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Contents

• Introduction
• Power Amplifier Definitions
• Power Supply Requirements
• Transformers
• Transformer Series Resistance
• VA Versus Watts
• Further Analysis
• Voltages and Currents
• Increasing Capacitance / Transformer Size
• Rectifier Types
• Full Wave
• Temperature
• Capacitor Value
• Rectifier Diodes
• DC Sound
• Summary
• Fusing and Protection
• Disclaimer
• References

Having searched the Web for reference material (and found very little!), this would appear to be the definitive article on the design of a "simple" linear power supply for a power amplifier.  Power supplies are needed for every type of amplifier we will ever use.  I do not intend to deal with "esoteric" designs with interesting names, but the simple, linear power supply that is still the mainstay of audio.

These supplies should not create any problems for anyone, because they are so simple, right?  Wrong!  They appear simple, but there are many inter-related factors that should be considered before just embarking on your next masterpiece.  The purpose of this article is to explain the terminology used, traps and pitfalls, and give some insight by way of a few practical examples.

Power supplies themselves require several definitions (these are discussed later in this article), but the requirements of the amplifier that is to be connected need to be understood before we start.  This makes a very big difference to the way the supply performs.

Poor earthing practices, such as connecting components to the nearest available ground reference can (and do) also create problems, and these can introduce hum, or more usually a "buzz" into the signal circuits.  This applies equally to Class-AB and Class-A amplifiers, but is usually more apparent with Class-A since the maximum current is drawn on a continuous basis.

Class-AB Amplifiers
I shall refer to the standard power amplifier as Class-AB - of all the amplifier types, these are the most common. Any amp that draws a quiescent current through the output devices is by definition, Class-AB. For true Class-B, there is no quiescent at all, and the output devices will conduct for exactly 180 degrees - this is rare.

Class-AB amps have a very widely varying current drain, which may be only 100mA or less with no signal, but rising to many amps when driven. The main problem is the revolting waveshape of the current on each supply lead, typically half-wave pulses, in sympathy with the program content.

These waveforms - this is current, not voltage - have sharply defined transitions, and as such will generate a magnetic field which varies with the current. Since a sharp transition equates to high order harmonics, care must be taken to ensure that voltages are not induced into the input stages of the amp from the supply lines. Because of the low inductance of the wiring of an amp, these problems are going to create distortion components which will tend to be worse at higher frequencies.

It is not only the amplifier which creates current pulses, but the rectifier / filter capacitor combination as well. The power supply rectifier diodes usually conduct for only a short time during each AC cycle - this may be as little as 3 or 4 degrees at idle, but both the angle of conduction and the amplitude of the current pulse will increase as more power is drawn from the supply.

Class-A Amplifiers
The other common amplifier type is Class-A.  These amps draw a large current on a continuous basis, and place a completely different loading on the supply.  The current pulses are gone from the supply leads, but the rectifier and filter now must handle the maximum current on a continuous basis.

The continuous load creates a new set of constraints on the design of a power supply, and the use of a Class-A amp implies that the builder already wants the very lowest noise.  Although the noise of the power supply DC output (hum/ ripple) will normally be low because of extensive filtering, regulation or a capacitance multiplier, the switching noise of the diodes in the rectifier can become more than a nuisance if proper care is not taken.

Somewhat surprisingly perhaps, the fundamental requirements of the final design are not greatly influenced by the different loading presented by the two amp types described above.  The continuous rating of a Class-A amp means that you must design the supply for a continuous (rather than transient) current, but since we are discussing properly designed, quality power supplies, the final result will often be quite similar.

When a power supply is used with an amplifier, the basic things we need to know before starting are as follows

Power output and minimum impedance, or ...
Peak / average current
Acceptable power supply ripple voltage

The first component of the power supply is the transformer.  Using magnetic coupling between windings, the transformer is used to isolate the amplifier (and the users) from the mains voltage, and to reduce (for solid state equipment at least) the voltage to something the amplifier can tolerate.  The primary winding will be rated at 240, 220 or 120V AC depending on where you live, and the secondary will be a more user friendly (or less user hostile) voltage to suit the amplifier.

DC Output:  The DC output is approximately equal to the secondary voltage multiplied by 1.414, but as we shall see, this is a rather simplistic calculation, and does not take the many variables into consideration.  At light loading, this rule can be applied without fear, and it will be accurate enough for most applications.  When an appreciable current is drawn, this simple approach falls flat on its face.

Mains variations:  These occur in all situations, and the mains voltage at any point in time will usually be somewhat different from the nominal voltage quoted by the supplier.  Any variation of 10% or less can be considered "normal", and greater variations are not at all uncommon.  In nearly all cases, an amplifier is rated at a certain power output into a specified load impedance, and at the nominal mains voltage.  For those who live close to a sub-station or pole transformer, expect the voltage (and power output) to be higher than quoted - the rest of us can expect a lower mains voltage and less power, especially during peak electricity usage times.

