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1970s Design Indulgence

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Post Options Post Options   Thanks (0) Thanks(0)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 27 Nov 2020 at 8:39pm
1 watt per channel 20kHz THD, showing 2nd, 3rd, 4th, 5th and 6th harmonics, as well as high frequency noise with spurious glitch noises. All noise -90dBu and below.

1 WPC 20kHz THD FFT



40 watts per channel 1kHz THD, showing falling high harmonics. The prominent "sidebands" indicate the near onset of clipping, where the FFT takes on a hedgehog shape, aka massive distortion.

40 WPC 1kHz THD



10 watts per channel 1kHz THD. This is the quarter-power amplifier torture, as it stress tests the output stage. Here it tells the tale of the less than perfect layout due to the several hack and bridge modification operations. Still, at 0.033%, it isn't bad!

Quarter-power 1kHz THD stress test



All tests use 20Hz high pass and 300kHz low pass filters.

A note regarding the words "it isn't bad" above: this depends on what is regarded acceptable, and for this retro 6 transistor configuration, it is acceptable. It would not be acceptable to Mr Doug Self!
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 28 Nov 2020 at 11:05am
I am not going to criticise anybody for not challenging me about the frequency equivalents of the time periods. Instead, I will chastise myself by pointing out my error.

We have been discussing conduction angle in the form of time t in relation to the rectified mains waveform.

By simply inverting t, I gave the equivalent frequency, now obviously, in a rectified system, that is wrong (but also right as we will come to).

A t of 100uS inverted gives the number 10,000. That being a positive pulse, when we add its negative pulse to produce a full cycle, it must be 5,000 cycles per second. Therefore, when I stated 5800Hz, it should have been 2900Hz.

Now, a capacitor might be ripple current rated for 100Hz, but that ripple current is from a rectified 50Hz. Surely the rating should be for 50Hz?

And therefore, a rating at 20kHz should mean a rectified 10kHz. Therefore, I have just stated that it was right too.

Capacitor ripple current ratings are useless to a degree. Their only use is where the ripple voltage is close or equal to the HT voltage, where the conduction angle is close to half the frequency being rectified (do you follow?).

Can we therefore establish a smoother capacitor's lifetime? I fear the answer is no.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Sylvain Quote  Post ReplyReply Direct Link To This Post Posted: 28 Nov 2020 at 5:41pm
A note regarding the words "it isn't bad" above: this depends on what is regarded acceptable, and for this retro 6 transistor configuration, it is acceptable. It would not be acceptable to Mr Doug Self!

I shall try understand the Graph and your description at teh various OUTPUT something you say in your notes reminds me of What I read this week about Astra Zeneca Oxford Vaccinne  BUT I shall reserve judgement.

Sylvain Jacques 
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 28 Nov 2020 at 6:34pm
Assuming we now have a "perfect" amplifier through the DC being good at all signal frequencies, we discover what should be obvious - that the whole thing gets hot!

After reading this, you might think I'm making it all up to sell more Proprius amps, but I'm afraid to say I'm not at all that clever.

Amplifier thermals need to be carefully considered. Quiescent conditions are established to obtain the amplifier's best performance, not for the amplifier's quiescent conditions to suit the amplifier's temperature.

If we put a heater of some sort in the case with the amplifier, the case will become hotter than if it only contained the amplifier. The amplifier will then run hotter. Because it is a bipolar junction transistor amplifier, its output stage draws more current due to thermal runaway behaviour.

Even if the heatsinks are large enough to survive apocalyptic thermal runaway, for the case/heatsink assembly to operate at a safe temperature, the temperature rise must be countered.

Ambient temperatures of operation are a nightmare of sorts for the designer to try and workaround, but what if we put a heater in the case as well?

The amplifier uses thermal "feedback," in which a heatsink mounted sensor opposes the increasing current, which the positive temperature coefficient of the output stage gives rise to.

This is supposed to ensure the optimal quiescent setting, which dictates the low volume distortion is maintained so that performance doesn't differ.

Suppose we install a heater that doesn't have any thermostatic control. In that case, the amplifier's thermal see-saw will need to cater for that as well.

Perhaps you cannot think of an occasion of putting a heater inside the amplifier case, here is a suggestion: its power supply!

The transformer is made of iron (or particular steel to be more accurate), and iron has low thermal conductivity. At switch-on, the transformer is cold (at ambient room temperature), but it produces heat with work.

The heat eventually spreads out to all parts of the case, including the amplifier output stage, mounted on the heatsink(s).

Now, what is the optimum quiescent setting of output stage bias? Should it be set at switch-on, or should it be set once the entire case has reached its thermal equilibrium?

But what if the case cannot take the temperature rise of both the heater and the output stage? The amplifier quiescent bias must then be fudged to assist the "equilibrium."

If we also added another heater to bleed-off current to widen the power supply reservoir capacitor conduction angle, we have worsened that problem.

Perhaps we must consider an outboard power supply (as per the Proprius). Otherwise, we must consider a thing called ventilation?

Below is an image of a correctly ventilated case (I assume this amplifier is a Self design). It can be seen that the chassis tray is formed from a poor heat conductor, steel, but is of a thin gauge, which helps. You can see the ventilation slots which decouple the "heater" (the transformer) from the thermally "fed back" power amplifier heatsink. What isn't shown is the raised "dimple" beneath the transformer, which has become such a feature to aid ventilation and convenient mounting, that large enclosure manufacturers keep a range of forming dies for the purpose.

case ventilation

Here, the do-it-yourselfer is disadvantaged, as the amount of casework required becomes prohibitive.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 28 Nov 2020 at 9:02pm
What might be fun is to research the idling current of a design to find how hot the amplifier runs and then view the finished product to see how it copes.

