• Introduction
• Amplifier Power Losses
• Class-B
• Class-A
• Supply Losses
• Further Reading
• References

Amplifier efficiency is a topic that seems to be poorly understood, judging from the comments and questions I have had from readers, so I thought that I would put something together to explain away some of the mystery.  As I progressed, it became obvious that I was going to have a lot more work to do than originally intended, since many of the basic concepts will also require explanation.

Efficiency in itself is not hard - power out versus power in.  No electronic device can ever be 100% efficient, and all lost power in an amplifier is converted to heat.  It makes some sense to try to make an amp as efficient as possible, but if this causes an unacceptable degradation of the sound quality, then it matters not how many efficiency stars it has, no-one will want to listen to it.

At the other end of the scale, there are Class-A amplifiers, some of which are so inefficient that it almost defies belief.  This waste of power does not necessarily translate into better sound, although reading some material would lead you to think that the lower the efficiency, the better it must sound.  This is (of course) not true.

This article describes a single channel amplifier, using Class-A and Class-B.  No allowance has been made for reactive loads (as occur in loudspeakers), which can increase power dissipation dramatically.

So, where do the power losses come into play?  We will need to look at a basic amplifier design first, so that the mechanism can be seen clearly.  I shall start with a Class-B stage, as these are the most common (although very few are truly Class-B, most are actually Class-AB - but more on this later).  For all calculations, a sinewave is the signal source, and for those who might be ready to claim that music is much more complex, music - and indeed any waveform - is derived entirely from sinewaves of differing frequencies and amplitudes.

Class-B

We will start with the assumption that the transistors in Figure 1 are "ideal", in that there is no base current needed (they have infinite gain), and no voltage at all is lost when fully conducting.  The bias voltage is adjusted so that the transistors are exactly on the verge of conduction, but no quiescent current flows.  Given that the supply voltage is +/-20V, this allows a peak swing into an ideal 8 Ohm resistive load of 20 Volts, which is an RMS voltage of 14.14 V.  The peak current is 20/8, or 2.5 Amps, and this equates to 1.768 Amps RMS.  The power to the load is 25 Watts.

P = V * I      .... or
P = V² / R     .... or
P = I² * R

The DC current waveform is shown on the diagram, and it can be seen that for each half cycle, one supply or the other supplies the current to the load.  This can safely be ignored, as RMS (with a sinewave signal) implies that  the waveform is symmetrical, and therefore the added individual DC currents make a "whole" RMS current.

The RMS input current (see note, below) must therefore equal the RMS output current, and we can now calculate the input power using the first equation above.  The DC applied is 20V, so the input power is 20 * 1.768A, or 35.36W.  The 10.36W difference is dissipated as heat in the output devices (5.18W each), and is removed using a heatsink.  We do not need to take the entire supply voltage (40V) into account, since only half of it is used at any point in time.

Efficiency may now be calculated as

Eff = P_out / P_in * 100 = 25 / 35.36 * 100 = 70.7%

Note:  It is worth pointing out that using RMS current for the input power calculation is convenient, but not strictly correct.  It can be shown that the average value is more accurate, and for a sinewave (the waveform of choice for all power tests) this is equivalent to 0.636 (rather than 0.707) of the peak input current.  The difference is actually not at all great, especially since the amplifier should spend virtually none of its time at maximum output power (since the likelihood of clipping is very great indeed).  It is in the interests of accuracy that I let you know that ...

Average current (using the same example as above) is 2.5A (peak) * 0.636 = 1.59A, so maximum efficiency is

Eff = P_out / P_in * 100 = 25 / 31.8 * 100 = 78.6%

My thanks to Robert Milby for pointing out that average current is correct for the purposes of calculation.  He also sent a spreadsheet that can be used to work this out very accurately - please contact ESP for a copy if desired.

At 1/2 power (12.5W in this case), the dissipation is at a maximum, and the dissipated power exactly equals the output power, giving an efficiency of 50%.  It is also apparent that at lower powers efficiency falls even more, but output stage dissipation also falls, so heating would not seem to be such a problem.  However, since we are working with music in the real world, we need to consider the dynamic range (or, to be more precise, the peak to average ratio) of a typical signal.  This is between 10dB and 20dB, so average power will be somewhere between 2.5 and 8 Watts.  This implies from the table that the amplifier will be working fairly consistently at an average dissipation of 8 to 12 Watts at maximum power output before clipping.