Losses:  Since all transformers have losses, these cannot be ignored in the design phase.  Magnetising loss (AKA iron loss) is the current that is required to maintain the design value of magnetic flux in the transformer core.  There is nothing you can do to affect this loss, as it is dependent on the size of the core and the design criteria of the manufacturer.  Large transformers will have a larger magnetising loss than small ones, but will be less affected by it due to the larger surface area which allows the transformer to remain cool at no load.  Small transformers have a greater loss per VA than bigger ones, and this is one of the reasons that small transformers run quite warm even when unloaded.

The iron losses are greatest at no-load and fall as more current is drawn from the transformer.  Copper losses are caused by the resistance of the winding, and are negligible at no load, and rise with increasing output current.  There is a fine balance between iron and copper losses during transformer design.  A relatively high iron loss means that copper losses will be reduced (thus improving regulation), but if too high, the transformer will overheat with no load.  A full description of the magnetising current and its effect on regulation is outside the scope of this article, and since there is little you can do about it, it shall be discussed no further.

Mains noise:  Noise can easily get through a transformer, both in transverse and common modes.  Transverse noise is any noise or waveform distortion that is effectively superimposed on the incoming AC waveform, and this is coupled through the transformer along with the wanted signal - the mains.

Common mode noise is any noise signal that is common to both the active (hot) and neutral mains leads.  This is not coupled through the transformer magnetically, but capacitively.  The higher the capacitance between primary and secondary windings, the more common mode noise will get through to the amplifier.  The much loved toroidal transformer is much worse than conventional "EI" (Ee-Eye) lamination transformers in this respect because of the large inter-winding capacitance.  An electrostatic shield will help, but these are uncommon in mass produced toroidal transformers.  The conventional transformer is usually better, and by using side-by-side windings instead of the more common (and cheaper) concentric windings, common mode noise can be reduced by an order of magnitude.

Input mains filters can remove either form of high frequency noise component to some degree, and large spikes can be removed using Metal Oxide Varistors (MOVs) that effectively short circuit the noise pulse, reducing it to a level that is (hopefully) inaudible.  Contrary to the beliefs of some, there is no panacea for noise, and it is best attacked in the equipment, rather than the now popular (but mainly misconceived) notion that an expensive mains lead will cure all.

Regulation:  The "regulation" of typical power transformers seems to be rather poor when some measurements are taken on a complete power supply.  When specified, regulation is based upon a resistive load over the full cycle, but when used in a capacitor input filter (99.9% of all amplifier power supplies), the quoted and measured figures will never match.

Since the applied AC spends so much of its time at a voltage lower than that of the capacitor, there is no diode conduction. During the brief periods when the diode conducts, the transformer has to replace all energy drained from the capacitor in the intervening 7 to 8ms or so (assuming 50Hz and full wave rectification).

Consider a power supply as shown in Figure 1. This is a completely conventional full-wave capacitor input filter (it is shown as single polarity for convenience). The circuit is assumed to have a total effective series resistance of 1 Ohm - this includes transformer winding resistances (primary and secondary) and diode losses. The capacitor C1 has an initial value of 4,700uF.

The transformer is rated at 60VA and has a primary resistance of 15 Ohms, and a secondary resistance of 0.5 Ohms. This calculates to an internal "copper loss" of 0.75 Ohm.

With a 20 Ohm load as shown, and at an output current of about 1.7 amps (1.69A actually), diode conduction is about 2.8ms, and the peak value of the current flowing into the capacitor is 6A - 100 times per second (10ms interval). Diode conduction is therefore 28% of the cycle. (To check that these figures are valid, we can take the peak capacitor current and the duty cycle, and calculate the average

Ia = (Ip / 100) * duty cycle so, Ia = (6 / 100) * 28 = 1.68

(close enough for our purposes)

Ripple across the load is 2.4V peak-peak (or about 740mV RMS), and is the traditional sawtooth waveform. Average DC load voltage is 33.8V. The no-load voltage of this supply is 41V, so at a mere 1.7A load, the regulation is

Reg (%) = ((Vn - Vl) / Vn) * 100 = 17.5%

Where V_n is the no-load voltage, and V_l is the loaded voltage

For this example, this works out to 17.5% which is terrible. By comparison, the actual transformer regulation would be in the order of 5% for a load current of 2A at 30V.  Note that the RMS current in the secondary of the transformer will be just over 3 amps AC for a load current of 1.7A DC - this must be so, since otherwise we would be getting something for nothing - a practice frowned upon by physics and the taxman.

Output power is 33.8V * 1.7A = 57.46W, and input power is 240V * 0.25A = 60VA.  The losses are caused by the diode voltage drop and winding resistance of the transformer.

As described above, I assumed a total equivalent series resistance for the transformer of 1 Ohm, which is about typical for a 60VA transformer as used here. Larger transformers will have lower series resistance (and vice versa), and the equivalent may be calculated - this is easier than actually measuring it under load.

If the secondary resistance is (say) 0.5 Ohm for a 240V to 30V transformer, it will be found that the primary resistance is (or should be) in the order of 15 Ohms. The actual figure will vary from one transformer type to another (e.g. "conventional" EI (ee-eye) laminations versus toroidal).

