Laughing even louder at the "engineers", thanks Old Hack Re: He who laugh last, laughs loudest, was Re: Wet your pants funny-- A.J. as an EE

Good golly, Old Hack, I am absolutely mortified to discover your list
of errors of which I read almost half. Especially as we have on RAT two
"engineers" (they say they are anyway) called Stewart Pinkerton aka
Psycho and Graham Stevenson aka Poopie whose express purpose in life is
to ride my ass every time I make an electrical error. How could they
have missed all those gross, dangerous, destructive errors for a whole
year now?

I think we'd better give the "engineers" Psycho Pinkerton and Poopie
Stevenson an opportunity to explain that your collation is a list of
out-of-context nits and and malicious mispresentations of what I said
and outright lies and wishful thinking and ignorance of the gaincard
movement, all the quicker to salvage what shreds remain of their
reputations.

And perhaps you would care to explain the miracle, in the presence of
so much incompetent supervision by those two clowns PinkoStinko and the
Poopster, of the amp actually working rather well -- and precisely as
intended.

ROTHF. Thanks, Old Hack, you've made my day. Who are you again?

Andre Jute
Visit Jute on Amps at http://members.lycos.co.uk/fiultra/
"wonderfully well written and reasoned information
for the tube audio constructor"
John Broskie TubeCAD & GlassWare
"an unbelievably comprehensive web site
containing vital gems of wisdom"
Stuart Perry Hi-Fi News & Record Review


Ancient_Hacker wrote:
Okay, here's a short list of the funnies in the web pages:

Good GOing, AJ, you've set up the rules so you win either way!

But here's a few quibbles:

I picked a 50K pot so that there can always be at least 10K to ground, the minimum load my Quad CD player expects to see.

Huh? Where does he get this 10K number? Your typical op-amp has many
megohms of input resistance, so there's no way to get 10K from a 50K
pot and many megohms.


Most opamps require voltage gain of at least 10 to work at all.

Um, no. Op-amps get a bit more sensitive to load impedance at very low
and very high voltage gains. But they tend to work perfectly at any
gain, especially if you follow the compensation recommended in the app
notes.

An opamp isn't a linear device like a tube.

??? tubes are not linear at all, they follow a 3/2 power rule.

Op amps are about 99.5% of the time used with negative feedback, making
them very linear indeed.

An opamp requires a reasonably stiff supply

Um, if you look at the LM675 spec *** it shows 90db of power supply
rejection. That's a whole BILLION. If the power supply voltage varies
a whole volt, the effect on the audio output will be sqrt(billion), or
about a millivolt out of 30 volts. Miniscule, so stiffness is the least
thing needed.

... even in class C.

One never operates an op-amp to process audio in class C. Class C is
only used in RF amplifiers, never in audio.

The 675 is rated 60V peak to peak, 3A, 30W.

Huh? It's rated at 60V DC input, 3A max DC input. I don't see how one
can get 30W out of that.
Even if you use misuse the wattage calculation rules, and think you're
going to get 60 volts peak-to-peak out, which is about 20 volts RMS
into 8 ohms, that's 50 watts. And you can't multiply 60 by 3 to get
180 watts, as the voltage and amp limits are individual limits, not
possible to use both at once! That would be 180 watts, and the
amplifier chip is only good for about 30 to 40 watts dissipation, even
with an infinite heat sink.

36Vct 120VA is overkill for me because I own horns among several pairs of sensitive speakers, so that a handful of watts is plenty for me.

120 watts is about 3 or 4 times more power than the IC can ever handle.

But everyone else should use at least a 40Vct 150VA power transformer. And if you intend to use Class A with insensitive speakers, 300VA will look right handy.

That's even more ridiculous.

If you are obsessive you can get a surface mount 12K resistor and solder it right across the opamp pins. It makes very little difference because in any event even a semi-'fast' power supply will be filthy.

The placement of this resistor has nothing to do with the quality of
the sound. And as we've explained above, the quality of the power
supply is a miniscule factor, about one part per billion. The color of
the paint on the case makes a bigger difference, as it affects the chip
temperature by several degrees C.

