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Series LC Circuits

The Circuit

A series LC circuit.

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 XC and XL.

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 vL = 43v and vC = 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 Vectors

The vectors in a series LC circuit.

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 vL at a phase angle of +90°. We also know that voltage lags current in a capacitance, so we show vC at -90°. And this gives us our first clue as to what is happening in this circuit and how we can get both vL and vC 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 XL > XC, the combined circuit looks purely inductive to the source. If XC > XL, the circuit looks capacitive. The only real question left is how to determine the lengths of the two vectors for vL and vC. Of course, we can always set up the circuit and measure the voltages, but we really need a way to calculate them as well.



The Mathematics

Our calculations for this circuit are based on Ohm's Law, just as they have been for other circuits. In general terms,

iL  =  vL

XL
 
iC  =  vC

XC
 
i  =  v

Z

Since this is a series circuit, iL = iC = i and we can use i as the reference value for all of our calculations. Also, since both XL and XC require us to determine the value of 2πf (= ω), let's calculate that now. For a frequency of 1 MHz,

ω = 2πf = 6.2831853 × 1,000,000 = 6283185.3

Now we can complete our calculations, starting with XL, XC, and Z:

XL  =  ωL    
 
   =  6283185.3 × 0.000150
 
   =  942.4778 Ω
 
 
XC  =  1  

ωC
 
   =  1

6283185.3 × 220 × 10-12
 
   =  1  

0.0013823008
 
   =  723.43156 Ω
 
 
Z  =  ((XL - XC)²)½
 
   =  | XL - XC |
 
   =  | 942.4778 - 723.43156 |
 
   =  219.04624 Ω
 
 
i  =  v  

Z
 
   =  10  

219.04624
 
   =  0.045652461
 
   =  45.652461 mA
 
 
vL  =  i × XL
 
   =  0.045652461 × 942.4778
 
   =  43.026431 v
 
 
vC  =  i × XC
 
   =  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 vL and vC 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 vL and vC: these voltages are always opposite in phase.

Since the total impedance of this circuit is the difference between XL and XC, 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?



When XL = XC

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 XL will increase while XC decreases. At lower frequencies, of course, XL will decrease while XC increases.

The question that begs to be asked is, How does the circuit behave at the frequency where XL = XC? Also, how can we determine what that frequency will be?

Finding the "magic" frequency is not a problem. Since both XL and XC depend on frequency, we simply set the two to be equal to each other, and solve the resulting expression for frequency:

XL  =  XC
 
ωL  =  1

ωC
 
ω²  =  1

LC
 
ω  =  1

 
f  =  1

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 XL = XC, it is also true that XL - XC = 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 short-circuited.

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.


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