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Band pass filter

The circuit above is called a band pass filter, because olnly a band of frquencies appears at the output. The capacitor stops all low frequencies and the inductor all high frequencies. The current flow in C and L is the same. The phase shift in the voltage across the capacitor is +90 degrees; the phase shift in the voltage across the inductor is -90 graden. The phase shift between both voltages is therefore 180 graden, and are thus in anti-phase! The output voltage reaches its maximum value if the voltages across C and L are equal; they simply cancel each other, and the full input voltage will appear across R. If the voltages are equal, XC must be equal to XL. The previous section already showed us that in that case:

We have also found a way to determine the quality factor Q of the inductor. We already know that Q=XL/rL. And rL can easily determined, because we have a simple voltage devider here, since the total impedance of C and L at resonance frequency is rL. The full procedure is:

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

  2. Adjust C to maximum output voltage;

  3. Calculate rL and Q.

Example: we want to determine the Q of an inductor at 159kHz.

  1. Adjust the source frequency to 159kHz;

  2. Adjust C to maximum output voltage. It appears to be 0.83 times the input voltage;

  3. The voltage across rL is 0.17 times the input voltage. So the ratio rL:R is 0.17:0.83. This means that rL is about 20 ohms. So: Q=1k/20=50.

Please note that we asumed that the internal resistance of the source is 0 ohms. In reality, that will never be the case. Additionally, we need very accurate voltage meters and rL is usually very small causing high current flows. Later in this lesson we will discover a better method.

But let's first determine the bandwith of our filter.

The bandwith is determined by the frequencies at which the output voltage is 3dB less than its maximum value. The maximum voltage is at -1.6dB, so we need to read the frequencies at which the output voltage is -4.6dB. The lowest frequency is about 150kHz and the highest 169kHz. So the bandwidth is about 169kHz-150kHz=19kHz. Fortunalely, we can also calculate the bandwidth:


In our case: B=159kHz∙120/1k=19.1kHz.

(R+rL)/XL looks very much like 1/Q. XL/Rtotal is therefore also called the quality factor of the circuit. So the bandwidth can also be noted as:


In our example Qc=1k/120=8.3. So B=159kHz/8.3=19.1kHz.

When we swap R and C, we get a low pass filter:

At low frequencies, the output voltage is equal to the input voltage (so Vout/Vin=0dB). We may expect it remains this way until the cut-off frequency (159kHz) at which point it will decay by 40dB per decade. However, we first see the capacitor voltage increasing to about 18.4dB. At this point, the voltage is 8.3 times higher than the input voltage! This ratio is eual to Qc. By making R 0 ohms, the voltage ratio will be equal to the inductor's Q! This is the second method to determine the Q of an inductor. We don't need very accurate voltage meters anymore, but the source's input resistance and the (possible) high currents are still bothering us.

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