Discussing audio amplifier design -- BJT, discrete

[David Eather]

Jon Kirwan wrote:

On Wed, 27 Jan 2010 17:31:00 +1000, David Eather
eather@tpg.com.au> wrote:

Jon Kirwan wrote:
On Tue, 26 Jan 2010 16:15:45 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 26 Jan 2010 12:57:13 -0800, Jon Kirwan
jonk@infinitefactors.org> wrote:

I'd like to take a crack at thinking through a design of an
audio amplifier made up of discrete BJTs and other discrete
parts as an educational process.

I imagine this will be broken up into three sections; input
transconductance, transimpedance VAS, and output driver. But
other arrangements (such as combining the VAS and output
driver using a signal splitting BJT) would work for me, in
learning.

I said "BJT" and "discrete" but I'm also open to the idea of
using BJT pairs, such as the BCV61 and BCV62. In the case of
current mirrors, that may make sense. But not high-priced,
elite and/or hard to get, or obsolete. And no FETs. This is
to be about learning to design with BJTs.

SMT vs through-hole isn't an issue for learning about a
design, I suppose. If I need to build up some section and
test it with a signal, I'll probably want to do it quickly
and without having to buy services every step of the way. So
I may 'dead bug' SMT parts to get there. (The basic idea
here is to learn, not to make something tiny.)

Although I have some other applications, right now I'd like
the target use to be as a computer speaker system (not unlike
those dirt cheap, sadly almost all of them 10% THD, systems
sold today into this market. Except that I'd like to work
through the design on my own, from start to end.

Given what I understand right now from a very short search on
the topic, the input should be taken as a maximum of 1.0Vrms
and the input's load should appear to be something like 10k
ohms. If someone knows different from that, I'll accept the
criticism and change that spec.

I'd like to consider a tone control and a volume control to
be included.

Output is to be into a small 8 ohm speaker. With that
maximum 1.0Vrms at the input and the volume control set to
maximum the wattage into 8 ohms should be around 10 watts.
Since human hearing won't tell much difference between 8
watts and 12 watts, this is a bit of a sloppy spec and I'm
open to anything in the area of 5-20 watts... though I'm
really wanting to keep the rail voltages down to something
modest and the BJTs not having to tolerate hugish Vce.

Now that I say this, an odd idea comes to mind because the
CFL light bulbs include two TO-220 BJTs that can handle quite
a high Vce on them. I could cannibalize those. But to be
honest, I'm still not needing high watt outputs. So there's
no reason to think about scavenging such parts.

I would like to design it to work into 4 ohms as a margin
bound and not as a design goal, but even 5.6 ohms would be
acceptable.

I'm not looking for this to be done quickly, either. If it
takes months of only occasional back-and-forth, I'm fine with
that. Also, I expect to do my work and don't expect someone
else to hand-hold me from complete ignorante to complete
enlightenment. I just need someone to slap my face when
I say something terribly stupid and/or point in a truely
useful direction when I need it. Or else someone who is
wanting to explore this with me and willing to work for it.

Is anyone here willing to consider a sincere discussion?

Jon
Back when transistors were young, and transistor manuals (GE, RCA)
were published, there were tons of such circuits around. They all
pretty much converged to a few forms, and haven't changed much since.
I remember reading in popular electronics about some audio
amps that I couldn't even come close to following at the
time. The series of them with the name 'tiger' in them.

I'm not so much interested in _circuits_, per se, as I am in
learning about topologies, various ideas built upon them, and
then the specific details of designing towards a specific
implementation. For example, I enjoyed learning about
bootstrapping as a general idea _and_ as it applies to a
couple of specific areas. Having both theory _and_ specific
practice helps firm up the ideas better.

I could post some circuits from the old manuals, it that wouldn't
spoil what you want to do.
It may serve as a point of discussion. Would you be willing
to discuss their details and the broader theories as applied?

10 watts into computer speakers sounds like a lot. Most AM radios
didn't make one watt. You might experiment first to see how much power
you really need.
Oh, I figure one watt is enough, too. As a practical matter
and as a consumer using a device and not as a designer trying
to learn something. That's what... 3Vrms? Into 8 ohms? A
voltage gain of 3, given 1Vrms input? I'm wanting to learn
some things, not place one BJT (okay, not really, but it
almost seems like that) down as an emitter follower and then
calling it good.

Up front, I thought I'd like to deal with perhaps something
on the order of about 10Vrms into 8 ohms. I figured that is
enough 'bad' that I'd have to cope with some interesting
corners along the way; but not enough 'bad' that I'd have to
deal with too much all at once.

For example, at around 10 watts or so, it's enough that I may
need to seriously consider avoiding class-A operation of the
output stage and move to class-B, instead. But it is low
enough that there is some room to discuss each, as well as
class-AB biasing, too. More power and I'm almost certain I'm
pushed into class-B. Less power and.. well, who cares that
much? At one watt or so, just class-A and be done with it? I
won't learn the reasoning behind trade-offs that way.

There's more. I just figured at about 10 watts I'm likely to
learn some things but not be forced to learn so much that I'm
overwhelmed.

I'm open to specific advice about all this, of course.

Jon
I like 10 watts as a starting size - at this size you have to start
doing things the way the big amps do, but it is not so big as to be
outrageously expensive, for example you still use a relatively small
power supply, heatsinks, and inexpensive transistors,

You appear to confirm my instincts.

and at the end you
can use it with your PC and really blow those 320 watt PFPO (peak
fantasy power output)speakers away.

Well, mostly I'm just trying to learn... not impress others
about the results.

I like your hesitation on class A. You want an amp with some power
output and class A is very inefficient, never more than 25% and often
way less. This would add greatly to the cost - a 40 watt power supply,
heatsinks capable of getting rid of the same as heat while keeping the
transistor junction temperature low, and beefier transistors. You also
get to put up with a shorter service life from all that heat. The "big
thing" with class A is there is no crossover distortion, which can lead
to better overall distortion figures, but the cost is huge - a kit for 2
x 20 watt class A sells for $600.

Egads. My instincts said class-A would add a lot to weight
and cost, but no idea a mere 20W kit could sell for $300!

Well it would cost something less at 10 watts.

My particular bias for an amp this size is to go class AB with a split
power supply. The majority of quality audio amps follow this topology
and this is, I think, I great reason to go down this design path (what
you learn is applicable in the most number of situations). I should hunt
down a schematics of what I'm seeing in the distance (which can/will
change as decisions are made) - some of the justifications will have to
wait

I'm fine with taking things as they come.

As far as the class, I guessed that at 10 watts class-A would
be too power-hungry and probably not worth its weight but
that class-AB might be okay.

I have to warn you, though, that I'm not focused upon some
20ppm THD. I'd like to learn, not design something whose
distortion (or noise, for that matter) is around a bit on a
16-bit DAC or less. I figure winding up close to class-B
operation in the end. But I'd like to take the walk along
the way, so to speak.

10 watts / PPM thd? Mmmm... maybe more like .1 - .05 % are realistic and
a few detours to see what would help or harm that.

The first step is to think about the output. The basic equations are

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

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

If you don't mind, I'd like to discuss this more closely. Not
just have it tossed out. So, P=V*I; or P=Vrms^2/R with AC.
Using Vpeak=SQRT(2)*Vrms, I get your Vpeak=SQRT(2*P*R)
equation. Which suggests the +/-12.7V swing. Which further
suggests, taking Vce drops and any small amounts emitter
resistor drops into account, something along the lines of +/-
14-15V rails?

Or should the rails be cut a lot closer to the edge here to
improve efficiency. What bothers me is saturation as Vce on
the final output BJTs goes well below 1V each and beta goes
away, as well, rapidly soaking up remaining drive compliance.

(2).....Imax = sqrt(2*P/R)

This comes out to 1.6 amps. You should probably also consider the case
when R speaker = 4 ohms when initially selecting a transistor for the
output 2.2 amps - remember this is max output current. The power supply
voltage will have to be somewhat higher than Vout to take into account
circuit drive requirements, ripple on the power supply and transformer
regulation etc.

Okay. I missed reading this when writing the above. Rather
than correct myself, I'll leave my thinking in place.

So yes, the rails will need to be a bit higher. Agreed. On
this subject, I'm curious about the need to _isolate_, just a
little, the rails used by the input stage vs the output stage
rails. I'm thinking an RC (or LC for another pole?) for
isolation. But I honestly don't know if that's helpful, or
not.

Mostly not needed, if you use a long tailed pair for the input / error
amplifier, but you might prefer some other arrangement so keep it in
mind if your circuit "motorboats"

Are you OK with connecting mains to a transformer? or would you rather
use an AC plug pack (10 watts is about the biggest amp a plugpack can be
used for)? The "cost" for using an AC plug pack is you will need larger
filter capacitors.

