# Amplifier Clipping

What is is and why it is bad

An audio amplifier will have two voltage supplies, one positive and one negative, and they will be of equal value.

The negative pole (side, or terminal) of the positive supply and the positive pole of the negative supply will be connected together and is the circuit "common" line, or 0V, is usually connected to the amplifier's chassis and often called "ground" - even when it isn't connected to ground.  (See here)

This is illustrated in the diagram using batteries, although most amplifiers' supplies will be derived from the mains. The principle is exactly the same. The line at the top is the positive supply, that at the bottom is the negative supply and the one in the middle is the 0V connection.  Called 0V because it is the reference from which the positive and negative supplies are measured.

With supplies (rails) of ±9V the output power can be no more than a few watts.

Let's consider something a little more powerful.  We will assume power rails of ±30V.  The red lines in the following diagrams represent the two 30V rails.

With no signal input to the amplifier the output of the amplifier, measured from, (or with respect to), the 0V line will be zero volts.

When a signal is applied to its input, the output from the amplifier will be a voltage which will vary in polarity, swinging positive and negative with respect to 0V, changing its amplitude (size) and shape in accordance with the input signal.

The output signal will be a larger copy of the input signal.  That's what amplification is - making a larger copy.  The size of this copy will be dependent upon three things, the amplitude of the input signal, the gain of the amplifier, and the setting of the volume control.

It is useful to illustrate the following with diagrams assuming the application of a sine wave signal, as is normally used for testing.

In normal operation, with a reasonably sized input signal and volume control setting, the maximum amplitude of the output will be at some value between 0 and almost 30 volts (it can't quite go to 30.) In this diagram the peak output voltage is about half the supply value, i.e. about 15V.  This is the voltage which is applied to the loudspeaker.

How much power this value of voltage develops in the loudspeaker depends on the 'speaker's impedance and can be calculated simply with Ohm's law. where P = power in watts,  E = the r.m.s. value of the voltage  and  R = the resistance in ohms (Ω)

The value of the peak voltage in the above diagram is about 15.  To obtain the r.m.s. value, the peak value of a sine wave must be multiplied by 0.707.  This gives us 10.6V.

Using a value of 8Ω for the 'speaker's impedance the equation is Now let's turn the volume up so that the output voltage swings so that its peaks almost reach the supply rails, as illustrated in the next diagram. The peak voltage is almost 30.  Let's call it 30 anyway.  The r.m.s. value is 21.2.  Using the same formula, gives 56W.

Note that because power is proportional to the square of voltage and the voltage has increased by a factor of 2, the power has increased by a factor of 4.

And now let's turn the volume up even further.

Although the amplitude of the voltage applied to the output stage of the amplifier has increased, it should be obvious that the output voltage can't rise above the supply rails.  It had already reached that value in the last example.  The result is that the first, and steepest, part of the signal, where it crosses the zero line on both its positive and negative excursions is copied well enough but it soon reaches the supply rails and can't increase any more.  It will just stay at the maximum voltage until the next opposite-polarity excursion, as is illustrated in this diagram. It looks as if the tops and bottoms of the sine wave have been cut off.  Or clipped.

What power will be developed in the 'speaker?  We can still use the same formula.  This time, however, the output waveform is no longer a sine wave.

It is more like a square wave.  If the amplifier were to be driven hard enough it would be difficult to distinguish the output waveform from a square wave, so let's do the figures on that assumption.

The r.m.s. value of a square wave is the same as its peak value, in this case 30 volts.  Putting the figures into the formula gives 112W.

If the loudspeaker were designed to accept 50, 60 or even 70 watts, this is clearly not a clever thing to do.

In practice not many amplifiers will have power supplies which will deliver twice their design power when driven into clipping.  The voltage rails will drop but the power still increases dramatically and that is why clipping is bad.

It is often said that a loudspeaker driven from an underrated and overdriven amplifier is at much more risk than one driven from a higher power amplifier which is not driven into clipping.

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