Home  www.playhookey.com  Sat, 06242017 

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  
Series LC Circuits 

The schematic to the right shows an ideal series circuit containing inductance and capacitance but no resistance. As with the RC and LC circuits we have already examined, capacitor C and inductor L form a voltage divider across the voltage source. In this case, however, we have no resistance to provide an absolute limit on the current that can flow through the circuit — we have only X_{C} and X_{L}.
For this example, we'll assume a frequency of 1 MHz (1,000,000 Hz). We'll also set L = 150 µh and C = 220 pf. These are reasonable values for radio frequencies. The source voltage is 10 vrms.
This all seems quite reasonable. However, when we measure the ac voltages across L and C, we find v_{L} = 43v and v_{C} = 33v. Yet our source voltage is still just 10v. What's going on here? How can we get 76v across two components in series across a 10v source?
The answer is clear when we look at the voltage vectors in this circuit. They are shown to the right.
Since this is a series circuit, the current is the same throughout the circuit. With no circuit resistance, there is no resistive voltage, so we simply show the current vector in red, at the reference phase angle of 0°.
We know that voltage leads current in an inductance, so we show v_{L} at a phase angle of +90°. We also know that voltage lags current in a capacitance, so we show v_{C} at 90°. And this gives us our first clue as to what is happening in this circuit and how we can get both v_{L} and v_{C} to be higher than the source voltage: they oppose each other, and at least partially cancel each other out. It is the difference between these two voltages that must match the source voltage, and sure enough, 43v  33v = 10v.
If X_{L} > X_{C}, the combined circuit looks purely inductive to the source. If X_{C} > X_{L}, the circuit looks capacitive. The only real question left is how to determine the lengths of the two vectors for v_{L} and v_{C}. Of course, we can always set up the circuit and measure the voltages, but we really need a way to calculate them as well.
Our calculations for this circuit are based on Ohm's Law, just as they have been for other circuits. In general terms,
i_{L}  =  v_{L} 
X_{L}  
i_{C}  =  v_{C} 
X_{C}  
i  =  v 
Z 
Since this is a series circuit, i_{L} = i_{C} = i and we can use i as the reference value for all of our calculations. Also, since both X_{L} and X_{C} require us to determine the value of 2πf (= ω), let's calculate that now. For a frequency of 1 MHz,
Now we can complete our calculations, starting with X_{L}, X_{C}, and Z:
X_{L}  =  ωL  
=  6283185.3 × 0.000150  
=  942.4778 Ω  
X_{C}  =  1  
ωC  
=  1  
6283185.3 × 220 × 10^{12}  
=  1  
0.0013823008  
=  723.43156 Ω  
Z  =  ((X_{L}  X_{C})²)^{½}  
=   X_{L}  X_{C}   
=   942.4778  723.43156   
=  219.04624 Ω  
i  =  v  
Z  
=  10  
219.04624  
=  0.045652461  
=  45.652461 mA  
v_{L}  =  i × X_{L}  
=  0.045652461 × 942.4778  
=  43.026431 v  
v_{C}  =  i × X_{C}  
=  0.045652461 × 723.43156  
=  33.026431 v 
All calculations were taken to the limit of accuracy of a calculator program, which is why so many significant digits appear for each value. Note that the difference between v_{L} and v_{C} is precisely 10 volts — the value of the voltage source. Nevertheless, the voltages across L and C each exceed the source voltage by several times. This is only possible because of the phase relationship between v_{L} and v_{C}: these voltages are always opposite in phase.
Since the total impedance of this circuit is the difference between X_{L} and X_{C}, what happens if these two values are equal? There will be some frequency at which this will be the case, so we need to consider it. How will the circuit behave in this case?
We have examined one specific case of a series LC circuit and its behavior at one particular frequency. We also know that at higher frequencies X_{L} will increase while X_{C} decreases. At lower frequencies, of course, X_{L} will decrease while X_{C} increases.
The question that begs to be asked is, How does the circuit behave at the frequency where X_{L} = X_{C}? Also, how can we determine what that frequency will be?
Finding the "magic" frequency is not a problem. Since both X_{L} and X_{C} depend on frequency, we simply set the two to be equal to each other, and solve the resulting expression for frequency:
X_{L}  =  X_{C} 
ωL  =  1 
ωC  
ω²  =  1 
LC  
ω  =  1 
f  =  1 
2π 
If this calculation looks familiar, it should. This is the resonant frequency of an LC circuit, as we discovered when we first looked at LC circuits with direct current. The resonant frequency is the same with an ac voltage applied, but the continuous ac input makes up for any losses in the circuit so the oscillations never die out.
At resonance, since X_{L} = X_{C}, it is also true that X_{L}  X_{C} = 0. Therefore, there is absolutely no reactive component to Z at the resonant frequency. In the absence of any resistance, current i rises without limit, and becomes theoretically infinite. The voltage source behaves as if it were directly shortcircuited.
In reality, no circuit is completely without resistance, and the resistance present will serve to limit the current from the source. However, if the resistance is very small, the current will still be high. In some cases, a resistor is deliberately added, to set a minimum impedance and maximum current at resonance.


 
All pages on www.playhookey.com copyright © 1996, 20002015 by
Ken Bigelow Please address queries and suggestions to: webmaster@playhookey.com 