# RLC circuits - Part 1

RLC circuits - Part 2 Resonance/Notches/Bandpass Cartoon from Agilent, www.educatorscorner.com Inductors - how do they work? R V0 dI VL = L dt Start with no current in the circuit. When the battery is connected, the inductor is resistant to the flow of L current. Gradually the current increases to the fixed value V0/R, meaning that the voltage across the inductor goes to zero. In reality the inductor has a finite resistance since it is a long wire so it will then be more like a pair of series resistances. Inductors - time constant L/R VR V0 VL Again the behavior of an inductor is seen by analysis with Kirchoffs laws. Suppose we start with no current. dI V0 =VR + VL = IR + L dt

V0 R then I = Rt 1 exp L and Rt Rt VL =V0 exp VR =V0 1 exp L L There is a fundamental time scale set by L/R, which has units of seconds (=Henry/Ohm) RLC circuits with sinusoidal sources The AC analysis of circuits with inductors is also easy, with the effective resistance (impedance) of an inductor equal to iL. From a phasor point of view this means that the inductor leads the resistor by 90 degrees. High pass and low pass filters can be made from inductors as well. However the inductors are usually bulkier and relatively expensive compared to capacitors (and more difficult to make in an integrated circuit) so are not used as commonly.

Another limitation is that they are far from ideal. The impedance is usually RL+iL, which means that in order to find the breakpoint you use f = L/(2(RL+R)). Notch and Bandpass A filter can also serve to select or eliminate a narrow band of frequencies. Examples are radio (select) and parental control channel eliminator circuits. Vout Vout Log10 f Log10 f Mathematical analysis of a series LRC circuit - bandpass filter Vin Z L =iL L 1 i C i = C ZC = C R Vout

First find the total impedance of the circuit 1 Z = R + i L C Using a voltage divider Vout = Vin R 1 R + i L C The phase shift goes from 90to -90. =tan 1 R 1 L

C Mathematical analysis of a series LRC circuit - bandpass filter (2) The magnitude of the gain, Av, is Vin Av = L C R Vout V out = V in R 1 2 R + L C 2 Note that for high frequencies L is dominant and the gain is

R/ L or small. At low frequencies the gain is RC because the impedance of the capacitor is dominant. At 2 = 1/ LC the gain is one (assuming ideal components). Graphing for a series LRC circuit RLC bandpass filter 1.2 f peak = 1 1 2 LC Gain 0.8 0.6 0.4 0.2 0 0 2 4 Log10 Freq 6 Although the gain falls off at 20 dB/decade at high and low frequencies (this

means that it is proportional to ) it is more typical to plot it as shown on a semi-log graph, since this emphasizes the peak. Q factor for a Series LRC circuit RLC bandpass filter 1.2 f peak Q= f 3dB L = R LC 1 Gain 0.8 0.6 0.4 0.2 0 0 Solve 2 1 = 2 4

Log10 Freq 6 R 2 1 R + L C 2 The quality factor or Q is defined as the energy stored divided by the energy loss/cycle. For an electronic bandpass it is the peak frequency divided by the width of the peak or bandwidth (defined by the frequencies where the gain is 3 dB lower than the maximum). f 3dB R = L Parallel LRC circuit 1 L Z = R +i

= R+i 1 ( 1 2 LC ) C L R Vout R A = v = 2 L L Vin R + i 2 R + 2 1 2 LC 1 LC Vin

C R f peak Q= f 3dB RC = LC Vout f valley = 1 2 LC Measured across the resistor, this circuit is a notch filter, that is it attenuates a small band of frequencies. The bandwidth in this case is defined by 3dB from the lowest point on the graph. Transients in a series LRC circuit - Ringing Vin L C R Vout Suppose instead of a sinusoidal source we had a slowly varying square waveform or a sharp turn on of

voltage. How would a LRC circuit behave? We can start by using Kirchoffs laws again. dI Q V0 =VR + VL + VC = IR + L + dt C 2 d Q dQ Q =L 2 + R+ dt dt C This is a second order differential equation that can be solve for the general and particular solutions. Transients in a series LRC circuit - Ringing (2) 1 1 x + x+ =0 RC LC The solutions to the quadratic above determine the form of the solutions. We will just state the solutions for different value of R, L and C. 2 Vin L

C R Vout 1 > RC 1 = RC 1 < RC 2 : Overdamped LC 2 : Critically damped LC 2 : Underdamped LC Transients in a series LRC circuit - Ringing (3) Vin L C R Vout

The last, underdamped, results in an exponentially decaying envelope and a sinusoidal oscillations. This ringing is commonly observed. It can be thought of as two parts: a loss of energy related to R and an oscillation related to the product LC. This not exact so lets look at the mathematical solution. 1 1 t VR (t ) = K1 cos 2 2 t exp LC 4 R C 2 RC When (RC)2>>LC the cos will oscillate several times at a frequency almost equal to the resonant frequency.

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