Behaviour of a passive filter

Question

^{simulate this circuit – Schematic created using CircuitLab}

In this circuit how is it possible that Vout is ten times greater then Vin? Vout's measure is made by an oscilloscope 1M ohm internal resistance.

Answer 1

It can be helpful to look at this problem using voltage division.

Lets look at one of the simplest cases: an inductor and a capacitor in a voltage divider.

^{simulate this circuit – Schematic created using CircuitLab}

Lets forget about the actual values of the inductor and capacitor. Lets assume that we are looking at a particular frequency and we have already converted the inductance and capacitance values into positive and negative reactances.

According the the rule of voltage division,

\$\frac{V_{out}}{V_{in}} = \frac{Z_1}{Z_1 + Z_2}\$

In this case,

\$Z_1 = \frac{1}{j2\pi fC} = -j50Ω\$

\$Z_2 = j2\pi fL = j40Ω\$

Again: I'm going to worry only about the reactances. I'm going to ignore the values of frequency, inductance, and capacitance in this example.

We can insert our values of impedance into the voltage divider equation like so:

\$\frac{V_{out}}{V_{in}} = \frac{-j50Ω}{-j50Ω + j40Ω} = \frac{-j50Ω}{-j10Ω} = 5\$

So this circuit has a voltage gain of 5 at the particular frequency we are concerned with.

If the inductive reactance and the capacitive reactance get closer and closer to being the same magnitude, then the voltage gain of the circuit will increase without bound. This is seen in the denominator of the voltage division equation: it tends toward zero when the reactances are made closer and closer to being equal and opposite.

In real life, the voltage gain is limited by resistances in the inductor and capacitor. To illustrate this, imagine what would happen to the voltage division equation if the reactances were +j49 and -j50 ohms. The denominator would be much smaller, making the voltage gain much higher.

p.s. You can do the same thing with positive and negative resistances. ;)

Cheers

Answer 2

Indeed, it is the second part of your circuit that has a problem. In this circuit, by voltage division, the transfer function is $$ T(s) = \frac{V_o(s)}{V_1(s)} = \frac{R_2\|sL_2}{\frac{1}{C_2s}+R_2\|sL_2}=\frac{R_2L_2s}{\frac{R_2+L_2 s}{sC_2}+R_2L_2 s}=\frac{L_2C_2R_2\;s^2}{L_2C_2R_2s^2+L_2s+R_2} = \frac{as^2}{as^2+bs+c}$$ where $$ a = L_2C_2R_2 = (50\cdot 10^{-3})\cdot(10\cdot10^{-9})\cdot(83\cdot 10^{3}) = 41.5\cdot10^{-6} \\ b = L_2 = 50\cdot10^{-3} \\ c = R_2 = 83\cdot 10^3$$

Solving for the poles, $$a s^2 + b s + c = 0,$$ gives $$s = -120.5 \pm j2\pi\cdot 7,117.6,$$ which shows the natural frequency response of \$7,117.6\;Hz.\$ Remember that a slight insertion of energy in the input in the form of a unit pulse will result in a response with this frequency, so it really is a characteristic of the system.

Further, notice as \$s = j\omega\$ goes to infinity, that \$\frac{as^2}{as^2+bs+c}\rightarrow\frac{as^2}{as^2} = 1\$, and that it it goes to 0 as frequency goes to 0. In other words, it is a high-pass filter.

However, at the natural frequency, if we just plot in the numbers, then $$ \left|\frac{V_o(s)}{V_1(s)}\right|_{s=j2\pi(7117.6)} = 31.4\;\text{dB} = 37.1$$

So at the natural frequency, it amplifies your voltage by a factor of 37, thus explaining what you see.

Answer 3

The 10 nF (C2) and 50 mH (L2) form a fairly resonant (aka peaking) high pass filter at a frequency of 7117.6 Hz. That is what gives you voltage amplification. Here's a quick simulation of just those parts: -

As can be seen the peak is at 7.119 kHz (near enough to my calculation) and has a magnitude of over 31 dB.