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| Elliott Sound Products | Output Capacitor (Single-Supply) Power Amplifiers |
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In theory, capacitor-coupled output stages are completely straightforward, and there's no uncertainty about how they work. We all know that a capacitor passes AC and blocks DC, but with a single-supply power amplifier (or any other Class-AB single-supply circuit for that matter), current is only drawn from the power supply with positive half-cycles. When 'at rest' (no signal), the amplifier's DC output voltage sits at ½ the supply voltage. During positive half-cycles, current to the load is provided through the upper transistors (typically a Darlington pair). It passes through the capacitor to the load as we would expect.
However, things aren't quite so clear for negative half-cycles. We know that the lower transistors pass current, because we see a negative voltage across the load. However, there's no matching current drawn from the power supply. It's almost like magic, but the only reasonable explanation is that the current is delivered from the output capacitor. But - how does the capacitor charge and discharge when the current through the upper transistors, the output capacitor and load is identical? Surely the current should be greater to 're-charge' the capacitor after it's been partially discharged through the load and lower transistors.
This article came about after a number of emails back and forth with a well-regarded supplier of 'high-end' equipment. Not being one to reject a challenge, I decided to look into this, because it is not immediately apparent. While no-one gives this a second thought (or so it seems), it does require some explanation. It can be proven without too much difficulty, but it remains a little mysterious.
Audio amps require local decoupling to minimise interactions between the power supply wiring and the amp itself. Cables have inductance, and this can cause instability or increased distortion. These caps are shown in all of the drawings, and are assumed to be 4,700µF. In reality they may be more or less, and if the amplifier is located very close to the supply filter cap(s) the amount of decoupling needed is usually minimal.
A simplified version of the 'standard' single supply amplifier is shown below. The output capacitor is 1,000µF for convenience, and the load is 8Ω (resistive). I've used a 30V supply (equivalent to a ±15V dual supply). The performance of each is analysed. The power output is immaterial, as the same principles affect all single-supply amplifiers equally, with the only variable being the peak output voltage and current. The topology of the amplifier is not relevant, since everything 'interesting' happens in the output stage. In all descriptions I've assumed a Class-AB amplifier, and while the behaviour of the output cap is the same for Class-A, DC supply current always flows, so the supply current waveform is completely different.
For testing, I used the Project 217 low-power amplifier, as it's the only one I have that uses a single supply. Testing shows that without a doubt, there is some degree of infrasonic disturbance with an unregulated supply. However, it's only very low-level, showing a shift in the DC operating point of less than 20mV at the amp's input. The DC input filter has a -3dB frequency of less than 0.5Hz, but some of the power supply variations with programme material do get through the filter. The amp has unity gain at DC, so any DC disturbance at the input is not amplified, but simply buffered.
With a 1,000µF output cap, that has a -3dB frequency of 20Hz. This further reduces any infrasonic disturbances prior to the resistive load. A speaker is not resistive though, so at resonance the impedance may be 40Ω or more. Even so, 20mV of infrasonic energy will not cause significant cone movement. Indeed, it's likely to be negligible with even the most sensitive speaker.
Figure 1.1 - Schematic Of The Test Amplifier
In many ways it's no accident that many early single-supply amplifiers used a regulated supply. The regulator was pretty crude, but it served two purposes. It all but eliminated ripple which could be easily reduced to less than 20mV, and also kept the supply voltage reasonably stable as the load changed. This meant that the relatively poor power supply rejection ratio (PSRR) didn't cause hum and noise at the amp's output, and it all but eliminated the likelihood of infrasonic disturbance. The latter effect is almost certainly 'incidental', as I've never seen a reference to infrasonic disturbances for capacitor-coupled amplifiers.
Note that half of the AC feedback is taken from after the output capacitor (via R12). This connection has a very minor effect on the generation of infrasonic signals, but was a common trick in the days when single-supply amplifiers were common. Because the capacitor is inside a feedback loop, low frequency response is improved, and damping factor is somewhat better than a design that doesn't include the cap in the feedback loop. However, most of the time there will be little audible difference one way or the other.
