Hi,
I have a circuit which I want to analyse using phasor at its natural frequency.
The circuit consists of a parallel inductor-capacitor \((L || C_1)\) in series with a parallel resistor-capacitor \((R || C_2)\) which is connected to an AC voltage source. So we have parallel inductor-capacitor impedence, call it \(Z_1\), and parallel resistor-capacitor impendence, call it \(Z_2\).
Therefore their impedences are:
\(Z_1 = \frac{1}{\frac{1}{j \omega L} - \frac{\omega C_1}{j}} = j\frac{\omega L}{1 - \omega^2 L C_1}\)
and,
\(Z_2 = \frac{1}{\frac{1}{R} - \frac{\omega C_2}{j}} = \frac{\frac{1}{R} - j \omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2}.\)
Since these impendences are in series the total impedence of the circuit is:
\(Z_1 + Z_2= j\frac{\omega L}{1 - \omega^2 L C_1} + \frac{\frac{1}{R} - j \omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} = \frac{\frac{1}{R}}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} + j \left[ \frac{\omega L}{1 - \omega^2 L C_1} - \frac{\omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} \right].\)
The natural frequency of this circuit occurs when the reactance of the series is zero so,
\(\frac{\omega L}{1 - \omega^2 L C_1} - \frac{\omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} = 0.\)
Rearranging this we get,
\(\omega_o = \sqrt{\frac{C_2 -\frac{L}{R^2}}{LC_2^2 + L C_1 C_2}}.\)
Let the voltage source be denoted by \(V_s\). At natural frequency, \(\omega_o\), the voltage across the impedence \(Z_1\) is,
\(V_1 = \frac{Z_1}{Z_1 + Z_2} \times V_s = j \frac{\frac{\omega_o L}{1 - \omega_o^2 L C_1}}{\frac{1/R}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}} \times V_s\)
and the voltage across the impedence \(Z_2\) is,
\(V_2 = \frac{Z_2}{Z_1 + Z_2} \times V_s = \left[ 1 - j \left( \frac{ \frac{\omega_o C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}}{\frac{1/R}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}} \right) \right] \times V_s = \left[ 1 - j \omega_o C_2 R \right] \times V_s\)
Okay, now let \(V_s = 12, \ L = 0.1mH, \ C_1 = 100nF, \ C_2 = 80nF, \ R = 100 \Omega\). Using the formulae above I got,
\(\omega_o = 220479.275\)Hz, \(V_1 = 12 \times (j 1.04364 \times 10^{3})\), and \(V_2 = 12 \times (1 - j 1.04364 \times 10^{3})\).
These mean \(|V_1| = 12.52368 \times 10^3\) and \(|V_2| = 12.52368 \times 10^3\). But my simulation using SPICE, I got both \(|V_1|\) and \(|V_2|\) roughly around 12 volts only! Can someone tell me where I am wrong?
I have a circuit which I want to analyse using phasor at its natural frequency.
The circuit consists of a parallel inductor-capacitor \((L || C_1)\) in series with a parallel resistor-capacitor \((R || C_2)\) which is connected to an AC voltage source. So we have parallel inductor-capacitor impedence, call it \(Z_1\), and parallel resistor-capacitor impendence, call it \(Z_2\).
Therefore their impedences are:
\(Z_1 = \frac{1}{\frac{1}{j \omega L} - \frac{\omega C_1}{j}} = j\frac{\omega L}{1 - \omega^2 L C_1}\)
and,
\(Z_2 = \frac{1}{\frac{1}{R} - \frac{\omega C_2}{j}} = \frac{\frac{1}{R} - j \omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2}.\)
Since these impendences are in series the total impedence of the circuit is:
\(Z_1 + Z_2= j\frac{\omega L}{1 - \omega^2 L C_1} + \frac{\frac{1}{R} - j \omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} = \frac{\frac{1}{R}}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} + j \left[ \frac{\omega L}{1 - \omega^2 L C_1} - \frac{\omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} \right].\)
The natural frequency of this circuit occurs when the reactance of the series is zero so,
\(\frac{\omega L}{1 - \omega^2 L C_1} - \frac{\omega C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega C_2 \right)^2} = 0.\)
Rearranging this we get,
\(\omega_o = \sqrt{\frac{C_2 -\frac{L}{R^2}}{LC_2^2 + L C_1 C_2}}.\)
Let the voltage source be denoted by \(V_s\). At natural frequency, \(\omega_o\), the voltage across the impedence \(Z_1\) is,
\(V_1 = \frac{Z_1}{Z_1 + Z_2} \times V_s = j \frac{\frac{\omega_o L}{1 - \omega_o^2 L C_1}}{\frac{1/R}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}} \times V_s\)
and the voltage across the impedence \(Z_2\) is,
\(V_2 = \frac{Z_2}{Z_1 + Z_2} \times V_s = \left[ 1 - j \left( \frac{ \frac{\omega_o C_2}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}}{\frac{1/R}{\left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2}} \right) \right] \times V_s = \left[ 1 - j \omega_o C_2 R \right] \times V_s\)
Okay, now let \(V_s = 12, \ L = 0.1mH, \ C_1 = 100nF, \ C_2 = 80nF, \ R = 100 \Omega\). Using the formulae above I got,
\(\omega_o = 220479.275\)Hz, \(V_1 = 12 \times (j 1.04364 \times 10^{3})\), and \(V_2 = 12 \times (1 - j 1.04364 \times 10^{3})\).
These mean \(|V_1| = 12.52368 \times 10^3\) and \(|V_2| = 12.52368 \times 10^3\). But my simulation using SPICE, I got both \(|V_1|\) and \(|V_2|\) roughly around 12 volts only! Can someone tell me where I am wrong?
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