| 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 RLC Circuits|
When we add a resistance to a series LC circuit, as shown in the schematic diagram to the right, the behavior of the circuit is similar to the behavior of the LC circuit with no resistance, but there are some variations. To see how the added resistance affects the operation of the circuit, we'll use the same parameters as with the Series LC Circuit, plus the resistor:
This time, our measured voltages come out as follows:
Does this make sense? We now know we must deal with the difference between vL and vC, which is just over 9 volts. But then we have a vR of over 4 volts. Did we somehow convert that 10 vrms input voltage to a 13 volt drop? Or have we overlooked something else?
As before, we must take into account the different phase angles between voltage and current for each of the three components in the circuit. The vector diagram to the right, while not to scale, illustrates this concept.
Since this is a series circuit, the current is the same through all components and is therefore our reference at a phase angle of 0°. This is shown in red in the diagram. The resistor's voltage, vR, is in phase with the current and is shown in green. The blue vector shows vL at +90°, while the gold vector represents vC, at -90°. Since they oppose each other diametrically, the total reactive voltage is vL - vC. It is this difference vector that is combined with vR to find vT (shown in cyan in the diagram).
We already know that vT = 10 vrms. Now we can see that vT is also the vector sum of (vL - vC) and vR. In addition, because of the presence of R, the phase angle between vT and i will be arctan((vL-vC)/vR), and can vary from -90° to +90°.
As always, voltage and current calculations for this circuit are based on Ohm's Law. Our basic expressions are:
Since this is a series circuit, the value of i in each expression is the same. That is, iL = iC = iR = i, and we can use i as our reference value for our calculations. We will also need to know the value of ω (= 2πf) to determine XL and XC. For f = 1 MHz,
Now we can complete our calculations, starting with XL, XC, and Z:
|=||6283185.3 × 0.000150|
|6283185.3 × 220 × 10-12|
|Z||=||((XL - XC)² + R²)½|
|=||((942.4778 - 723.43156)² + 100²)½|
|=||((219.04624)² + 100²)½|
|=||((47981.255) + 10000)½|
|vL||=||i × XL|
|=||0.041529452 × 942.4778|
|vC||=||i × XC|
|=||0.041529452 × 723.43156|
|vR||=||i × R|
|=||0.041529452 × 100|
|v||=||((vL - vC)² + vR²)½|
|=||((39.140586 - 30.043716)² + 4.1529452²)½|
|=||((9.09687)² + 4.1529452²)½|
|=||(82.753044 + 17.246954)½|
Allowing for calculator round-off errors through all these calculations, the total voltage v is exactly 10 volts, which is what we initially specified. Therefore our calculations check out, and our results are valid.
One effect of adding R to the circuit has been to reduce the current through the circuit, and increase the circuit impedance. How will this affect the circuit at resonance, when XL = XC?
At very low frequencies, capacitor C will be an open circuit, and virtually no current will flow through the circuit. At very high frequencies, inductor L will be an open circuit, and again no current will flow. However, at intermediate frequencies, both XC and XL will be moderate, and the difference between them will be small. At resonance, that difference will be zero, and only R will limit the current flowing in the circuit.
The graph to the right shows normalized values of current through a series RLC circuit at frequencies ranging from 0.01 times the resonant frequency, to 100 times that frequency. Beyond that range, as you can see from the graph, no significant current will flow at all. Within that range, current depends primarily on the value of R.
In order to get a graph with a resonant frequency at 1, we temporarily assign values of 1 henry to L and 1 farad to C, and use frequency in radians/second (ω, where ω = 2πf). We will also assume a normalized signal voltage of 1 volt (rms). Now we can set the current at resonance by selecting the value of R.
(Note that these values are used specifically to obtain a normalized graph. Once we have the graph, we can change the component values and know that even when the resonant frequency and relative value of R change, the shape of the curve will always be the same, so long as the ratio L/C remains constant. We'll see shortly what happens as that ratio changes.)
In a fully normalized circuit, R = 1Ω. This will permit a current of 1 ampere to flow at resonance, as indicated by the green curve on the graph. Similarly, if we set R to 2Ω the current will be 0.5 ampere at resonance. The blue curve shows this.
The remaining curves show what happens if we reduce R. The yellow curve shows current when R = 0.5Ω, while the red curve is for R = 0.1Ω.
Note that for low values of R, the current at resonance peaks much higher, but falls off very quickly as frequency changes. For higher values of R, the curve is much broader, and maximum current remains much lower. This is the standard trade-off between bandwidth and maximum current, and the value of R is critical for controlling this factor.
When we change the ratio of L/C, we change the reactances of both components at any given frequency without changing the resonant frequency. This is accomplished by making sure that the product of L and C remains constant even when we change their ratio. Thus, if L = 1H and C = 1f, LC = 1 and L/C = 1. However, if L = 2H and C = 0.5f, we still have LC = 1, but now L/C = 4. Or, if L = 0.5H and C = 2f, L/C = 0.25.
By changing L and C in this manner, we change the values of XL and XC at and near the resonant frequency without changing the resonant frequency itself. This controls the overall impedance of the circuit at frequencies near resonance, and gives the resistor, R, either more or less control over current at those frequencies. The result is a change in the frequency range over which this circuit will conduct significant amounts of current. The following three graphs illustrate this:
|L/C = 4||L/C = 1||L/C = 0.25|
When we plot the curves this way, it becomes clear that as we increase the L/C ratio, we limit the circuit to passing current over an increasingly narrow band of frequencies. However, as we reduce the L/C ratio, we widen the frequency band over which this circuit will pass significant amounts of current. This becomes very important when we deal with certain types of filters, and especially with tuned circuits.
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