I'd much prefer to __avoid__ using someone else's "pack" for
the supply. All discrete parts should be on the table, so to
speak, in plain view. And I don't imagine _any_ conceptual
difficulties for this portion of the design. I'm reasonably
familiar with transformers, rectifiers, ripple calculations,
and how to consider peak charging currents vs averge load
currents as they relate to the phase angles available for
charging the caps. So on this part, I may need less help
than elsewhere. In other words, I'm somewhat comfortable
here.

Ah, then there are questions of what voltage and VA for a transformer.
So there are questions of usage (music, PA, PA with an emergency alert
siren tied in etc) and rectifier arrangement and capacitor size /
voltage to get your required voltage output at full load.

I should also ask if you have a multi meter, oscilloscope (not necessary
but useful)and how is your soldering? But it would be wise to keep this
whole thing as a paper exercise before you commit to anything.

I have a 6 1/2 digit HP multimeter, a Tek DMM916 true RMS
handheld, two oscilloscopes (TEK 2245 with voltmeter option
and an HP 54645D), three triple-output power supplies with
two of them GPIB drivable, the usual not-too-expensive signal
generator, and a fair bunch of other stuff on the shelves.
Lots of probes, clips, and so on. For soldering, I'm limited
to a Weller WTCPT and some 0.4mm round, 0.8mm spade, and
somewhat wider spade tips in the 1.5mm area. I have tubs and
jars of various types of fluxes, as well, and wire wrap tools
and wire wrap wire, as well. I also have a room set aside
for this kind of stuff, when I get time to play.

OK. Next serious project, I'm coming around to your place! Your gear is
better than mine. I had to ask, rather than just assume just in case my
assumptions got you building something you didn't want to, and got you
splattered all over the place from the mains, or suggesting you choose
the miller cap by watching the phase shift of the feedback circuit - I
don't read a lot of the posts so I didn't know what you could do.

Jon

Have a look at
http://en.wikipedia.org/wiki/Electronic_amplifier
The bits on class A might be interesting as it says 25% efficiency and
50% obtainable with inductive output coupling (i.e. with a transformer)
which is what I said, not what blow hard Phil said.

 

[Phil Allison]

"David Eather is a Fuckwit LIAR"

Have a look at
http://en.wikipedia.org/wiki/Electronic_amplifier
The bits on class A might be interesting as it says 25% efficiency and 50%
obtainable with inductive output coupling (i.e. with a transformer)
which is what I said, not what blow hard Phil said.

** ROTFLMAO !!!

So this Eather WANKER gets his WRONG info from bloody Wiki !!!!

Here is the actual quote:

" Class A amplifiers are the usual means of implementing small-signal
amplifiers. They are not very efficient; a theoretical maximum of 50% is
obtainable with inductive output coupling and only 25% with capacitive
coupling."

So, the para is clearly about "small signal" class A stages

- ie RC coupled pre-amp stages !!!!!!!

Not class A ** POWER AMPS ** !!!!!

Wot a fucking FUCKWIT !!

BTW

Inductive coupling also refers to the use of chokes as the collector or
plate loads for ( single ended ) class A operation.

Way over this Eather Google Monkey's pointy head.



..... Phil

 

[David Eather]

Phil Allison wrote:

"David Eather = a LIAR and a Wanker "


** Learn to trim - wanker!!

I like your hesitation on class A. You want an amp with some power
output and class A is very inefficient, never more than 25% and often
way less.
** Class A amplifiers are up to 50 % efficient.
Yes, if you use a matching transformer,

** 100%, absolute BULLSHIT !!!!!!!

YOU KNOW NOTHING MORON !!!!!!!

http://en.wikipedia.org/wiki/Electronic_amplifier
agrees with me and not with you. As does every textbook - they agree
with me about the efficiency of class A. If you are up to it prove it -
work an example where you get better than 25% - or show a text book
making your claim 50% (without using a transformer, which you said
wasn't needed)

This would add greatly to the cost - a 40 watt power supply,
** 20 watts is all that is needed.
IBID

** Fuck you - you damn imbecile.

Sorry it is you

heatsinks capable of getting rid of the same as heat while keeping the
transistor junction temperature low,
** Junction temps can settle at 125 C with no problems.
Most transistors are spec'ed at 125 C for 1000 or 2000 hours.


** More, 100% absolute BULLSHIT !!!!!!!

This RATBAG just keeps piling on the LIES !!!!!

How long do you think they are rated for Phil?

people would like there devices to run a bit longer than this. At least
one semiconductor manufacturer believes that for each lowering of the
junction temperature by 10 C doubles the life of transistor


** MORE 100% absolute BULLSHIT !!!!!!!

This RATBAG just keeps piling on the damn LIES !!

RCA Power Transistors Application Note AN-6688 "A Practical Approach to
Audio Amplifier Design" page 5

Would you like me to post you a copy?

You also get to put up with a shorter service life from all that heat.
** Bull.
I think just about everyone will disagree with you on that point.


** What some fuckwit like you choses to "think" is of zero consequence.


heat dries out electrolytics

** ROTFLMAO !!!

Shame BJTs are not electrolytic.

You FUCKING MORON !!!!!!!!


and heat stress cycling is a major failure mode of semiconductors


** No heat cycling occurs with class A.

Huge amounts occur with all class B and AB designs.

You FUCKING KNOW NOTHING MORON !!!!!!!!

If no heat cycling occurs then no signal output occurs either.

The "big thing" with class A is there is no crossover distortion, which
can lead to better overall distortion figures, but the cost is huge - a
kit for 2 x 20 watt class A sells for $600.
** Irrelevant what some unspecified kit sells for.

No, it is an example of how costly class A can become.

** Absolute fucking CRAP !!!!!!!!

YOU are an example of how STUPID people can become.

An extreme example of an damn LIAR too.

Please Phil, show us all how to build a cheap 2 x 20 watt class A amp

A 20 watt class A power stage ONLY needs bigger heatsinks compared to
usual low bias, class AB operation.

** No big expense there.



My particular bias for an amp this size is to go class AB with a split
power supply. The majority of quality audio amps follow this topology
** Bollocks.
Well, somewhat recently the trend has gone to mono block IC amps so I'll
take that on the chin

** Go drop dead - you pig ignorant LYING WANKER !


Only some very high powered hi-fi and pro-audio amps use additional DC
rails.

So what.


** Proves you WRONG - you stinking LIAR !!!!!!

No, it shows you can't construct a logical argument.

The power supply voltage will have to be somewhat higher than Vout to
take into account circuit drive requirements, ripple on the power supply
and transformer regulation etc.

** There is no variation in the DC current draw from a PSU with class A
amplifiers - in fact, this is the very definition of class A operation
of an audio amp. So the DC rails will remain steady from no drive to
full output.
True!

** Proves you WRONG AGAIN - you stinking LIAR !!!!!!

I agreed with you and you said that proves you wrong and makes you a
stinking liar. Correct!, you're wrong and you are a stinking liar

Also, the heatsink will cool considerably when the amp is operated at
full
sine wave power.

I thought there was no heat stress cycling, so why would the heatsink be
"cooler" wouldn't it be at a constant temperature?.

Not relevant at the moment.

** Course it is - you DAMN LIAR !!!!!!!!!
How is the heat sink (currently) relevant when there is no spec for the

power supply, no spec for transistor de-rating, no spec for heat
dissipation?

Please tell us all what is the correct size of heatsink and how did you
arrive at that magical number?

GO DROP FUCKING DEAD

- YOU BULLSHITTING IMBECILE !!!!!!!!!




..... Phil

Phil,

Is it hard to walk around with so much malice and vitriol? What does
your wife think of that? How do you treat her and others close to you?

You are simply a stain on humanity. When you die, as we all one day
must, those few who remember you will remember you as a negative
influence and the few people you would have influenced will be better
off without you.

That is a sad monument to your life. The best you are is an internet
troll and a stain on life.

 

[David Eather]

Phil Allison wrote:

"David Eather is a Fuckwit LIAR"

Have a look at
http://en.wikipedia.org/wiki/Electronic_amplifier
The bits on class A might be interesting as it says 25% efficiency and 50%
obtainable with inductive output coupling (i.e. with a transformer)
which is what I said, not what blow hard Phil said.


** ROTFLMAO !!!

So this Eather WANKER gets his WRONG info from bloody Wiki !!!!

Here is the actual quote:

" Class A amplifiers are the usual means of implementing small-signal
amplifiers. They are not very efficient; a theoretical maximum of 50% is
obtainable with inductive output coupling and only 25% with capacitive
coupling."

So, the para is clearly about "small signal" class A stages

- ie RC coupled pre-amp stages !!!!!!!

Not class A ** POWER AMPS ** !!!!!

Wot a fucking FUCKWIT !!

BTW

Inductive coupling also refers to the use of chokes as the collector or
plate loads for ( single ended ) class A operation.

Way over this Eather Google Monkey's pointy head.



