Discussing audio amplifier design -- BJT, discrete

[John Larkin]

On Sun, 7 Feb 2010 16:10:14 +0530, "pimpom" <pimpom@invalid.invalid>
wrote:

Jon Kirwan wrote:
On Sat, 06 Feb 2010 10:48:16 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


A few percent distortion at power levels is essentially
inaudible.
Speakers do that already. Low-level crossover distortion is
obvious
and obnoxious.

Yes, I think that's now much clearer to me now than it was
say two weeks ago -- without needing my ears to say so. Just
on understanding better _what_ crossover distortion is and
does.

Jon, to see a graphical illustration of JL's point, see this
screenshot of some simple simulations I just ran:
http://img694.imageshack.us/img694/8967/crossoverdistortion.png

On the right, the complementary output stage is driven without
any bias, . The upper trace shows the output when the input
amplitude is +/-1V peak. The transistors are operating in Class C
and manage to conduct for less than half of each half cycle. Now
that's going to sound awful by any standard. I _know_ it sounds
awful because, when I was doing a lot of repairing work on
consumer products in the 80s, I came across some amps whose
biasing circuit had developed a fault.

Do a Fourier analysis and you'll get lots of harmonics. Reduce
the input amplitude even further and there won't be any output at
all below a certain amplitude.

The lower trace shows the output with +/-9V input. Crossover
distortion is much reduced, though still evident. This may or may
not be acceptable depending on the application. For anything that
needs good audio quality, any waveform distortion that can be
clearly seen in graphical form is still too high, especially in a
low-resolution bitmap trace like this.

On the left, we have the same amp with diode biasing added.
Visible distortion of the waveshape has disappeared. The slight
irregularities in the sinusoidal shape are due to limitations of
the low-res, non-antialiased bitmap image.

In your "unbiased" circuit, try adding a 1K resistor from the bases to
the output. Now the flat spots in the output waveform become slopes.
But the transfer function is now continuous, so negative feedback can
reduce distortion without ugly slewing problems. And with zero bias,
there's no idle power dissipation and no possibility of thermal
runaway.

The second circuit, with the bias diodes, is a likely firebomb.

John

 

[John Larkin]

On Sun, 07 Feb 2010 13:29:07 GMT, N0Spam@daqarta.com (Bob Masta)
wrote:

On Fri, 05 Feb 2010 11:32:16 -0800, Jon Kirwan
jonk@infinitefactors.org> wrote:

snip

Interesting to me that you say that crossover distortion
might not be such an issue for a musical instrument
amplifier, though. I take it you must mean for stage work
where the power is going to be set pretty high, generally?


Yes. There might be a *very* quite venue
somewhere, where (say) the final note of a song
might trail away into audible crossover
distortion, but I'm skeptical.

However, note that the previously-mentioned
quantization distortion was first discovered as a
problem on early CDs where a final piano note
decayed into silence. When it got very soft, it
also got very "chunky" since it only used a few
active bits, and it sounded gritty just before it
became inaudible. The fix was to add dither
(noise) when recording, which effectively gave
PWM that miraculously eliminated the distortion at
the cost of a small increase in noise.

(You can get a fairly dramatic demo of this using
Daqarta. See <www.daqarta.com/dw_0gbb.htm>. This
is done using just the Generator, so it is totally
free.)

Alas, there is nothing like dither to "fix"
crossover distortion!

Mag tapes had inherently gross crossover distortion. The fix was to
add a pretty high-level "bias" oscillator to the record path to smear
it out. The bias voltage might be 20 volts at 60 KHz, way bigger than
the record signal going into the head.


John

 

[Jon Kirwan]

On Mon, 08 Feb 2010 11:54:42 -0800, I wrote:

snip
(Also, I suppose, the Early effect will add yet another
slight modification, since the Vce is slightly changed so is
the Ic for the same Vbe. The higher required Ic (assuming
the current source or resistor is supplying more current,
instead of less) requires a higher Vbe, as stated. So the
multiplied voltage at Vce is higher. But that multipled
voltage also slides over on the Vce axis for whatever Vbe
that has become and that suggests still more Ic due to Early
effect, so it is a positive feedback contributing to the
already existing problem, I think. I haven't tried to work
out just what percent it contributes, though.)
snip

I'm rethinking this. I think the Early effect counters the
effect, slightly. The higher Ic does increase Vbe. The
increased Vbe does increase Vce. So Vce is higher and we are
also on a different Vbe curve in the Vce vs Ic graph where
the Early effect shows up clearly. But Ic is 'given' because
it is being divided by the structure, so I should have
rotated the chart in my mind and using Ic as the independent
variable. Had I done that, I would have 'seen' that a given
rise in Ic would have suggested (holding Vbe constant for a
moment) a certain change in Vce. But that a now slightly
higher Vbe would have chosen a new Vbe curve that (before
rotating the chart) is _above_ the earlier Vbe curve, which
will cut the new Vbe curve where that same Ic intersects it a
little sooner on the Vce axis. Thus, it acts against.

My gut was telling me that nature works to oppose change and
I should have gone with that instinct, I think.

The effect is small. On the order of about 0.1%, roughly.

Jon

 

[Jon Kirwan]

On Mon, 08 Feb 2010 11:54:42 -0800, Jon Kirwan
<jonk@infinitefactors.org> wrote:

On Tue, 2 Feb 2010 22:30:31 +0530, "pimpom"
pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Sat, 30 Jan 2010 01:11:28 +0530, "pimpom" wrote:

snip
I've seen this as a modification. In ASCII form:

A
|
,---+---,
| |
| \
| / R3
\ \
/ R2 /
\ |
/ +--- C
| |
| |
| |/c Q1
+-----|
| |>e
\ |
/ R1 |
\ |
/ |
| |
'---+---'
|
B

We've already decided that R1 might be both a simple resistor
plus a variable pot to allow adjustment. The usual case I
see on the web does NOT include R3, though. However, I've
seen a few examples where R3 (small-valued) exists and one of
the two output BJTs' base is connected at C and not at A.

The above circuit is a somewhat different version of the Vbe
multiplier/rubber diode thing. The difference being R3,
which I'm still grappling with.

I've seen R3 used in that position too, but never gave it much
thought until you brought it up. Offhand I still can't see a
reason for it either. Maybe for stability against a local
oscillation? Perhaps taking some time to think about it will
bring some revelation. Or someone else can save us the trouble
and enlighten us.
snip

I had earlier said I thought you might be right about this R3
value. Now, I don't. I think it deals with something else
-- unregulated rail voltage variations.

In this thread, you've posted circuits with a resistor on one
side and the VAS on the other of this structure. The VAS
yanks one end around while the other side mostly follows it
around. However, the current through the resistor varies, of
course. Even if one of those BJTs+2 diodes thingies is used
for a current source instead of the resistor, which does
improve things, it still isn't very constant. Using the 2
BJT structure doesn't change that fact, though it does impact
variations. No matter how you arrange it, resistor or
current source, the fact is that the current into the Vbe
multiplier device changes around as the VAS yanks around one
side of it.

This variation means that Q1's Ic varies. To accomodate that
variation, Vbe varies. Since Vbe varies, so does the
multiplied value. And for no _other_ reason than variations
in signal. That changes the bias. Changing the bias changes
the quiescent current. Etc.

