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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:


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:


The image below shows the frequency response.

The bandwidth is about 160.7kHz-157.5kHz=3.2kHz.

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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