger tuned circuit is a factor of the output impedance. The circuit can be thought of as an impedance transformer. But the input and output impedance are different from the low values of series resistance that reduce the Q of resonant circuits. One resistance carries the resonant circulating current, the other is the ratio of voltage to current at the ports of the 4 terminal block. Every network has an input and an output impedance. Each impedance is the other, but transformed by the network.

 

[brainbaby]

It seems that you might mean the input impedance is high

.why at non resonant frequencies resistance don't get coupled to the primary circuit as at resonant frequencies only pure resistance get coupled from the secondary and increases the primary resistance...

First I am having no clue of how the input impedance loads the tuned circuit acc to your quote since the text dosen't mention it ...
you may be right ...
rather I am much more interested in neglecting the loading effect caused by the input impedance as it may make the analysis much more complex...

Second thing is that I believe that you didn't got my point...as I said that..
let us assume that resistance is present in a circuit...now an ac current flows through it...so as by the nature of resistance it will oppose the flow of current ...the same will happen when there are other components present in the circuit as well like inductor or capacitor..the current still flows in the circuit but now L and C will show some of their effect in the circuit..causing the output to lag or lead the input which depends upon the frequency however the resistance is still present which will show its effect while the output leads or lag the input..but this resistance got completely neglected at non resonance frequencies where only reactive impedance gets coupled...unlike at resonant frequency only resistance get coupled to the primary and over all increase the resistance of the circuit...

Please clarify?

 

[Baluncore]

let us assume that resistance is present in a circuit...now an ac current flows through it...so as by the nature of resistance it will oppose the flow of current

An inductor opposes changes in current, a capacitor opposes changes in voltage. The voltage drop across a resistor, divided by the current through that resistor, is a ratio called resistance, R = V / I.
A resistor dissipates energy, W = V ∙ I = I² ∙ R = V² / R.

At resonance, energy E = ½ ∙ L ∙ I² = ½ ∙ C ∙ V² is circulating between L and C. Any resistance present has many, = Q, opportunities to dissipate that energy. When at resonance, XL+XC = 0, and so R will dominate the circuit impedance. That is why you need to consider R when parallel LC is resonant, but can usually ignore R when LC is far from resonance. At low frequencies XL will dominate while at high frequencies XC will dominate.

Every network has an input and an output impedance. Ideally, they will be matched by the network. It is a reciprocal process for a network, you can work out the optimum input impedance given the output impedance, or the optimum output impedance given the input impedance. The network transforms the port impedances.

Do you understand how to use vectors or complex numbers to calculate impedance?

 

[Merlin3189]

Just to help show Baluncores point perhaps, I made this table of impedances for an LCR circuit around its resonance at about 1MHz.
You can see that at resonance, ZT = R, but as you move away from resonance, it is swamped by size of ZC or ZL and R is irrelevant to the net impedance. Only if R were very large - a highly damped circuit - would it have much effect on the total impedance.

 

[brainbaby]

That is why you need to consider R when parallel LC is resonant, but can usually ignore R when LC is far from resonance

but as you move away from resonance, it is swamped by size of ZC or ZL and R is irrelevant to the net impedance.

At frequency 0.955376 MHz the net impedance(reactive) is 167.97 and suppose we have taken the value of resistance to be 100 ohm for study..now according to claim if we moves away from resonance then the value of resistance should have no effect as it is subsided for non resonant frequencies ..but according to the data you have presented…the net total impedance is 167.97 and the value of resistance is 100 ohm which is though quite comparable to the net impedance as it is 59.53% of ZT…..then how can you say to neglect the resistance..

 

[Baluncore]

@brainbaby

Do you understand how to use vectors or complex numbers to calculate impedance?

https://en.wikipedia.org/wiki/Phasor

 

[brainbaby]

Do you understand how to use vectors or complex numbers to calculate impedance?

I don't have much knowledge in vectors ..but I ve once done it at university...all those polar/rectangular forms ..their conversion...etc...I don't remember it well...
Rather I am looking to understand this phenomenon in general sense with minimum involvement of mathematics..

Like vector will involve imaginary number into existence which lead and lag of output caused by inductance and capacitance present in the resonant circuit..so technically its the same thing...

 

[Baluncore]

Rather I am looking to understand this phenomenon in general sense with minimum involvement of mathematics..

I don't think it is possible to understand LCR circuits without realising that reactance is perpendicular to resistance, and that inductive reactance is positive and capacitive reactance is negative. At resonance they cancel, the total reactance is zero, but the resistance is still R. The further from resonance the greater the total reactance and the less important is the fixed resistance.

