The Principle and Realization of Medium Wave Q Table
Principle: Ultra low impedance signal voltage source is output across R4, which is connected in series in the LC resonant circuit. When the circuit resonates, the inductive and capacitive reactances of L and C cancel each other out, leaving only R4 in series with the loss resistance r of the LC resonator. And the voltage across R4 is the voltage across r. In this way, we only measure the voltage U1 across R4, we can get the voltage across r is also U1. Then measure the voltage U2 across C3, and finally Q = U2 / U1.
Why R4 is designed so small. In fact, it can be larger, even if it is 1 ohm, then R3 should be increased to reduce the excitation current, so that the voltage across R4 will not be too high. One disadvantage is that the voltage across R4 is too high. That is, in a high-Q circuit, the voltage U2 across C3 is too high and will exceed the dynamic range of T2. In addition, looking at the two ends of L7, R4 and r are connected in parallel, then R4 uses 1 ohm has a disadvantage, that is, the high-Q circuit r is very small, the current mainly flows through r instead of R4, so it flows through the inductance under test The current of L is relatively constant, then the voltage across C3 is relatively stable near the resonance frequency. At this time, we have to measure the voltage across R4 in reverse to know whether it is accurately resonant, and the voltage across R4 is relatively small, which is not easy to measure. So it is recommended that R4 be smaller. However, R4 cannot be too small, otherwise the excitation is not enough. If the Q value of the large loop antenna is measured, the radio signal induced by the large loop may be stronger than the excitation signal and interfere with the measurement.
L4 has only 1 turn, and the proportional relationship with L1 is 1:14, so L1 is connected to a load of R3 of 100 ohms, which is only equivalent to a load of 14 * 14 * 100 = 19.6 kiloohms at both ends of L1, which will not affect the oscillator. normal work. The voltage across L4 is 0.4 volts, which refers to the peak-to-peak voltage, which is equivalent to the AC peak voltage of 0.4 / 2 = 0.2V. The AC signal refers to the peak-to-peak voltage or current unless otherwise specified. R4 is equivalently converted to L6, which is equivalent to 0.1 * 20 * 20 = 40 ohms, then the current flowing through R3 is 0.4V / (100 + 40) = 0.0028mA. After passing through the current transformer, the current on L7 is about 0.0028 * 20 = 56mA, then the voltage across R4 is 56mA * 0.1 ohm = 5.6mV. Of course, when actually measuring the Q value of the coil, the voltage across R4 is subject to the actual measurement. The current in the L7 and R4 circuits is very large, which will adversely affect the circuit. Therefore, the leads of the circuit should be shorter. If the leads are made long, the inductance of the wire circuit will also affect the current conversion effect of the transformer.
T2 is an impedance converter, and the related circuit should be far away from the oscillator and transformer to avoid interference. This part of the circuit is connected overhead with the exception of the positive and negative connections of the power supply, so that the distribution parameters are minimal.
In the power amplifier circuit of the audio system, some resistance of a few zero ohms will be used, and it is enough to find those that are not in the form of coils. I used two 0.22 ohm resistors in parallel to get a 0.11 ohm non-inductive resistor.
Then the current flowing through L6 is about if it is a high-Q coil, the pure part of the load impedance of C4 should not be ignored, and it is best to correct it in the calculation. This pure resistance is equivalent to about 8M ohms across C3. In addition, the loss of the ceramic capacitor may also be estimated. I have not found out the loss of the ceramic capacitor.
A small amount of waveform distortion: Because the load is relatively heavy, the output waveform is somewhat distorted, but it is relatively slight and does not affect the accuracy of the voltage measurement across R4.
The problem of non-inductive resistance: adjust C2 to change the signal frequency. If the voltage at both ends of R4 changes, it means that R4 is not a non-inductive resistance and should be replaced.
The primary inductance of transformer L6 is about 400uH. The magnetic core uses the magnetic ring inside the energy-saving lamp, which is a high-frequency ring with high permeability.
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