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,
X_{C} must be equal to X_{L}. 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=X_{L}/r_{L}. And
r_{L} can easily determined, because we have a simple
voltage devider here, since the total impedance of C and L at resonance
frequency is r_{L}. The full procedure is:

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

Adjust C to maximum output voltage;

Calculate r

_{L}and Q.

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

Adjust the source frequency to 159kHz;

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

The voltage across r

_{L}is 0.17 times the input voltage. So the ratio r_{L}:R is 0.17:0.83. This means that r_{L}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 r_{L} 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:

**B=f∙(R+r _{L})/X_{L}**

In our case: B=159kHz∙120/1k=19.1kHz.

(R+r_{L})/X_{L} looks very
much like 1/Q. X_{L}/R_{total} is
therefore also called the quality factor of the circuit. So the bandwidth
can also be noted as:

**B=f/Q _{c}**

In our example Q_{c}=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 V_{out}/V_{in}=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
Q_{c}. 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.