Once you put two stagger tuned LCR circuits together to get a band pass filter you will need to use complex numbers. Complex numbers are really simple, they are a 'complex' of two ordinary numbers that travel together. There is nothing imaginary about perpendicular x and y axes. The j in (x + jy) is only there to keep the x and the y 90° apart, it stops them being simply added.

You can have more fun with simple complex numbers than you can with tuned circuits, put them together and you can have a ball.

 

[brainbaby]

The further from resonance the greater the total reactance and the less important is the fixed resistance.

What ever you said I agree with it completely ...ya the further we go from resonance the more resistance becomes "less significance"...but what should be the importance of this less significance?? ...i mean that resistance is not frequency dependent and hence it exist the same value for all frequencies..If resistance is 100 ohm...it will be there for resonant frequency as well as for non resonant frequency...At resonance since reactive impedance is zero so only resistance is counted ..when at non resonance reactive impedance will be there in the circuit ...but still there would be 100 ohm resistance in the circuit which will be of some value in comparison to reactive impedance just as I quoted in my previous post discussion example quoted by Merlin...

There is nothing imaginary about perpendicular x and y axes. The j in (x + jy) is only there to keep the x and the y 90° apart, it stops them being simply added.

And regarding imaginary complex numbers I have a good explanation regarding this...
have a look..

so I am just trying to understand the point this way...

 

[Baluncore]

And regarding imaginary complex numbers I have a good explanation regarding this...

It is quite unnecessary to invoke the confusion of the square root of a negative number. That is a mathematical fascination that gets in the way of vectors and phase.

Replace the word 'Real' with the word 'Resistive', R.
Replace the word 'Imaginary' with the word 'Reactive', X.
R and X are at right angles to each other, so they sum to form a vector:
The sum of R and X is Impedance, Z = R + jX.

Multiplying by j just means turn 90° left. If you are a unit vector heading east along the +x axis and turn left, you are then going north, = +y. Turn left again and you are going west. That is why multiplying by j twice reverses the x direction, x * j * j = x * j² = –x, (which is why j² = –1). Multiply by j again and you turn left again, so you are then going south, (x * j³ = –y ). Finally multiply by j again and you are going east, which is the way you started, x * j⁴ = +x.

That might explain why energy is said to 'circulate' in an LC circuit.

 

[brainbaby]

R is irrelevant to the net impedance

@Baluncore

I appreciate your way to link physical understanding of a phenomenon with mathematical tools, its a great approach for boosting your understanding over a topic...but currently I feel my self to be very close to understand a point...and that is to neglect R at non resonance frequencies ... but the confusion is that R is a static quantity at all frequencies and suppose at a particular non resonant frequencies reactive impedance is dominant then since R has some value, it is still making some contribution to the REAL PART of the total impedance thereby effecting the value of the impedance...so is it right to neglect R completely...?

 

[Baluncore]

...and that is to neglect R at non resonance frequencies ... but the confusion is that R is a static quantity at all frequencies and suppose at a particular non resonant frequencies reactive impedance is dominant then since R has some value, it is still making some contribution to the REAL PART of the total impedance thereby effecting the value of the impedance...so is it right to neglect R completely...?

It is never right to neglect anything, especially the resistance which determines Q, and so the optimum frequency separation of the stagger tuned elements.

That is why you need to consider R when parallel LC is resonant, but can usually ignore R when LC is far from resonance.

It comes down to language. Neglect = fail to consider in any way. Ignore = disregard intentionally. Far = well outside the combined pass-band, where the phase shift will be close to 90° and not sensitive to the resistance.

It must be made clear that this analysis is for stagger tuned LCR circuits. They must be lightly coupled so they do not resonate as one, at one frequency. The impedance changes with frequency of the two resonators tend to cancel between the two resonant frequencies. Outside that range the impedance changes tend to sum.

 

[The Electrician]

Hi friends,
My question...
At non resonance frequencies the resistance of primary is minimum or maximum?