..... Phil

So how are class A POWER AMPS more efficient? A little math please..
Oh, and you sniped or forgot to mention that the textbooks agree with me
too..

 

[Phil Allison]

"David Eather is a Fuckwit LIAR"

Have a look at
http://en.wikipedia.org/wiki/Electronic_amplifier
The bits on class A might be interesting as it says 25% efficiency and
50% obtainable with inductive output coupling (i.e. with a transformer)
which is what I said, not what blow hard Phil said.


** ROTFLMAO !!!

So this Eather WANKER gets his WRONG info from bloody Wiki !!!!

Here is the actual quote:

" Class A amplifiers are the usual means of implementing small-signal
amplifiers. They are not very efficient; a theoretical maximum of 50% is
obtainable with inductive output coupling and only 25% with capacitive
coupling."

So, the para is clearly about "small signal" class A stages

- ie RC coupled pre-amp stages !!!!!!!

Not class A ** POWER AMPS ** !!!!!

Wot a fucking FUCKWIT !!

BTW

Inductive coupling also refers to the use of chokes as the collector or
plate loads for ( single ended ) class A operation.

Way over this Eather Google Monkey's pointy head.



So how are class A POWER AMPS more efficient?

** Cos they operate in push pull - you IMBECILE !!

Look it up - you pig ignorant, arrogant, shit for brains LIAR !!

Forget stupid bloody Wiki cos it is full of missing info and mistakes.





..... Phil

 

[Phil Allison]

"David Eather = a LIAR and a Wanker "



** ABSOLUTELY IGNORE this

KNOW NOTHING IDIOT !!!!!!!!!!

He has ZERO tech knowledge.

He has ZERO actual experience with any audio electronics at all.

He is another pathetic GOOGLE MONKEY !!!

ALL his advice is WRONG !!!




..... Phil

 

[Jon Kirwan]

On Thu, 28 Jan 2010 11:17:02 +1000, David Eather
<eather@tpg.com.au> wrote:

Jon Kirwan wrote:

On Wed, 27 Jan 2010 17:31:00 +1000, David Eather
eather@tpg.com.au> wrote:
snip

My particular bias for an amp this size is to go class AB with a split
power supply. The majority of quality audio amps follow this topology
and this is, I think, I great reason to go down this design path (what
you learn is applicable in the most number of situations). I should hunt
down a schematics of what I'm seeing in the distance (which can/will
change as decisions are made) - some of the justifications will have to
wait

I'm fine with taking things as they come.

As far as the class, I guessed that at 10 watts class-A would
be too power-hungry and probably not worth its weight but
that class-AB might be okay.

I have to warn you, though, that I'm not focused upon some
20ppm THD. I'd like to learn, not design something whose
distortion (or noise, for that matter) is around a bit on a
16-bit DAC or less. I figure winding up close to class-B
operation in the end. But I'd like to take the walk along
the way, so to speak.

10 watts / PPM thd? Mmmm... maybe more like .1 - .05 % are realistic and
a few detours to see what would help or harm that.

Hehe. I'm thinking of some numbers I saw in the area of
..002% THD. I hate percentages and immediately convert them.
In this case, it is 20e-6 or 20 ppm. Which is darned close
to a bit on a 16-bit dac. That's why I wrote that way. I
just don't like using % figures. They annoy me just a tiny
bit.

Regarding .1% to .05%, I'm _very_ good with that. Of course,
I'm going to have to learn about how to estimate it from
theory as well as measure it both via simulation before
construction and from actual testing afterwards. More stuff
I might _think_ I have a feel for, but I'm sure I will
discover I don't as I get more into it.

But speaking from ignorance, I'm good shooting for the range
you mentioned. It was about what I had in mind, in fact,
figuring I could always learn as I go.

The first step is to think about the output. The basic equations are

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

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

If you don't mind, I'd like to discuss this more closely. Not
just have it tossed out. So, P=V*I; or P=Vrms^2/R with AC.
Using Vpeak=SQRT(2)*Vrms, I get your Vpeak=SQRT(2*P*R)
equation. Which suggests the +/-12.7V swing. Which further
suggests, taking Vce drops and any small amounts emitter
resistor drops into account, something along the lines of +/-
14-15V rails?

Or should the rails be cut a lot closer to the edge here to
improve efficiency. What bothers me is saturation as Vce on
the final output BJTs goes well below 1V each and beta goes
away, as well, rapidly soaking up remaining drive compliance.

(2).....Imax = sqrt(2*P/R)

This comes out to 1.6 amps. You should probably also consider the case
when R speaker = 4 ohms when initially selecting a transistor for the
output 2.2 amps - remember this is max output current. The power supply
voltage will have to be somewhat higher than Vout to take into account
circuit drive requirements, ripple on the power supply and transformer
regulation etc.

Okay. I missed reading this when writing the above. Rather
than correct myself, I'll leave my thinking in place.

So yes, the rails will need to be a bit higher. Agreed. On
this subject, I'm curious about the need to _isolate_, just a
little, the rails used by the input stage vs the output stage
rails. I'm thinking an RC (or LC for another pole?) for
isolation. But I honestly don't know if that's helpful, or
not.

Mostly not needed, if you use a long tailed pair for the input / error
amplifier, but you might prefer some other arrangement so keep it in
mind if your circuit "motorboats"

Okay. I've _zero_ experience for audio. It just crossed my
mind from other cases. I isolate the analog supply from the
digital -- sometimes with as many as four caps and three
inductor beads. There, it _does_ help.

Are you OK with connecting mains to a transformer? or would you rather
use an AC plug pack (10 watts is about the biggest amp a plugpack can be
used for)? The "cost" for using an AC plug pack is you will need larger
filter capacitors.

I'd much prefer to __avoid__ using someone else's "pack" for
the supply. All discrete parts should be on the table, so to
speak, in plain view. And I don't imagine _any_ conceptual
difficulties for this portion of the design. I'm reasonably
familiar with transformers, rectifiers, ripple calculations,
and how to consider peak charging currents vs averge load
currents as they relate to the phase angles available for
charging the caps. So on this part, I may need less help
than elsewhere. In other words, I'm somewhat comfortable
here.

Ah, then there are questions of what voltage and VA for a transformer.
So there are questions of usage (music, PA, PA with an emergency alert
siren tied in etc) and rectifier arrangement and capacitor size /
voltage to get your required voltage output at full load.

I figure on working out the design of the amplifier and then
going back, once that is determined and hashed out, with the
actual required figures for the power supply and design that
part as the near-end of the process. Earlier on, I'd expect
to have some rough idea of how "bad" it needs to be -- if the
initial guesses don't raise alarms, then I wouldn't dig into
the power supply design until later on. The amplifier, it
seems to me, dictates the parameters. So that comes later,
doesn't it?

I should also ask if you have a multi meter, oscilloscope (not necessary
but useful)and how is your soldering? But it would be wise to keep this
whole thing as a paper exercise before you commit to anything.

I have a 6 1/2 digit HP multimeter, a Tek DMM916 true RMS
handheld, two oscilloscopes (TEK 2245 with voltmeter option
and an HP 54645D), three triple-output power supplies with
two of them GPIB drivable, the usual not-too-expensive signal
generator, and a fair bunch of other stuff on the shelves.
Lots of probes, clips, and so on. For soldering, I'm limited
to a Weller WTCPT and some 0.4mm round, 0.8mm spade, and
somewhat wider spade tips in the 1.5mm area. I have tubs and
jars of various types of fluxes, as well, and wire wrap tools
and wire wrap wire, as well. I also have a room set aside
for this kind of stuff, when I get time to play.

OK. Next serious project, I'm coming around to your place!

You come to the west coast of the US and I'll have a room for
you!

Your gear is
better than mine. I had to ask, rather than just assume just in case my
assumptions got you building something you didn't want to, and got you
splattered all over the place from the mains, or suggesting you choose
the miller cap by watching the phase shift of the feedback circuit - I
don't read a lot of the posts so I didn't know what you could do.

To be honest, I can do a few things but I'm really not very
practiced. My oscilloscope knowledge is lacking in some
areas -- which becomes all too painfully obvious to me when I
watch a pro using my equipment. And I'm still learning to
solder better. It's one of a few hobbies.

Jon

Have a look at
http://en.wikipedia.org/wiki/Electronic_amplifier

Done.

The bits on class A might be interesting as it says 25% efficiency and
50% obtainable with inductive output coupling (i.e. with a transformer)
which is what I said, not what blow hard Phil said.

What I first see there is the amplifier sketch at the top of
the page (I don't really care too much about arguing about
efficiencies right now -- I'm more concerned about learning.)
The input stage shown is a voltage-in, current-out bog
standard diff-pair. First thing I remember about is that R4
shouldn't be there and better still both R3 and R4 should be
replaced with a current mirror. R5 should be a replaced with
a BJT, as well. I assume the input impedance of that example
is basically the parallel resistance of R1 and R2, but if we
use split supplies I'd imagine replacing the two of them with
a single resistor to the center-ground point. There's no
miller cap on Q3, I'd probably replace the two diodes with
one of those BJT and a few resistor constructions I can't
remember the name of (which allows me to adjust the drop.)
The feedback ... well, I need to think about that a little
more. There's no degen resistors in the emitters of Q4 and
Q5.