The effective primary series resistance is calculated (approximately) by

Re = Rp / (Tr)2

Where R_e is equivalent primary resistance, R_p is measured primary resistance, and T_r is the transformer turns ratio (in this case, 240 / 30 = 8

Re = 15 / 64 = .234 Ohm

This value is now added to the secondary resistance to calculate total series resistance. Please don't bother to e-mail to tell me that these figures are not correct - this is intended as a rough approximation - calculating actual values for transformers is worthy of an article in itself (which I am not about to write! :-)

The sad truth is that accuracy here is completely unimportant, as there is also series resistance in the mains power wiring from the power generation plant all the way to the power transformer primary winding. This is going to vary from one outlet to another and from one house to the next. Although it can be measured, this is a completely pointless exercise since it will only be relevant for one household. Other factors are the actual supply voltage (nominal 115V, 220V, 240V, etc.) which varies widely from day to day and hour to hour. (For what its worth, the actual supply voltage was 233V when I measured the mains impedance at approximately 0.8 Ohms at my workbench - does this help?). We shall now do what everyone else does, and ignore it completely.

However - again for what its worth - your 100W / 8 Ohm amplifier will be reduced to just over 90W, simply by connecting a 2400W heater to an adjacent power outlet, based on 0.8 Ohms mains wiring impedance and a genuine 240V supply voltage (before connection of the heater). The situation is likely to be slightly worse in the US, because the much lower AC supply voltage means that all currents are doubled for the same power.

An important distinction must be made between power (Watts) and VA.  Power is a measure of work, and it is quite possible (common, actually) to have a situation where there is voltage and current, but little or no work.  The product of voltage and current is Volts * Amps, or VA, and there is commonly a wide variance between VA and Watts.

Various loads (capacitive or inductive) will draw current from the output of a transformer or the mains supply.  If the load is purely inductive or capacitive, there is no power (work) at all, even though the current may be quite high.  Fluorescent lighting fixtures are renowned for this, where the current can be several times what was expected based on the power rating of the tubes.

This phenomenon is called "power factor", and a power factor of 1 means that there are no power losses due to inductance or capacitance.  Likewise, a power factor (PF) of 0 means that there is lots of voltage and current, but no power.  In the case of fluorescent lighting, power factor correction capacitors are used to try to maintain the PF at as close to unity as possible.  If this were not done, the wiring to the fittings (especially in large commercial buildings) will overheat, and a much greater load than necessary is placed on the local power sub-station, and indeed on the entire power grid.  Electricity supply companies worldwide have the same problems, and in most countries, there is legislation that determines the minimum acceptable power factor for any installation.

The switch mode power supplies used in computers have a very poor PF, and there are many new designs that improve this.  These can be expected to become mandatory in the not too distant future, as a poor power factor makes electricity more expensive to supply, and therefore more expensive for the consumer.

Note that I do not propose to cover the topic of power factor in depth (in fact that was it!), but a basic understanding is useful, and will make some of the following information more sensible.

A power transformer does not care if work is being done at the output or not.  It has internal resistance and inductive losses, and cares only about the output voltage and current.  A power transformer can be overloaded and destroyed by a large capacitance directly across the output terminals.  The capacitor does not even get warm, since it dissipates no power and does no work.  The transformer "sees" only the load current, and heats up proportionally - if the VA rating is exceeded consistently, the transformer will eventually overheat and die.  Equally, a transformer may be operated at 500% of its ratings for a short period, and as long as it has enough time to cool down between overloads, will be unaffected by the ordeal.  Unfortunately, this otherwise useful characteristic is pointless in audio, since the voltage will fall too far with the load, and amplifier power output suffers badly.  Having said this, most "mainstream" power amps will economise on the transformer, and rely on the duty cycle of typical programme material to provide an adequate supply voltage for normal music signals.  The continuous (erroneously called "RMS") power will be lower, sometimes significantly.

The term "dynamic headroom" used to be used to describe the difference between continuous and peak output power.  A large figure (2dB or more) indicates that the transformer is too small for the job, since the supply voltage collapses under a sustained load.

Because we are going to use the transformer in an unfriendly manner, with a rectifier and large capacitance as the load, the VA rating is much higher than the power rating of the amplifier may indicate.  There are some basic rules of thumb for the most common rectifier types, and these are shown below.

To properly see the effects of the losses and currents involved, a simpler circuit will be used from this point.  This consists of a 25V RMS "ideal" generator, and the copper losses are simulated by a resistance.  Since the full-wave bridge rectifier is a very common configuration, this is what shall be used for the detailed analyses that follow.  There are variations and exceptions to everything, but simulations and real-life testing on these simple circuits are very close, so this is what shall be used.

A simple resistance is the load, and we shall see the vast differences in peak AC current, capacitor ripple current and output voltage as the various parameters are changed.  A solid understanding of the behaviour of the transformer, rectifier and filter capacitor is essential if worthwhile power supplies are to be designed.

Figure 2 shows the voltages and currents present in a typical supply.  The waveforms will be examined shortly - for now we are interested in the average current and voltage in each section of the supply.  The generator voltage is 25V RMS, and for this supply I have used a winding resistance of 0.75 Ohm - roughly equivalent to a 120VA transformer.  The voltages and currents are all RMS - although in practice very few RMS meters can give an accurate reading of the spiky current waveform.

Note the big difference between DC output current, capacitor ripple current and AC input current.  The important parameters are listed in the table below ...
 

