1970s Design Indulgence

Hi Sylvain, yes, you're right about cooling the output transistors, but it helps to understand the interaction between them and other hot items attached to and within the case.

The heat should be purely produced by the output transistors. The "thermal feedback," that is, the heat fed back to the Vbe multiplier responsible for setting the output stage current, works to control that exclusively.

However, the transformer also produces heat, and if allowed to warm the Vbe multiplier, it impedes its function.

The heatsinks get warmer than the amplifier alone should get. The Vbe multiplier is then correcting the output transistors for something they're not doing. Therefore, the optimum bias isn't optimum.

The image shows an attempt to decouple the warming effect of the transformer from the heatsink by admitting airflow between the two. A chimney or flue effect is set in motion because heat rises.

My prototype is at the point where I must now think about how I'm going to achieve the required thermal decoupling. One would expect that having my heatsinks on either end of the case would be sufficient. However, the transformer heat is coupled into the baseplate, coupled to the heatsinks, front, back, and top panels, and round it goes.

If I "waste" too much current, the case gets warmer because the "current wasting" power resistors also have to be cooled by being "heatsinked" to the case. If I waste less, then the case doesn't get as hot.

The interconnected reason for the current wasting resistors (or load) is to increase the conduction angle on the reservoir capacitor(s) to lengthen their lifespan. Rather than go over that subject again, I will cut to the quick by throwing-in another consideration.

If this were to be an integrated amplifier (hopefully), what powers the preamp stage?

At this moment in time, I am trying 30% of current waste, which is about 60mA. Would a two-channel preamp use that amount of current? If that preamp were to be made of discrete transistors, there is a possibility of using a proportion of, if not all, the 60mA. That is because the transistor preamp is always class-A (op-amps are class-B).

If you follow my reasoning - which I know is difficult - if one made a power amp from an integrated amp by merely removing its preamp, would the power amp perform the same?

In this investigation, I am daring to ask questions which have not been answered before, but I think they are evidence based. One gleans from the texts written over the years that similar problems exist, and the texts show the solutions employed.

There are also what I call "accidental" solutions, such as the "tree-trunk" layouts of early commercial transistor amplifiers, which prevented stereo loop distortion. While on that subject, we find the fixes, which if we study the Dinsdale stereo amplifier and his 10-ohm ground connection resistor, we can see the same in the Quad 303 schematic - 2R2 in that case. We also see the "most satisfactory" method adopted by the Decca Radio Laboratory, which, by now, most amplifiers employ.

From what I've been reading on the internet, it looks as if I'm a few slates short of a full roof (bungalow Graham perhaps)

For all my years, I've been getting it wrong. Ripple current is a function of transformer VA and nothing to do with the current drawn by the load.

However, ripple voltage is due to the current drawn by the load, given by Q (charge) = CV = IT, and if not, then a lot of engineers have greatly screwed-up over several years.

Current is drawn like water is drawn. The pressure might be there to fill a beaker instantly, but we control the flow by the tap. In electronics, the flow is determined by the load.

Then again, if the current capability isn't there, then the ripple cannot be exceeded - perhaps?

Relying on myself, and to a degree, Mr. Self, it looks like we could both be wrong. I thought we needed what the transformer/rectifier relationships say (for a capacitive input filter)?

If we need 2.3A for max (40W into 8 ohms) output power, then the AC current for an FWBR used to be 1.62x. That would make it 3.7 amps, and for both channels to do that simultaneously, 7.4 amps.

Wanting 72 volts, the AC is 0.7 x DC, and so the AC input needs to be 50V.

Therefore, the transformer should have a rating of 370VA (AC watts), and Mr. Self says I can get away with 70% of that, making 260VA.

Initially, I wanted 50WPC, and so chose a 300VA transformer. It seems we're wrong! It seems we only need a 100VA transformer for 50WPC!

That would limit the ripple current, I guess? Only 2.4 amps available before the voltage dips, or should that be only 1.5 amps due to the long-held belief of current division due to a capacitively input-filtered rectifier?

Imagine only 0.75 amps producing 50W into 8 ohms. But maybe we can use our imagination quite a lot here?

Looking at just one channel, if at full output, using say a 4700uF reservoir, 2.5 amps is required, then the ripple voltage is 5.3V. The cos-1 of that is 0.386, and the time constant of 50Hz is 0.00318, so the charge timing is 0.0012 seconds.

There being 8.8 seconds left, with no charge, can't we use it to inflate the current? What's 1+(0.01/0.0012)? Why it's 9.3!

So, if we only have 0.75 amps, we can make it 7 amps!

Perhaps that seems a bit pie in the sky, so let's take the average and call it 3.5 amps.

Or perhaps we should simply forget it and go back to being thick?

I have built the prototype using one 100VA transformer for each channel, and the most I can push each amplifier to is 40 watts. That is approximately 2.5 times the amplifier power, and not "A 50W amp, therefore, needs a 50VA transformer", as I just read.

When I first started building power amplifiers, it was more than 40 years ago, and I thought, "how can you get bass while the power's being chopped-up?"

And

"If the power is being delivered into 8 ohms, what resistance has the smoothing cap?"

I had this mad idea that 1/2pi.F.C might hold the answer because at 50Hz, using a 4700u cap, the answer is 0.67 ohms. I saw it as being in series with the output, but it isn't if we ignore Kirchhoff's laws - which means it is!

