Home  www.playhookey.com  Thu, 07242014 

Direct Current

Alternating Current

Semiconductors

Digital

Logic Families

Digital Experiments

Computers

 Analog  Analog Experiments  Oscillators  Optics  HTML Test  

 The Fundamentals  Resistance and Reactance  Filter Concepts  Power Supplies   

Series RC Circuits

Series RL Circuits

Parallel RC Circuits

Parallel RL Circuits

 Series LC Circuits  Series RLC Circuits  Parallel LC Circuits  Parallel RLC Circuits   AC Applications of the Wheatstone Bridge  
Parallel LC Circuits 

In the schematic diagram shown to the right, we show a parallel circuit containing an ideal inductance and an ideal capacitance connected in parallel with each other and with an ideal signal voltage source. For consistency, we will use the same example values that we used when examining the series LC circuit. Thus,
According to Ohm's Law:
If we measure the current provided by the source, we find that it is 3.2126777 mA — the difference between i_{L} and i_{C}.
The question to ask about this circuit, then, is, "Where does the extra current in both L and C come from, and where does it go?"
The vectors that apply to this circuit imply the answer, as shown to the right. Here, the voltage is the same everywhere in a parallel circuit, so we use it as the reference. There is no resistance, so we have no current component in phase with the applied voltage.
We already know that current lags voltage by 90° in an inductance, so we draw the vector for i_{L} at 90°. Similarly, we know that current leads voltage by 90° in a capacitance. Therefore, we draw the vector for i_{C} at +90°.
Combining these two opposed vectors, we note that the vector sum is in fact the difference between the two vectors. This matches the measured current drawn from the source.
The remaining current in L and C represents energy that was obtained from the source when it was first turned on. This energy, and the current it produces, simply gets transferred back and forth between the inductor and the capacitor.
If we begin at a voltage peak, C is fully charged. Since current is 90° out of phase with voltage, the current at this instant is zero. But C now discharges through L, causing voltage to decrease as current increases. When C is fully discharged, voltage is zero and current through L is at its peak. This current has caused the magnetic field surrounding L to increase to a maximum value. This completes ¼ cycle.
The second quartercycle sees the magnetic field collapsing as it tries to maintain the current flowing through L. This current now charges C, but with the opposite polarity from the original charge. As current drops to zero and the voltage on C reaches its peak, the second ¼ cycle is complete.
The other half of the cycle sees the same behavior, except that the current flows through L in the opposite direction, so the magnetic field likewise is in the opposite direction from before. At the conclusion of the second halfcycle, C is once again charged to the same voltage at which it started, with the same polarity. Now, a new cycle begins and repeats the actions of the old one.
The calculation for the combined impedance of L and C is the standard productoversum calculation for any two impedances in parallel, keeping in mind that we must include our "j" factor to account for the phase shifts in both components. Thus,
Z  =  (jX_{L})(jX_{C}) 
(jX_{L}) + (jX_{C})  
=  j²(X_{L})(X_{C})  
j(X_{L}  X_{C})  
=  j(X_{L})(X_{C})  
(X_{L}  X_{C})  
This equation tells us two things about the parallel combination of L and C:
Because the denominator specifies the difference between X_{L} and X_{C}, we have an obvious question: What happens if X_{L} = X_{C} — the condition that will exist at the resonant frequency of this circuit? Clearly there's a problem with a zero in the denominator of a fraction, so we need to find out what actually happens in this case.
At the resonant frequency of the parallel LC circuit, we know that X_{L} = X_{C}. At this frequency, according to the equation above, the effective impedance of the LC combination should be infinitely large. In fact, this is indeed the case for this theoretical circuit using theoretically ideal components.
The currents flowing through L and C may be determined by Ohm's Law, as we stated earlier on this page. The current drawn from the source is the difference between i_{L} and i_{C}. However, when X_{L} = X_{C} and the same voltage is applied to both components, their currents are equal as well. Therefore the difference is zero, and no current is drawn from the source. This corresponds to an infinite impedance, or an open circuit.
This doesn't mean that no current flows through L and C. Rather, all of the current flowing through these components is simply circulating back and forth between them without involving the source at all. The currents calculated with Ohm's Law still flow through L and C, but remain confined to these two components alone. As a result of this behavior, the parallel LC circuit is often called a "tank" circuit, because it holds this circulating current without releasing it.
There is one other factor to consider when working with an LC tank circuit: the magnitude of the circulating current. We can use many different values of L and C to set any given resonant frequency. Keep in mind that at resonance:
X_{L}  =  X_{C} 
ωL  =  1 
ωC  
ω²  =  1 
LC  
ω  =  1 
As long as the product L × C remains the same, the resonant frequency is the same. However, if we use a large value of L and a small value of C, their reactances will be high and the amount of current circulating in the tank will be small. If we reverse that and use a low value of L and a high value of C, their reactances will be low and the amount of current circulating in the tank will be much greater. Many applications of this type of circuit depend on the amount of circulating current as well as the resonant frequency, so you need to be aware of this factor. In fact, in realworld circuits that cannot avoid having some resistance (especially in L), it is possible to have such a high circulating current that the energy lost in R (p = i²R) is sufficient to cause L to burn up!


 
All pages on www.playhookey.com copyright © 1996, 20002013 by
Ken Bigelow Please address queries and suggestions to: webmaster@playhookey.com 