The 675 is a push-pull amp operating on the verge of Class C all the time.

As we noted before, not true. Not even remotely possible. Class C
generates huge amounts of distortion. The only reason Class C is used
is in narrow-bandwidth RF amplifiers, where the terrible harmonics can
be easily filtered away.

For the tube-hobbyists still with us, a transistor (of which an opamp is a unified collation) has a transfer function precisely like a tube but with much uglier curves.

How can something be precisely like something else, but much uglier?
And a unified collection of transistors acts nothing like an individual
transistor.

The quiescent operating point of a transistor is Iq and it is found as the square root of the theoretical power divided by twice the load.

Hmm, well under quiescent conditions, there can be no power flowing to
the load, so the theoretical output power has nothing to do with it.
Plus the theoretical power is AC and reactive, and quiescent means DC
conditions, no reactance. So this analysis is so far off, it bends the
needle on the bogo-ometer.

The theoretical highest signal is the available input times the voltage gain or 22V, which is also the voltage we expect from the power supply.

Um no, if you look at the curves for the LM675, specifically the one
"Output voltage swing vs supply voltage", you'll see this op amp can't
reach the supply limits-- it poops out anywhere from two to five volts
from the extremes.

, so the bias resistor must be 16 ohm and it will dissipate 30W.

Right!. But a little thought will reveal that the op-amp must be
dissipating the same wattage. But 30 watts is the LM675's absolute
maximum rating. So this added resistor has used up all the IC's power
dissipating capacity, leaving nothing for our signal.

One watt into eight ohms is 2.83V and the required bias is 0A35 or 350mA.

Again with the completely wrong calculation. That's the RMS voltage,
whilst the bias has to handle the signal peaks. And at the signal
peaks, the resistor has to supply the current while it only has 22V
minus the peak signal voltage. Not to mention that most speakers are
not pure resistances, so it's going to take considerably extra current
to feed the reactances under signal conditions. So the math is all
foobared.

Notice a missing safety device on the circuit. I haven't put a bleeder across the power supply output but you may wish to consider that 10000uF caps can give you a nasty jolt

The voltages are only 25 volts on either side of ground. Most people
can't feel 25 volts, much less get any variety of nasty jolts.

.... or melt the tips of your pliers and screwdrivers.

10,000uF at 25 volts is only 0.25 of a joule. That might make a minor
pock mark, but it's totally unable to melt anything.

And if you look at the picture, you see AJ's charming idea of
insulation. On the AC leads he hasnt bothered to use wire nuts, or
crimp connectors, or even heat-shrink tubing. He's peeled the outer
PVC jacket off some telephone wire. Never mind that telecom wire is
only rated to 24 volts, room temperature only. One really should use
something less prone to arcing.


The transient peaks 10dB or 20dB higher can 'afford' much more distortion because they are so much louder. Our ear makes its judgment in the first 90dB where our one watt of Class A suffices.

If this was true .... aww what the heck, no need to explain, it's
obviously bogus to everyone.

The resistor we have so far used as a constant current source is poor at its job. You can improve it by splitting the resistance in two with bypass caps from the junction, which splits the load on the opamp into AC and DC components, one part of which also filters high frequency noise on the power rail.

A resistor is anything but a constant-current source-- the current is
directly and irrevocably proportional to the voltage across it.
Putting a capacitor at the junction doesnt help, quite the opposite--
it makes it exactly TWICE as poor a constant-current source, as you
then have twice the voltage swing as before across the second resistor,
therefore twice the current swing..

If you divide 50 by the temperature rise of a device in degrees Celsius per Watt and square the result,
you arrive at the required radiating area of the heatsink in square
inches.

Er, no. The temperature rise has nothing to do with the device, it's
the temperature rise of the heat sink above ambient.
A very different parameter.


So there you have it, just what you'd expect from someone with an
interest in the field, but no real idea what is going on.

.