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• • • ## Band pass filter with a smaller band This circuit is also a band pass filter. High frequencies are shorted by the capacitor and low frequencies by the inductor.

It can be proven that:

Zr=Q∙XL

where Zr is the impedance at resonance freqency and Q is the quality factor of the inductor. This equation is an approximation, but the Q of most inductors is so large that it can almost always be used.

Some complex math proves that the impedance reaches its maximum value at: Again, this is an approximation that assumes the series resistance of the inductor is so small that it doesn't affect the frequency.

In our case f is 159kHz again. If rL is 20 ohms, Q is 1k/20=50. Zr will be 50∙1k=50k. The current and the input voltage are in phase, so:

Vout,max/Vin=Zr/(R+Zr)=50k/1050k=0.0476.

The image below shows the frequency response. If R»Zr and Q is large enough, then: B=f/Q. In our case B=159kHz/50=3.18kHz.

And this is finally the best way to determine the Q of an inductor:

1. Adjust the source frequency to the frequency you want to know the Q of;

2. Adjust C to maximum output voltage;

3. Turn the source frequency down until the output voltage decreased by 3dB (=factor 1.41);

4. Increase the source frequency until the output voltage is again 3dB less than the maximum voltage;

5. Subtract both frequencies. This is the bandwidth;

6. Calculate Q: Q=f/B

With this method, the internal resistance of the source doesn't matter, and we don't run the risk of a high current flow.

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