Um.. okay, I need to sit down and think. Mind is spinning,
but I've not set a finger to paper yet and there is lots to
think about in that one. I could be way, way off base.

Jon

 

[Jon Kirwan]

On Wed, 27 Jan 2010 18:51:00 -0800, I wrote:

The amplifier, it
seems to me, dictates the parameters. So that comes later,
doesn't it?

By "that" I mean "the power supply." In case it isn't clear.

Jon

 

[pimpom]

Jon Kirwan wrote:

On Thu, 28 Jan 2010 11:17:02 +1000, David Eather
eather@tpg.com.au> wrote:


Have a look at
http://en.wikipedia.org/wiki/Electronic_amplifier

Done.

The bits on class A might be interesting as it says 25%
efficiency
and 50% obtainable with inductive output coupling (i.e. with a
transformer) which is what I said, not what blow hard Phil
said.

What I first see there is the amplifier sketch at the top of
the page (I don't really care too much about arguing about
efficiencies right now -- I'm more concerned about learning.)
The input stage shown is a voltage-in, current-out bog
standard diff-pair. First thing I remember about is that R4
shouldn't be there and better still both R3 and R4 should be
replaced with a current mirror. R5 should be a replaced with
a BJT, as well. I assume the input impedance of that example
is basically the parallel resistance of R1 and R2, but if we
use split supplies I'd imagine replacing the two of them with
a single resistor to the center-ground point. There's no
miller cap on Q3, I'd probably replace the two diodes with
one of those BJT and a few resistor constructions I can't
remember the name of (which allows me to adjust the drop.)
The feedback ... well, I need to think about that a little
more. There's no degen resistors in the emitters of Q4 and
Q5.

Either the circuit was designed by someone with a limited
experience or it was deliberately presented this way for clarity
as an illustration of the basic topology.

In a practical design using an unregulated power supply, R1
should be split into two with a capacitor from the split point to
ground. This is to decouple the input stage for stability as well
as for hum filtering. R6 should also be split and the split point
bootstrapped with a capacitor to the mid point of the output
stage.

Talking about RCA's 70W amp got me nostalgic about those days.
Here's a 1W amp using _germanium_ transistors:
http://img716.imageshack.us/img716/2583/1wamp.png
This is one of my early solid-state designs based, of course, on
topologies I'd learned by studying others' designs. It's no hi-fi
by any stretch of imagination, but I actually constructed a few
of these in the early 70s for myself and for friends. One of them
fed the input from an early Sony Walkman to drive an 8-inch
Philips dual-cone "Hi-Q" speaker and gushed over how good it
sounded!

 

[Jon Kirwan]

On Thu, 28 Jan 2010 14:00:18 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Thu, 28 Jan 2010 11:17:02 +1000, David Eather
eather@tpg.com.au> wrote:


Have a look at
http://en.wikipedia.org/wiki/Electronic_amplifier

Done.

The bits on class A might be interesting as it says 25%
efficiency
and 50% obtainable with inductive output coupling (i.e. with a
transformer) which is what I said, not what blow hard Phil
said.

What I first see there is the amplifier sketch at the top of
the page (I don't really care too much about arguing about
efficiencies right now -- I'm more concerned about learning.)
The input stage shown is a voltage-in, current-out bog
standard diff-pair. First thing I remember about is that R4
shouldn't be there and better still both R3 and R4 should be
replaced with a current mirror. R5 should be a replaced with
a BJT, as well. I assume the input impedance of that example
is basically the parallel resistance of R1 and R2, but if we
use split supplies I'd imagine replacing the two of them with
a single resistor to the center-ground point. There's no
miller cap on Q3, I'd probably replace the two diodes with
one of those BJT and a few resistor constructions I can't
remember the name of (which allows me to adjust the drop.)
The feedback ... well, I need to think about that a little
more. There's no degen resistors in the emitters of Q4 and
Q5.

Either the circuit was designed by someone with a limited
experience or it was deliberately presented this way for clarity
as an illustration of the basic topology.

Well, having looked a little more at the web site, I see them
talking about everything from opamps to servo amps so maybe
they just did the basics. But they missed the signal split
technique aka the old tube days, then.

In a practical design using an unregulated power supply, R1
should be split into two with a capacitor from the split point to
ground. This is to decouple the input stage for stability as well
as for hum filtering.

Now _that_ makes a lot of sense. I missed it.

R6 should also be split and the split point
bootstrapped with a capacitor to the mid point of the output
stage.

That one I really need to think about. This is what I wanted
to happen here. Throwing out things (I'm assuming correct
things, of course) that force me to consider and think.
Thanks.

Talking about RCA's 70W amp got me nostalgic about those days.
Here's a 1W amp using _germanium_ transistors:
http://img716.imageshack.us/img716/2583/1wamp.png
This is one of my early solid-state designs based, of course, on
topologies I'd learned by studying others' designs. It's no hi-fi
by any stretch of imagination, but I actually constructed a few
of these in the early 70s for myself and for friends. One of them
fed the input from an early Sony Walkman to drive an 8-inch
Philips dual-cone "Hi-Q" speaker and gushed over how good it
sounded!

Okay. So lets talk about some aspects. It'll expose my
terrible ignorance, but what the heck.

Input loading. I think I can ignore the R2 feedback as it is
10k. At least, for now. C1 will present about Z=800 at
20Hz, Z=160 at 100Hz, and Z goes down from there. R1 is 1k,
obviously in series with C1. Then there is R3=1k in parallel
with Q1's impedance, which maybe I can approximate as R4
times beta, or call it 50*33 or about 1500 ohms? So about
600 ohms counting that and R3 in parallel, that itself in
series with 1k and whatever C1 presents? So call it around
2k ohms loading, or so? (Which adds to the idea that the R2
feedback can be mostly ignored as a load.) Would that be an
okay, off-the-hip guess? Or how would you go about it?

D1 is, I guess, silicon and given that you said _germanium_,
I'll take that to suggest that the Vbe on those are about
half that of a silicon BJT. Which is why only one DR25 was
needed there.

DC bias point of Q1... hmm. Well, assuming no signal, SPK1
is roughly a dead short, so R5 is tied one side to a rail.
The other side moves Q2's base and Q2's emitter follows. As
Q2's emitter rises with it, R2 and R3 act to split that as
1/11th to the Q1 base. Q1's emitter follows up for a ways,
allowing DC current via R4 which must go through R5, dropping
Q2's base and thus Q2's emitter, lowering Q1's base voltage
in opposition. So there will be a middle point found.

Assuming Q1's Vbe should be something on the order of 300mV
(random guess), and I(R4) roughly equals I(R5), let's
establish where Q1's base will wind up. Call it Vb. The
value at Q2's emitter (which is also the other side of R2
from the Q1 base) will be 11 times higher because R2 and R3
split things that way. And Q2's base will be 300mV (same
random guess, again) higher than that. The difference
between there and the 9V battery voltage sets the current in
R5 and, by implication, in R4 as well. Of course, Q1's
emitter is 300mV away from that Vb value we are fussing over.
The equation looks like:

I(R5) = (9V - Vb*11 - 300mV) / 560
I(R4) = (Vb - 300mV) / 33
I(R4) = I(R5)
So,
(9V - Vb*11 - 300mV) / 560 = (Vb - 300mV) / 33
33/560 * (9V - Vb*11 - 300mV) = (Vb - 300mV)
Vb = 33/560 * (9V - Vb*11 - 300mV) + 300mV
Vb = 33/560*9V - 33/560*Vb*11 - 33/560*300mV + 300mV
Vb + 33/560*Vb*11 = 33/560*9V - 33/560*300mV + 300mV
Vb * (1 + 33/560*11) = (33/560*9V - 33/560*300mV + 300mV)
Vb = (33/560*9V - 33/560*300mV + 300mV) / (1 + 33/560*11)
or,
Vb = 493mV
and thus the current routing through R5, D1, Q1, and R4 is
about 193mV/33 or 5.85mA. That's not the total quiescent
current because D1 uses that 5.85mA to develop a voltage
across it that is probably on the order of 700mV. With that
between the Q2 and Q3 bases, both Q2 and Q3 are passing
collector currents, rail to rail. Hard to know how much
without data sheets, I suppose. But something. Their shared
emitter node would be on the order of 11*490mV or about 5.4V.

That neglected the base current for Q1 flowing via R2. As
I'm now guessing almost 6mA as Ic, and since we are talking
germanium here, I will pick a beta of about 60 and figure
about 100uA base current, then. That's about another 1V
across R2, less than that a little because that lowers Vb a
bit which lowers the 5.85mA figure a bit, which probably then
gets things very darned close to the midpoint of 4.5V one
might wish there.