(Also, I suppose, the Early effect will add yet another
slight modification, since the Vce is slightly changed so is
the Ic for the same Vbe. The higher required Ic (assuming
the current source or resistor is supplying more current,
instead of less) requires a higher Vbe, as stated. So the
multiplied voltage at Vce is higher. But that multipled
voltage also slides over on the Vce axis for whatever Vbe
that has become and that suggests still more Ic due to Early
effect, so it is a positive feedback contributing to the
already existing problem, I think. I haven't tried to work
out just what percent it contributes, though.)

So a cludge fix for this is to insert a resistor in the
collector, which will act in the opposite direction to some
degreee. I'd imagine this would create a second degree poly
curve, with a maximum somewhere but gentle 'arms' outward,
which means less variation of the Vbe-multiplied value with a
tweakable peak point based upon a nominal Ic.

I imagine this is NOT nearly as important for class-A
operation, though, since it is already "biased up" and
variation at that point of operation probably isn't so
important. It _would_ matter, I think, in other classes of
operation.

Thinking as I am that I don't want to go with class-A, I am
trying to think of still better ways of replacing R3 with (or
adding) an active device to further improve it. Anyone have
a suggestion there?

Jon

Side bar:

The small signal analysis of the Vbe multiplier, if I got it
right, is based squarely upon the small-re of the BJT which
is, itself: (kT/q)/Ic. There is, of course, also the value
of R2, but since its effect is only affected by the change in
base current, it's contribution is divided by beta. So the
actual equation is something like:

R = (1/Ic)*(kT/q)*(1+R2/R1) + R2/beta

For a 2X multiplier where R2=R1, this is:

R = (2/Ic)*(kT/q) + R2/beta

Note the variation on Ic! (We _want_ the variation on T but
do NOT want the variation on Ic.) With R2=1k, for example,
that part contributes 5 ohms. But with Ic=5ma, for example,
the other part of it is sitting at 10.4 ohms -- a total of
15.4 ohms. So a variation of half an mA in Ic suggests a
7.7mV change in the bias point, ignoring any residual Early
effect on it.

I mentioned the Early effect being on the order of 0.1%. It
took me a moment to think about it, but the figure works out
to something like this:

R_early = dV/dI = -Ic/VA*R^2

It's a negative resistance that adds to R. If R is 15.4 ohms
and Ic is around 5mA and VA=100V, for example, you get about
-10mOhms. Which is roughly a factor of 1000 less than 15.4
Ohms. Which is where I get the 0.1% as a ballpark estimate.

The fuller equation would be:

R = (2/Ic)*(kT/q) + R2/beta - Ic/Va*R^2

Which requires a quadratic solution to solve for R. I'm
comfortable, for now, that I can ignore the Early effect and
just focus on the broader R figure for the Vbe multiplier,
applying changes in Ic to it to see how the voltage shifts
around.

So, in the diagram, R3 now reduces the effect by that change
in Ic*R3. If R3 is on the order of the above computed R,
that generally sets things so that right at that Ic used to
compute R the effect of R3 will be just in the right amount
to compensate for nearby changes in Ic.

I think.

Jon

 

[pimpom]

John Larkin wrote:

On Sun, 7 Feb 2010 16:10:14 +0530, "pimpom"
pimpom@invalid.invalid
wrote:

Jon Kirwan wrote:
On Sat, 06 Feb 2010 10:48:16 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


A few percent distortion at power levels is essentially
inaudible.
Speakers do that already. Low-level crossover distortion is
obvious
and obnoxious.

Yes, I think that's now much clearer to me now than it was
say two weeks ago -- without needing my ears to say so. Just
on understanding better _what_ crossover distortion is and
does.

Jon, to see a graphical illustration of JL's point, see this
screenshot of some simple simulations I just ran:
http://img694.imageshack.us/img694/8967/crossoverdistortion.png

On the right, the complementary output stage is driven without
any bias, . The upper trace shows the output when the input
amplitude is +/-1V peak. The transistors are operating in
Class C
and manage to conduct for less than half of each half cycle.
Now
that's going to sound awful by any standard. I _know_ it
sounds
awful because, when I was doing a lot of repairing work on
consumer products in the 80s, I came across some amps whose
biasing circuit had developed a fault.

Do a Fourier analysis and you'll get lots of harmonics. Reduce
the input amplitude even further and there won't be any output
at
all below a certain amplitude.

The lower trace shows the output with +/-9V input. Crossover
distortion is much reduced, though still evident. This may or
may
not be acceptable depending on the application. For anything
that
needs good audio quality, any waveform distortion that can be
clearly seen in graphical form is still too high, especially
in a
low-resolution bitmap trace like this.

On the left, we have the same amp with diode biasing added.
Visible distortion of the waveshape has disappeared. The
slight
irregularities in the sinusoidal shape are due to limitations
of
the low-res, non-antialiased bitmap image.


In your "unbiased" circuit, try adding a 1K resistor from the
bases to
the output. Now the flat spots in the output waveform become
slopes.
But the transfer function is now continuous, so negative
feedback can
reduce distortion without ugly slewing problems. And with zero
bias,
there's no idle power dissipation and no possibility of thermal
runaway.

The second circuit, with the bias diodes, is a likely firebomb.

John

I thought it would be obvious that those were simplified
arrangements purely for illustration. No one in his right mind
would think of using them in a practical audio amp.

 

[pimpom]

John Larkin wrote:

Mag tapes had inherently gross crossover distortion. The fix
was to
add a pretty high-level "bias" oscillator to the record path to
smear
it out. The bias voltage might be 20 volts at 60 KHz, way
bigger than
the record signal going into the head.

DC bias of the recording head was also used in many cheap
portables. The erase head was a permanent magnet that swiveled
out of the way on playback. On recording, the erase head
magnetised the tape to saturation in one direction. The dc bias
is polarised in the opposite direction, with enough strength to
place the operating point in the linear region. The noise level
is higher than with AC erase and bias, but it works.

 

[Paul E. Schoen]

"pimpom" <pimpom@invalid.invalid> wrote in message
news:hkr77v$lvl$1@news.albasani.net...

John Larkin wrote:

Mag tapes had inherently gross crossover distortion. The fix was to
add a pretty high-level "bias" oscillator to the record path to smear
it out. The bias voltage might be 20 volts at 60 KHz, way bigger than
the record signal going into the head.


DC bias of the recording head was also used in many cheap portables. The
erase head was a permanent magnet that swiveled out of the way on
playback. On recording, the erase head magnetised the tape to saturation
in one direction. The dc bias is polarised in the opposite direction,
with enough strength to place the operating point in the linear region.
The noise level is higher than with AC erase and bias, but it works.

I had one of those cheap cassette recorders, and it worked OK. But I had an
8-track tape player in my car and I wanted to be able to record, so I built
an AC bias circuit (I think mine was 40 kHz), using a circuit that I found
in an old databook. It incorporated the RIAA non-linear amplitude curve as
well. I also made a device which used a cheap turntable and crystal pickup,
with two J-FET (2N3819) linear amplifiers and VU meters.

http://en.wikipedia.org/wiki/RIAA_equalization
http://freecircuitdiagram.com/2008/11/27/gramophone-pre-amp-a-pre-amplifier-with-riaa-response-curve/

This was in 1970, and the only decent piece of test equipment I had was a
refurbished HP140A scope, with 100 kHz bandwidth and a fast blue phosphor
CRT made for photography. I still have everything except the 8-track tape
player for the car. It was stolen, along with most of my tapes, when I
foolishly neglected to put them in the trunk when I parked in a marginally
bad neighborhood in Washington, DC in 1972.

