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I built a simple RLC circuit:

enter image description here

I'm using a signal generator as input, and I'm reading the voltage across the capacitor with an oscilloscope.

I've calculated that the resonant frequency (where I will also measure the highest voltage across the cap) is around 73kHz. This was verified to be approximately correct.

enter image description here

Now, I understand that the bandwidth of such a circuit is defined to be the difference between frequencies where the output voltage drops to 0.707 of the maximal value. Here the maximal value is at resonance, around 2.38V. The cut-off frequencies should be therefore where the voltage across the cap is 0.707*2.38=1.68V. I found the corresponding frequencies by turning the frequency higher and lower:

enter image description here

enter image description here

(the voltage peak-to-peak voltage given by the oscilloscope drifted around slightly, but stayed mostly at the 1.68V)

So, the high cutoff frequency was around 135kHz, and the lower one was around 45kHz. This is strange, should the cut-off frequencies be equal distance from the resonant frequency? But here 135kHz - 73kHz = 62kHz and 73kHz - 45kHz=28kHz. So the higher cut-off frequency seems to be over twice as far from the resonance frequency as the lower one! What could be the reason? What am I doing wrong?

I also calculated the Q factor using the formula:

$${\displaystyle {\begin{aligned}B_{\mathrm {f} }&={\frac {1}{\,R\,}}{\sqrt {{\frac {L}{C}}~}}\\Q&=R{\sqrt {{\frac {C}{L}}~}}\,~.\end{aligned}}}$$

With my values of resistance of 10kOhm, capacitance of 47nF and 100uH, I get a Q of 216. As Q is also defined as the ratio of resonant frequency and bandwidth, I can divide the resonant frequency by 216 to get a bandwidth of about 300, which also isn't anything close to reality. So what am I doing wrong?

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The upper and lower 3 dB points are not equi-distant in hertz from the resonant point. Basically, if the upper 3 dB point is 85% higher (73 kHz --> 135 kHz) then the lower 3 dB point will be 73 kHz/1.85 = 39.5 kHz.

Given measurement errors and impractical current sources I would say this ties in with what you see. What you can be sure of is that if the two 3 dB points are at 45 kHz and 135 kHz then the resonant point is mathematically this: -

$$F_{resonant} = \sqrt{F_{upper}\times F_{lower}}$$

This would make the resonant frequency more like 78 kHz.

So what am I doing wrong?

Are you sure you are using a good current source to stimulate this circuit? I've just simulated it and to get the numbers you have you must be using a 50 ohm signal generator output and not a current source. This makes a massive difference and totally explains why your predicted Q value is nothing like what you see in reality. Try putting the 10 k resistor in series with your signal generator output instead.

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  • \$\begingroup\$ Interesting! As English is not my first language I have to ask for a clarification: Are you saying that in general the upper and lower 3dB points are not even supposed to be equidistant from the resonant point? I've just gotten used to seeing illustrative graphs of frequency responses that always have the bandwidth marked, with the resonant point exactly in between the upper and lower points. And what is this formula, $F_{resonant} = \sqrt{F_{upper}\times F_{lower}}$? How is it derived? I will try your suggestion with the resistor. \$\endgroup\$ – S. Rotos Dec 13 '18 at 12:27
  • \$\begingroup\$ When Q is high the bandwidth is low and the difference is negligible but when Q is low the two 3 dB frequencies are NOT equi-distant. They are never equi-distant but high Q systems can make it seem that they are. The formula is that using geometric mean principles \$\endgroup\$ – Andy aka Dec 13 '18 at 12:35
  • \$\begingroup\$ I see. I repeated the test using a 10kOhm resistor as you suggested and now the upper cut-off frequency seems to be around 87kHz, and the lower at about 70kHz, and the resonance at 79kHz, so this would be much closer to what I expected. The exact values are hard to pinpoint since the output voltage is now very low because of the high voltage drop over the new resistor and there is a lot of noise. I will try to tweak around a bit and report back soon. \$\endgroup\$ – S. Rotos Dec 13 '18 at 12:45
  • \$\begingroup\$ Maybe you should consider using a simulator. I use micro-cap and the student edition is free. \$\endgroup\$ – Andy aka Dec 13 '18 at 12:53
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I believe your results don't match your calculations because of the output impedance of your signal generator. That output impedance is in parallel with your 10k ohms resistor. You didn't specify what generator you are using but chances are its output impedance is either 50 ohms or 600 ohms, both of which are much less than 10k ohms. A parallel RLC circuit should be driven with a high impedance source so as not to load the circuit. Essentially you should drive it with a current source, not a voltage source. You can approximate this by putting a large resistor in series with your signal generator, probably at least 100k ohms. Do this and repeat your measurements.

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Your signal generator has an output impedance \$ R_o \$. Depending on the model this will either be \$ 50 \Omega \$ or \$ 600 \Omega \$. I'd guess \$ 50 \Omega \$ is more likely in this case but check the manual. Without this impedance the output voltage would not change with frequency.

Your effective circuit is thus

schematic

simulate this circuit – Schematic created using CircuitLab

Now this can be simplified using Thevenin to:

schematic

simulate this circuit

Where \$ R_{th} \$ is equal to \$ R_o \$ in parallel with \$ R \$

The gain of this circuit is thus:

\$ \dfrac{V_o}{V_i} = \dfrac{\dfrac{s \cdot L \cdot \dfrac{1}{s \cdot C}}{s \cdot L + \dfrac{1}{s \cdot C}}}{R_{th} + \dfrac{s \cdot L \cdot \dfrac{1}{s \cdot C}}{s \cdot L + \dfrac{1}{s \cdot C}}} = \dfrac{\dfrac{s^2 \cdot L \cdot C}{1 + s^2 \cdot L \cdot C}}{R_{th} + \dfrac{s^2 \cdot L \cdot C}{1 + s^2 \cdot L \cdot C}} = \dfrac{s^2 \cdot L \cdot C}{s^2 \cdot L \cdot C \dot (1 + R_{th}) + R_{th}} \$

Note you can't measure \$ V_o \$ as this is internal to the signal generator.

None of this applies if your signal generator is actually a good current source but I would be surprised if it's not a voltage source with output impedance.

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