Not too bad given I have no idea about the BJTs and am using
a lot of random guesses as I go.

R2 is not only a DC divider but also NFB, I think. Can you
talk a little about how you figure on calculating both the
NFB you want _and_ the DC biasing of this thing, both of
which affect R2's value, I think? And although I've _seen_
miller feedbacks in the small nF range, could you talk a
little about how that was set at 2.2nF? Also, I think I
_almost_ get the idea of hooking one side of R5 to SPK1
instead of to the (-) side of 9V... but not quite sure. Can
you talk about that choice, as well?

Have at me. I probably got a lot wrong in the above, but
that's my thinking exposed like a soft worm to be crushed. If
I learn in the process, crush away!

Jon

 

[pimpom]

Jon Kirwan wrote:

On Thu, 28 Jan 2010 14:00:18 +0530, "pimpom"
pimpom@invalid.invalid> wrote:


Talking about RCA's 70W amp got me nostalgic about those days.
Here's a 1W amp using _germanium_ transistors:
http://img716.imageshack.us/img716/2583/1wamp.png
This is one of my early solid-state designs based, of course,
on
topologies I'd learned by studying others' designs. It's no
hi-fi
by any stretch of imagination, but I actually constructed a
few
of these in the early 70s for myself and for friends. One of
them
fed the input from an early Sony Walkman to drive an 8-inch
Philips dual-cone "Hi-Q" speaker and gushed over how good it
sounded!

Okay. So lets talk about some aspects. It'll expose my
terrible ignorance, but what the heck.

I'm no expert myself, but I'm willing to help where I can.

Input loading. I think I can ignore the R2 feedback as it is
10k. At least, for now.

R2 provides both dc and ac feedback. DC for bias stabilisation
and setting the emitters of the output transistors to about half
of Vcc (more about that later). For ac, it may be easier to think
of it as current feedback. Q2 needs about +/-50uA peak of base
current at full drive. At signal frequencies, R2 (plus the much
smaller input impedance of Q1) is effectively in parallel with
the output. The output swings by about 4V peak at max power,
which has 400uA of negative feedback current going back through
R2. The input current requirement goes up by a factor of 9. IOW,
a negative feedback of 19db. This is substantially better than
nothing and should significantly reduce distortion and improve
frequency response.

C1 will present about Z=800 at
20Hz, Z=160 at 100Hz, and Z goes down from there. R1 is 1k,
obviously in series with C1.

The already low input impedance of Q1 is further reduced by the
negative feedback, so R1 represents practically the whole input
impedance of the amp. The -3db cutoff frequency is 1/2*pi*C1*R1
which is about 16Hz.

Then there is R3=1k in parallel
with Q1's impedance, which maybe I can approximate as R4
times beta, or call it 50*33 or about 1500 ohms?

No. R4 is bypassed by C3 and has little effect on input impedance
except at very low frequencies.

So about
600 ohms counting that and R3 in parallel, that itself in
series with 1k and whatever C1 presents? So call it around
2k ohms loading, or so? (Which adds to the idea that the R2
feedback can be mostly ignored as a load.) Would that be an
okay, off-the-hip guess? Or how would you go about it?

It's mostly the internal dynamic emitter resistance that
determines Q1's input impedance. That resistance is 26/Ie at 20
deg C. Q2 is biased at about 7.7mA emitter current, giving about
3.4 ohms. Multiply that by hfe, add the ohmic base resistance and
you get Q1's basic input Z. I don't have my old data book handy,
but I think the AC126 had a typical hfe of about 150 and rbb of
maybe 100 ohms. This gives an input Z of about 600 ohms.

D1 is, I guess, silicon and given that you said _germanium_,
I'll take that to suggest that the Vbe on those are about
half that of a silicon BJT. Which is why only one DR25 was
needed there.

The output transistors need only about 0.1V each of Vbe to bias
them at a few mAs of Ic. D1 is germanium, but at the dc current
level flowing through it, two of them in series will have too
much voltage drop (I measured several samples). Ge transistors
have a more rounded knee than their Si counterparts in the Vbe
vs. Ic curve. So I felt that a single diode would present less
chance of thermal runaway for the output Trs and still cause a
reasonably low crossover distortion.

Oops. Have to go out for a while. Will take up the rest later.

 

[pimpom]

Jon Kirwan wrote:

On Thu, 28 Jan 2010 14:00:18 +0530, "pimpom"
pimpom@invalid.invalid> wrote:


R6 should also be split and the split point
bootstrapped with a capacitor to the mid point of the output
stage.

That one I really need to think about. This is what I wanted
to happen here. Throwing out things (I'm assuming correct
things, of course) that force me to consider and think.
Thanks.

This is what happens without a booststrap: When it's Q5's turn to

conduct on the negative half-cycle of the signal, the base drive
current has to come via R6. At the same time, Q5's current is
pulling its emitter - and therefore the base - down towards
ground, decreasing the voltage drop across R6. This decreases the
base drive current available just when it's needed.

Now look at it modified with a boostrap:
http://img715.imageshack.us/img715/4259/boostrap.png

For simplicity, let R6 = R7. At steady-state, C will be charged
to about a quarter of Vcc. When Q5 pulls its emitter (and
therefore the positive electrode of C) towards ground, the
voltage across C cannot change instantaneously and will push its
negative terminal down too. Beyond a certain level of drive, the
junction of C, R6 and R7 will even go down past oV and become
negative with respect to ground. This maintains the voltage
across R6 at an approximately constant level.

Oops again. Guests this time. Will be back when I can.

 

[pimpom]

Jon Kirwan wrote:

D1 is, I guess, silicon and given that you said _germanium_,
I'll take that to suggest that the Vbe on those are about
half that of a silicon BJT. Which is why only one DR25 was
needed there.

I think this is where I left off earlier.

DC bias point of Q1... hmm. Well, assuming no signal, SPK1
is roughly a dead short, so R5 is tied one side to a rail.
The other side moves Q2's base and Q2's emitter follows. As
Q2's emitter rises with it, R2 and R3 act to split that as
1/11th to the Q1 base. Q1's emitter follows up for a ways,
allowing DC current via R4 which must go through R5, dropping
Q2's base and thus Q2's emitter, lowering Q1's base voltage
in opposition. So there will be a middle point found.

Assuming Q1's Vbe should be something on the order of 300mV
(random guess), and I(R4) roughly equals I(R5), let's
establish where Q1's base will wind up. Call it Vb. The
value at Q2's emitter (which is also the other side of R2
from the Q1 base) will be 11 times higher because R2 and R3
split things that way. And Q2's base will be 300mV (same
random guess, again) higher than that. The difference
between there and the 9V battery voltage sets the current in
R5 and, by implication, in R4 as well. Of course, Q1's
emitter is 300mV away from that Vb value we are fussing over.
The equation looks like:

I(R5) = (9V - Vb*11 - 300mV) / 560
I(R4) = (Vb - 300mV) / 33
I(R4) = I(R5)
So,
(9V - Vb*11 - 300mV) / 560 = (Vb - 300mV) / 33
33/560 * (9V - Vb*11 - 300mV) = (Vb - 300mV)
Vb = 33/560 * (9V - Vb*11 - 300mV) + 300mV
Vb = 33/560*9V - 33/560*Vb*11 - 33/560*300mV + 300mV
Vb + 33/560*Vb*11 = 33/560*9V - 33/560*300mV + 300mV
Vb * (1 + 33/560*11) = (33/560*9V - 33/560*300mV + 300mV)
Vb = (33/560*9V - 33/560*300mV + 300mV) / (1 + 33/560*11)
or,
Vb = 493mV
and thus the current routing through R5, D1, Q1, and R4 is
about 193mV/33 or 5.85mA. That's not the total quiescent
current because D1 uses that 5.85mA to develop a voltage
across it that is probably on the order of 700mV. With that
between the Q2 and Q3 bases, both Q2 and Q3 are passing
collector currents, rail to rail. Hard to know how much
without data sheets, I suppose. But something. Their shared
emitter node would be on the order of 11*490mV or about 5.4V.

That neglected the base current for Q1 flowing via R2. As
I'm now guessing almost 6mA as Ic, and since we are talking
germanium here, I will pick a beta of about 60 and figure
about 100uA base current, then. That's about another 1V
across R2, less than that a little because that lowers Vb a
bit which lowers the 5.85mA figure a bit, which probably then
gets things very darned close to the midpoint of 4.5V one
might wish there.

Not too bad given I have no idea about the BJTs and am using
a lot of random guesses as I go.

Your reasoning is correct. However, the output transistors Q2 and
Q3 need only about 100mV each of Vbe for Class AB bias. IIRC,
beta of Q1 is about 150 and Vbe at that level of current is about
0.12V.