In Fig 1, it's assumed that the supply voltage will be unregulated. Almost all tests I carried out on the design used an unregulated supply, and the DC voltage must fall when current is drawn. The amount of voltage drop depends on the size of the transformer and filter capacitor, and the signal amplitude and load impedance. Normally, we can expect the voltage to fall by at least 10% at full power, but if the transformer is only just big enough (around 20VA for example) you'll lose somewhat deal more when the amp is driven hard. This could see the average DC voltage fall from a nominal 30V to perhaps 26-28V under load. This voltage variation will affect the bias point, as it's derived from a voltage divider (R1, R2 and R11).
Figure 2.1 - Infrasonic Disturbances Caused By Supply Voltage Variation
The above graph shows just what I'm talking about. From 'low power' (176mW) to 'high power' (8.56W), the average supply voltage fell by just over 1V. While it's doubtful that the disturbances seen would be audible on most systems, the possibility cannot be discounted. You would need a very revealing set of speakers and an excellent listening environment to hear anything, certainly far better than the speakers I have in my workshop. The peak-peak amplitude of the disturbance is just under 800mV, so it's not going to cause large speaker cone excursions. A power supply with worse regulation will make matters worse of course. The effect can be reduced by increasing the value of C6, which filters the bias voltage, but it can't be eliminated without using a regulator to supply bias.
A tone-burst is a brutal test for capacitor-coupled amplifiers, and fortunately, music is far less demanding. This does not mean that there are no disturbances, but they will generally be comparatively subdued. Very simple amplifiers with only one gain stage (such as the El Cheapo [Project 12A]) may be expected to be affected more than the example used here, although a simulation showed (surprisingly) less effect. Be aware that the output capacitor itself removes at least some of the disturbance, because it's a high-pass filter.
The infrasonic effects seen above are all but eliminated if the supply is regulated. However, this adds extra parts and means a bigger heatsink due to the power dissipated by the regulator. These results can be duplicated easily, either using the test amp described above, or any commercial amp from before ca. 1975. Most of these early designs used an output capacitor, and several used a simple regulated supply. You can see the advancements in power amp designs in the article Power Amp Development Over The Years.
You can also regulate the bias supply. In the case of the Figure 1 amp, a fixed voltage of +25V applied in place of C6 will do just that, assuming a 30V supply. This reduces the amount of disturbance, but it doesn't eliminate it. This is because the remainder of the amplifier still has a supply voltage that varies with load, and that changes the operating conditions. Almost without exception, modern power amps use a dual supply, and the reference is the amplifier's ground connection. This doesn't move around, and infrasonic disturbances are almost unheard of. This is covered in detail below.
This is something that you'll be hard-pressed to find any information about. I suspect that the likely search terms are partly to blame, because the major search-engines will prioritise other material that seems to fit the criteria. Enclosing 'suitable' searches in quotes doesn't appear to be very helpful, because there are thousands of pages that refer to capacitor coupling, but none that I found that describe the process in detail. It's possible that there may be something behind a 'paywall', but it's a risky business to pay for an article based only on a short excerpt. I consider this to be an abuse of the spirit of the internet.
The capacitor acquires a charge when the amp's output is positive (referenced to the quiescent voltage of 15V), equal to I × t (time in seconds) coulombs. By definition, if a current of 1A flows for 1 second, the charge is 1C. The charge with 1A for 0.5ms (e.g. a 1kHz squarewave) is 0.5mC. When the amplifier's output is below the quiescent voltage, this charge is reversed, and will provide (for example) 1A for 0.5ms, leaving the net charge across the output capacitor the same as it was after the amplifier stabilised after power-on. The quiescent charge for CC is about 15mC, obtained during power-on. A 1,000µF (1mF) cap with 15V across it has a charge (Q) of ...
