Thank you Professor Fiore for this great lesson on Parallel RLC circuit! Everything makes sense when the circuit is driven with a current sinusoidal source. I tried to analyze the same circuit with a voltage sinusoidal source. I ran into a big trouble when I tried to derive the bandwidth expression. Would it be possible if you post another lesson on this same topic with a voltage sinusoidal source. Thank you!
As a parallel circuit, the voltage is the same across each component. What you see instead is a change in the branch currents. You see the same sort of thing with the branch currents that you did with the voltages here (in reciprocal fashion), but generally, using a current source (e.g., a bipolar transistor) is the way to go. I suggest that you download my text "AC Electrical Circuit Analysis: A Practical Approach" from my web site (free) or grab the inexpensive paper version on Amazon as there are a large number of graphs that illustrate the impedance and voltage/current changes across frequency.
@@ElectronicswithProfessorFiore Thank you Professor Fiore for your kind reply! I will download your text to see if there is an answer for me there already. In case of current source, the dissipated power of circuit reduces when frequency deviates from the resonant frequency, so it makes sense to define the circuit bandwidth using side frequencies where the dissipated power is half of the power at the resonant frequency. In case of a voltage source, as long as the amplitude of the voltage source is fixed, then the dissipated power is not changing with frequency, then we are no longer able to use the dissipated power to calculate the bandwidth. This is the trouble I have with the voltage source.
@@yougoog1 The voltage is constant but the impedance isn't. At resonance, the impedance of the network is at maximum. As you move away from resonance, the network's impedance decreases, and thus, the source and branch currents change. Therefore, instead of concentrating on the change of voltage across frequency, consider the change in current across frequency. Instead of "square root of 2" change in voltage, we look at "square root of 2" change in current. Technically, with a constant voltage the true power (P) will not change although the apparent power (S) will. Using a current source, the increase in Z at resonance causes an increase in voltage. If the voltage is constant, Ohm's law dictates that the current will decrease at resonance. It can be shown that the branch currents through the capacitor and idealized inductor are Q times larger than the source current. This might sound impossible, but it is due to the fact that the L and C currents are 180 degrees out of phase and effectively cancel each other at resonance, leaving just the branch current flowing through the idealized R. This is similar to the situation in series resonance where the L and C voltages are Q times larger than the source voltage (again, cancelling due to phase).
@@ElectronicswithProfessorFiore Thank you Professor Fiore for your help! Because the voltage is constant, the apparent power at the frequency where the total current is “square root of 2” in current increases by the same factor of “square root of 2” instead by the factor of “2”. I understand the bandwidth is initially defined at side frequencies where the power gets changed by either half of or double of the power at the resonant frequency. The reason we could use the “square root of 2” change in current or voltage is because the power is proportional to the square of current or voltage, but in this case where voltage is held constant, the power is no longer proportional to the square of current, can we still use the principle of “square root of 2” change in current? Thank you for your insightful analysis from the current cancellation between the capacitor and inductor branches. I do understand the capacitor and inductor form a LC tank where the current oscillates back and forth between the L and C. Based upon the ratio of the maximum stored energy in L or C at resonance and the dissipated energy per cycle in resistor, the Q factor can be obtained to be R/(wL) where w is the resonant frequency and w= 1/sqrt(LC). From the bandwidth definition of Q factor, ie Q = w/bandwidth, the bandwidth can be derived to be 1/(RC). From the half power definition, we should end up with this same result of bandwidth, but I had trouble when I tried to do so. Yes. I did get the same result of bandwidth if I set the cutoff frequency at the “square root of 2” change in current, but should the cutoff frequency be set at the double of the apparent power?
@@yougoog1 You might be going down a rabbit hole here. Usually, the point of resonant circuit is to obtain frequency selectivity, for example, in a tuned amplifier. For that sort of application, you'd be using a current source (a BJT or FET) to drive the network. Because the impedance peaks at resonance, the gain peaks at resonance. That is, the gain curve mimics or follows the impedance curve. While we can certainly chase down what happens with a constant voltage source, it's an atypical application. In fact, to investigate parallel resonant response in a lab, we often mimic a constant current source by using a voltage source (i.e., a function generator) in series with a large resistance. This is because lab grade precision AC current sources are not very common and it's very easy to see an measure the Q/bandwidth effects using current drive.
Great video.
12:50 did you mean in parallel?
Yes, my apologies. I will make a note of that in the description. Thank you.