I don't know about others, but with low voltage circuits, I
usually try to fix the quiescent voltage at the output mid-point
at slightly more than half of Vcc. This is because Q1's Ve plus
its Vcesat reduces the available downward swing of Q3's base.

For this design, I tentatively chose a target of 4.6V at Q3's
emitter. Add Q2's Vbe and that leaves 4.3V for R5 plus the
speaker's dc resistance. The speaker's resistance has only a
minor effect but, just for the heck of it, let's take it as 6
ohms. So Q1's Ic = 4.3/566 = 7.6mA.

Q1's dc beta = 150, so Ib is about 50uA, and Ie = 7.65mA = I(R4).
7.65*33 places Q1's emitter at 252.45mV above ground. That plus
Vbe of 0.12V gives Vb = 372.45mV.
I(R3) = 372.45uA
I(R2) = I(R3) + Ib = 422.45uA
I(R2)*R2 = 4.2245V
V(R2) + Vb = (4.2245 + 0.37245)V = 4.59695V.

It just so happens that, in this case, common resistor values
produce almost exactly the desired quiescent bias level. If they
didn't, a slight departure from the target voltages would be
acceptable. In any case, tolerances on resistor values and
transistor characteristics could throw off actual values a bit.

R2 is not only a DC divider but also NFB, I think. Can you
talk a little about how you figure on calculating both the
NFB you want _and_ the DC biasing of this thing, both of
which affect R2's value, I think?

For such a simple design without a high level of audio quality as
the target, I wasn't too particular about the amount of signal
feedback as long as it's a reasonable amount. I chose a
compromise value for R3 first - low enough for bias stability so
that the current through it would be several times Ib, but not
too low to avoid excessive shunting of the signal input current.
Then I let the value of R2 be what it needs to be for correct
bias.

Then I calculate the amount of NFB as outlined in one of my
earlier replies and accept it if it's within reason. If I really
wanted more NFB, I'd parallel R2 with another resistor, but with
a capacitor in series to avoid upsetting the dc levels. BTW, that
can be used to provide some bass boost by choosing the proper
values of cap and resistor.

And although I've _seen_
miller feedbacks in the small nF range, could you talk a
little about how that was set at 2.2nF?

That's a guesstimated value, partly empirical and partly based on
observation of other people's designs. No PCs and simulation
software 40 years ago. For such a simple circuit, I didn't bother
with complex calculations for loops and phase shifts that
wouldn't be precise anyway due to wide tolerances in component
characteristics.

The reason for the relatively high capacitance is that this was a
low-Z low-gain circuit. But I might have made a mistake in
showing it now as 2.2nF. I might have used something like 1nF.

Also, I think I
_almost_ get the idea of hooking one side of R5 to SPK1
instead of to the (-) side of 9V... but not quite sure. Can
you talk about that choice, as well?

This is a variation of the bootstrap circuit I described in my
other post. R5 and the speaker serve the same functions as R6 and
R7 respectively in the other circuit.

 

[Jon Kirwan]

On Thu, 28 Jan 2010 19:19:06 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Thu, 28 Jan 2010 14:00:18 +0530, "pimpom"
pimpom@invalid.invalid> wrote:

Talking about RCA's 70W amp got me nostalgic about those days.
Here's a 1W amp using _germanium_ transistors:
http://img716.imageshack.us/img716/2583/1wamp.png
This is one of my early solid-state designs based, of course,
on
topologies I'd learned by studying others' designs. It's no
hi-fi
by any stretch of imagination, but I actually constructed a
few
of these in the early 70s for myself and for friends. One of
them
fed the input from an early Sony Walkman to drive an 8-inch
Philips dual-cone "Hi-Q" speaker and gushed over how good it
sounded!

Okay. So lets talk about some aspects. It'll expose my
terrible ignorance, but what the heck.

I'm no expert myself, but I'm willing to help where I can.

Input loading. I think I can ignore the R2 feedback as it is
10k. At least, for now.

R2 provides both dc and ac feedback. DC for bias stabilisation
and setting the emitters of the output transistors to about half
of Vcc (more about that later).

This much I can see, immediately. And thanks for dealing
with more, later on.

For ac, it may be easier to think of it as current feedback.

Okay.

Q2 needs about +/-50uA peak of base
current at full drive. At signal frequencies, R2 (plus the much
smaller input impedance of Q1) is effectively in parallel with
the output.

R2 is connected from the output to an input, which
effectively doesn't move much after arriving at it's DC bias
point. As you later point out, the _AC_ input impedance is
lowish (near 600 ohms), so the 10k is pretty close to one of
the rails at AC, anyway. Is that a different way of saying
what you just said? Or would you modify it?

The output swings by about 4V peak at max power,
which has 400uA of negative feedback current going back through
R2. The input current requirement goes up by a factor of 9. IOW,
a negative feedback of 19db. This is substantially better than
nothing and should significantly reduce distortion and improve
frequency response.

Okay. This goes past me a little (as if maybe the earlier
point didn't.) I'd like to try and get a handle on it.

Let's start with the 4V peak swing at max power.

Since you are discussing AC and converting it 400uA current
via the 10k, I would normally take this to mean 4Vrms AC.
Which in Vp-p terms would be 2*SQRT(2) larger, or 11.3V which
I know is impossible without accounting for the BJTs, given
the 9V supply. So this forces me to think in terms of
something else. But what? Did you mean 4Vpeak, which would
be 8Vp-p? If so, that would be about 2.8Vrms. In that case,
wouldn't a better "understanding" come from then saying that
the negative feedback is closer to 280uA?

The next point is on your use of "goes up by a factor of 9."
Can you elaborate more on this topic? Where the 9 comes
from? For volts, not power, I think I can gather the point
that 20*log(9) = 19.085), so I'm not talking about that
conventional formula. I'm asking about the 9, itself, and
also your thinking along the lines of concluding that it
significantly reduces distortion. How does one decide how
much is enough?

C1 will present about Z=800 at
20Hz, Z=160 at 100Hz, and Z goes down from there. R1 is 1k,
obviously in series with C1.

The already low input impedance of Q1 is further reduced by the
negative feedback, so R1 represents practically the whole input
impedance of the amp. The -3db cutoff frequency is 1/2*pi*C1*R1
which is about 16Hz.

I think I follow. Leading towards a comment you make soon
below (about C3), C3's very low impedance even after beta
multiplication almost bypasses R3 completely so the impedance
is as you say, mostly depending upon C1 and R1. I think.

Then there is R3=1k in parallel
with Q1's impedance, which maybe I can approximate as R4
times beta, or call it 50*33 or about 1500 ohms?

No. R4 is bypassed by C3 and has little effect on input impedance
except at very low frequencies.

Thanks for the knock in the head there. I had been ignoring
C3 for DC bais-point thinking and forgot to put it back in
when talking about AC loading. Your point is made.

So about
600 ohms counting that and R3 in parallel, that itself in
series with 1k and whatever C1 presents? So call it around
2k ohms loading, or so? (Which adds to the idea that the R2
feedback can be mostly ignored as a load.) Would that be an
okay, off-the-hip guess? Or how would you go about it?

It's mostly the internal dynamic emitter resistance that
determines Q1's input impedance. That resistance is 26/Ie at 20
deg C. Q2 is biased at about 7.7mA emitter current, giving about
3.4 ohms. Multiply that by hfe, add the ohmic base resistance and
you get Q1's basic input Z. I don't have my old data book handy,
but I think the AC126 had a typical hfe of about 150 and rbb of
maybe 100 ohms. This gives an input Z of about 600 ohms.

Okay. I get your point about small-case 're' based upon kT/q
and Ie. I'm mostly following here.

D1 is, I guess, silicon and given that you said _germanium_,
I'll take that to suggest that the Vbe on those are about
half that of a silicon BJT. Which is why only one DR25 was
needed there.

The output transistors need only about 0.1V each of Vbe to bias
them at a few mAs of Ic. D1 is germanium, but at the dc current
level flowing through it, two of them in series will have too
much voltage drop (I measured several samples). Ge transistors
have a more rounded knee than their Si counterparts in the Vbe
vs. Ic curve. So I felt that a single diode would present less
chance of thermal runaway for the output Trs and still cause a
reasonably low crossover distortion.

Thanks.

Oops. Have to go out for a while. Will take up the rest later.

Well, I think I'm roughly following so far. Please kick me
where I'm still off-track, if you feel you can afford the
moment to do it. I appreciate it very much.

Jon

 

[Jon Kirwan]

On Fri, 29 Jan 2010 00:20:15 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Thu, 28 Jan 2010 14:00:18 +0530, "pimpom"
pimpom@invalid.invalid> wrote:


R6 should also be split and the split point
bootstrapped with a capacitor to the mid point of the output
stage.

That one I really need to think about. This is what I wanted
to happen here. Throwing out things (I'm assuming correct
things, of course) that force me to consider and think.
Thanks.