Q = C × V
Q = 1m × 15 = 15mC (milli coulombs)
This initial charge is reached in (for example) 150ms with a constant current of 100mA. In reality, the charge curve is less well defined because there's a series resistance (the loudspeaker) and an uncontrolled charge current. For the case with a signal present, we can look at a 1kHz (1ms period) sinewave. We need to include the sinewave average constant of 0.637 to obtain the average current over time. We'll assume a peak output of 8V and an 8Ω load (1A peak). With a sinewave, the output cap will gain a charge (Q) of ...
Q = I × t
Q = 1A × 0.637 × 0.5ms = 0.3185mC = 318.5µC
The charge acquired/ released is obviously greater at lower frequencies and smaller at higher frequencies. On the negative half-cycle, this charge becomes a discharge. The charge on the capacitor increases and decreases by about ±0.5mC with a squarewave. Provided the charge/ discharge cycle is small compared to the total stored charge in the output capacitor, the frequency response is relatively unaffected. Using the same capacitor and load, the -3dB frequency is close enough to 20Hz, and the on/ off periods at that frequency are each 25ms. Under these conditions, the capacitor gains/ loses 14.6mC for each cycle, almost the total stored charge. This isn't easily calculated because the current waveform is differentiated due to the capacitor and load creating a high-pass filter. When Xc (capacitive reactance) is equal to the load impedance, the output level is reduced by 3dB. For a sinewave, we use the average value, which is 0.637 ...
Iavg = ( 1 / π × 2 )
Iavg = 0.636.62 ( 0.637 )
The capacitor gains its initial (quiescent) charge during power-up. The charge time is determined by the risetime of the bias network, the size of the output capacitor and the load impedance. If the amp's output voltage jumped to Vq (the quiescent output voltage) of 15V instantly, the initial current would be 1.875A for a 30V supply, tapering off to zero when the cap is fully charged. To measure the stored charge, you have to use the average current and the time period from power-on to where the charge current falls to (almost) zero. Again, this is not easily calculated, but it can be simulated easily enough. Alternately, just use the simple formula shown above.
Although no-one ever thinks about it, the exact same process applies with all capacitor-coupled circuits, from preamps (valve or transistor) to power amps.
Figure 3.1 - Voltage & Current For Symmetrical ±8V Output
In the drawing, I've shown a symmetrical ±8V sinewave output from the amplifier. For the positive half-cycle, current is drawn from the supply, controlled by Q1, through the capacitor (CC and then through the load to the ground return. As this is a series circuit, the current is identical at any point of the loop. For a negative half-cycle, current is drawn from the capacitor, controlled by the lower transistor (Q2), and passed through the load. Again, it's a series circuit with identical current at all points in the loop. The average level of a half-sinewave is 0.637, so the charge on CC increases by 318.5µC for the positive half-cycle, and releases 318.5µC for the negative half-cycle.
In each case, 8V must cause a peak current of 1A. There is a small voltage 'lost' across the capacitor due to ESR and capacitive reactance. With a 1kHz signal, it should be about ±100mV, partly due to the reactance of the cap itself (159mΩ at 1kHz) plus a small loss due to the cap's ESR (equivalent series resistance). ESR should be less than 100mΩ (0.1Ω). These losses are ignored in the following calculations because they have little effect on the outcome.
So, during the 'charge' period with a 1kHz sinewave (amp output 8V greater than 15V), the capacitor accumulates a 318.5µC charge described above. For negative outputs (15V - 8V), the cap loses 318.5µC of charge. Equilibrium is established quickly. If there were no state of equilibrium, the capacitor could charge or discharge in one direction until it reached the supply voltage or zero, but this doesn't happen over the long term. The small periods where equilibrium is not maintained perfectly represent the infrasonic disturbances seen in Figure 2.1.
The situation is more complex when a music signal is used, as there are always periods of asymmetry, and music is dynamic. This means that the DC voltage across the capacitor will change, but most of the asymmetry has been eliminated thanks to the input capacitor. This goes through the same process as the output cap, but of course the voltages, currents and amount of charge are all a great deal smaller. Any asymmetrical waveform will cause a DC shift, but most of it is removed by the capacitors throughout the circuit. Asymmetry can be re-created if transients (in particular) are allowed to clip. The clipping will often be inaudible due to the short duration, but the asymmetry created is very real. Capacitively-coupled asymmetrical signals can create a DC offset under some conditions, but a lab experiment and real-life are different.