Being the owner of a copy of Audio Power Amplifier Design by the venerable Self, who makes no bones about designing for Mr. Richer's Cambridge Audio, I am only passing on information publicly available.

Doug uses 0.1-ohm emitter resistors and tells us the voltage across them should be set to 25mV. Ohms law says the current is I = V/R, 0.025/0.1 = 0.25 amps. That is a lot more quiescent current than an early transistor amplifier, yet it is not high bias. It is on the edge of what might be called gm-doubling. It is class-B with a larger dose of A than most B's.

We also know how Doug treats his TO-92 transistors - he "likes" them hot - they are run at 10mA. Many of his designs use a 200-ohm charge sweeping resistor, and then there is the IPS supply current that might be another 5mA.

Output, driver, and voltage stages thus consume around 275mA per amplifier or 550mA per stereo pair.

So as you know, I am not breaking any copyright here; the same information can be gleaned from back issues of Wireless World.

The total of the two HT rails might be 90 volts for a 100 WPC design, and so standing power is 90 x 0.55 = 49.5 watts. Yes, I find that hard to believe too. But I can only go on the figures I have seen.

Let us now translate that into VA. 90 is derived from root 2 x Vac, and so 90 comes from 63Vac, ignoring diode losses. 0.55 entirely bridge rectified would be the multiple of the Fibonacci number or 0.9A, which is nearly 60VA. Further reading might show the uses of a 300VA transformer, and so 20% of the available power is used for idling.

It can safely be assumed that things run hot. It can also be seen by examination of internet images how the heat gets dissipated. The technique is that of flue technology, which uses heat to shift heat even faster. As I might be in trouble for pointing out the obvious, I will let you search the internet images. The transformer sits amid heatsink fins above a hole or slot matrix, which admits air, and the lid is equally pierced.

But why the extreme current? Self argues against voltage regulation for power amplifiers. Still, the pre-loading of a goodly portion of current has some relationship with a power zener shunt.

But, look back at my argument for increasing the reservoir capacitor's conduction angle. Draw a healthy current, and the conduction angle increases. Other designs might resort to voltage regulation, which counters any "damage" caused to the smoothers.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 29 Nov 2020 at 11:06am
To look at another design where all the information is available, I am studying Bob Cordell's Designing Audio Power Amplifiers book. He, too, argues for the 25mV emitter resistor voltage, which is the maximum allowable before gm-doubling. I note what seems to be a preference for 0.22-ohm emitter resistors - this sets an output stage current of 114mA.

The rest of the amplifier consumes around 20mA for the drivers, 10mA for the VAS, and let's guess 5mA for the IPS.

The total current is 298mA per stereo pair, and for the sake of roundness, we can say 300mA.

The Cordell power supply illustrations include bleeder resistors across reservoir capacitors, value 2k ohms, and on a voltage rail of 55 volts, consume 27.5mA.

However, when it comes to working-out the conduction angle, we are met with a surprise in that the value of capacitance he uses is huge! 50,000uF or 2 x 25,000uF are the values illustrated.

Running this through V ripple = I T / C , we find for 325mA and on 60Hz USA mains, 0.325 x 0.0083 / 0.05 = 0.054 V, or just 54mV ripple voltage.

To obtain the conduction angle time, we again use cos-1 (HT - V ripple / HT) x 60Hz time constant, which is 0.0443 x 0.00265 = 0.000117, or 117 uS.

Straight away, the opposite direction can be seen. The charge current pulses are of shorter duration than Self's and are half that of my design (excluding the bleed resistors). 117uS per 0.0083S is 1.4% of the time, where on 50Hz mains, I had 200uS per 0.01S, which is 2% of the time.

The current drawn on idle is 325mA, which means a 23 amp ripple spike. On my design, the current drawn was 140mA, which meant a 7 amp ripple spike.

I can only conclude that with 50,000uF capacitance, the ripple current might match that available in a capacitor's specification. The use of two Kemet 22,000uF ALF20 capacitors in parallel indicates 22.68 amps availability at 10kHz, with a minimum lifetime of 18,000 hours.

To date, I have not found any designer pay much attention to this phenomenon. Therefore, I can only deduce that Self and Cordell work on an "it's always been like that" basis, "so why change it?"

I have only found one designer who pays attention to the conduction angle, that being Morgan Jones.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 29 Nov 2020 at 7:42pm

 

Self

Cordell

Slee 30% bleed-off

Slee 60% bleed-off

HT smoother assumed

5250uF

25500uF

6000uF

6000uF

Idling current II

550mA

325mA

195mA

350mA

HT (+ to -)

90

90

72

72

Rectifier frequency

100Hz

120Hz

100Hz

100Hz

Rectifier period tr

0.01S

0.0083S

0.01S

0.01S

Vripple = IT/C

1.05V

0.106V

0.325V

0.583V

Supply tc = 1/ω

0.00318S

0.00265S

0.00318S

0.00318S

t = tc x cos-1 (HT-Vripple/HT)

0.486mS

0.129mS

0.302mS

0.405mS

Conduction = 100(t/tr)

4.86%

1.55%

3.02%

4.05%

Idle Iripple = 100/% x II

11.31A

20.97A

6.46A

8.64A

Fripple = 1/t

2057Hz

7751Hz

3311Hz

2469Hz

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