We said that class 'A' amplifiers were VERY inefficient. Class 'AB' amplifiers are also inefficient
are more more efficient than class 'A' amplifiers. Class 'AB' mobile amplifiers are generally 60%
efficient when driving a 4 ohm load at maximum power (just before clipping). The reason that these
amplifier configurations are inefficient is because there is a difference of potential (voltage) across
the output transistors and current flowing through the output transistors. When you have voltage
across the device and current flow through the device, there will be power dissipation in the form of
heat. The power needed to produce this heat is wasted power. When there is (virtually) no voltage
drop across a device (such as a large piece of wire or a transistor), there can be a significant
amount of CURRENT flow through the device with (virtually) no power dissipation. This means
that there is virtually no heat given off (highly efficient). The inverse is also true. If you have
significant amount of VOLTAGE across the device (transistor, wire...) but no current flow through
the device, again, there will be no wasted power.
OK, now to the point. A class 'D' amplifier, which may also be known as a switching amplifier or a
digital amplifier, utilizes output transistors which are either completely turned on or completely
turned off (they're operating in switch mode). This means that when the transistors are conducting
(switched on) there is virtually no voltage across the transistor and when there is a significant voltage
across the transistor (switched off), there is no current flowing through the transistor. This is very
similar to the operation of a switching power supply which is very efficient.
Output transistor switching:
In the diagram below, there are 3 waveforms. Point 'X' is the point where the transistor would be
fully 'on', it would be conducting the full power supply voltage.This is the same as taking a piece
wire and connected the rail voltage to the speaker output of the amplifier. Point 'Y' shows that the
transistor is fully 'off' and the output voltage is effectively 0 volts. This is the same as completely
disconnecting the speaker from the amplifier. The 'duty cycle' of waveform 'A' is approximately
10%. If you averaged the voltage over time, the effective output voltage would be approximately
10% of the power supply voltage. Waveform 'B' has a 50% duty cycle. Waveform 'C' has a 90%
This diagram (below) shows the effective D.C. output from the respective waveform. Please keep
in mind that the effective output voltage is represented as it would be while there is a 'load' on the
amplifier's output transistors. If there were no load, the output voltage would stay at the full power
supply voltage because nothing would pull the voltage back toward the reference (ground).
Averaged DC voltage:
You should notice that the longer the output transistors are 'on', the higher the effective output
voltage. From a purely mathematical stand point, the output of the top waveform is at 100% of the
supply's output voltage for 10% of the time and at 0% (0 volts) for 90% of the time. If we assume
that the amplifier has a rail voltage of 40 volts, the output is at 40 volts for 10% of the time and
volts for 90% of the time. This would give you ((10*40)+(90*0))/100 or 4 volts of average output
In the diagram below, waveform 'A' is the pulse width modulated portion of an audio signal. Notice
that it switches between the yellow reference line (ground) to the rail voltage (white line).
Waveform 'B' shows how the voltage is stepped up as the width (time on) of the pulse increases.
This part of the signal is not really present at any point in the amplifier. It is simply included for
clarity. You should notice how the effective voltage is stepped up as the pulse width increases.
Waveform 'C' is a portion of the sine wave output signal, after passing through a low pass filter.
Keep in mind that the pulses are a high frequency square wave, it may be as high as 500,000 hertz.
A low pass filter which will allow a 20,000 hertz signal to pass (the highest frequency needed to
reproduce the audio spectrum) will have virtually no effect on the audio output signal and will
completely filter out the switching pulses.
In the diagram below, notice the circle on the sine wave at the bottom of the drawing. It shows the
portion of the sinewave that is being reproduced in the example.