This is what happens without a booststrap: When it's Q5's turn to
conduct on the negative half-cycle of the signal, the base drive
current has to come via R6. At the same time, Q5's current is
pulling its emitter - and therefore the base - down towards
ground, decreasing the voltage drop across R6. This decreases the
base drive current available just when it's needed.

Now look at it modified with a boostrap:
http://img715.imageshack.us/img715/4259/boostrap.png

For simplicity, let R6 = R7. At steady-state, C will be charged
to about a quarter of Vcc. When Q5 pulls its emitter (and
therefore the positive electrode of C) towards ground, the
voltage across C cannot change instantaneously and will push its
negative terminal down too. Beyond a certain level of drive, the
junction of C, R6 and R7 will even go down past oV and become
negative with respect to ground. This maintains the voltage
across R6 at an approximately constant level.

Oops again. Guests this time. Will be back when I can.

I'm appreciating this very much. Printed the two and am
looking at both. I think I'm following, but need to sit down
and do a little paper calcs to make sure I burn it in a
little more. If I think of something useful, I'll post a
question or two. For now, I feel like I'm following you.

Jon

 

[sparky]

On Jan 27, 9:16 pm, "Phil Allison" <phi...@tpg.com.au> wrote:

 "David Eather is a Fuckwit LIAR"







Have a look at
http://en.wikipedia.org/wiki/Electronic_amplifier
The bits on class A might be interesting as it says 25% efficiency and
50% obtainable with inductive output coupling (i.e. with a transformer)
which is what I said, not what blow hard Phil said.

** ROTFLMAO !!!

So this Eather WANKER  gets his WRONG info from bloody Wiki  !!!!

Here is the actual quote:

" Class A amplifiers are the usual means of implementing small-signal
amplifiers. They are not very efficient; a theoretical maximum of 50% is
obtainable with inductive output coupling and only 25% with capacitive
coupling."

So, the para is clearly about "small signal" class A stages

 -  ie RC coupled pre-amp stages  !!!!!!!

Not class A  ** POWER AMPS **   !!!!!

Wot a fucking FUCKWIT !!

BTW

Inductive coupling also refers to the use of chokes as the collector or
plate loads for ( single ended ) class A operation.

Way over this Eather Google Monkey's pointy head.

So how are class A POWER AMPS more efficient?

** Cos they operate in push pull -  you  IMBECILE !!

 Look it up -   you pig ignorant, arrogant, shit for brains  LIAR !!

 Forget stupid bloody Wiki cos it is full of missing info and mistakes.

....  Phil- Hide quoted text -

- Show quoted text -

Phils panties are all in a twist ! Again !

 

[Jon Kirwan]

On Fri, 29 Jan 2010 02:09:16 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:

D1 is, I guess, silicon and given that you said _germanium_,
I'll take that to suggest that the Vbe on those are about
half that of a silicon BJT. Which is why only one DR25 was
needed there.

I think this is where I left off earlier.

DC bias point of Q1... hmm. Well, assuming no signal, SPK1
is roughly a dead short, so R5 is tied one side to a rail.
The other side moves Q2's base and Q2's emitter follows. As
Q2's emitter rises with it, R2 and R3 act to split that as
1/11th to the Q1 base. Q1's emitter follows up for a ways,
allowing DC current via R4 which must go through R5, dropping
Q2's base and thus Q2's emitter, lowering Q1's base voltage
in opposition. So there will be a middle point found.

Assuming Q1's Vbe should be something on the order of 300mV
(random guess), and I(R4) roughly equals I(R5), let's
establish where Q1's base will wind up. Call it Vb. The
value at Q2's emitter (which is also the other side of R2
from the Q1 base) will be 11 times higher because R2 and R3
split things that way. And Q2's base will be 300mV (same
random guess, again) higher than that. The difference
between there and the 9V battery voltage sets the current in
R5 and, by implication, in R4 as well. Of course, Q1's
emitter is 300mV away from that Vb value we are fussing over.
The equation looks like:

I(R5) = (9V - Vb*11 - 300mV) / 560
I(R4) = (Vb - 300mV) / 33
I(R4) = I(R5)
So,
(9V - Vb*11 - 300mV) / 560 = (Vb - 300mV) / 33
33/560 * (9V - Vb*11 - 300mV) = (Vb - 300mV)
Vb = 33/560 * (9V - Vb*11 - 300mV) + 300mV
Vb = 33/560*9V - 33/560*Vb*11 - 33/560*300mV + 300mV
Vb + 33/560*Vb*11 = 33/560*9V - 33/560*300mV + 300mV
Vb * (1 + 33/560*11) = (33/560*9V - 33/560*300mV + 300mV)
Vb = (33/560*9V - 33/560*300mV + 300mV) / (1 + 33/560*11)
or,
Vb = 493mV
and thus the current routing through R5, D1, Q1, and R4 is
about 193mV/33 or 5.85mA. That's not the total quiescent
current because D1 uses that 5.85mA to develop a voltage
across it that is probably on the order of 700mV. With that
between the Q2 and Q3 bases, both Q2 and Q3 are passing
collector currents, rail to rail. Hard to know how much
without data sheets, I suppose. But something. Their shared
emitter node would be on the order of 11*490mV or about 5.4V.

That neglected the base current for Q1 flowing via R2. As
I'm now guessing almost 6mA as Ic, and since we are talking
germanium here, I will pick a beta of about 60 and figure
about 100uA base current, then. That's about another 1V
across R2, less than that a little because that lowers Vb a
bit which lowers the 5.85mA figure a bit, which probably then
gets things very darned close to the midpoint of 4.5V one
might wish there.

Not too bad given I have no idea about the BJTs and am using
a lot of random guesses as I go.

Your reasoning is correct. However, the output transistors Q2 and
Q3 need only about 100mV each of Vbe for Class AB bias. IIRC,
beta of Q1 is about 150 and Vbe at that level of current is about
0.12V.

Thanks for the adjustments. At least, I didn't wander too
far afield, for once. It's nice to hear I'm not totally out
of my depth.

I don't know about others, but with low voltage circuits, I
usually try to fix the quiescent voltage at the output mid-point
at slightly more than half of Vcc. This is because Q1's Ve plus
its Vcesat reduces the available downward swing of Q3's base.

Okay, that I follow.

For this design, I tentatively chose a target of 4.6V at Q3's
emitter. Add Q2's Vbe and that leaves 4.3V for R5 plus the
speaker's dc resistance. The speaker's resistance has only a
minor effect but, just for the heck of it, let's take it as 6
ohms. So Q1's Ic = 4.3/566 = 7.6mA.

I follow. Ve(Q2)=4.6V, Vbe(Q2)=0.1V, so Vb(Q2)=4.7V.
Assuming a solid 9V power source (not what a 9V battery is,
if that were used, but...), that leaves 9V-4.7V = 4.3V across
R5 and SPK1. Which sets a current to the Vb node of Q2 that
needs to be disposed of via D1 and then Q1. 7.6mA it is.

Q1's dc beta = 150,

That high? I had anticipated germaniums were lower. Okay.

so Ib is about 50uA,

Followed.

and Ie = 7.65mA = I(R4).

Yes. 7.6mA+0.05mA = 7.65mA that must be dumped into R4.
There will be some Q1 base current (another 50uA?) added to
Ie(Q1) that is ignored here. No problem. That would only
mean 7.7mA instead of 7.65mA for your calcs below.

7.65*33 places Q1's emitter at 252.45mV above ground.

or 7.7mA*33 = 254.1mV, if you add Q1's base drive?

That plus
Vbe of 0.12V gives Vb = 372.45mV.

Okay.

I(R3) = 372.45uA
I(R2) = I(R3) + Ib = 422.45uA
I(R2)*R2 = 4.2245V
V(R2) + Vb = (4.2245 + 0.37245)V = 4.59695V.

It just so happens that, in this case, common resistor values
produce almost exactly the desired quiescent bias level. If they
didn't, a slight departure from the target voltages would be
acceptable. In any case, tolerances on resistor values and
transistor characteristics could throw off actual values a bit.

I'm with you. Thanks for the care, here.

R2 is not only a DC divider but also NFB, I think. Can you
talk a little about how you figure on calculating both the
NFB you want _and_ the DC biasing of this thing, both of
which affect R2's value, I think?

For such a simple design without a high level of audio quality as
the target, I wasn't too particular about the amount of signal
feedback as long as it's a reasonable amount. I chose a
compromise value for R3 first - low enough for bias stability so
that the current through it would be several times Ib, but not
too low to avoid excessive shunting of the signal input current.
Then I let the value of R2 be what it needs to be for correct
bias.

Okay. So I remember in an earlier post you talking about
400uA being a factor of 9 higher. Guessing that Q1's base
current is around the same area of 50uA, this would be a
factor of 8... not the 9 you mentioned before. But at least
I'm starting to see where that number 9 came from?