Paul

 

[Michael A. Terrell]

"Paul E. Schoen" wrote:

"pimpom" <pimpom@invalid.invalid> wrote in message
news:hkr77v$lvl$1@news.albasani.net...
John Larkin wrote:

Mag tapes had inherently gross crossover distortion. The fix was to
add a pretty high-level "bias" oscillator to the record path to smear
it out. The bias voltage might be 20 volts at 60 KHz, way bigger than
the record signal going into the head.


DC bias of the recording head was also used in many cheap portables. The
erase head was a permanent magnet that swiveled out of the way on
playback. On recording, the erase head magnetised the tape to saturation
in one direction. The dc bias is polarised in the opposite direction,
with enough strength to place the operating point in the linear region.
The noise level is higher than with AC erase and bias, but it works.

I had one of those cheap cassette recorders, and it worked OK. But I had an
8-track tape player in my car and I wanted to be able to record, so I built
an AC bias circuit (I think mine was 40 kHz), using a circuit that I found
in an old databook. It incorporated the RIAA non-linear amplitude curve as
well. I also made a device which used a cheap turntable and crystal pickup,
with two J-FET (2N3819) linear amplifiers and VU meters.

http://en.wikipedia.org/wiki/RIAA_equalization
http://freecircuitdiagram.com/2008/11/27/gramophone-pre-amp-a-pre-amplifier-with-riaa-response-curve/

This was in 1970, and the only decent piece of test equipment I had was a
refurbished HP140A scope, with 100 kHz bandwidth and a fast blue phosphor
CRT made for photography. I still have everything except the 8-track tape
player for the car. It was stolen, along with most of my tapes, when I
foolishly neglected to put them in the trunk when I parked in a marginally
bad neighborhood in Washington, DC in 1972.

Paul

RIAA is for records. Tape used NAB equalization.


--
Greed is the root of all eBay.

 

[Paul E. Schoen]

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:4B723882.898DEB80@earthlink.net...

"Paul E. Schoen" wrote:

"pimpom" <pimpom@invalid.invalid> wrote in message
news:hkr77v$lvl$1@news.albasani.net...
John Larkin wrote:

Mag tapes had inherently gross crossover distortion. The fix was to
add a pretty high-level "bias" oscillator to the record path to smear
it out. The bias voltage might be 20 volts at 60 KHz, way bigger than
the record signal going into the head.


DC bias of the recording head was also used in many cheap portables.
The
erase head was a permanent magnet that swiveled out of the way on
playback. On recording, the erase head magnetised the tape to
saturation
in one direction. The dc bias is polarised in the opposite direction,
with enough strength to place the operating point in the linear
region.
The noise level is higher than with AC erase and bias, but it works.

I had one of those cheap cassette recorders, and it worked OK. But I had
an
8-track tape player in my car and I wanted to be able to record, so I
built
an AC bias circuit (I think mine was 40 kHz), using a circuit that I
found
in an old databook. It incorporated the RIAA non-linear amplitude curve
as
well. I also made a device which used a cheap turntable and crystal
pickup,
with two J-FET (2N3819) linear amplifiers and VU meters.

http://en.wikipedia.org/wiki/RIAA_equalization
http://freecircuitdiagram.com/2008/11/27/gramophone-pre-amp-a-pre-amplifier-with-riaa-response-curve/

This was in 1970, and the only decent piece of test equipment I had was
a
refurbished HP140A scope, with 100 kHz bandwidth and a fast blue
phosphor
CRT made for photography. I still have everything except the 8-track
tape
player for the car. It was stolen, along with most of my tapes, when I
foolishly neglected to put them in the trunk when I parked in a
marginally
bad neighborhood in Washington, DC in 1972.

Paul


RIAA is for records. Tape used NAB equalization.

Thanks for clearing that up. I distinctly recall the RIAA curve, but it was
probably because I was recording from records. I may have also included an
NAB curve, but IIRC it was a rather simple circuit. I remember
special-ordering a choke or a transformer, but that may have been for the
bias. This was forty years ago! I don't feel like trudging through the snow
to my other house (which is unheated) just to open up the old equipment or
possibly find the databook which has the circuit. I was just happy to get
the thing to work and I wasn't too critical about the sound quality. I
found an interesting discussion of NAB equalization:
http://home.comcast.net/~mrltapes/equaliz.html

That was for 15 in/sec, which was popular for high quality reel-to-reel. I
think 8 track tapes were 7.5 or 3.75 in/sec. And cassettes were (and
probably still are) 1-7/8 in/sec.

Paul

 

[David Eather]

Jon Kirwan wrote:

On Sat, 06 Feb 2010 06:22:37 +1000, David Eather
eather@tpg.com.au> wrote:

Jon Kirwan wrote:
On Thu, 04 Feb 2010 05:46:33 +1000, David Eather
eather@tpg.com.au> wrote:

A 30 volt CT transformer with 15% regulation and 7% mains over-voltage,
less voltage drop for the diode bride would give rails of +/- 25.
I'm not entirely familiar with the formal terms,
Four..mal terms? ....I don't think I have used one of those.

I probably should have written, "I may not be familiar with
your use of this term, which may have a convention I'm
unaware of." I assumed your use had a formalized or
conventional meaning.

Regulation
is just 1 - (no load AC voltage / full load AC voltage) expressed as a
percentage. A first order approximation of a real transformer would be
an ideal transformer with a resistor in series. PS pretty much nailed it
in his response.

Yes, Paul helped me understand your meaning better.

There is no ripple to consider as there is no (significant) load on the
transformer and hence no (significant)load on the power supply. Also,
there is no (significant) current flowing through the diode bridge.

Well, I had been thinking about the loaded case perhaps too
much and hadn't broken out of that thinking. Seemed to me
that the unloaded situation never really occurs and that I
needed to maintain a fully loaded mindset at all times.

But I'd also wrongly imagined something that Paul caused to
change for the better. The bridge rectifier's peak current
occurs _before_ the top of the AC rise and has nearly gone
away by the time the rise hits the peak point. So the peak
voltage on the caps gets closer than I'd earlier guessed it
would. I knew that the slope (dv/dt) of the AC determines
the charging current on the caps and that the slope was zero
at the peak point, but a firm, simultaneous grasp hadn't yet
solidified in mind until he wrote.

I think this, taken together with your comment above, I must
also realize that the transformer itself will only be loaded
down for a small fraction of each cycle. If it has 15%
regulation, as you say, then for a short time while this peak
load is taking place the secondary voltage will be depressed.
But the fact remains that by the time the peak of the AC
arrives, the loading will be very light and thus the voltage
on the output will be at the higher, nearly unloaded value.
And thus, also, the caps (and BJTs) will experience this
value.

Did I get this about right?

Yes.

It was a 30 volt CT transformer with 15% regulation and 7% mains
over-voltage, less voltage drop for the diode bride would give rails of
ą 25.

Vsupply(max no load) = SQRT(2) * 15 * 1.15 * 1.07 - voltage lost from
the diode bridge. I rounded of the answer to +/- 25 volts. If I had
already selected a bridge I would have used whatever the data sheet
called the minimum voltage drop. That +/- 25 is the fully rectified and
filtered output of the power supply which is supplied by a relatively
cheap 30 volt CT transformer.

