Understanding LC Tank Q, Impedance, and Losses

David Wagner 2008/04/20 20:05

A resonant inductor-capacitor (LC) tank is the beating heart of a crystal radio, filtering the signal of interest from the sea of radiofrequency (RF) energy received by an antenna. Properly loaded, a high-quality (high-Q) tank will give a crystal radio greater sensitivity and selectivity.

For a parallel tank, a smaller inductor and larger capacitor can result in higher Q.

Tank Impedance

When tuned to the station of interest, the crystal radio's LC tuning tank between the antenna and detector presents a very high resistive impedance to ground. Estimating the magnitude of this effective shunt resistance is not difficult, but it is important to keep in mind how the tank impedance is in parallel with both the source and the detector/load impedance.

If the source and load impedances are low in comparison to the tank's shunt resistance, this resistance may be neglected, the tank loss will be fairly low, and the antenna can be matched directly to the serial combination of the detector and the audio load. However, passing a low impedance signal significantly loads the tank and results in broader tuning. To minimize tank loading, the signal impedance will approach the tank's shunt resistance, and this resistance must be considered as part of the driven load. Estimating the parallel LC tank's effective shunt resistance can be done by measuring or assuming the tank's unloaded Q.

  • R_tank = 2 pi f L_tank Q_unloaded

Tank Loading

For efficient energy transfer, the antenna impedance should be matched to the parallel combination of the load and shunt impedances. In effect, the tank must be kept fully 'powered up' to maintain its narrow frequency response while reflecting the rest of the signal into the load.

  • R_src = 1/{ {1/{R_load}} + {1/{R_tank}} }{1/{R_load}} + {1/{R_tank}} = 1/{R_src}

The LC tank shunt resistance can then be used to estimate upper limits on the tank's loaded Q and the matched source and load impedances.

  • R_P = 1/{ {1/{R_src}} + {1/{R_load}} + {1/{R_tank}} } = 1/{ {1/{R_src}} + {1/{R_src} } } = {1/2}R_src
    • Q_loaded =  {R_P}/{2 pi f L_tank} =  {R_src}/{4 pi f L_tank}
    • R_src = 2 R_P

This is for very small signals.

  • R_load = R_det + R_audio = 2 R_det

This may not be correct since R_load(RF) may be only about half of R_load(audio) and R_load(DC).

Assume all loss is through the LC tank shunt resistance (R_tank); losses due to circulating current resistance are small in comparison to primarily capacitive losses to ground.

  • R_src = 1/{ {1/{2 R_det}} + {1/{R_tank}} }
  • R_P = {1/2}R_src = 1/{ {1/{R_det}} + {2/{R_tank}} }
  • Q_loaded = 1/{ ({{2 pi f L_tank}/{R_det}} + {2/{Q_unloaded}}) } = { {Q_unloaded R_det}/{2 pi f L_tank Q_unloaded + 2 R_det} }2)
  • Q_loaded = 1/{ ({{4 pi f L_tank}/{R_load}} + {2/{Q_unloaded}}) } = { {Q_unloaded R_load}/{4 pi f L_tank Q_unloaded + 2 R_load} }

The tank loss can now be estimated.

  • Efficiency (%): {R_tank}/{R_tank + 2 R_det} = {pi f L_tank Q_unloaded}/{pi f L_tank Q_unloaded + R_det}= {1}/{ {{R_det}/{pi f L_tank Q_unloaded}} +1}
  • Tank Loss (dB): 10 log{ {R_tank}/{R_tank + 2 R_det} } dB
  • Tank Loss (%): {R_det}/{R_tank + 2 R_det} = {R_det}/{2 pi f L_tank Q_unloaded + 2 R_det}

Circulating Current Losses

The preceding should be valid so long as the tank's Q is limited primarily by its shunt resistance. To check this assumption, verify the tank's circulating current series resistance (R_circ) is much lower than what it would need to be to be solely responsible for the LC tank's unloaded Q.

  • R_circ << {2 pi f L_tank}/Q_unloaded = 1/{2 pi f C Q_unloaded}

For the 'worst case' broadcast band values, this series resistance is small, but not impossible to beat.

  • R_circ << {2*pi*f*90*10^{-6}}/1000 approx0.3 Ω at 520 kHz, and 1 Ω at 1720 kHz.

For comparison, the 30' (10 m) of 12 AWG (2 mm) solid wire sufficient to make a stout cylindrical air-core inductor of this value is about 0.05 Ω.

One thing to note is how much circulating current resistance by itself is equivalent to the tank loading resistance (again by itself) resulting in the same tank Q.

  • Q = {R_tank}/{2 pi f L_tank} = {2 pi f L_tank}/{R_circ}R_tank R_circ = (2 pi f L_tank)^2

Thus, at 1 MHz, one ohm of circulating current resistance is equivalent to loading a 100 µH tank with about 400 kΩ, while it would load a 200 µH tank only one-quarter as heavily, about 1.6 MΩ. Since parallel loading dominates most crystal radio tank designs, it seems worthwhile to use tanks with smaller inductors. However, parallel loading decreases Q with increasing frequency, and BCB reception requires an increasing Q to maintain constant bandwidth, so some trade-offs may be worth considering.

Now, if it were true that the unloaded tank Q can be modeled accurately from only the inductance (L_tnk) and parallel tank resistance (R_tnk), then the Q should always decrease with frequency, but this is not the case. Perhaps a combination of parallel and serial (circulating) resistance can be used.

  • Q_unloaded = 1/{ {2 pi f L_tank}/{R_tank} + {R_circ}/{2 pi f L_tank} }

Tank Design

  • Typical ferrite toroid inductor Q is 300.4)
  • Extremely good tanks have R_tank increasing from 1 MOhm to 2 MOhm across the broadcast band.5)

Personal Tools