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schematic

simulate this circuit – Schematic created using CircuitLab

I'm designing a 2nd Order passive RC band-pass filter and found that the quality factor is given by:

\$Q=\frac{\sqrt{C_1C_2R_1R_2}}{C_1R_1+C_2R_1+C_2R_2}\$

When i make Q = (value > ~0.5) and solve for \$C_1\$, for example, Wolfram Mathematica always give me complex or negative answers. So, my question is: are there any limits for Q factor in passive circuits or I just done something wrong?

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  • \$\begingroup\$ Nearly correct. Would you consider passive LC bandpass? The limits of Q are mostly constrained by the source resistance and load resistance when compared to L & C reactance. Now Q can be much larger. In your RC example, source R and load R constrain Q to even lower limits (you haven't yet considered these constraints). \$\endgroup\$ – glen_geek May 8 '18 at 2:20
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    \$\begingroup\$ The Q factor of a passive RC circuit can never exceed Qmax=0.5. This is because the poles are always negative-real (and cannot be complex) \$\endgroup\$ – LvW May 8 '18 at 7:02
  • \$\begingroup\$ @Lvw But how can i proof that? I came upon this value by simply putting numbers in the formula. \$\endgroup\$ – Vinicius ACP May 8 '18 at 13:03
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    \$\begingroup\$ Vinicius, find the transfer function and identify the zeros of the denominator (poles of the transfer function). You will see that the well known formula for finding the roots of a quadratic function will always led to REAL solutions. Now - applying the definition of Q expressed through pole parameters you will see that always q<0.5. Examples: For R1=R2 and C1=C2 we have Q=1/3 and for R1C1=R2C2 but with R2>>R1 and C2<<C1 we have Q very close to 0.5. \$\endgroup\$ – LvW May 8 '18 at 14:23
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    \$\begingroup\$ Vinicius - why do you think we have "active filters" based on amplifiers with feedback? Answer: Because passive RC-filters do not allow quality factors >0.5. In your circuit, we have Q<0.5. Only in case of decoupling both parts with a buffer a value of Q=0.5 is possible. If you want pasive filters with Q>0.5 you are required to use LC-resonance. However, inductors have a lot of disadvantages and limitations - that`s the case for active filters which "emulate" the inductor effect (creating conjugate-complex poles with Q>0.5. \$\endgroup\$ – LvW Jun 7 '18 at 9:39
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It is possible to prove mathematically that the Q factor of this band-pass RC circuit is limited to: $$ 0<Q<0.5 $$


First, we need to remember that the standard form of the transfer function for a second-order band-pass filter is: $$H(s)=\frac{K(2ζω_n)s}{(s^2+2ζω_ns+ω_n^2 )}$$ Where:

  • \$K\$ is the maximum gain of the filter
  • \$ζ\$ is the damping ratio
  • \$ω_n\$ is the undamped natural frequency

And the transfer function based on component values is: $$H(s)=\frac{sC_1 R_1}{C_1 C_2 R_1 R_2 \cdot s^2+(C_1 R_1+C_2 R_1+C_2 R_2)s+1}$$


To prove that 0 < Q < 0.5, I will start by setting up an equation that, when solved, give the poles (roots of the denominator) of the above transfer function: $$C_1 C_2 R_1 R_2 s^2+(C_1 R_1+C_2 R_1+C_2 R_2)s+1=0$$

Note that this expression is a quadratic equation in the form \$ as^2+bs+c=0\$. Its discriminant is given by \$Δ=b^2-4ac\$. So, we have:

$$Δ=(C_1 R_1+C_2 R_1+C_2 R_2)^2-4⋅C_1 C_2 R_1 R_2$$ Assuming that \$ C_1>0\$, \$ C_2>0\$, \$R_1>0\$ and \$R_2>0\$, it follows that \$Δ>0\$ for any choice of components.

Now, doing the same process but using the standard form this time, we have: $$s^2+2ζω_n s+ω_n^2=0$$ $$Δ=(2ζω_n )^2-4ω_n^2=4ζ^2 ω_n^2-4ω_n^2=4ω_n^2⋅(ζ^2-1)$$ But we already know that \$Δ>0\$. So, using the standard form:

$$4ω_n^2⋅(ζ^2-1)>0$$

The term \$4ω_n^2\$ is surely positive. Then, to satisfy the above inequality, its necessary that $$(ζ^2-1)>0$$ Before proceeding, it's important to remember one more thing: $$Q≜\frac{1}{2ζ} \implies ζ=\frac{1}{2Q}$$ Then, by making the substitution, we have:

$$\left(\frac{1}{2Q}\right)^2-1>0 \implies \left|\frac{1}{2Q}\right|>1$$ Analyzing the expression of Q factor for the filter, we have, doing the same assumptions made above with \$Δ\$: $$Q=\frac{\sqrt{C_1C_2R_1R_2}}{C_1R_1+C_2R_1+C_2R_2}>0$$ Thus, we can finally conclude that: $$\frac{1}{2Q}>1 \implies 0<Q<0.5$$

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    \$\begingroup\$ 100% correct. Gratulations! \$\endgroup\$ – LvW Jun 7 '18 at 9:40

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