Yes, I am better gathering the use of the term "regulation"
now. And maybe _why_ you think that way about it. You are,
I imagine, thinking about worst case voltage stresses that
need to be held off. The sqrt(2)*15 is the specified peak
stress, but one needs to be aware that under no load the
transformer will likely produce higher voltages on its output
which _will_ be passed along to the caps and BJTs.

You use this voltage when you check that all transistors, caps etc are
adequately rated for the job and for the worst case situations you can
imagine - like that killer of speakers, the "1812 overture" driven into
heavy clipping.

Yes, now that I read this it sounds almost like what I just
wrote. So maybe I am "getting a clue," now. Thanks very
much for taking the time for me.

Welcome

but I figure
the 15% regulation you mentioned above must mean that the
ripple voltage goes from 100% to 85%. In general, the
equation for the angle would be something like:

angle = arcsin( 1 - Rf*(1-Vd/Vpk) )

This is with Rf being the ripple factor (not in percent
terms, obviously) and Vd being the sum of the diode drops
(full wave would be something like 2V) and Vpk being the
sqrt(2)*Vrms. At least, that's what the equation works out
for me on paper. (I can develop out here, if needed.)

The peak diode current happens just as the first moment of
conduction (which is neatly defined by the ripple factor, if
I understand you) and would be something like:

Ipk = 2*pi*f*C*Vpk*cos( angle )

Since the cos(arcsin(x)) is just sqrt(1-x^2), the computation
looks like:

Ipk = 2*pi*f*C*Vpk*SQRT(1-(1-Rf*(1-Vd/Vpk))^2)

There's probably some other adjustments to nail it, but that
probably gets somewhat close.

For a full wave bridge (I know, I think you were talking
about a half wave, but let's go with this for a moment) with
Vpk=25.2*SQRT(2) and Vd=2V (for two diodes in conduction in
the bridge) and Rf=.15 and f=60 (US-centric) and let's say a
C=2200uF, that gets something like Ipk = 15.2A.

None of this takes into account the average load current or
peak load currents. It just assumes that the ripple factor
is somehow known to be correct. I started out assuming that
the average load current would be defined entirely by the
droop (which is, of course, based upon the ripple factor
assumption) and the time from the peak of a previous cycle to
the point at which the above angle occurs and conduction
again begins. But it is complicated slightly by the fact
that the transformer supplies the entire current draw (if it
can) for a short part of the cycle _after_ the peak, as its
slope is less than the droop slope of the cap. I played with
accounting for all that but then decided that in most
practical cases the droop is hopefully not too excessive and
if not, then the angle after the peak isn't that much.. maybe
5 to 8 degrees or so... So I decided to ignore it and just
rely upon the capacitor's droop only:

Iave = C*Vpk*Rf/((pi/2+angle)/(2*pi*f))

So in the case just mentioned, I get Iave = 1.7A. I take it
that Ipk can easily be a factor of 9 or 10 greater. Also, I
note that it would probably be helpful to have nifty charts
of some kind to help pick off details like these.

Can you expand a little on what you were talking about,
though? Was that a half wave suggestion? I'm not sure I can
make sense of the rails, if so. If not, then I'd still
appreciate some of the calculations so that I can sure I
follow all of it.


I'm guessing that if a rail is to have a minimum of 25V on it
at the bottom of the ripple, and you are talking about 15%
regulation, the peak is going to be 29.4V -- not counting the
diode drops. Add 1V for that and it's 30.4V. Another 7% on
top would be 32.7V for the peak. RMS would be that figure
divided by sqrt(2), wouldn't it? Or 23.1Vrms or so?
No. 25 volt max. Under load less and also subtract the ripple.

Right. I've been corrected and I think I may gather the why
of it.

So would that suggest two of the cheap 25.2Vrms transformers
and plan on rails still slightly higher?
No. 1 CT transformer might be OK. The PSU unloaded voltage might be

V supply(max no load) = SQRT(2) * 12.6 * 1.15 * 1.07 - bridge

which I would round off to +/- 21 volts. In my mind I rejected this
option because I thought it would be difficult to make work in a worst
case, such as when the mains is 7% under voltage and the transformer is
at it's rated load

V supply(max load) = SQRT(2) * 12.6 * 1.0 * 0.93 - bridge (say 2 volts
at 1.6 amps) equals 14.6 volts

For 10 watts of output the peak output voltage is 12.7 volts. You also
need some (configuration dependent) voltage to keep the output in linear
range and even before you make any allowance for ripple voltage you are
in trouble.

I see your thinking much better.

You could deal with this situation a few ways, like saying it is not
your fault if the mains voltage is under. Use a bigger 12.6 transformer
so it is never fully loaded and hence the voltage is somewhat higher,
try to pass of an 8 watt design as 10 watts music power (which it
probably is), use massive filter caps (infinite in this case) or use a
higher voltage transformer. My experiance suggests that no one
complains if their 10 watt amp can actually put out 11 watts, but lots
of them complain when their 10 watt amp only puts out 9 watts (even
though you really can't hear the difference)

Well, this is for learning and eventually for building and
testing here. Complaining to myself won't get far. But
I am very glad for the broader thinking process. Thanks.

********
(1).....Vout(to speaker) = sqrt(2*P*R)

With R as 8 ohms for a common speaker and 10 watts that is 12.7 volts -
actually ą 12.7 volts with a split power supply.

(2).....Imax = sqrt(2*P/R) This comes out to 1.6 amps.
*********

Yes, I think we've calculated that one a few times, already.

Oh, crap. The VA rating. That's another one to consider.
Later, I guess.
- always one more thing... later is OK. Just start thinking about how
the amp might be used (domestic music, PA, emergency announcement,
musician's band etc)

I'm going to hold off on that until I get more grasped well.
For now, I will imagine in my head that the main thing here
will be to look at the volts*amps right around the point
nearby the peak recharging current each cycle, except that
such a transformer would then be spec'd for a worst case peak
and would likely be overly large for realistic needs. Or
something like that.

A dual 12.6 volt transformer would give a minimum (worst case with
transformer at full load and mains 7% under voltage) of 16.something
volts meaning big filter caps if you were serious at getting 10 watts.
One of the reasons to go PSU first I think. (Also I live in a tiny
jerk-water town where no one knows what a custom made transformer is let
alone where you can get one wound)
I suppose the old filament-type 12.6VAC transformers must be
common everywhere. I also see that Radio Shack (yes, I'm
holding my nose for a moment) still carries some "commodity"
type 25.2VAC CT 2.0A rated transformers for about US$10.
Their 12.6VAC CT 3.0A transformers are priced identically.

So what's considered to be generally available?
Job dependent. For a one off, whatever you can get. For high spec
commercial audio, whatever you can get custom made for the application.
For something laid out as a set of instructions what ever is widely
available - like 12.6v. In OZ we have a couple of hobbiest suppliers
that stock essentially the same range 6.3, 12.6, 15, 18, 20, 25, 30, 35..

Okay. 25.2V is very common here, so if they are similarly
common elsewhere perhaps it's better to leave the final
output specification as an output of the end of the design
process, rather than an input at the front end of it. I
don't actually _need_ 10 watts. I just figured that would be
a good place to target to learn enough things to be useful.
In terms of Radio Shack, I've no idea what the regulation
spec on their transformers would be. But I have to imagine
that it's about as cheaply made as they can get away with.

I tracked down a very nice transformer in my box which may be
okay. It has two secondaries and was intended for 60Hz use.
It weighs in at 2.8 lbs (1.25 kg.)

