Discrepancy in calculating Q Factor using Energy instead of R/wL

[phantomvommand]

I have 2 questions, the first has been asked (unfortunately not directly answered) below:
https://electronics.stackexchange.c...or-of-parallel-rlc-with-series-resistive-load
I have reposted the question below for ease of reference.

For the given circuit, calculating Q factor via the energy method seems to give a wrong answer. What was wrong with the energy method?

My second question involves the converted Norton Equivalent with R || L || C (see picture above).
1. How is Q = R/wL derived for this Norton equivalent?
2. In the norton equivalent, is there current flowing through R? I ask this because converting the norton back into its equivalent thevenin will result in R being in series with a tank circuit, which is a set up with infinite impedance. Therefore, there should be 0 current flowing through R. Contradictorily, applying the current divider rule to the Norton equivalent suggests that current through R is equal to the norton source current.
3. If there is 0 current through R, is Q even defined for a norton R||L||C circuit?

 

[Baluncore]

If the current source is AC, at the frequency of the LC resonance.
Energy will circulate between L and C, with an AC voltage appearing across the circuit.
That AC voltage will cause an AC current to flow through the resistor.
The energy delivered by the current source will be dissipated in the resistor.
The Q will be related to R / XL, since the proportion of the current that flows through the resistor is in that ratio.

Draw the resonant LC together so it is obviously resonant.
Draw the parallel resistor outside the LC tank circuit, to show it is external to the tank.

 

[phantomvommand]

If the current source is AC, at the frequency of the LC resonance.
Energy will circulate between L and C, with an AC voltage appearing across the circuit.
That AC voltage will cause an AC current to flow through the resistor.
The energy delivered by the current source will be dissipated in the resistor.
The Q will be related to R / XL, since the proportion of the current that flows through the resistor is in that ratio.

Draw the resonant LC together so it is obviously resonant.
Draw the parallel resistor outside the LC tank circuit, to show it is external to the tank.

Thanks, this makes sense.

Regarding my 2nd qn: converting the norton back into its equivalent thevenin will result in R being in series with a tank circuit, which is a set up with infinite impedance. Therefore, there should be 0 current flowing through R. Contradictorily, applying the current divider rule to the Norton equivalent suggests that current through R is equal to the norton source current.

I now realise that there is no contradiction here as this is exactly what the Thevenin <-> Norton conversion should give.

Would you be able to answer the first part on energy? Thank you.

 

[Baluncore]

Regarding my 2nd qn: converting the norton back into its equivalent thevenin will result in R being in series with a tank circuit, which is a set up with infinite impedance. Therefore, there should be 0 current flowing through R.

When energy circulates in the tank, current flows through any series resistance. The tuned circuit only appears to have high impedance, at the resonant frequency, when there are no losses due to the series resistance of the inductance, or leakage resistance through the dielectric of the capacitor.

Would you be able to answer the first part on energy? Thank you.

Only when V1 = 0, does the loss resistance = (R || Rload). That may also be considered the case when the tank is not resonant with V1, or V1 is a broadband signal.

If the circuit is resonant and stable, then V1 ≅ V2, and the value of dV/Rload determines the tank drive current.
You need to avoid extreme assumptions.

 

[phantomvommand]

Only when V1 = 0, does the loss resistance = (R || Rload). That may also be considered the case when the tank is not resonant with V1, or V1 is a broadband signal.

If the circuit is resonant and stable, then V1 ≅ V2, and the value of dV/Rload determines the tank drive current.
You need to avoid extreme assumptions.

Sorry I did not understand this. Exactly which line in the energy approach to derving Q is wrong? Thanks

 

[Baluncore]

Exactly which line in the energy approach to derving Q is wrong?

The assumptions.
Assume V1 = 0.
Combine resistors before the derivation; Rp = (Rload || R).
Assume a 1 volt sinusoidal tank signal.

 

[tech99]

View attachment 354092
My second question involves the converted Norton Equivalent with R || L || C (see picture above).
1. How is Q = R/wL derived for this Norton equivalent?
2. In the norton equivalent, is there current flowing through R? I ask this because converting the norton back into its equivalent thevenin will result in R being in series with a tank circuit, which is a set up with infinite impedance. Therefore, there should be 0 current flowing through R. Contradictorily, applying the current divider rule to the Norton equivalent suggests that current through R is equal to the norton source current.
3. If there is 0 current through R, is Q even defined for a norton R||L||C circuit?

This diagram shows a parallel circuit, so I suggest you assume a generator voltage of 1 volt. Then the current in each of the three arms will be V/Z, so current flows in all three arms. You can see this with a simple phasor diagram. Current in R is in phase with the generator, current in C leads by 90 degs and current in L lags by 90 degs.

 

[Tom.G]

A short cut would be to realize that a voltage source is generally assumed to have zero or negligible internal impedance.

This leads to your series resistance being a parallel resistance.

Cheers,
Tom

 

[phantomvommand]

The assumptions.
Assume V1 = 0.
Combine resistors before the derivation; Rp = (Rload || R).
Assume a 1 volt sinusoidal tank signal.

This would explain why R load can be (approximately) combined in parallel with R, but does not explain why the energy derivation is wrong?

 

[phantomvommand]

View attachment 354120

A short cut would be to realize that a voltage source is generally assumed to have zero or negligible internal impedance.

This leads to your series resistance being a parallel resistance.

Cheers,
Tom

Yes, that will explain why R_load can be approximately combined in parallel with R, but does not explain why the energy derivation is incorrect.

 

[tech99]

View attachment 354093
For the given circuit, calculating Q factor via the energy method seems to give a wrong answer. What was wrong with the energy method?

When you calculate the Energy Lost Per Cycle, using V^2/R, have you placed the two resistors in series instead of parallel?

 

[phantomvommand]

When you calculate the Energy Lost Per Cycle, using V^2/R, have you placed the two resistors in series instead of parallel?

Yes (see denominator)

 

[phantomvommand]

View attachment 354120

A short cut would be to realize that a voltage source is generally assumed to have zero or negligible internal impedance.

This leads to your series resistance being a parallel resistance.

Cheers,
Tom

While I can see why this is true, this does not explain why the energy approach is incorrect. Is there any direct reason for why the energy approach is wrong?

 

[tech99]

When you calculate Q from energy relations, you are using the parallel model. So I think you need to use v2 in calculating both the stored energy and the energy lost per cycle.

 

[phantomvommand]

When you calculate Q from energy relations, you are using the parallel model. So I think you need to use v2 in calculating both the stored energy and the energy lost per cycle.

This contradicts the textbook method of combining R_load with R, since you would effectively have ignored R_load.

BTW, I can understand the textbook method of combining R_load with R -- they are simply taking the Norton equivalent and applying Q = 2pi * energy stored / energy loss per cycle. However, there is clearly a discrepancy, since applying Q = 2pi * energy stored / energy loss per cycle, on the circuit and its Norton equivalent, appears to give different Q factors.