** This figure is somewhat misleading, since there is both a charge and discharge cycle.  During the discharge, there is a relatively constant current of -1.44A (the negative means the current is flowing out of the capacitor).  During the charge period, the rectifier takes over the supply of current to the load, then re-charges the capacitor at the peak current shown.

About 0.9V is lost across each diode during conduction, but this will vary in practice, based on the current capacity of the rectifier diodes.  Since this is a bridge rectifier, there are two diodes conducting at the +ve and -ve peaks of the waveform, so the total voltage loss is 1.8V - so the output DC should be around 32V.  The measured values of 23.6V AC and 28.8V DC is a direct result of the measurement inaccuracy!  Because current is drawn only at the peak of the AC waveform, the input to the rectifiers is not a sinewave.  The voltage and current waveforms are shown below, and it can be seen that the voltage waveform has been "flattened" at the peaks.  This is due to the high peak current drawn during this time, and no voltmeter will give the correct value - you must use an oscilloscope to be able to measure the peak-to-peak value of the waveform.

This reveals very different information from the voltage and current measurements taken before.  Both are essential in understanding the rectification process.  The peak AC input is only 31V, where we would normally expect 25 * 1.414 = 35V or (allowing for the voltage drop within the transformer) 33.4V.  We have some missing voltage (31 - 1.8 is 29.2, not the 28.8V measured) if we simply use normal multimeters.  Examination with the oscilloscope shows that the peak DC voltage is higher than the average value shown by the meter, and is 29.2V, so everything really does fall into place.

It must be pointed out that these waveforms were taken at different times, and are not exactly in phase.  The positive going part of the output ripple voltage, the peaks of the AC current, the positive peaks in capacitor ripple current and the flattening of the AC input voltage all occur at exactly the same time.

It is well known that bigger transformers have better efficiency that small ones, so it is a common practice to use a transformer that is over-rated for the application.  This can improve the effective regulation considerably, but also places greater stresses on the filter capacitor due to higher ripple current.  This is quoted in manufacturer data for capacitors intended for use in power supplies, and must not be exceeded.  Excessive ripple current will cause overheating and eventual failure of the capacitor.

Capacitor ripple current ratings can be ignored at your peril.

Large capacitors usually have a higher ripple current current rating than small ones (both physical size and capacitance).  It is useful to know that two 4,700uF caps will usually have a higher combined ripple current than a single 10,000uF cap, and will also show a lower ESR (equivalent series resistance).  The combination will generally be cheaper as well - one of the very few instances where you really can get something for nothing.  Using ten 1,000uF caps will generally give even better overall figures again, but the cost (in time and effort) of assembling them into a proper filter bank may not be felt worthwhile.

Above a specific value, as the capacitance is increased, the peak charging current will remain much the same for the same sized transformer, but the capacitor retains more of its charge between cycles.  The switch-on current will be very much higher, and the surge will last longer as the capacitor charges.  At capacitor values below optimum, the peak charge current will decrease somewhat, but there will be far greater output ripple.

There is no hard and fast rule for determining the optimum value for the filter cap, but in general I would suggest that the value should be at least that required to give a full load ripple voltage of less than 5V peak to peak.  Based on this, my recommendation is that the absolute minimum value is 2,000uF per amp DC, so a 5A (continuous) power supply will have an absolute minimum of 10,000uF capacitance.

What is achieved by increasing the capacitance, is the ability of the capacitors to retain more of their charge between AC cycles. Since the current demands of a Class-AB amplifier vary so widely - with the majority of the time at very low average currents - the actual operating voltage will be closer to the no-load voltage.

With large capacitors, the momentary current peaks created by the program material will not be of sufficient duration to discharge the caps to the full load voltage levels, so there is more voltage available on a more or less consistent basis. This equates to more power for transient signals, and lower ripple voltages the rest of the time.

With a 4,700uF capacitor and a peak current of 5A (equivalent to the peak current of a 100W amp into 8 Ohms), the capacitor will lose voltage at the rate of 1V / ms between "charges". As the capacitance is increased, this discharge rate naturally falls proportional to the capacitance. Doubling the capacitance halves the discharge rate and the ripple voltage for a given current, but increases the capacitor ripple current and the peak AC current - although the average value remains much the same.  There are some small variations, but these are eventually accounted for if we analyse the waveforms critically - again, this is a relatively pointless exercise, and will not be undertaken.

What we will do, is see what happens in each individual case when ...

• The capacitance is increased
• The transformer is made bigger

Table 3 is the same data, with the original 4,700uF capacitor, but now with a transformer having 0.5 of the original total winding resistance (0.375 ohms) - this is equivalent to a transformer of about 4 times the 120VA rating used before (or about 500 VA).
 

The increases in both average and peak currents are quite substantial, and the output voltage is higher by a small (but not especially useful) amount.  The DC current is higher, only because the voltage is greater, and this is the sole reason for the increase in ripple voltage.  The big test is to use the 10,000uF filter cap with this very much larger transformer, and see what increases occur.
 

In case anyone is wondering why I used a load resistance of 20 Ohms, this was to simulate one half of a 55W Class-AB amplifier operating at the onset of clipping into an 8 Ohm load, with a steady sinewave input.  Any dynamic analysis is very difficult, and the results are not particularly meaningful unless the exact signal source is known, along with the specifics of the power amplifier that is connected to the supply.

Actually, the current of 1.44A is slightly on the high side, since the RMS current will only be about 1.375A, which is 1/2 the speaker current (the other half of the speaker current is provided by the negative power supply).

So far we have looked at the full wave bridge rectifier, and this is but one of several different configurations.  The most common rectifiers are ...

• Full Wave Bridge
• Full Wave
• Dual Full Wave (Full Wave Centre Tap)

The full wave voltage doubler is still common in valve amplifier circuits and for preamp supplies, and shall be covered briefly in an update to this article (at a later date).

Of the major types, the generally accepted voltage and current ratios are as follows ...
 

*   Choke input filters have not been included, since although they provide superior filtering and lower noise, they are very expensive to produce because of the sheer size of the inductance.
** This figure is for each winding of the transformer.

Unlike the bridge rectifier which uses 100% of the transformer winding at all times, the full wave rectifier uses only 1/2 of the winding on each half cycle of the AC waveform.  This leads to some additional losses, since the winding must be double the number of turns of that for a bridge rectifier.  This usually means that the winding resistance is more than double that for a bridge rectifier.

Figure 4 - Full Wave Rectifier

I have reverted to a transformer for this section, rather than the simulated version used before.  This makes the drawings clearer, and the same depth of analysis will not be performed again this time.  The capacitor ripple current is unaffected in principle, except that it will be slightly lower for a given VA rating since it is directly related to transformer winding resistance.

For the example in Figure 4, the AC current in each winding is 1.82A for a 1.47A DC load.  This is quite close to the ratio of 1.2:1 shown in Table 5, and the difference is readily explained by measurement inaccuracy.

The biggest difference between a bridge rectifier and a full wave for a single output voltage is the transformer VA rating.  Using the data from Table 5, and staying with a 25V (or 25-0-25 for the full wave) transformer and a nominal load of 1.5A ...

Bridge:  Voltage = 25V, Current = 1.5 * 1.8 or 2.7A, so 25 * 2.7 = 67VA
Full wave: Voltage = 50V, Current = 1.5 * 1.2 or 1.8A, so 50 * 1.8 = 90VA

This version now uses the entire winding all the time - each winding is used for both positive and negative supplies.  Full winding utilisation means that the AC current is now the same as for a bridge rectifier, at 1.8 times the DC current, but only for a common mode load (i.e. between the supplies, rather than from one supply or the other to ground - this is identical to equal current from each supply to ground).  Where the load is from only one supply or the other, the 1.2 rule applies, but power amplifiers will draw from both supplies (more or less) equally.  The frequency of the signal waveform is mostly above the power supply input frequency, so the supply will effectively be loaded as common mode.  All dual power supply designs must assume this load, or the result will be most unsatisfactory.

In all power supply designs where the power is significant, the transformer temperature rise must be accounted for.  Apart from the transformer radiating its heat to nearby components, any temperature rise will increase the copper losses, leading to reduced performance and even more heat.  Use of a larger than required transformer will help considerably, but at the expense of capacitor ripple current.

Naturally, an increase in ripple current will cause the capacitors to become hotter, and as always, an increase in temperature causes increased losses and shortened component life expectancy.

For similar reasons, it is unwise to mount the filter capacitors anywhere near a significant heat source, such as large wirewound resistors, heatsinks or other heat generating components.  Valve amplifiers are the natural enemy of electrolytic caps, due to the often elevated temperatures within the chassis.  Some manufacturers have resorted to mounting the filter caps in a separate metal enclosure on the outside of the chassis (and reasonably well away from output valves) in an attempt to keep temperatures down.

The required capacitance for a given load current and ripple voltage is determined (approximately) by the formula [1] ...

I_L = 1.44, ripple = 2V p-p, therefore C = 5,040uF

I have seen it advocated that 100,000uF is the minimum that should be used with a powerful amp (say 200W / channel or so), but I find this difficult to justify.  The "law of diminishing returns" comes into play quite quickly, and with both capacitance and transformer VA rating, this law becomes significant once you have doubled each of these values.  With Class-AB amps, a ripple voltage of 2V P-P at full power will do nothing more than reduce the power by a few watts at the onset of clipping.  Even reducing the capacitance further to (say) 1,500uF per amp will only reduce the continuous power output by a small amount.  Normal music signals with their dynamic range will allow an amp with a relatively small capacitance to still provide the same maximum power for short transients.

As an example, a 100W/ 8 Ohm power amp will have a maximum output current of about 3.5A RMS into a resistive load.  Since we know that a loudspeaker is not resistive, this figure should be doubled, to 7A.  This is not an absolute rule - you can multiply by three if that makes you any happier.  The supply current for each supply (+ve and -ve) will therefore have a peak value of about 10A (7 * 1.414), and the average will be 1/2 of the speaker current, or 3.5A.  Based on the 3,500uF / amp figure above (assuming a 50Hz supply), 12,250uF per side is enough to ensure that ripple voltage never exceeds 2V P-P, but since this is a non-standard value, we would use 10,000uF.

In reality, it is probably going to be quite OK with 4,700uF, and the loss of power is negligible in real terms.  As stated above, continuous power will be reduced, but normal music signals will not have transients of sufficient duration to discharge the filter caps appreciably.  Note that this does not apply to amplifiers that are used for sub-woofers, nor for Class-A amps.  These will place a greater load on the supply, and on a more consistent basis.

If we were to increase the capacitance to infinity (big cap!), the ripple voltage will be 0V, and we will get an extra volt (peak) from the amp before the onset of clipping with a continuous signal.  This represents about 4W increase, which is completely insignificant - especially when we remember that the mains voltage could (will - at some stage) fall by up to 10%, which represents a power drop of around 20W - the 100W amp will only be capable of 80W with 10% low mains.  Considering that an amplifier should not be operated at or near clipping anyway, the difference is inconsequential.

Use of a small (e.g. 1uF) polyester or polycarbonate capacitor across the DC output is a common practice.  Electrolytics all exhibit a small inductance, and this causes their impedance to rise at higher frequencies.  This is dependent on the size of the cap - bigger caps usually have greater inductance.  Again, the use of a paralleled bank of small (1,000uF) electros will be better in this respect than a single large can type, and will also be easier to mount and cheaper.  I have never found the necessity to add a bypass cap to maintain amp stability, but it cannot hurt (some amplifiers will oscillate if the supply impedance is allowed to exceed some specific (low) impedance at high frequencies).  Essentially it is a good idea, and in the greater scheme of things, is inexpensive.

The manufacturers' ripple current rating is the maximum continuous ripple if the quoted life expectancy of the capacitor is to be achieved (usually 2000 hours, but 12000 to 26000 hours for some manufacturers). The ripple current rating is for the capacitor operating in an ambient of 85^o C (or higher for high temperature types) and that the maximum ripple current can be increased by up to 2.5 times as the operating temperature is reduced (2.5 times at 30^o C), though going above about 1.5 times is risky because the ESR increases as the capacitor ages and causes more heat for the same ripple current. [3]

Capacitors in power supplies feeding Class-A amps should be operated well within their ripple current rating.  In a Class-AB amp, the maximum ripple is at maximum output which only occurs occasionally (if at all!).  Occasional excursions up to or even above the maximum ripple current will not significantly affect the life of the capacitor.  In a Class-A, however, the ripple is at a maximum whenever the amp is switched on.  If the ripple current is at the maximum for the capacitor, the life expectancy would be 2000 hours (for the normal types).  This equates to a life of less than 2 years if the amp is used for 3 hours a day.

A formula for calculating ripple current would be very useful, but unfortunately (despite claims made in some articles I have read), it is almost entirely dependent on the series resistance provided by the incoming mains and the power transformer.  Any formulae that do exist are only true for "sub-optimal" values of capacitance (in other words, the cap is too small to be useful).  The summary (below) has some guidelines that may be useful, but be aware that these are guidelines only - the final outcome has so many variables that it is impossible to give an accurate prediction of capacitor ripple current.

Remember that large capacitor values will have a smaller surface area per unit capacitance than smaller ones, so the use of multiple small caps instead of a single large component can be beneficial.  There is more surface area, the ESR will be lower, ripple current rating higher, and the combination will most often be cheaper as well.  This is an "all win" situation - rarely achieved in any form of engineering.  An example is worth showing  - the following details are from an Australian electronics retailer's catalogue:
 

For the sake of the exercise, we will assume that we need 8,000uF at a minimum of 50V, and with a ripple current rating of 7A - this is more than adequate for a 100W amp, and meets all the design criteria.

• A single 8,000uF 80V cap could be used, with a price of $18.95, ripple current of 3.46A and a surface area of 8467 mm².  Overall, a simple solution but ripple current rating is 1/2 what I want, so it is excluded.
• Two 4,000uF 75V caps will cost $29, but have a ripple current of 9.2A and a surface area of 15,268 mm².  Considerably more expensive, but very good performance.
• Four 2,200uF 50V caps cost $11.40, ripple current is 7.6A, and surface area is 7,240 mm².  The most economical, but there is a minor performance deficiency by comparison with the previous and next options (but you do get some extra capacitance).  This would be my choice for most systems, as it meets or exceeds all requirements.
• Eight 1,000uF 63V caps will cost $15.60.  The ripple current is 11.2A, and surface area is 13,272 mm².  For performance vs. price, there really is no contest.  More effort is required to mount them, though.
• Naturally, you could also use the 10,000uF cap, but why would you do that?

It is worth pointing out that historically, filter capacitors are the number one cause of power supply failure.  This is almost always because of the effects of temperature and ripple current, and close attention to this is very much worth your while.

One thing I strongly recommend is the use of 35A chassis mounted bridge rectifiers.  Because of the size of the diode junctions, these exhibit a lower forward voltage drop than smaller diodes, and they are much easier to keep cool, since they will be mounted to the chassis which acts as a heatsink.  As always, lower temperatures mean longer life, and as was demonstrated above, the peak currents are quite high, so the use of a bigger than normal rectifier does no harm at all.

Even given the above, I have had to replace bridge rectifiers on a number of occasions - like any other component, they can (and do) fail.  The bigger transformers increase the risk of failure, due to the enormous current that flows at power-on, since the capacitors are completely discharged and act as a momentary short circuit.

Diodes used in a FWCT (Full Wave Centre Tapped) or single Full Wave supply rectifier must be rated at a minimum of double the worst case peak AC voltage.  So for example, a 25V RMS transformer will have a peak AC voltage of 35V when loaded, but may be as high as 40V unloaded, and double this is 80V.  100V Peak Inverse Voltage (PIV) diodes would be the minimum acceptable for this application.

For a single bridge rectifier, PIV needs to be greater than the peak AC voltage, since there are effectively two diodes in series.

In the case of a dual supply (using a 25-0-25V transformer), the worst case peak AC voltage is 80V, so the diodes must be rated at 200V PIV.  The most common 35A chassis mounted bridge rectifiers are rated at 400V, and this is sufficient for all supplies commonly used for power amplifiers of any normal (i.e. < 500W into 8 ohms).  Beyond this, the voltage rating is fine, but the current rating is inadequate, and a higher current bridge should be used.

There is currently a trend towards using fast recovery diodes in power supplies, since these supposedly "sound better".  This is basically a bad idea, and there is absolutely no requirement for them.  The purpose of a fast recovery (or any other fast diode) is to be able to switch off quickly when the voltage across the diode is reversed.  All diodes will tend to remain in a conducting state for a brief period when they are suddenly reverse biased.  This is extremely important for switched supplies, since they operate at high frequency, and have a squarewave input.  Standard diodes will fail in seconds with the reverse current, since it causes a huge power loss in the diode.

At 50 or 60Hz, and with a sinewave input, the slowest diodes in the universe are still faster than they need to be.  Indeed, Nelson Pass suggests that even the standard diodes should be slowed down with paralleled capacitors [2].  This can be an excellent idea, as it reduces the radiated harmonics from the diode switching.  Typically, 100nF capacitors (optionally with a small series resistance) are wired in parallel with each diode in the bridge, and this is quite common with some high end equipment and test gear where minimum noise is essential.

For reasons that I find completely obscure, there are some who claim that there are audible differences between power supply filter caps, diodes and mains leads.  The net result of all the transforming, rectifying and filtering described above, is to produce DC.  OK, it is not pure DC, in that it has some superimposed AC, as ripple voltage.  Very few power amplifiers are so intolerant of ripple or other signals on their supply lines that a couple of volts will be audible.  If this is the case, then a regulated supply, or at least a capacitance multiplier, is essential.

Very small amounts of noise that manage to get through the supply also should have no effect on an amp, and if the amp is so afflicted, the remedy is the same as for ripple.  Remember that at all audio frequencies, the reactance of the capacitor is very low, and will act as a short circuit for any stray noise signals.  A 10,000uF cap has a theoretical reactance (impedance) of 1.6 milliohms at 10kHz.  This will never be achieved in practice - the wire has more resistance than that.  Suffice to say that the impedance is pretty low, and it is extremely hard for any appreciable signal to get past the filter capacitors.  The addition of one or more 1uF film type bypass caps ensures that impedance stays low even as the inductance of the electrolytics starts to have an influence.

The above assumes that the problem is noise, but this is rarely stated as the "improvement" - it is more likely to be bass "authority" or extension, or perhaps the "veil" is lifted from the highs.  It can readily be demonstrated that as long as a power supply is well engineered in the first place, the "quality" of the DC output is unaffected by any of the "remedies".  Ripple voltage will remain the same, amplifier power output (at all frequencies) will be unchanged, and mostly the signal to noise ratio will be unaffected.  It is very easy to monitor the supply rails with an oscilloscope and a monitor loudspeaker (capacitively coupled of course), and from this it is possible to directly assess any difference should it exist.

I have yet to hear from anyone who can give a plausible explanation or send me a sound file that demonstrates the difference between any two mains leads, all else being equal.

In conclusion, there are some rules of thumb that can be applied to save calculations and test measurements.  These should not be considered as gospel - they are merely my suggestions on acceptable minimum requirements for a power supply.

Diode Ratings

• For most amplifiers or 50W or greater, use a 35A 400V rectifier bridge.  Smaller amps can naturally use something with lower ratings.  Current should be based on 10A per 100W for an 8 ohm load, or 20A per 100W for 4 ohms.
• Voltage rating must be a minimum of 200V for 100W into 8 ohms, or 100V for 100W for 4 ohms.  This is not linear, so direct extrapolation is not advised.  400V diodes are adequate for amplifiers up to 2000W/ 8 ohms.

• Capacitance
• Class-AB:  The filter cap should be a minimum of 4,700uF per 100W into 8 ohms, and 10,000uF per 100W into 4 ohms.  The actual value can be extrapolated from here.
• Class-A:  The cap needs to be selected on the requirements of the amplifier, but a minimum of 4,700uF for each 10W into 8 ohms is recommended.

• Voltage rating is as required by the maximum DC supply, and will usually be 50, 75 or 100V

 
• Ripple current ranges (from the examples shown here) from 3.3 times the load current up to 4.6 times load current, depending largely on the transformer size.
• Class-AB:  Because the amplifier will rarely draw maximum current for prolonged periods, a ripple current rating equal to double the peak output current will usually be enough.  Thus a 100W/ 8 ohm amp having a peak output current of 3.5A will usually be OK with a ripple current rating of 7A. (Note:  guitar amplifiers are specifically excluded from this! - I suggest double the above value as a minimum.)
• Class-A:  For the worst case possibility, I suggest that the ripple current rating should be 5 times the load current for these amplifiers.  A 20W/ 8 ohm Class-A amp will draw a continuous 2.5A (typical), so ripple current rating for the capacitors needs to be 12.5A.

• The voltage is determined by the power that you need from the amplifier.  Calculate power from the formula

P = Va² / RWhere P is power in Watts, Va is RMS speaker voltage, and R is speaker nominal impedance
 
• The supply voltage (allowing only for basic losses) is calculated as follows

V_RMS = Va * 1.1WhereV_RMS is the transformer secondary voltage (for each supply rail)

VA Rating - Class-AB

The minimum VA rating suggested is twice the amplifier power.  A 50W amp therefore needs a 100VA transformer, or 200VA for stereo 50W amps.  Larger transformers will provide a "stiffer" power supply, and this may be beneficial.  For continuous operation at full power (never actually needed in audio), the transformer should have a VA rating of 4 to 5 times the amplifier power.

It is suggested by some transformer manufacturers (and no doubt gleefully adhered to by many amplifier makers) that the VA rating needs only to be 0.7 of the maximum amplifier power.  While this will work, you will not have a "stiff" power supply - I would suggest "soggy" as an appropriate term - the voltage will waffle about like a limp biscuit.

VA Rating - Class-A

The minimum VA rating suggested is 5 times the amplifier power.  A 20W Class-A amp therefore needs an 100VA transformer, or 200VA for stereo 20W Class-A amps.  Larger transformers will again provide a "stiffer" power supply, and this will be beneficial.  Nelson Pass suggests the ratio should be 7.5 times the amp power [2].

Since the power supply is connected to the mains, it is necessary to protect the building wiring and the equipment from any failure that may occur.  To this end, fuses are the most common form of protection, and if properly sized will generally prevent catastrophic damage should a component fail.

Toroidal transformers have a very high "inrush" current at power-on, and slow-blow fuses are highly recommended to prevent nuisance blowing.  In the case of any toroid over 500VA, a slow-start circuit is very useful to ensure that the initial currents are limited to a safe value.  An example of such a circuit is presented on the Project Pages, and represents excellent insurance against surge damage to rectifiers and capacitors.

Calculating the correct value for a mains fuse is not easy, since there are many variables, but a few basic rules may help.
First, determine the maximum operating current of the system, based on continuous maximum power.  The calculations done previously will help ...

The mains current is determined by the turns ratio of the transformer, calculated by

Tr = Vpri / VsecTr is turns ratio, Vpri is primary (mains) voltage, and Vsec is secondary voltage

Ipri = Isec / TrIpri is primary current, Isec is secondary current

Thermal protection (often by way of a once-only thermal fuse) is often included in transformers.  If the "one-time" thermal fuse has been used, should the transformer overheat it must be discarded, since the fuse is buried inside the windings and cannot be replaced.  All the more reason to ensure that the transformer is properly protected at the outset.
 

Note that primary fuse or circuit breaker protection does not protect the amplifier against overload or shorted speaker leads.  If this happens, or should the amplifier fail, the primary fuse offers no protection against catastrophic failure and possibly fire.  For this reason, secondary DC fuses should always be used - no exceptions.

Multi-tapped primaries (e.g. 120, 220, 240V) create additional problems with fusing, and often a compromise value will be used.  The transformer protection is then not as good as it could be, but will generally still provide protection against shorted diodes or filter caps.

Additionally, it may be an advantage to fit Metal Oxide Varistors (MOVs) to the mains - between the active and neutral leads.  These will absorb any spikes on the mains, and help to prevent clicks and pops coming through the amplifier.

The information presented in this article is intended as a guide only, and the author takes no responsibility for any damage disfigurement or injury to persons (including but not limited to loss of life) or property that results from the use or misuse of the data or formulae presented herein.  It is the readers' responsibility absolutely to assess the suitability of a design or any part thereof for the intended purpose, and to take all necessary precautions to ensure the safety of himself/ herself and others.

The reader is warned that the secondary voltages present in nearly all power supplies for amplifiers are potentially lethal, and to observe all applicable laws, statutory requirements and other restrictions or requirements that may exist where you live.
 

WARNING:  All mains wiring should be performed by suitably qualified persons, and it may be an offence to perform such wiring unless so qualified.  Severe penalties may apply.

All power supplies must be fused or protected by an approved circuit breaker, and all mains wiring must be suitably insulated and protected against accidental contact to the specifications that apply in your country.

1. Voltage Regulator Handbook - National Semiconductor Corp. - 1981 Edition
2. Manufacturers Report - "The Importance of the Power Supply" - May, 1997  (By Nelson Pass)
3. Information supplied by Geoff Moss, my unofficial editor from the UK.  As always, thanks Geoff!