Thank you Professor Fiore for this great lesson on Parallel RLC circuit! Everything makes sense when the circuit is driven with a current sinusoidal source. I tried to analyze the same circuit with a voltage sinusoidal source. I ran into a big trouble when I tried to derive the bandwidth expression. Would it be possible if you post another lesson on this same topic with a voltage sinusoidal source. Thank you!
As a parallel circuit, the voltage is the same across each component. What you see instead is a change in the branch currents. You see the same sort of thing with the branch currents that you did with the voltages here (in reciprocal fashion), but generally, using a current source (e.g., a bipolar transistor) is the way to go. I suggest that you download my text "AC Electrical Circuit Analysis: A Practical Approach" from my web site (free) or grab the inexpensive paper version on Amazon as there are a large number of graphs that illustrate the impedance and voltage/current changes across frequency.
@@ElectronicswithProfessorFiore Thank you Professor Fiore for your kind reply! I will download your text to see if there is an answer for me there already. In case of current source, the dissipated power of circuit reduces when frequency deviates from the resonant frequency, so it makes sense to define the circuit bandwidth using side frequencies where the dissipated power is half of the power at the resonant frequency. In case of a voltage source, as long as the amplitude of the voltage source is fixed, then the dissipated power is not changing with frequency, then we are no longer able to use the dissipated power to calculate the bandwidth. This is the trouble I have with the voltage source.
@@yougoog1 The voltage is constant but the impedance isn't. At resonance, the impedance of the network is at maximum. As you move away from resonance, the network's impedance decreases, and thus, the source and branch currents change. Therefore, instead of concentrating on the change of voltage across frequency, consider the change in current across frequency. Instead of "square root of 2" change in voltage, we look at "square root of 2" change in current. Technically, with a constant voltage the true power (P) will not change although the apparent power (S) will.
Using a current source, the increase in Z at resonance causes an increase in voltage. If the voltage is constant, Ohm's law dictates that the current will decrease at resonance. It can be shown that the branch currents through the capacitor and idealized inductor are Q times larger than the source current. This might sound impossible, but it is due to the fact that the L and C currents are 180 degrees out of phase and effectively cancel each other at resonance, leaving just the branch current flowing through the idealized R. This is similar to the situation in series resonance where the L and C voltages are Q times larger than the source voltage (again, cancelling due to phase).
@@ElectronicswithProfessorFiore Thank you Professor Fiore for your help!
Because the voltage is constant, the apparent power at the frequency where the total current is “square root of 2” in current increases by the same factor of “square root of 2” instead by the factor of “2”. I understand the bandwidth is initially defined at side frequencies where the power gets changed by either half of or double of the power at the resonant frequency. The reason we could use the “square root of 2” change in current or voltage is because the power is proportional to the square of current or voltage, but in this case where voltage is held constant, the power is no longer proportional to the square of current, can we still use the principle of “square root of 2” change in current?
Thank you for your insightful analysis from the current cancellation between the capacitor and inductor branches. I do understand the capacitor and inductor form a LC tank where the current oscillates back and forth between the L and C. Based upon the ratio of the maximum stored energy in L or C at resonance and the dissipated energy per cycle in resistor, the Q factor can be obtained to be R/(wL) where w is the resonant frequency and w= 1/sqrt(LC). From the bandwidth definition of Q factor, ie Q = w/bandwidth, the bandwidth can be derived to be 1/(RC). From the half power definition, we should end up with this same result of bandwidth, but I had trouble when I tried to do so. Yes. I did get the same result of bandwidth if I set the cutoff frequency at the “square root of 2” change in current, but should the cutoff frequency be set at the double of the apparent power?
@@yougoog1 You might be going down a rabbit hole here. Usually, the point of resonant circuit is to obtain frequency selectivity, for example, in a tuned amplifier. For that sort of application, you'd be using a current source (a BJT or FET) to drive the network. Because the impedance peaks at resonance, the gain peaks at resonance. That is, the gain curve mimics or follows the impedance curve. While we can certainly chase down what happens with a constant voltage source, it's an atypical application. In fact, to investigate parallel resonant response in a lab, we often mimic a constant current source by using a voltage source (i.e., a function generator) in series with a large resistance. This is because lab grade precision AC current sources are not very common and it's very easy to see an measure the Q/bandwidth effects using current drive.
19:30 yup this is like where magic and science meet; it makes sense but it shouldnt make sense