Electronic – RC/RL circuits and frequency selection

impedancetheory

We have a lab that wants us to simulate an RC and RL circuit, and then a CR and LR circuit (series). From reading it is my understanding that reversing order of components results in reversing the action (ie frequency selection curve). In other words

LR=RC
CR=RL

Why is this?

Best Answer

There is a lot you implied or are assuming that you didn't state. The order of two components in series only matters if you access the node between them. It doesn't matter either if they are in parallel. You really should be more careful in explaining what you are talking about than just "RC and RL circuit". There are many ways to hook up resistors, capacitors, and inductors.

You apparently mean the special case of a RC and RL circuit where the two components are used as a frequency-dependent voltage divider. In that case the order matters just like it would if two different resistors are used in a voltage divider. Clearly these two circuits are not the same:

If the 10 kΩ resistors were replaced by capacitors, it would be even more obviously not the same. For the purpose of analyzing R-C filters, you can think of the C as a resistor that goes down with frequency. If R1 were replaced with a capacitor, the voltage divider would attenuate less at high frequencies and more at low frequencies, which we call a high pass filter. At DC it would attenuate infinitely, meaning it blocks DC. Similarly, if R3 were replaced by a capacitor, the voltage divider would attenuate more at high frquencies and less at low frequencies, which we call a low pass filter. At DC the capacitor looks like a open circuit, so it wouldn't attenuate at all at DC.

Related Solutions

Electronic – Resonance frequency of an RLC circuit

KCL: (\$v_o\$ over your output, \$v_i\$ over voltage source and \$v\$ over the internal node:

$$ \frac{v_o}{R_3} + \frac{v_o-v}{Z_{C1}}+\frac{v_o-v_i}{R_2} = 0 \\ \frac{v}{Z_L} + \frac{v-v_i}{Z_{C2}} + \frac{v-v_o}{Z_{C1}} = 0 $$

can be solved to get rid of the internal node \$v\$, setting \$Z_{C i} = \frac{1}{sC_i}\$ and \$Z_L = sL\$:

$$ \frac{v_o}{v_i} = \frac{L C_1 C_2 R_2 R_3 s^3 + L (C_1 + C_2) R_3 s^2+R_3}{ L C_1 C_2 R_2 R_3 s^3 + L(R_2+R_3)(C_1+C_2) s^2 + (R_2 R_3 C_1) s + (R_2 + R_3)} $$ which is in the form $$ \frac{s^3+n_2s^2+n_0}{s^3+d_2 s^2 + d_1 s + d_0}$$ with $$ n_2 = \frac{(C_1+C_2)}{C_1 C_2 R_2} = \frac{1}{R_2C_*}$$ $$ n_0 = \frac{1}{LC_1 C_2 R_2} $$ $$ d_2 = \frac{(R_2+R_3)(C_1+C_2)}{C_1 C_2 R_2 R_3} = \frac{1}{R_\| C_*}$$ $$ d_1 = \frac{1}{LC_2} $$ $$ d_0 = \frac{R_2+R_3}{LC_1 C_2 R_2 R_3} =\frac{1}{LC_1 C_2 R_\|}$$ where I introduced the parallel resistance and serial capacitances $$ \frac{1}{R_\|} = \frac{1}{R_2} + \frac{1}{R_3}\\ \frac{1}{C_*} = \frac{1}{C_1} + \frac{1}{C_2} $$

This is a third-order high-pass filter, so general second order LRC intuition does not apply. You have to plug in your values and show them on a Bode-plot in order to find oscillation problems.

Appendix

Maxima script:

sol  : solve([v_o/R3+(v_o-v)*s*C1+(v_o-v_i)/R2=0, 
              v/(s*L)+(v-v_i)*s*C2 + (v-v_o)*s*C1 = 0], [v_i, v_o]) $
trf  : ev(v_o / v_i, sol) $
res  : rat(trf, s);

Electrical – Why is Capacitance Effect Modeled as Admittance Branch and Not as Series Impedance in Transmission Line Circuits

Because the capacitance models electric field between the signal and return conductors.

And the electric field in a capacitor is directed from one plate to the other. Therefore in our model, one plate of the capacitor must be connected to the signal wire and the other to the return wire.

An obvious way to think about why the capacitor isn't in series with the signal wire: What is the behavior of a capacitor at low frequency? What is the behavior of a transmission line at low frequency?

Best Answer

Related Solutions

Electronic – Resonance frequency of an RLC circuit

Appendix

Electrical – Why is Capacitance Effect Modeled as Admittance Branch and Not as Series Impedance in Transmission Line Circuits

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