Then I calculate the amount of NFB as outlined in one of my
earlier replies and accept it if it's within reason. If I really
wanted more NFB, I'd parallel R2 with another resistor, but with
a capacitor in series to avoid upsetting the dc levels.

Got it. That would provide additional AC feedback but keep
the DC biasing. So you don't have to screw around balancing
R2 against two different considerations, you just "fix it"
with a patch like that. Makes total sense.

BTW, that
can be used to provide some bass boost by choosing the proper
values of cap and resistor.

Hmm... Lower frequencies would have less NFB, higher
frequencies more. Okay. There is also other areas where
higher frequencies are going to see less gain in this
design... Now I'm starting to wonder about phase shift not
exactly 180 degrees in the NFB over frequency. But I need to
sit down and think more.

And although I've _seen_
miller feedbacks in the small nF range, could you talk a
little about how that was set at 2.2nF?

That's a guesstimated value, partly empirical and partly based on
observation of other people's designs. No PCs and simulation
software 40 years ago. For such a simple circuit, I didn't bother
with complex calculations for loops and phase shifts that
wouldn't be precise anyway due to wide tolerances in component
characteristics.

The reason for the relatively high capacitance is that this was a
low-Z low-gain circuit. But I might have made a mistake in
showing it now as 2.2nF. I might have used something like 1nF.

Okay. Nuff said. I'll leave that for later thinking.

Also, I think I
_almost_ get the idea of hooking one side of R5 to SPK1
instead of to the (-) side of 9V... but not quite sure. Can
you talk about that choice, as well?

This is a variation of the bootstrap circuit I described in my
other post. R5 and the speaker serve the same functions as R6 and
R7 respectively in the other circuit.

Hmm. I generally get the thrust. I need to think more
closely about the value of it. But I'll take it as something
to explore more.

Jon

 

[Paul E. Schoen]

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:vcm2m5lhelb2t4imsik31fjuchkf4d931m@4ax.com...

Have at me. I probably got a lot wrong in the above, but
that's my thinking exposed like a soft worm to be crushed. If
I learn in the process, crush away!

Have you used LTSpice (AKA SwitcherCAD)? Some time ago I tried some ideas
for linear amplifiers, and I came up with a design that is DC coupled and
uses just three MOSFETs. You can trim the quiescent current to trade
efficiency for crossover distortion. I originally made it with capacitor
coupling and a single supply, but I redid it with a dual supply and DC
coupling.

As a practical matter, the amplifier may be a bit unstable and prone to
overheating and even self-destruction if the bias current is set high
enough to eliminate crossover distortion. If you can tolerate a few percent
(tens of thousands of PPM, if you must), then it should be fairly stable.

This design is basically an output stage only, and has no voltage
amplification. That can be easily achieved with a simple class A input
stage.

When the output is driven just about to the rails, it puts out 5.5 watts
into 8 ohms, with input power of 8.9 watts, or 62% efficiency. I'm using
+/- 12 VDC rails and 7.7 VRMS input. There are two IRL3915 NMOS output
transistors (STD30NF06 also work), and one IRF7205 PMOS. I have not
actually built this circuit, and there are probably a number of problems
with making it work using actual components, but I think it's worth a try.
Of course, this is a MOSFET design and not BJT, but a similar circuit could
be built using BJTs if that is a requirement.

The ASCII file follows.

Paul

---------------------------------------------------------------

Version 4
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SYMATTR Value 10k
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SYMATTR Value 1N4148
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SYMATTR Value 1N4148
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SYMATTR Value 1N4148
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SYMATTR InstName D4
SYMATTR Value 1N4148
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SYMATTR Value 12
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SYMATTR Value 220
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SYMATTR InstName R5
SYMATTR Value 220
TEXT 264 704 Left 0 !.tran .5
TEXT 416 704 Left 0 !;ac oct 5 20 20000

 

[Jon Kirwan]

On Thu, 28 Jan 2010 22:14:12 -0500, "Paul E. Schoen"
<paul@peschoen.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:vcm2m5lhelb2t4imsik31fjuchkf4d931m@4ax.com...

Have at me. I probably got a lot wrong in the above, but
that's my thinking exposed like a soft worm to be crushed. If
I learn in the process, crush away!

Have you used LTSpice (AKA SwitcherCAD)?

Oh, yes. However, my interest is in _learning_. I use
LTspice to explore my own ignorance, at times. And I use it
to test what I think I've learned. But I don't "hack" with
it. Much.

Some time ago I tried some ideas
for linear amplifiers, and I came up with a design that is DC coupled and
uses just three MOSFETs. You can trim the quiescent current to trade
efficiency for crossover distortion. I originally made it with capacitor
coupling and a single supply, but I redid it with a dual supply and DC
coupling.

No FETs. I think I stated that in the beginning. There are
a variety of reasons why. But suffice it that I don't want
to go there... for now.

As a practical matter, the amplifier may be a bit unstable and prone to
overheating and even self-destruction if the bias current is set high
enough to eliminate crossover distortion. If you can tolerate a few percent
(tens of thousands of PPM, if you must), then it should be fairly stable.

This design is basically an output stage only, and has no voltage
amplification. That can be easily achieved with a simple class A input
stage.

When the output is driven just about to the rails, it puts out 5.5 watts
into 8 ohms, with input power of 8.9 watts, or 62% efficiency. I'm using
+/- 12 VDC rails and 7.7 VRMS input. There are two IRL3915 NMOS output
transistors (STD30NF06 also work), and one IRF7205 PMOS. I have not
actually built this circuit, and there are probably a number of problems
with making it work using actual components, but I think it's worth a try.
Of course, this is a MOSFET design and not BJT, but a similar circuit could
be built using BJTs if that is a requirement.

Yeah. That's a requirement. I've still some learning ahead
of me. But I'll take a look at the schematic and tuck it
away, at least. Thanks.

Jon

The ASCII file follows.

Paul
snip of LTspice .ASC file

 

[Paul E. Schoen]

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:thl4m5l8u5vk38tnmu5imr4hqvd829n693@4ax.com...

On Thu, 28 Jan 2010 22:14:12 -0500, "Paul E. Schoen"
paul@peschoen.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:vcm2m5lhelb2t4imsik31fjuchkf4d931m@4ax.com...

Have at me. I probably got a lot wrong in the above, but
that's my thinking exposed like a soft worm to be crushed. If
I learn in the process, crush away!

Have you used LTSpice (AKA SwitcherCAD)?

Oh, yes. However, my interest is in _learning_. I use
LTspice to explore my own ignorance, at times. And I use it
to test what I think I've learned. But I don't "hack" with
it. Much.

Some time ago I tried some ideas
for linear amplifiers, and I came up with a design that is DC coupled and
uses just three MOSFETs. You can trim the quiescent current to trade
efficiency for crossover distortion. I originally made it with capacitor
coupling and a single supply, but I redid it with a dual supply and DC
coupling.

No FETs. I think I stated that in the beginning. There are
a variety of reasons why. But suffice it that I don't want
to go there... for now.

As a practical matter, the amplifier may be a bit unstable and prone to
overheating and even self-destruction if the bias current is set high
enough to eliminate crossover distortion. If you can tolerate a few
percent
(tens of thousands of PPM, if you must), then it should be fairly stable.

This design is basically an output stage only, and has no voltage
amplification. That can be easily achieved with a simple class A input
stage.

When the output is driven just about to the rails, it puts out 5.5 watts
into 8 ohms, with input power of 8.9 watts, or 62% efficiency. I'm using
+/- 12 VDC rails and 7.7 VRMS input. There are two IRL3915 NMOS output
transistors (STD30NF06 also work), and one IRF7205 PMOS. I have not
actually built this circuit, and there are probably a number of problems
with making it work using actual components, but I think it's worth a
try.
Of course, this is a MOSFET design and not BJT, but a similar circuit
could
be built using BJTs if that is a requirement.

Yeah. That's a requirement. I've still some learning ahead
of me. But I'll take a look at the schematic and tuck it
away, at least. Thanks.

OK. So I made a similar amplifier output stage using two 2N3055s, and a
2N3904 and a 2N3906. With 8.4 VRMS input the output is 7.3 VRMS into 8 ohms
for 6.66 watts. Input power is 9.97 watts, efficiency is 67%. Some very
slight crossover distortion. 6 mA drive current (about 1.2k input
impedance). Looks good 20 Hz to 20 kHz. I added an output inductor which
affects output at higher frequencies. LTSpice ASCII follows.

Paul

-----------------------------------------------------------

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SYMATTR Value 2N3906
SYMBOL npn 800 -96 R0
SYMATTR InstName Q4
SYMATTR Value 2N3904
SYMBOL ind 1040 240 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L1
SYMATTR Value 100ľ
SYMBOL diode 720 32 R0
SYMATTR InstName D2
SYMATTR Value 1N4148
TEXT 264 704 Left 0 !.tran .5
TEXT 416 704 Left 0 !;ac oct 5 20 20000