Primary:
115VAC, 5.0 Ohms, 16 gauge
Secondaries: (Tested using 120.5VAC RMS on primary)
16VAC RMS CT, 0.05 Ohms, 14 gauge
30.4VAC RMS CT, 2.6 ohms, 22 gauge

The 16VAC RMS outer winding across a 56 ohm resistor yields
15.88VAC RMS. (I don't have a large wattage resistor with
lower values of resistance, so that needs to suffice.) Half
of the 30.4VAC RMS winding (CT to one side) yields 14.75VAC
RMS loaded with the same 56 ohm resistor. Across the entire
30.4VAC windings it is 28.9VAC RMS. (The poor thing is just
a 5W, so I didn't measure for longer than a few seconds.)

The 30.4VAC secondary looks reasonable, I think, for the two
amplifier rails and ground.

Yes - I think it's Ok

The 16VAC might make another

supply for some other reason or, perhaps, provide another
pair of rails to use for a 2 ohm speaker?

Could use it for a preamp. Just from inspection I think a little bit
high for 2 ohms, would be good for 4 ohms (but calculations talk,
opinions walk)

I hadn't thought about that aspect, but as you earlier
pointed out the 25.2VAC CT standard transformer might be a
little light for a 10W amplifier... unless I spec'd a 4 ohm
speaker, I suppose. Then it might be fine.

Yes. But it is only a little light. If you are happy with 8-9 watts RMS
and 10 watt peaks I think it is OK.

Anyway, it looks like it may be a reasonable choice as
something I have available and ready from the junk box.

Jon
Yes.

 

[David Eather]

Jon Kirwan wrote:

On Fri, 05 Feb 2010 16:46:22 -0800, I wrote:

snip
I tracked down a very nice transformer in my box which may be
okay. It has two secondaries and was intended for 60Hz use.
It weighs in at 2.8 lbs (1.25 kg.)

Primary:
115VAC, 5.0 Ohms, 16 gauge
Secondaries: (Tested using 120.5VAC RMS on primary)
16VAC RMS CT, 0.05 Ohms, 14 gauge
30.4VAC RMS CT, 2.6 ohms, 22 gauge

The 16VAC RMS outer winding across a 56 ohm resistor yields
15.88VAC RMS. (I don't have a large wattage resistor with
lower values of resistance, so that needs to suffice.) Half
of the 30.4VAC RMS winding (CT to one side) yields 14.75VAC
RMS loaded with the same 56 ohm resistor. Across the entire
30.4VAC windings it is 28.9VAC RMS. (The poor thing is just
a 5W, so I didn't measure for longer than a few seconds.)

The 30.4VAC secondary looks reasonable, I think, for the two
amplifier rails and ground. The 16VAC might make another
supply for some other reason or, perhaps, provide another
pair of rails to use for a 2 ohm speaker?

I hadn't thought about that aspect, but as you earlier
pointed out the 25.2VAC CT standard transformer might be a
little light for a 10W amplifier... unless I spec'd a 4 ohm
speaker, I suppose. Then it might be fine.

Anyway, it looks like it may be a reasonable choice as
something I have available and ready from the junk box.

Second thoughts. The 30.4VAC RMS CT secondary shows 2.6 ohms
and is 22 gauge. That's 1.3 ohms per half. I believe from
calculation that the peak diode current _might_ be 8-10 times
the load current in the ideal case (0 ohms.) Taking into
account the winding resistance, I may need to think more
closely about using this transformer in this application. The
winding resistance will limit the current and thus the energy
per unit time that can be transferred to the caps and that
will very likely lower the achievable rail voltage on the
other side of the bridge since the bridge itself simply won't
ever see the idealized peak voltage even right up to the
moment of peak where the dv/dt goes to zero. By the time
that happens, the cycle will already be on a decline again
while the resistance continues to limit inflow of charge.
Cripes.

Darn it. Back to monster caps to get a slight decent rail
voltage there.

Jon

It is not a big stress. You can always use the junk-box transformer and
if it really isn't suitable replace it latter. For your consideration -
the RMS power of even compressed samples of music is only about 20% of
the peak.

There are a few variations on that figure. RCA did a lot of research in
the area and found that Radio broadcasts of compressed FM signals of
"Rock Music" - an undefined term, was the most demanding at 15%. Some
companies are trying to redefine that. IRF who call the same figure 1/8
of max power (12.5%) - which just happens to make their newest audio
mosfets look really good. It might be the other way around. They may
really believe it, and designed the mosfets to match. I forget where but
some group stated the 20% figure with respect to new modern music
styles. IIRC they were regarded as technically competent in the area and
had no axe to grind or wheelbarrow to push - so I filed the info away.
In any case an overestimate leads to a more conservative design and 5%
is not much. I'd be wary of definition of "modern music" too - badly
played organ music can be a stream of full amplitude waveforms that only
change in frequency at random intervals.

I'd use the junk box transformer and forget about allowing for the
electricity company slackness and just choose good sized caps that are a
reasonable price. I think a learning experience allows for a little
compromise.

 

[David Eather]

Jon Kirwan wrote:

On Fri, 05 Feb 2010 17:06:54 -0800, I wrote:

On Fri, 05 Feb 2010 16:46:22 -0800, I wrote:

snip
I tracked down a very nice transformer in my box which may be
okay. It has two secondaries and was intended for 60Hz use.
It weighs in at 2.8 lbs (1.25 kg.)

Primary:
115VAC, 5.0 Ohms, 16 gauge
Secondaries: (Tested using 120.5VAC RMS on primary)
16VAC RMS CT, 0.05 Ohms, 14 gauge
30.4VAC RMS CT, 2.6 ohms, 22 gauge

The 16VAC RMS outer winding across a 56 ohm resistor yields
15.88VAC RMS. (I don't have a large wattage resistor with
lower values of resistance, so that needs to suffice.) Half
of the 30.4VAC RMS winding (CT to one side) yields 14.75VAC
RMS loaded with the same 56 ohm resistor. Across the entire
30.4VAC windings it is 28.9VAC RMS. (The poor thing is just
a 5W, so I didn't measure for longer than a few seconds.)

The 30.4VAC secondary looks reasonable, I think, for the two
amplifier rails and ground. The 16VAC might make another
supply for some other reason or, perhaps, provide another
pair of rails to use for a 2 ohm speaker?

I hadn't thought about that aspect, but as you earlier
pointed out the 25.2VAC CT standard transformer might be a
little light for a 10W amplifier... unless I spec'd a 4 ohm
speaker, I suppose. Then it might be fine.

Anyway, it looks like it may be a reasonable choice as
something I have available and ready from the junk box.
Second thoughts. The 30.4VAC RMS CT secondary shows 2.6 ohms
and is 22 gauge. That's 1.3 ohms per half. I believe from
calculation that the peak diode current _might_ be 8-10 times
the load current in the ideal case (0 ohms.) Taking into
account the winding resistance, I may need to think more
closely about using this transformer in this application. The
winding resistance will limit the current and thus the energy
per unit time that can be transferred to the caps and that
will very likely lower the achievable rail voltage on the
other side of the bridge since the bridge itself simply won't
ever see the idealized peak voltage even right up to the
moment of peak where the dv/dt goes to zero. By the time
that happens, the cycle will already be on a decline again
while the resistance continues to limit inflow of charge.
Cripes.

Darn it. Back to monster caps to get a slight decent rail
voltage there.