My analysis...
There are two tuned circuit tuned to same frequency . In primary circuit at low frequency XL is low and large amount of current will flow at the inductor L1, now due to this same current is generated in the secondary circuit because both the circuit are tuned to same resonant frequency. The impedance coupled from the secondary will be of same magnitude but opposite in sign and will cancel the inductive reactance of the primary circuit as discussed per theory. Now since XL is canceled in the primary so no inductive reactance will present which technically means that there will be no driving force which will cause large current to flow in the primary circuit as earlier. This can be thought as an "occurrence of large impedance which cause little current flowing in the inductive branch of primary circuit." So at non resonant frequency the presence of a tuned circuit in the vicinity of other tuned circuit fixed at same resonant frequency will cause high primary impedance to persist which force me to assume that at non resonant frequencies the primary resistance is high..

but according to theory its wrong. The text states that at non resonant frequency the primary resistance is low which causes large current to flow causing occurrence of two humps in the output curve.

Trying to figure out what is happening in coupled circuits by just thinking about it with words is not a fruitful way to proceed. Using mathematics is the way to go. Steinmetz showed that AC circuits can be easily analyzed with complex arithmetic rather than the old way of solving differential equations.

Consider the first image shown in post #5. Let the left side be driven with a 50 ohm source impedance. Choose some component values:

C1 = 300 pF
C2 = 300 pF
L1 = 100 uH (L1 is the inductor connected to C1)
L2 = 100 uH

Those values will give a resonance frequency for the two tanks of about 919 kHz, fairly near 1 MHz.

The two parallel resonant tanks are coupled; the amount of coupling is described by the mutual inductance (M) between the two. A mutual inductance of 100 uH would be 100% coupling. Let's start with fairly tight coupling, but not 100%--choose M = 90 uH.

R2 = 900 ohms (R2 is the load on the right tank circuit)

Now, given those values we can calculate the input impedance magnitude, input resistance (real part of the input impedance) and input reactance (imaginary part of the input impedance)

Here's a plot of the three quantities vs. frequency for those starting values. The impedance magnitude is in black, real part in blue and imaginary part in red:

We don't see any resonance at 1 MHz--the tight coupling has resulted in two resonances fairly separated from 1 MHz.

Now let's reduce the mutual inductance from 90 uH to 45 uH which decreases the coupling. Here's the plot resulting from that change:

The previously widely separated resonance frequencies have come much closer to 1 MHz.

Now, let's change C2 to 265 pF. Here's the resulting input parameter plot:

Now we see the desired double humped input impedance magnitude although the input resistance (real part of the impedance) is a little skewed. The input impedance is near 1500 ohms, and is fairly constant over the passband. Away from the passband the impedance becomes lower than 1500 ohms. The real part of the input impedance drops much more rapidly than does the impedance magnitude; this shows that the input impedance is mostly resistive over the passband, but becomes reactive away from the passband.

These results would be very difficult to determine by just thinking about it--one needs to use mathematics.

 

[brainbaby]

It comes down to language. Neglect = fail to consider in any way. Ignore = disregard intentionally. Far = well outside the combined pass-band, where the phase shift will be close to 90° and not sensitive to the resistance.

Thanks for correcting my speech...

Away from the passband the impedance becomes lower than 1500 ohms. The real part of the input impedance drops much more rapidly than does the impedance magnitude; this shows that the input impedance is mostly resistive over the passband, but becomes reactive away from the passband.

Yeah I was looking for this type of plot since a long time...thanks you provided it for me...but I want to draw your attention here...as for frequency x1 in the plot the value of the real part of the impedance is a1 i.e near 500 ohms and the value of imaginary part of impedance(reactive) is a4 which is approx near to 1000 ohms...(both a1 and a4 are significant values) similar is with x2 and x3...and now I have to be cautious enough about my language as quoted by Baluncore.. here is that how should I assume R to be really far away from the "passband region" since a1 and a4 are quite noteworthy...

So can I conclude like this that the "resistance do show its effect forward from point Y in synchronization with the reactive impedance"

Moving towards left of point Y, I fully agree to what you have said since the value of R is quite trival ...I don't have any problem in it...

Further could you please provide me the output voltage curve for the passband region, reactive and real voltage in sync ?

 

[The Electrician]

Sorry I couldn't get back to you sooner.

The plots in post #22 are of the input impedance of the two parallel tank circuit. The only components in the circuit are L1, C1, L2, C2 and R2. R2 is the 900 ohm load on the second tank. The impedance is what is seen looking at the top node of the left tank. There is no 50 ohm resistor in series with this input for the plots in question.

Thus the only losses in the circuit are R2 and the variation in the real part of the input impedance is an example of the impedance transforming property of coupled resonators.

