Practical Electronics/Parallel RC

Parallel RC

Circuit Impedance

$\frac{1}{Z}=\frac{1}{Z_R}+\frac{1}{Z_C}$

$\frac{1}{Z}=\frac{1}{R}+j\omega C=\frac{j\omega CR+1}{R}$

$Z=R\frac{1}{j\omega CR+1}$

Circuit Response

$I=I_R+I_C$

$I=\frac{V}{R}+C\frac{dV}{dt}$

$V=(IR-RC\frac{dV}{dt})$

Parallel RL

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Circuit Impedance

$\frac{1}{Z}=\frac{1}{Z_R}+\frac{1}{Z_L}$

$\frac{1}{Z}=\frac{1}{R}+\frac{1}{j\omega L}=\frac{R+j\omega L}{j\omega RL}$

$Z=\frac{j\omega RL}{R+j\omega L}=j\omega L\frac{1}{1+j\omega \frac{L}{R}}$

Circuit Response

$I=I_R+I_L$

$I=\frac{V}{R}+\frac{1}{L}\int Vdt$

$V=IR-\frac{R}{L}\int Vdt$

Parallel LC

Circuit Impedance

$\frac{1}{Z}=\frac{1}{Z_L}+\frac{1}{Z_C}$

$\frac{1}{Z}=\frac{1}{j\omega L}+j\omega C=\frac{(j\omega {)}^{2}LC+1}{j\omega L}$

$Z=\frac{j\omega L}{(j\omega {)}^{2}LC+1}$

Circuit response

$I=I_L+I_C$

$I=\frac{1}{L}\int Vdt+C\frac{dV}{dt}$

Parallel RLC

Circuit Impedance

$\frac{1}{Z}=\frac{1}{Z_R}+\frac{1}{Z_L}+\frac{1}{Z_C}$

$\frac{1}{Z}=\frac{1}{R}+\frac{1}{j\omega L}+j\omega C$

$\frac{1}{Z}=\frac{(j\omega {)}^{2}RLC+j\omega L+R}{j\omega RL}$

$\frac{1}{Z}=\frac{(j\omega {)}^{2}LC+j\omega \frac{L}{R}+1}{j\omega L}$

Circuit response

$I=I_R+I_L+I_C$

$I=\frac{V}{R}+\frac{1}{L}\int Vdt+C\frac{dV}{dt}$

$I=\frac{V}{R}+\frac{1}{L}\int Vdt+C\frac{dV}{dt}$

$V=IR-\frac{R}{L}\int Vdt-CR\frac{dV}{dt}$

Natural Respond

$0=\frac{V}{R}+\frac{1}{L}\int Vdt+C\frac{dV}{dt}$

Forced Respond

$I_t=IR+L\frac{dI}{dt}+\frac{1}{C}\int Idt$

Second ordered equation that has two roots

ω = -α ± $\sqrt{{\alpha }^2-{\beta }^2}$

Where

$\alpha =\frac{R}{2L}$

$\beta =\frac{1}{\sqrt{LC}}$

The current of the network is given by

A e^{ω1 t} + B e^{ω2 t}

From above

When ${\alpha }^2={\beta }^2$, there is only one real root

ω = -α

When ${\alpha }^2>{\beta }^2$, there are two real roots

ω = -α ± $\sqrt{{\alpha }^2-{\beta }^2}$

When ${\alpha }^2<{\beta }^2$, there are two complex roots

ω = -α ± j$\sqrt{{\beta }^2-{\alpha }^2}$

Resonance Response

At resonance, the impedance of the frequency dependent components cancel out . Therefore the net voltage of the circui is zero

$Z_L-Z_C=0$ and $V_L+V_C=0$

$\omega L=\frac{1}{\omega C}$

$\omega =\sqrt{\frac{1}{LC}}$

$Z=Z_R+(Z_L-Z_C)=Z_R=R$

$I=\frac{V}{R}$

At Resonance Frequency

$\omega =\sqrt{\frac{1}{LC}}$ .

$I=\frac{V}{R}$ . Current is at its maximum value

Further analyse the circuit

At ω = 0, Capacitor Opened circuit . Therefore, I = 0 .

At ω = 00, Inductor Opened circuit . Therefore, I = 0 .

With the values of Current at three ω = 0 , $\sqrt{\frac{1}{LC}}$ , 00 we have the plot of I versus ω . From the plot If current is reduced to halved of the value of peak current $I=\frac{V}{2R}$ , this current value is stable over a Frequency Band ω₁ - ω₂ where ω₁ = ω_o - Δω, ω₂ = ω_o + Δω

• In RLC series, it is possible to have a band of frequencies where current is stable, ie. current does not change with frequency . For a wide band of frequencies respond, current must be reduced from it's peak value . The more current is reduced, the wider the bandwidth . Therefore, this network can be used as Tuned Selected Band Pass Filter . If tune either L or C to the resonance frequency $\omega =\sqrt{\frac{1}{LC}}$ . Current is at its maximum value $I=\frac{V}{R}$ . Then, adjust the value of R to have a value less than the peak current $I=\frac{V}{R}$ by increasing R to have a desired frequency band .

• If R is increased from R to 2R then the current now is $I=\frac{V}{2R}$ which is stable over a band of frequency

ω₁ - ω₂ where

ω₁ = ω_o - Δω

ω₂ = ω_o + Δω

For value of I < $I=\frac{V}{2R}$ . The circuit respond to Wide Band of frequencies . For value of $I=\frac{V}{R}$ < I > $I=\frac{V}{2R}$ . The circuit respond to Narrow Band of frequencies

Summary

Circuit	Symbol	Series	Parallel
RC			A parallel RC Circuit
Impedance	Z	$Z_t=R+\frac{1}{\omega C}=\frac{\omega CR+1}{\omega C}$	$\frac{1}{Z_t}=\frac{1}{Z_R}+\frac{1}{Z_C}=\frac{1}{R}+\omega C=\frac{R}{\omega CR+1}$
Frequency	${\omega }_o=2f_o$	$Z_R=Z_C$ $R=\frac{1}{\omega C}$ $\omega =\frac{1}{CR}$	$\frac{1}{R}=\frac{1}{\omega C}$ $\frac{1}{R}=\omega C$ $\omega =\frac{1}{CR}$
Voltage	V	$V=IR+\frac{1}{C}\int Idt$	$I=\frac{V}{R}+C\frac{dV}{dt}$
Current	I	$\int Idt=C(V-IR)$	$\frac{dV}{dt}=\frac{1}{C}(I-\frac{V}{R})$
Phase Angle		Tan θ = 1/2πf RC f = 1/2π Tan CR t = 2π Tan CR	Tan θ = 1/2πf RC f = 1/2π Tan CR t = 2π Tan CR