Jon

An addition or two. The transformer mentioned above uses 18
gauge, not 16 gauge, for the primary. And since I'm still
wrestling with why there are 5 ohms, measured, I think it's
likely that is merely the wiring to the outside and that
perhaps even smaller diameter wire was used to wrap around
the core. So externally labeled gauge probably is NOT a
precise indication of what was used in the core.

I took a look at the weight of some similarly shaped 60Hz
power transformers, available from Stancor. It seems that
similar weights are on the order of 80 VA. (I've seen a few
rated 100 VA, but I'm betting less.)

For now, I'll assume that transformer is too small, not
because of my guessed-at VA rating but because of the
measured resistance in the 30.4VAC secondary.

I want to get back to the power supply design and finish
that.

I understant that at 10W, the 8 ohm speaker will experience
sqrt(2*10W*8 ohms) or 12.65Vrms and 1.58Arms. I could always
modify this to fit what I have available but rather than do
that, I think it's better to "stay on target" and see where
that goes.

OK. I'm good with either approach

So that's what is expected "at the speaker." The peak figure
required will be rounded up to 18V. To reach that peak, the
output BJTs will need some headroom of their own. Staying
out of saturation and assuming the output stage might use two
BJTs on either side, requiring perhaps two diode drops if
either of these quadrants uses both NPN or both PNP, I would
best figure another 4V of headroom on each side. So 22V
minimum there, __under load.__

4v is not excessive but you could probably trim it back a little if push
came to shove. (Any "brown out" would coincided closely with clipping
and would be difficult to detect with an ear) - but in general it is
always better to do it the right way in design stage.

And I begin to see why 25V isn't a bad target.

Which brings in the question about a linear regulator. It's
my vague feeling that there is NO need for one.

Good intuition.

I should be

able to arrange the circuity (current sources, etc.) so that
they are sufficiently immune to modest ripple that the 60Hz
(and other components due to loading causing cap voltage
changes, as well) can be rejected well enough.

Yes.

Besides, a

linear regulator would mean just that much more headroom and
wasted power/heat. So unless something very difficult is
shown to me, I'd like to take the position that a linear
regulator is a lot extra trouble without worthwhile payback.

In almost every case here your correct. I'm trying to think of an
exception. The only thing I can think of is a specific configuration of
a calls D amp - and then if that is a problem they should have used a
different configuration.

(And dealing with the added poles/zeros would seem to make my
worries compounded, as if the rest weren't enough.)

The filter capacitors will probably have to be spec'd at 50V
given what I've read here. It seems 35V wouldn't be entirely
safe, given the comments about regulation at 15% and another
7% margin, as well.

25 volts included those "extras" - I'd be perfectly happy with 35v caps
for a device in intermittent use with an intended lifespan of 3 - 5
years - but that is a total guess of intended use and lifespan.

And something else that is bothering me.

Charging only takes place for short bursts and happen
_before_ the windings reach peak voltage

before and until the winding reach peak voltage.


.. So there is a

small duty cycle during which usable energy is transferred.
Does this suggest that one might _under specify_ the VA
rating for the transformer to save cost and weight and get
away with it?

Yes, but.... I would attribute that more to the low duty cycle of the
music signal.

I actually find it frustrating to see kits designed for home use with a
power supply (and heatsinks) for full time 100% output. The only time it
will ever be needed is in (long) prototype testing. Just a waste of
money IMO)

Jon

 

[David Eather]

Paul E. Schoen wrote:

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:4B723882.898DEB80@earthlink.net...
"Paul E. Schoen" wrote:
"pimpom" <pimpom@invalid.invalid> wrote in message
news:hkr77v$lvl$1@news.albasani.net...
John Larkin wrote:
Mag tapes had inherently gross crossover distortion. The fix was to
add a pretty high-level "bias" oscillator to the record path to smear
it out. The bias voltage might be 20 volts at 60 KHz, way bigger than
the record signal going into the head.

DC bias of the recording head was also used in many cheap portables.
The
erase head was a permanent magnet that swiveled out of the way on
playback. On recording, the erase head magnetised the tape to
saturation
in one direction. The dc bias is polarised in the opposite direction,
with enough strength to place the operating point in the linear
region.
The noise level is higher than with AC erase and bias, but it works.
I had one of those cheap cassette recorders, and it worked OK. But I had
an
8-track tape player in my car and I wanted to be able to record, so I
built
an AC bias circuit (I think mine was 40 kHz), using a circuit that I
found
in an old databook. It incorporated the RIAA non-linear amplitude curve
as
well. I also made a device which used a cheap turntable and crystal
pickup,
with two J-FET (2N3819) linear amplifiers and VU meters.

http://en.wikipedia.org/wiki/RIAA_equalization
http://freecircuitdiagram.com/2008/11/27/gramophone-pre-amp-a-pre-amplifier-with-riaa-response-curve/

This was in 1970, and the only decent piece of test equipment I had was
a
refurbished HP140A scope, with 100 kHz bandwidth and a fast blue
phosphor
CRT made for photography. I still have everything except the 8-track
tape
player for the car. It was stolen, along with most of my tapes, when I
foolishly neglected to put them in the trunk when I parked in a
marginally
bad neighborhood in Washington, DC in 1972.

Paul

RIAA is for records. Tape used NAB equalization.

Thanks for clearing that up. I distinctly recall the RIAA curve, but it was
probably because I was recording from records. I may have also included an
NAB curve, but IIRC it was a rather simple circuit. I remember
special-ordering a choke or a transformer, but that may have been for the
bias. This was forty years ago! I don't feel like trudging through the snow
to my other house (which is unheated) just to open up the old equipment or
possibly find the databook which has the circuit. I was just happy to get
the thing to work and I wasn't too critical about the sound quality. I
found an interesting discussion of NAB equalization:
http://home.comcast.net/~mrltapes/equaliz.html

That was for 15 in/sec, which was popular for high quality reel-to-reel. I
think 8 track tapes were 7.5 or 3.75 in/sec. And cassettes were (and
probably still are) 1-7/8 in/sec.

Paul


The RIAA curve would have been used to "straighten out" the signal from

the record player before recording onto tape.

 

[David Eather]

Paul E. Schoen wrote:

"pimpom" <pimpom@invalid.invalid> wrote in message
news:hkr77v$lvl$1@news.albasani.net...
John Larkin wrote:
Mag tapes had inherently gross crossover distortion. The fix was to
add a pretty high-level "bias" oscillator to the record path to smear
it out. The bias voltage might be 20 volts at 60 KHz, way bigger than
the record signal going into the head.

DC bias of the recording head was also used in many cheap portables. The
erase head was a permanent magnet that swiveled out of the way on
playback. On recording, the erase head magnetised the tape to saturation
in one direction. The dc bias is polarised in the opposite direction,
with enough strength to place the operating point in the linear region.
The noise level is higher than with AC erase and bias, but it works.

I had one of those cheap cassette recorders, and it worked OK. But I had an
8-track tape player in my car and I wanted to be able to record, so I built
an AC bias circuit (I think mine was 40 kHz), using a circuit that I found
in an old databook. It incorporated the RIAA non-linear amplitude curve as
well. I also made a device which used a cheap turntable and crystal pickup,
with two J-FET (2N3819) linear amplifiers and VU meters.