Note: Fully DC coupled amplifiers might seem like a good idea, but consider the fact that any DC offset will cause speaker cones to shift relative to their rest position. This can cause distortion because the voicecoil is no longer centred within the magnetic circuit. You have a choice - either allow all asymmetrical signals to pass through the amp to the speaker (including any DC component), or use one (or more) capacitors to remove the DC component. If you choose the latter, there will be some infrasonic disturbance, but it's a great deal less than the effective DC component. Everything you listen to has passed through multiple capacitors, so the idea of eliminating 'evil' capacitors is just silly and isn't worth discussion.
It should be obvious from the above that load power is drawn from the supply only during positive (greater than Vq) signal excursions. As there is no negative supply, the negative portion of the output waveform is derived from the charge stored in the output capacitor. For a perfectly symmetrical signal, the two balance out, leaving the net charge on Cc (the output capacitor) unchanged. At first glance it may seem that we are getting something for nothing, as the negative half-cycle is 'free'. Naturally, nothing of the sort happens.
When you look at the current distribution in a single-ended (capacitor coupled) amplifier, it's apparent that current is drawn from the power supply only during positive-going signals, when the output voltage is greater than the quiescent state. That's +15V for the example here, but it can be up to +35V with a +70V supply. You might imagine that this means that the negative-going signals get 'free' power, because it's supplied by the output capacitor. Getting something for nothing is frowned upon by the laws of physics (and the Taxman), so we have to assume that there is no 'free' power involved.
The easiest way to demonstrate the power used is to examine both input and output power. The current drawn by the remainder of the amp is ignored. Using the same waveforms as shown in Figure 3, we can examine the input power, delivered from the power supply. The single supply is 30V, and the average output power is 4W (8V peak is 5.66V RMS). The input current averages 364mA, so with a 30V supply the input power is 10.92W. It's immediately apparent that we don't get that free lunch after all - the input power is 2.72 times greater than the output power with the conditions described.
With a dual supply amplifier (±15V) it's obvious that the speaker current and therefore the supply current for each half-cycle must be equal. Each part is a series circuit, so if 1A peak flows from the supply to the speaker via the transistor, the current in each part of the circuit has to be identical. With each half-cycle, the peak current is again 1A, and the average is also 346mA. With half the voltage, the power delivered from each supply (one positive, one negative) is 5.46W, exactly half that of the single 30V supply. Because there are two supplies, the total is 10.92W.
The laws of physics are satisfied, and the input power is identical for single-supply and dual-supply amplifiers under the same conditions - total supply voltage, signal amplitude and load impedance. It's somewhat counter-intuitive at first, but examination of input vs. output power is by far the easiest way to work out what happens. You can also measure the mains current, but if you were to do that the circuits must be identical other than the power supply configuration. If you're unwilling to build the amps and supplies to take measurements, the results can be simulated.
A dual supply amplifier uses ground as its reference, with a positive and a negative power supply. So I could use the same amplifier (both for display here and for simulations), a -2.5V bias was used at the input to obtain zero voltage at the output with no signal. Otherwise there's no difference in the circuit, other than changing from a single +30V supply to a ±15V supply.
Figure 5.1 - Dual Supply Test Amplifier
This arrangement doesn't require a great deal of comment, as the dual supply is the defacto standard today. This doesn't mean that capacitor coupling is not used though, and there are a surprisingly large number of amplifiers that still use an output capacitor. These are primarily low-power designs, and they are used in many consumer products because they are cheaper to build than a dual supply.