This is not often taken into consideration, and many amplifiers have insufficient heatsink capacity to operate at anywhere near full power for prolonged periods.  Thankfully, most music is listened to at fairly moderate levels most of the time, so the problem is rarely as bad as it looks.  Sinewave testing is a most arduous test of an amplifier, quite contrary to the beliefs of some, who claim that it is too simple, and cannot give any reliable indication of the amp's performance on complex music.  It is rare music indeed that will stress an amp like this, as there is a constant variation of power, allowing the heatsink time to dissipate the heat generated during loud passages.

If the bias voltage of the amp in Figure 1 is increased to ensure that the transistor's quiescent current is about equal to the peak speaker current, then we have a Class-A amplifier.  The efficiency of this amp can also be easily calculated by the same method.  Since a Class-A amp conducts at all times, the DC current must be at least equal to the peak output current.  A 20V peak signal will still create a 2.5A peak current in the load, so the DC quiescent current must be equal to (or greater than) this figure.  The output to the load will still be 25W, but the DC input power is now 100W (2.5A * 40V).  Since all of the supply voltage is in use all of the time, we must perform calculations based on this.  Efficiency now (from equation 4) is

Eff = 25 / 100 * 100 = 25%

But wait, there's more!  The power supply is expected to convert it's input sinewave (at 50 or 60Hz) into a nice smooth DC that the amplifier can use.  This also has losses, since the laws of physics are omnipresent.  Since we used +/-20V as the power for the amp, we shall work with the same voltage output from the power supply, and will assume "typical" components.

A 14.14V RMS sinewave will provide 20V when rectified using "ideal" diodes.  In real life, there will be a voltage drop of about 1V (or more, depending on current) across each diode, and also losses in the transformer, which can be expected to have an internal resistance of about 1.0 Ohm (typical of a 120VA transformer).  The situation here is complicated by the fact that diode conduction is for only as very short period, so the current waveform is difficult to measure accurately.  We can still use basic theory to work out what happens, and Figure 2 shows a typical supply circuit and waveforms.  We will assume that the capacitance is so great that output ripple (although shown) can be considered negligible (bad news for the diodes though, since the peak curent will be very high).

As a matter of interest, the only meter that will give a passably accurate reading of the current is a moving coil (average responding) type.  True RMS meters will get it (sometimes horribly) wrong.  The best is to use an oscilloscope, and calculate the true average from the waveform shape and amplitude.

With the input of 14.14V AC, we will have a DC output of 20V (less the diode voltage drops), since the capacitors charge to the peak value of the RMS voltage.  If the peak output current (on each supply rail) is 2.5A as demonstrated above, then the input power must be at least equal to the output power plus any losses.  Even this configuration is open to some degree of interpretation.  Is the circuit of Figure 2 a full-wave centre-tap, or a bridge.  I consider it to be a dual full wave centre tap, but I could easily be wrong!.

In reality, the AC transformer current can only be accurately determined using complex analysis, but the table below is useful for passably accurate results.  There are several other variations (such as choke input filters), but these are not commonly used due to the excessive cost and weight penalty, and are not shown.
 
 

Assuming a Class-B amp first, 20V peak output (14.14V RMS) at an AC current of 1.768A.  I have indicated that the DC current must be equal to the AC load current, so using the above table we can determine the transformer secondary current using a full wave centre tap configuration as shown above

Transformer secondary current = DC current * 1.2 = 1.768A * 1.2 = 2.12A
Transformer secondary voltage = 28.28 + 2 x 1V diode voltage + 2.12V (transformer loss)
                                                      = 28.28 + 2 + 2.12 = 33V (approx)

P = V * I = 33 * 2.12 = 70VA

The power lost by the transformer winding resistance and diodes is (typically)

(2V + 2.12V) * 2.12A = 8.73W

Transformers also have "iron loss".  This is power that is needed to maintain the magnetic flux in the transformer core, and is always present.  As the load is increased, the magnetising current becomes less significant, and at full power, the resistive losses in the windings are usually far greater than the iron loss.  It is the magnetising current that causes a transformer to get warm even with no connection to the secondary winding.  Typically, small transformers have greater iron and copper losses than large transformers per volt-amp of output (i.e. they are less efficient).

VA = V * A = 38 * 3 = 114 VA

P = 5V * 2.5A = 12.5W (each supply, +ve and -ve)

From the above, you can now see where the power goes to.  There are losses everywhere, and it is essential that the amplifier builder has a reasonable knowledge of losses and dissipation requirements for an amplifier before starting construction.

For further reading, see Amp Design and Heatsink Design, both of which will assist in the final determination of the supply and heatsinking requirements for your next project.

1. Voltage Regulator Handbook - National Semiconductor 1982
2. Amplifier Design (ESP web page)
3. Heatsink Design (ESP web page)