Didn't all that stuff make the car hard to drive?

http://en.wikipedia.org/wiki/RIAA_equalization
http://freecircuitdiagram.com/2008/11/27/gramophone-pre-amp-a-pre-amplifier-with-riaa-response-curve/

This was in 1970, and the only decent piece of test equipment I had was a
refurbished HP140A scope, with 100 kHz bandwidth and a fast blue phosphor
CRT made for photography. I still have everything except the 8-track tape
player for the car. It was stolen, along with most of my tapes, when I
foolishly neglected to put them in the trunk when I parked in a marginally
bad neighborhood in Washington, DC in 1972.

Paul

 

[David Eather]

Jon Kirwan wrote:

On Fri, 05 Feb 2010 17:06:54 -0800, I wrote:

On Fri, 05 Feb 2010 16:46:22 -0800, I wrote:

snip
I tracked down a very nice transformer in my box which may be
okay. It has two secondaries and was intended for 60Hz use.
It weighs in at 2.8 lbs (1.25 kg.)

Primary:
115VAC, 5.0 Ohms, 16 gauge
Secondaries: (Tested using 120.5VAC RMS on primary)
16VAC RMS CT, 0.05 Ohms, 14 gauge
30.4VAC RMS CT, 2.6 ohms, 22 gauge

The 16VAC RMS outer winding across a 56 ohm resistor yields
15.88VAC RMS. (I don't have a large wattage resistor with
lower values of resistance, so that needs to suffice.) Half
of the 30.4VAC RMS winding (CT to one side) yields 14.75VAC
RMS loaded with the same 56 ohm resistor. Across the entire
30.4VAC windings it is 28.9VAC RMS. (The poor thing is just
a 5W, so I didn't measure for longer than a few seconds.)

The 30.4VAC secondary looks reasonable, I think, for the two
amplifier rails and ground. The 16VAC might make another
supply for some other reason or, perhaps, provide another
pair of rails to use for a 2 ohm speaker?

I hadn't thought about that aspect, but as you earlier
pointed out the 25.2VAC CT standard transformer might be a
little light for a 10W amplifier... unless I spec'd a 4 ohm
speaker, I suppose. Then it might be fine.

Anyway, it looks like it may be a reasonable choice as
something I have available and ready from the junk box.
Second thoughts. The 30.4VAC RMS CT secondary shows 2.6 ohms
and is 22 gauge. That's 1.3 ohms per half. I believe from
calculation that the peak diode current _might_ be 8-10 times
the load current in the ideal case (0 ohms.) Taking into
account the winding resistance, I may need to think more
closely about using this transformer in this application. The
winding resistance will limit the current and thus the energy
per unit time that can be transferred to the caps and that
will very likely lower the achievable rail voltage on the
other side of the bridge since the bridge itself simply won't
ever see the idealized peak voltage even right up to the
moment of peak where the dv/dt goes to zero. By the time
that happens, the cycle will already be on a decline again
while the resistance continues to limit inflow of charge.
Cripes.

Darn it. Back to monster caps to get a slight decent rail
voltage there.

Jon

An addition or two. The transformer mentioned above uses 18
gauge, not 16 gauge, for the primary. And since I'm still
wrestling with why there are 5 ohms, measured, I think it's
likely that is merely the wiring to the outside and that
perhaps even smaller diameter wire was used to wrap around
the core. So externally labeled gauge probably is NOT a
precise indication of what was used in the core.

I took a look at the weight of some similarly shaped 60Hz
power transformers, available from Stancor. It seems that
similar weights are on the order of 80 VA. (I've seen a few
rated 100 VA, but I'm betting less.)

For now, I'll assume that transformer is too small, not
because of my guessed-at VA rating but because of the
measured resistance in the 30.4VAC secondary.

I want to get back to the power supply design and finish
that.

I understant that at 10W, the 8 ohm speaker will experience
sqrt(2*10W*8 ohms) or 12.65Vrms and 1.58Arms. I could always
modify this to fit what I have available but rather than do
that, I think it's better to "stay on target" and see where
that goes.

So that's what is expected "at the speaker." The peak figure
required will be rounded up to 18V. To reach that peak, the
output BJTs will need some headroom of their own. Staying
out of saturation and assuming the output stage might use two
BJTs on either side, requiring perhaps two diode drops if
either of these quadrants uses both NPN or both PNP, I would
best figure another 4V of headroom on each side. So 22V
minimum there, __under load.__

And I begin to see why 25V isn't a bad target.

Which brings in the question about a linear regulator. It's
my vague feeling that there is NO need for one. I should be
able to arrange the circuity (current sources, etc.) so that
they are sufficiently immune to modest ripple that the 60Hz
(and other components due to loading causing cap voltage
changes, as well) can be rejected well enough. Besides, a
linear regulator would mean just that much more headroom and
wasted power/heat. So unless something very difficult is
shown to me, I'd like to take the position that a linear
regulator is a lot extra trouble without worthwhile payback.
(And dealing with the added poles/zeros would seem to make my
worries compounded, as if the rest weren't enough.)

The filter capacitors will probably have to be spec'd at 50V
given what I've read here. It seems 35V wouldn't be entirely
safe, given the comments about regulation at 15% and another
7% margin, as well. And something else that is bothering me.
Charging only takes place for short bursts and happen
_before_ the windings reach peak voltage. So there is a
small duty cycle during which usable energy is transferred.
Does this suggest that one might _under specify_ the VA
rating for the transformer to save cost and weight and get
away with it?

Jon

I didn't give a full answer. The small duty cycle (and large I**2R
losses) mean you have to over specify the VA of the transformer - but
because of the low duty cycle (for a domestic application) the net
result can be a transformer smaller than the amplifier output wattage.

There are common figures used for various configurations. If I ever knew
how they were derived I don't remember now. Can someone step in here?

 

[Paul E. Schoen]

"David Eather" <eather@tpg.com.au> wrote in message
news:VMOdnQ-MerwJIe7WnZ2dnUVZ_tP_fwAA@supernews.com...

Paul E. Schoen wrote:

I had one of those cheap cassette recorders, and it worked OK. But I had
an 8-track tape player in my car and I wanted to be able to record, so I
built an AC bias circuit (I think mine was 40 kHz), using a circuit that
I found in an old databook. It incorporated the RIAA non-linear
amplitude curve as well. I also made a device which used a cheap
turntable and crystal pickup, with two J-FET (2N3819) linear amplifiers
and VU meters.


Didn't all that stuff make the car hard to drive?

Not at all. My car was a 1965 Chevy Malibu that was good on the curves.
Most importantly, it had a Class A driver

Paul

 

[pimpom]

Paul E. Schoen wrote:

"David Eather" <eather@tpg.com.au> wrote in message
news:VMOdnQ-MerwJIe7WnZ2dnUVZ_tP_fwAA@supernews.com...
Paul E. Schoen wrote:

I had one of those cheap cassette recorders, and it worked
OK. But
I had an 8-track tape player in my car and I wanted to be
able to
record, so I built an AC bias circuit (I think mine was 40
kHz),
using a circuit that I found in an old databook. It
incorporated
the RIAA non-linear amplitude curve as well. I also made a
device
which used a cheap turntable and crystal pickup, with two
J-FET
(2N3819) linear amplifiers and VU meters.


Didn't all that stuff make the car hard to drive?

Not at all. My car was a 1965 Chevy Malibu that was good on the
curves. Most importantly, it had a Class A driver

Was the driver good on curves _inside_ the car?

 

[Jon Kirwan]

On Thu, 11 Feb 2010 17:51:27 +1000, David Eather
<eather@tpg.com.au> wrote:

snip
It is not a big stress. You can always use the junk-box transformer and
if it really isn't suitable replace it latter. For your consideration -
the RMS power of even compressed samples of music is only about 20% of
the peak.

There are a few variations on that figure. RCA did a lot of research in
the area and found that Radio broadcasts of compressed FM signals of
"Rock Music" - an undefined term, was the most demanding at 15%. Some
companies are trying to redefine that. IRF who call the same figure 1/8
of max power (12.5%) - which just happens to make their newest audio
mosfets look really good. It might be the other way around. They may
really believe it, and designed the mosfets to match. I forget where but
some group stated the 20% figure with respect to new modern music
styles. IIRC they were regarded as technically competent in the area and
had no axe to grind or wheelbarrow to push - so I filed the info away.
In any case an overestimate leads to a more conservative design and 5%
is not much. I'd be wary of definition of "modern music" too - badly
played organ music can be a stream of full amplitude waveforms that only
change in frequency at random intervals.

I'd use the junk box transformer and forget about allowing for the
electricity company slackness and just choose good sized caps that are a
reasonable price. I think a learning experience allows for a little
compromise.

Okay. I'm back to the power supply, again. (I'm convinced
that my junkbox unit will work fine -- I think it can hold
maybe 18V minimum under load on each rail. Which seems more
than enough headroom for 12.7V, plus output stage overhead.

I take a little issue with your use of terms in this phrase,
"RMS power of even compressed samples of music is only about
20% of the peak." Power is average and I don't think RMS
applies to power. Volts-to-power is a squared-phenomenon. So
are amps-to-power. RMS makes sense for those two. But power
is an average (integrated Joules divided by time.)

So I believe I have to interpret your meaning as suggesting
that the short-term power required (also an average of some
ill-defined kind, I suppose) when playing music can be a
factor of 5 times more than its long-term average power. You
also mentioned a figure as low as 12.5%, which would suggest
a factor of 8 used as a margin instead of 5.

But a requirement to support short-term power levels is
really just a compliance requirement on the power supply
rails, isn't it?

So put another way, if I wanted a long-term average of 10W
output and I wanted the extra margins required to support the
worst case estimate of a factor of 8 for short-term power
bursts, then I'd need to design rails that support a voltage
compliance level substantially higher. The parts would need
to withstand it, too. And because of the much higher rail
voltages that need to be dropped most of the time, the output
BJTs would need to have just that much more capacity to
dissipate.

Or put still another way, assuming that my output swing at
the output stage emitters cannot exceed a magnitude of 15V
and that everything is sized for dissipating 10W, does this
mean the amplifier is a 10W amplifier that can support a peak
of 14W=(15^2/(2*8))? (Which isn't so good, considering your
comments above regarding "music?")

What is meant when one says, '10 watts?'

This gets worse when I consider the class of operation,
doesn't it? I mean, class-B might be specified as 10W into 8
ohms, but wouldn't that be 20W into 4 ohms? But if class-A,
it's pretty much 10W no matter what?

I'm beginning to imagine amplifiers should be specified as to
their peak output voltage compliance into 8, 6, and 4 ohms;
instantaneous and sustained without damage to the unit. For
example, 35V into 8 ohms instantaneous, 15V sustained. Or
80W instantaneous, 15W sustained. That way, someone might
have some knowledge about how well it might handle _their_
music at, say, 15W average power. And could compare that
against another unit specified as 20V into 8 ohms, 15V
sustained.

How does one know what they are buying? What a headache.

Jon

 

[Jon Kirwan]

On Tue, 16 Feb 2010 20:28:30 -0800, I wrote:

20V into 8 ohms, 15V sustained.

I mean "20V instantaneous into 8 ohms, 15V sustained."

Jon

 

[Paul E. Schoen]

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:kvlmn5dhciv4c51a0au32qi52rjnl67sro@4ax.com...

Okay. I'm back to the power supply, again. (I'm convinced
that my junkbox unit will work fine -- I think it can hold
maybe 18V minimum under load on each rail. Which seems more
than enough headroom for 12.7V, plus output stage overhead.

I take a little issue with your use of terms in this phrase,
"RMS power of even compressed samples of music is only about
20% of the peak." Power is average and I don't think RMS
applies to power. Volts-to-power is a squared-phenomenon. So
are amps-to-power. RMS makes sense for those two. But power
is an average (integrated Joules divided by time.)

So I believe I have to interpret your meaning as suggesting
that the short-term power required (also an average of some
ill-defined kind, I suppose) when playing music can be a
factor of 5 times more than its long-term average power. You
also mentioned a figure as low as 12.5%, which would suggest
a factor of 8 used as a margin instead of 5.

But a requirement to support short-term power levels is
really just a compliance requirement on the power supply
rails, isn't it?

So put another way, if I wanted a long-term average of 10W
output and I wanted the extra margins required to support the
worst case estimate of a factor of 8 for short-term power
bursts, then I'd need to design rails that support a voltage
compliance level substantially higher. The parts would need
to withstand it, too. And because of the much higher rail
voltages that need to be dropped most of the time, the output
BJTs would need to have just that much more capacity to
dissipate.

Or put still another way, assuming that my output swing at
the output stage emitters cannot exceed a magnitude of 15V
and that everything is sized for dissipating 10W, does this
mean the amplifier is a 10W amplifier that can support a peak
of 14W=(15^2/(2*8))? (Which isn't so good, considering your
comments above regarding "music?")

What is meant when one says, '10 watts?'

This gets worse when I consider the class of operation,
doesn't it? I mean, class-B might be specified as 10W into 8
ohms, but wouldn't that be 20W into 4 ohms? But if class-A,
it's pretty much 10W no matter what?

I'm beginning to imagine amplifiers should be specified as to
their peak output voltage compliance into 8, 6, and 4 ohms;
instantaneous and sustained without damage to the unit. For
example, 35V into 8 ohms instantaneous, 15V sustained. Or
80W instantaneous, 15W sustained. That way, someone might
have some knowledge about how well it might handle _their_
music at, say, 15W average power. And could compare that
against another unit specified as 20V into 8 ohms, 15V
sustained.

How does one know what they are buying? What a headache.

Yes, as an extension of what (I think) Mark Twain said, there are lies,
damn lies, statistics, and specifications. Then there is the matter of
testing. An amplifier is a complex entity and its performance depends on
the power supply, the load, its components, environmental conditions, and
the nature of the signal being applied. So it may seem fair to level the
playing field by testing with a pure sine wave at certain frequencies and
determining that it maintains a certain level of maximum distortion without
overheating or shutting down over an extended period of time in a
controlled environment.

But in real life there are many more factors involved, and the actual
performance in an individual situation may vary widely. Power is indeed an
average function, but the ability to provide power involves efficiency and
a duty-cycle rated function of maximum temperature of components, and also
the ability of the power supply to maintain a certain voltage level for
long enough to "ride out" brief peaks in the signal of typical music.

The power that can be supplied to various loads depends largely on
impedance matching. But most solid state amplifiers are capable of
supplying a certain amount of current, so if it is optimized for eight
ohms, it may be able to provide even less continuous power at 4 ohms, but
possibly more peak power.

You have brought up some good points. But for most purposes, an amplifier
rated conservatively at 10W continuous power should be plenty for home
music listening. When pushed beyond its normal limits, much depends on how
the amplifier handles overloads, and your personal threshold of annoyance
when the inevitable distortion occurs.

Paul