Tutorial 13Filter Circuits


At the end of this topic you should be able to:

  • Calculate the reactance of a capacitor with the formula Xc = 1/(2pfC).

  • Draw and explain passive filters using RC circuits

  • Draw and explain first order active filters including treble cut, treble boost, bass cut and bass boost.

  • Calculate the break point of active filter circuits.

This tutorial has two pages.  Click next using the button at the bottom of the page to look at active filters.


The Reactance of a Capacitor

The following demonstration shows how a capacitor “blocks” direct current, but “allows” alternating current to flow.


Question 1

Use your knowledge of capacitor action to explain these observations.



It seems that in Circuit A the current is being blocked, while in Circuit B the current is being allowed to pass.


We see this because the capacitor is being continually charged and discharged.  No current is flowing through the capacitor.  It cannot because there is an insulating layer.  However we will treat the capacitor as if it were allowing a current to flow.


If we replace the 1000 mF capacitor with one of 100 mF, the charging and discharging currents are less. Therefore the bulb will glow less brightly.  We can make the bulb brighter again by increasing the frequency.


From this we can conclude:

  We can sum this up in the next diagram:



The reactance of a capacitor is the ratio of the voltage to the current.  The symbol for reactance is XC and the units are ohms (W).  It is the “effective” resistance of the capacitor at a particular frequency.  The reactance is linked to the capacitance, C, and the frequency, f, in the following relationship:


                        XC = __1__




The capacitance must be in farads.  Watch out for this bear trap.





We can also define the reactance in terms of the rms voltage and the rms current of a sinusoidal waveform:

                                    XC = Vrms



Question 2

Why do we use the term reactance to describe the ratio of voltage to current, in preference to resistance?



Notice that in calculations we treat reactance like resistance as in Ohm’s Law.


Worked Example

A 400 mF capacitor is connected to a 56 Hz supply.  If the supply voltage is 14 V rms, what is the rms current through the capacitor?


Use XC = __1__ to work out the capacitor reactance.



XC = __1__ = __________1_____________ = 7.11 W

         2pfC       2 ´ p ´ 56 Hz ´ 400 ´ 10-6 F


Current = 14 V ¸ 7.11 W = 1.97 A



Question 3

What is the reactance of a 1000 mF capacitor at a frequency of 1000 Hz?



We can plot a graph of how the reactance varies as frequency.  As XC varies as 1/f, the graph will be a hyperbola.


To get a straight line, we would need to plot XC against 1/f, the reciprocal of the frequency, which is T, the period.  The gradient will be 2pC.


The essential difference between a reactive and a resistive circuit is that:

  At the very simplest level:


Filter Circuits

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 Vs 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:

XC = __1__  and that VC = I XC



We also know that VR = 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 discussed in my electrical engineering notes, 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 2.0 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?

(c)    Work out the current.

(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.


XC = __1__ = ____________1____________      = 1590 W

         2pfC      2 × p × 1000 Hz × 0.1 × 10-6 F


(b) Now work out the impedance.

                  Z2 = R2 + XC2

                                                     = (2200)2 + (1590)2 =  7.36 × 106 W2

                                             => Z = (7.36 × 106)1/2 = 2714 W

( Note: I have had to use the notation X1/2 to denote the square root)


(c) The current can be worked out by I = V/Z

                        I = 2.0 V ÷ 2714 W = 7.4 × 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. 

VR = IR and  VC = IXC. 

It is a series circuit so the current is the same all the way round.   

                         V = IR = 7.4 × 10-4 A × 2000 W = 1.63 V


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

VC = IXC  = 7.4 × 10-4 A × 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 a Passive Filter

There are occasions where a filter circuit is useful:

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?



The RC circuit is like a potential divider. Formula:

This circuit acts as a voltage balance:

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

To analyse the circuit, we need the equations:

V = IR


XC =    1_





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


Question 5

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



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:

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:


Question 6

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



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.



Active filters are discussed in the next part of the Tutorial.





Self Test