Students are always taught using the "magic battery," where the current is miraculously created at the pos terminal and then disappears into oblivion at the negative terminal. It is born and dies in just one circuit pass! Being DC, though, I guess it doesn't move, so where does the electromotive part of EMF come from?

A rectifier squirts current at a smoother to create charge. If that smoother weren't there, the current would flow into the load and look like a series of loops above a reference we could call zero. The squirts happen only because the smoother is there. If we drag more out, the squirts get longer and bigger. How can it be that if the smoother cannot take the squirt's size, it merely passes the smoother by?

Does it, therefore, squirt into a resistive load? How can it? It only squirts because there's a capacitive load!

Time also has something to do with it. The diodes only conduct when required by the current demanded by the load. They switch every 0.01 of a second, and they'll do the full 0.01 of a second period into a resistive load. A reactive load does what it says - it reacts, and therefore current flows in by squirty jerks.

To comply with energy conservation, the current is lost by the load, as it is turned into another form of energy. Power, which is volts times amps, is transformed into movement energy.

Therefore current is lost, and so the charge is lost, and so the voltage drops (because power is volts times amps), and each 0.01 of a second, some amount of "squirt" gets drawn from the rectifier, for as long as it is required to replace the charge.

We can apply formulas to behaviours through the joy of mathematical modeling, and these formulas are so rarely wrong. They can only ever be wrong if we don't input the correct data. This is the worrying bit.

If you're taking out five amps continuously, you have to put in 5 amps continually, but what happens if the input isn't available continually?

The rectifier will only conduct if there is a voltage drop greater than its own - isn't that how diodes work?

The voltage drop isn't detected until well into the charge cycle, at which time the rectifier conducts. Does it only put five amps in?

If it only puts five amps in for a proportion of the cycle, how can you get five amps out continuously? Answer: you cannot!

Charge has a time component, so if the charge is half of the time, then current is twice what's being drawn. My table, which I drew earlier, shows a lot more than five amps being squirted in to do around 200mA out. However, this is an angular thing, and so it doesn't grow linearly. To obtain 4.5 amps on full load, the current squirt needs only be 27 amps!

Ah but, I hear you say, "how can you get 27 amps from a 300VA transformer?" Good question! I find it funny that by the time things get so complicated, the chapter, or article, or white paper, has finished.

Now, I'm not much good at transformers, but aren't they an energy store too? They have inductance, don't they? Mind you, the transformer's measured inductance is only 45uH, with the "short circuit" of the mains' stiffness on the primary. Perhaps the transformer voltage doesn't go down as we exceed its current rating? Maybe it is so fast that it doesn't notice?

Then again, perhaps it's a bit of both. Does the transformer output 11 volts for 1/6th of a half cycle and offload voltage (greater than 50V) for the other 5/6ths? Damn it! I have been unable to measure this. What else could there be that could give up the struggle, such that the transformer still delivers 50V? Maybe we should ask Kirchhoff? Perhaps it is heat?

Actually, due to transformer regulation, the voltage does fall a little. Also, the capacitor cannot give up its charge so fast. And that is how a steady DC voltage at a given current, works. At maximum current, the transformer voltage is trying to fall. The ripple voltage is at the maximum allowable, but within the mathematics we know about, it works and can be, and has been, measured.

The other thing that is measured is the temperature. It rises until the metal box it is in reaches equilibrium, and that's where the lost energy goes, because work in Joules, which is related to charge in C (coulombs) represented by Q, because C here, stands for capacitance, is ... do we have to do this stuff?!

But what if the load current is transitory? That is to say, all over the place all the time. Large and small current, then small and large current, then small and small current, and large and large current. These changes happening rapidly at the speed of?

The capacitor charge current is the average taken over the period of a mains half-cycle for that half cycle, which differs from the next and to the previous. But, the current being drawn is changing more rapidly than that.

Can the capacitor offer the stiffness of supply as it loses charge to the load in less than 0.01 seconds before getting a squirt of current to top it up on the next 0.01 second period? Is it, therefore, passive between charge pulses? If it is, and the charge is current flow, then signal flows through it.

So, if the reservoir capacitor is C and the frequency F, then the series resistance must be 1/2pi.C.F, surely?

But when the rectifier conducts for a portion of 0.01S, there is no series resistance. And now for the matter of sub charge period frequencies. Well, "nearing DC," it is still going to be the same picture of squirting-in current, with perhaps one subtle difference being the transformer voltage possibly fluctuating a little.

Unfortunately, for the engineering, we have low bass frequencies and all the rest of the frequencies happening together (puls their harmonics, and mix of harmonics producing more harmonics).

The smoother's instantaneous leakage current might be sufficiently high at switch-on to muddy the picture, and the dissipation factor so poor as to make the changes slow right down, and all this might sound like music - for a while.

Eventually, however, the leakage abates. The heat produced inside the capacitor and conducted into the capacitor might change the separator water's ionic capabilities - like an overcharged car battery fizzes - such that a roving change in capacitance happens dynamically.

So, is that what happens between switch-on and 48 hours, or 72 hours, or 168 hours, seemingly depending on the model of the capacitor(s) used?

If so, then how can an amplifier sound the same over a three hour listening period - between switching it on, then off afterward?