RC Filters

A capacitor in series or parallel with a resistor can be used to make a filter circuit that allows us to select frequencies. A filter circuit consisting of a capacitor in series with a resistor can be made to act as a voltage divider, in the same way as two series resistors form a potential divider.

A series circuit with a capacitor C and a resistor R is connected to an AC supply of rms voltage V_s and frequency f.

We know that for a series circuit, the current is the same all the way round. For any frequency we know that:

X_C = __1__ and that V_C = I X_C

2pfC

We also know that V_R = IR

However, if we add the voltages across the resistor and the capacitor, they DO NOT add up to the supply voltage. For example, if the voltages across the capacitor and the resistor were 3 V and 4 V respectively, we would find that the supply voltage was not 7 V, but 5 V.

Why does this happen? The answer is that in a capacitor, the current and charge are not in phase. Phase relationships are not on the syllabus, so we won't consider them any further here.

Worked Example

A series circuit consisting of a 0.1 mF capacitor and a 2000 W resistor is connected to a 2 V supply set at a frequency of 1000 Hz.

(a) What is the reactance of the capacitor?

(b) What is the impedance of the circuit?

(d) Work out the voltage across and the resistor.

(e) Work out the voltage across the capacitor.

(a) First work out the reactance of the capacitor.

X_C = __1__ = ____________1____________ = 1590 W.

2pfC 2 x p x 1000 Hz x 0.1 x 10^-6 F

(b) Now work out the impedance.

Z² = R² + X_C²

= (2200)² + (1590)² = 7.36 x 10⁶ W²

=> Z = (7.36 X 10⁶)^1/2 = 2714 W

( Note: I have had to use the notation X^1/2 to denote the square root)

I = 2 V/2714 W = 7.4 x 10^-4 A = 0.74 mA.

(d) Now work out the voltage across the resistor:

We can use Ohm’s Law to calculate the voltages. V_R = IR and V_C = IX_C. It is a series circuit so the current is the same all the way round.

V = IR = 7.4 x 10^-4 A x 2000 W = 1.63 V

(e) Now do a similar calculation for the voltage across the capacitor:

V_C = IX_C = 7.4 x 10^-4 A x 1590 W = 1.77 V

If we add these two voltages up, we find that they don’t add up to 2 volts, but their squares do add up to 4 V2.

The Function of an RC Circuit

There are occasions where a filter circuit is useful:

To boost low frequency sounds when music is played at low level. The ear does not detect low frequency sounds very well.
Gets rid of high frequency noise such as a tape hiss
Compensates for imperfection in sound sources.

The RC circuit is a passive filter circuit. It cuts treble or bass frequencies, but cannot boost them. To boost the treble or bass, we need an amplifier in the circuit; this is an active circuit.

Question 4. What is the difference between an active and a passive circuit? ANSWER

The RC circuit is like a potential divider. Formula: Vout = Vin (R2/[R1 + R2])

This circuit acts as a voltage balance:

If R1 is smaller than R2, then the voltage across R2 (Vout) will be larger than the voltage across R1, and vice versa.
The voltages across both resistors add up to the supply voltage.
We can alter the voltage balance by changing either of the resistances.

We can lay out the RC circuit in exactly the same way.

To analyse the circuit, we need the equations:

V = IR
Xc = 1_
2pfC
Vc = IXc

In this case the circuit is acting as a treble cut filter:

At low frequencies, the reactance of the capacitor is high. This means that the voltage across the capacitor is high, since the current is the same throughout the circuit.
The voltage across the resistor is low at low frequencies.

Question 5. Explain what happens to the voltage across the capacitor when the frequency is high. ANSWER

· There is a frequency at which the voltage across the capacitor is the same as the voltage across the resistor. This is called the break frequency. Therefore the reactance and the resistance are the same:

R = 1_
2pfC

This is an important result.

We can measure the voltage across the capacitor at different frequencies and plot the log of the voltage against the log of the frequency.

A logarithm is a number expressed as a power of 10. So 10 000 = 10⁴. Therefore log₁₀ 10000 = 4. Logarithmic scales allow for a much larger range of values to be displayed than a linear scale.

Note that:

The plot is almost level until the break frequency. There is a change in the voltage balance, but remember that the effect will be less marked as the voltages add up as vectors. The effect is made even less by the logarithmic scale.
The plot passes under the frequency axis at 1 volt, since log₁₀ 1 = 0. Voltages of less than 1 volt will show up as negative values on the log₁₀ V axis.

Question 6. A filter circuit has a 10 mF capacitor in series with a 1 kilohm resistor. What is the break frequency? ANSWER

If we place the input to an amplifier at Vout, the amplifier would amplify preferentially the lower frequencies, since the voltage produced by higher frequencies would be very small.

If we turn the RC circuit upside down, we get a passive bass cut filter.

If we increase the frequency, the reactance of the capacitor goes down. Therefore the voltage across the capacitor must go down as well. So the voltage balance shifts and the voltage across the resistor goes up. Vout will be high when the frequency is high. A graph can be plotted of log10 voltage against log10 frequency:

Question 7. Explain why the bass cut filter behaves in this way. ANSWER

Now move on to ACTIVE FILTERS

Back to REACTANCE