Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Circuit Analysis

Status
Not open for further replies.

kaosad

New Member
Hi,

I have a circuit which I want to analyse using phasor at its natural frequency.

The circuit consists of a parallel inductor-capacitor [latex](L || C_1)[/latex] in series with a parallel resistor-capacitor [latex](R || C_2)[/latex] which is connected to an AC voltage source. So we have parallel inductor-capacitor impedence, call it [latex] Z_1 [/latex], and parallel resistor-capacitor impendence, call it [latex] Z_2 [/latex].

Therefore their impedences are:
[latex]Z_1 = \frac{1}{\frac{1}{j \omega L} - \frac{\omega C_1}{j}} = j\frac{\omega L}{1 - \omega^2 L C_1}[/latex]
and,
[latex]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}.[/latex]

Since these impendences are in series the total impedence of the circuit is:
[latex]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].[/latex]

The natural frequency of this circuit occurs when the reactance of the series is zero so,
[latex]\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.[/latex]
Rearranging this we get,
[latex]\omega_o = \sqrt{\frac{C_2 -\frac{L}{R^2}}{LC_2^2 + L C_1 C_2}}.[/latex]

Let the voltage source be denoted by [latex]V_s[/latex]. At natural frequency, [latex]\omega_o[/latex], the voltage across the impedence [latex]Z_1[/latex] is,
[latex]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[/latex]
and the voltage across the impedence [latex]Z_2[/latex] is,
[latex]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 \left( \frac{\omega_o C_2}{R \left[ \left( \frac{1}{R} \right)^2 + \left(\omega_o C_2 \right)^2 \right]^2} \right) \right] \times V_s[/latex]

Okay, now let [latex]V_s = 12, \ L = 0.1mH, \ C_1 = 100nF, \ C_2 = 80nF, \ R = 100 \Omega[/latex]. Using the formulae above I got,
[latex]\omega_o = 220479.275[/latex]Hz, [latex]V_1 = 12 \times (j 1.04364 \times 10^{3})[/latex], and [latex]V_2 = 12 \times (1 - j 1.04364 \times 10^{3})[/latex].

These mean [latex]|V_1| = 12.52368 \times 10^3[/latex] and [latex]|V_2| = 12.52368 \times 10^3[/latex]. But my simulation using SPICE, I got both [latex]|V_1|[/latex] and [latex]|V_2|[/latex] roughly around 12 volts only! Can someone tell me where I am wrong?
 
Last edited:
Ooppss! I made a mistake for [latex]V_2[/latex], here is the correct one (someone pointed it out to me):

[latex]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_2R \right] \times V_s[/latex].

My calculations were wrong too!! The correct one should be [latex]V_1 \approx V_2 \approx 21.1654[/latex]. Funnily, when I rerun the simulation I got [latex]V_1 \approx 21.4169[/latex] and [latex]V_2 \approx 23.4816[/latex].
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top