Figure 5.2 - Voltage & Current For Symmetrical ±8V Output
The current paths are also exactly what you'd expect. Positive output current flows from the positive supply, through Q1, the load and back to the power supply common (ground). Negative half-cycles are provided from the negative supply, through Q2 and the load back to the supply's common. This is all very easy to follow. The load current is controlled by the transistors, which are within a feedback loop to ensure that the output signal is an accurate (but larger) image of the input signal.
A point that's generally missed is that the power supply filter capacitors form part of the audio circuit, both for single and dual supplies. The supply doesn't exist in some fugue state, divorced from the 'real world' and acting as a separate entity with no association with the amplifier. The filter capacitors supply the current for positive transitions (single supply) or both positive and negative half-cycles (dual supply), with the job of the transformer and rectifier being only to maintain the required voltages at the current being drawn. I expect that this may not 'sit well' with some people who claim to abhor capacitors in the audio path, but it should be obvious that they are there whether you like it or not.
There will always be a (small) voltage dropped across the output capacitor. The voltage difference you can measure easily is due to the ESR of the capacitor which is in-phase but almost always slightly non-linear! This is the reason that output capacitors nearly always cause increased distortion, particularly at low frequencies where the reactance is greater, and more voltage is developed across the capacitor. This gives us an additional voltage component across the capacitor that's harder to measure, due to the reactance of the capacitor. This varies with frequency, and is 90° out of phase. It's this voltage component that's created as the capacitor gains or loses the charge (in coulombs). Capacitive reactance and the charge are directly related, so as the reactance is reduced (e.g. with increasing frequency) so too is the stored charge (and vice versa of course).
You'll see many capacitor coupled amplifiers (including the one shown here that I used for testing) that derive at least part of their negative feedback signal from after the output capacitor. This helps to minimise distortion created by the capacitor. The other method is to use a capacitor with a higher value, as this reduces both ESR and capacitive reactance. The 1,000µF cap shown is actually too small for very good performance.
In the interests of completeness, I've included the conversion factors so coulombs (charge) can be converted to joules (energy), along with other useful conversions. While these aren't necessary to understand the processes involved with capacitor coupled amplifiers (or other applications using capacitive coupling), they may come in handy some day.
Energy in capacitor = Q × V / 2
Q² / 2 × C
C × V² / 2 (also written as ½ × C × V²)
Where the energy is in joules, Q is the charge in coulombs, V is the voltage in volts, and C is the capacitance in farads.
I suggest that if you intend to work a lot with capacitors (something we can't escape with electronics), you should make a note of these formulae. You won't need them for most activities, but there will come a time when you'll want to know, either for interest's sake or because not knowing will leave you in the dark as to what happens within a circuit.
All of the results shown here were simulated, but a bench-test using the P217 amplifier was also performed. This was done both with music and a tone-burst, and the infrasonic disturbance was visible, but not very pronounced. This was because the unregulated supply I used has better regulation than expected (at least at the modest power drawn by the test amplifier). In the majority of cases, any infrasonic disturbance will be quite small, and audibility (or otherwise) isn't something I'm willing to comment upon. The effects are real and easily simulated, but are probably less easily measured.
Nevertheless, the way a coupling capacitor works in a circuit isn't something I've seen described anywhere else. Mostly, we just know it works because we can see it in operation. We know that both power transistors dissipate much the same power, and probably don't give a great deal of thought to the processes involved. As it turns out, there is more to it than we imagined, particularly the charge existing on the coupling capacitor itself.
Most people don't worry about the charge (in coulombs) lurking on a capacitor, or it being increased and decreased with the signal. Indeed, it's not something that I've discussed other than in passing for any of the many articles on the ESP site. Mostly, it's pretty much irrelevant to the majority of audio circuits, and although the principles explained here apply for all coupling capacitors, it's never necessary to go into any detail.
There are no references for the specific topic, but the formula for Coulombs was obtained from Wikipedia and verified elsewhere. The specifics of this topic seem to have escaped attention over the years, largely I suspect because not many people actually care, as long as it works.
One link you may find useful is Capacitor Charging Equation (Hyperphysics), along with Lumen Learning (Energy Stored in Capacitors)
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