To show the output voltage Vo on the same frequency range, I used a 1500 ohm resistor in series with an ideal voltage source to drive the input since the calculated input impedance magnitude was about 1500 ohms. This plot shows the three parameters: magnitude of Vo (black), real part of Vo (blue) and imaginary part of Vo (red). The vertical scale is linear rather than log because of showing the real and imaginary parts which are negative at certain frequencies.

Normally the phase of Vo would be shown rather than the real and imaginary parts, but you asked for the two parts:

This isn't exactly a high performance bandpass filter.

 

[brainbaby]

@The Electrician
Our main discussion was the analyse the role of resistance R near resonance and far from resonance...

Far from resonance the reactive impedance is dominant over R and effects the output...

I have considered a frequency x1 which is quite far from resonance, and according to the curve, R has a value of a1 and the reactive impedance value is a4...
and if we compare a1 with a4, a1 is near about 50% of a4...so according to me R still have some "significant " i repeat significant effect over the response ..but majoity of textbooks which I have read so far states that at non resonance frequencies the imaginary part(reactive impedance) is dominant over R.. i.e the output voltage would be dominant by reactive impeance...
So I am just in an effort to synchonize my observation with the texts...

 

[The Electrician]

I suppose that how one interprets the statements from the textbooks depends somewhat on how one understands words like "significant" and "dominant".

Be aware that the way the real part and imaginary parts of impedance combine to give the magnitude is not just arithmetically, but rather "vectorially".

For example, at frequency X1, the real part in post #23 is about 400 and the imaginary part is about 860. These give an impedance magnitude of sqrt((400)^2+(860^2)) = 948. The impedance magnitude is only about 10% greater than the imaginary part, not 50% greater as it would be if the two parts combined arithmetically. Because of this way the real and imaginary parts combine to give the impedance magnitude, it is reasonable to say that the imaginary part "dominates".

 

[brainbaby]

It seems I nearly got it..since the imaginary impedance is just a virtual concept which do not contribute real voltage drop. It just contribute to the direction of the voltages which you can define it by vectors or time delay.

The mistake which I probably made was that I confused myself with the idea that at frequency x1 the real voltage drop is signified by the resistance as well as imaginary impedance. However at x1 the voltage drop i.e 860 is imaginary..i.e it do not contribute to the real voltage drop as it only signifies the direction (phase relationship) of the output voltage to the input voltage..

The example screen shot clearly shows that at resonance , j has some value which gets canceled out due to equal an opposite value j provided by other component.

However at non resonance both the j have some arbitrary values which gets subtracted and yields some net value, and that net value is either inductive or capacitive only signifying how much degrees input voltage leads or lags the output voltage..

Therefore near x2 we are just concerned on phase shifts of the voltages rather than real voltage drop and hence we may call imaginary part to be dominant on the real part as baluncore in his post 17 said

”The further from resonance the greater the total reactance and the less important is the fixed resistance.”

 

[Baluncore]

The example screen shot clearly shows that at resonance , j has some value which gets canceled out due to equal an opposite value j provided by other component.
However at non resonance both the j have some arbitrary values which gets subtracted and yields some net value, and that net value is either inductive or capacitive only signifying how much degrees input voltage leads or lags the output voltage..

The operator j means “perpendicular to”, you should instead be using the symbol X and the term “reactance”. So long as you confuse the terminology you will confuse yourself and others.

The mathematical term 'imaginary' should be avoided when discussing currents. The term 'reactive' is more meaningful when used in conjunction with resistance.

When in series, do inductive and capacitive reactances add, subtract or cancel?
X_C is negative, X_L is positive, so they simply sum, to zero at resonance.
X = X_L + X_C.

In series, impedances add. In parallel, admittances add.
Impedance, Z = ( R +j X ); where R is resistance and X is reactance.
Admittance, Y = ( G +j B ); where G is conductance and B is susceptance.

Admittance and impedance are the reciprocal of each other.
Y = 1 / Z; To numerically convert or transform between Z and Y use;
Impedance to Admittance; G = +R / (R²+X²); B = –X / (R²+X²).
Admittance to Impedance; R = +G / (G²+B²); X = –B / (G²+B²).

Congratulations, you have come all the way back from a stagger tuned pair of coupled parallel LC circuits with resistance, to a single series LC circuit without resistance.

 

[brainbaby]

I got it...but just need to confirm that at point x1, x2, x3 the phase relationship between input and output voltage will be something out of phase ..i.e shown in example below..

 

[The Electrician]

Here is a plot showing the phase of Vout/Vin. The phase is shown in degrees, and the magnitude has been multiplied by 500 so it will fit on the same plot: