Tutorial: RLC Circuits and Q
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This week, we started considering new kinds of circuits involving a resistor, capacitor, and inductor all in one circuit. For this problem, let's consider the following circuits, one with the RLC all in series and another with them all in parallel.
If we wanted to solve these circuits for the values in question using differential equations, the differential equations for the two would have very similar forms. Let's take a look at those.
Series Version: Differential Equation
For the series version of the circuit, we'll have a differential equation like the following (considering only the homogeneous part of the answer):
{\displaystyle {{\rm d}^2 \over {\rm d}t^2}v_o(t) + 2\alpha {{\rm d}\over {\rm d}t}v_o(t) + \omega_o^2v_o(t) = 0}
R
, L
, and C
? Enter your answer as a simple Python expression:
\alpha =~
R
, L
, and C
? Enter your answer as a simple Python expression:
\omega_o =~
Parallel Version: Differential Equation
For the parallel version of the circuit, we'll have a differential equation like the following, considering only the homogeneous part of the answer):
{\displaystyle {{\rm d}^2 \over {\rm d}t^2}i_o(t) + 2\alpha {{\rm d}\over {\rm d}t}i_o(t) + \omega_o^2i_o(t) = 0}
R
, L
, and C
? Enter your answer as a simple Python expression:
\alpha =~
R
, L
, and C
? Enter your answer as a simple Python expression:
\omega_o =~
Solutions
OK, so these differential equations (and, in fact, the ones we could write for other voltages and currents in those circuits, too) have very similar forms. Let's go ahead and solve things out. For now, let's focus on the homogeneous part of the solution, which is instead given by the following (with x replaced by i_o or v_o depending on what we care about):
If we assume a form for our answer of x(t) = Ae^{st}, we end up with the following:
Since this equation is quadratic in s, there will, in general, be two solutions, so our output variable will actually have a form like x(t) = Ae^{s_1t} + Be^{s_2t}, where s_1 and s_2 are given by:
Depending on the particular values in question, we can end up with a number of different behaviors:
-
If \alpha > \omega_o, we say that the system is overdamped. In this case, the solutions for s are purely real, and our output looks like the sum of two exponential decays: x(t) = Ae^{s_1t} + Be^{s_2t}
-
If \alpha = \omega_o, we say that the system is critically-damped. In this case, s_1 = s_2 = -\alpha, and our output looks like x(t) = (A + Bt)\left(e^{-\alpha t}\right)
-
Perhaps the most interesting case is when \alpha < \omega_o, in which case we say that the system is underdamped.
The Underdamped Response
If the system is underdamped, s_1 and s_2 are complex: s = -\alpha \pm j\omega_d.
omega_o
and alpha
?
\omega_d =~
Solving out, we arrive at a solution like the following:
where A and \phi depend on initial conditions.
Quality Factor (Q)
In electrical engineering, it's common to use characterize RLC circuits by their "quality factor" Q, which is a single number that tells us how underdamped our system is. Other disciplines sometimes use other related quantities (for example, the damping ratio \zeta is common in mechanical systems), but for EE's, Q is usually the way.
We define the quality factor as follows: Q \equiv {\displaystyle \omega_o\over 2\alpha}. Physically, Q tells us about how much energy is lost per cycle of oscillation.
We can use this parameter Q to characterize whether the system is overdamped, critically-damped, or underdamped.
Q in terms of Circuit Component Values
In an underdamped system, our output will be of the form x(t) = Ae^{-\alpha t}\cos(\omega_d + \phi), i.e., a sinusoid with an exponential envelope. Q tells us about how quickly the response decays, compared to how quickly it oscillates.
Another way of thinking about this is: the higher the Q value, the closer we are to a perfect LC oscillator like we saw before. Theoretically, as Q\to\infty, our system becomes a perfect oscillator.
Of course, we usually think about Q as a way of thinking about the values in our circuit. So let's go ahead and relate these values back to our original circuit.
R
, L
, and C
?
Q_{\rm series} =~
R
, L
, and C
?
Q_{\rm parallel} =~
Q in Underdamped Systems: Time Domain
One important physical approximation is that, after Q cycles of oscillation, the amplitude of the sinusoid will be about 4% (1/25) of whatever value it started at, since e^{-\alpha (2\pi / \omega_o) Q} = e^{-\pi} \approx 1/25.
Q\approx ~