9V 1A schematic needed

[Fred Bloggs]

Larry Brasfield wrote:

I have patiently explained the rationale for that,
Fred.

We already shot down your so-called rationale as a typical troll dodge.

Is there any specific part of it you do not
understand or believe to be irrelevant?

Not really- we understand you all too well.

[...snip evasion and dodging...]

You are a fake- that much is clear. If you could come close to finishing
the detector easily, it would have require far less than the 10,000
words of bs excuse you have since posted. You can't because you are a fake.

 

[Larry Brasfield]

"John Fields" <jfields@austininstruments.com> wrote in message
news:lqer31pv3l2b7828o4iatqeebcv674pr1f@4ax.com...

On Sun, 20 Mar 2005 08:38:38 -0800, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

If you search my posts on rec.audio.tech, you could
find that analysis already done, if you're not doing
this for fun or education.
---
So, instead of referring the OP to a particular post or thread
containing the salient information, (if, indeed, such a post or thread
exists)
[More on this later.]
you would have him wallow in your offal. To what end?

I thought, since he is doing an analysis that I went
to some trouble awhile back to complete and put
into moderately clear form, that Mac might be
interested enough to do a simple Google news
search. It appears that you are unfamilier with
the power of that tool because I was able to
find that thread in less than a minute using the
obvious name and group search criteria plus
body containing ("transformer" and "capacitor"
and ("rectifier" or "bridge")). I had no need of
any special knowledge to spec that search.
Here is the link:
http://groups-beta.google.com/groups?as_q=capacitor+transformer&num=10&scoring=r&hl=en&ie=UTF-8&as_epq=&as_oq=rectifier+bridge&as_eq=&as_ugroup=rec.audio.tech&as_usubject=&as_uauthors=Larry+Brasfield&lr=&as_drrb=q&as_qdr=&as_mind=1&as_minm=1&as_miny=1981&as_maxd=20&as_maxm=3&as_maxy=2005&safe=off
(watch for word wrap)

You might want to take a look at it. There
are examples of both rational discourse and
dealing with a spew emitting crackpot who
has achieved some fame. There is even an
example of "ass handed" if you like.

[Insulting speculation cut.]
Just for your edification: I was not ready to
go spend the same minute Mac could spend
if he wanted without some reason to believe
he was interested. It was a pure efficiency
consideration, nothing as awful as you would
like to believe and proclaim.

And John, before you elect to insinuate that I
am a liar, would not common decency tempt
you to attempt a search?

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Clarence_A]

"fuclarence" <nobody@wateoftime.com> wrote in message
news:423D610C.6080805@wateoftime.com...

PLONK Again!

 

[Fred Bloggs]

Larry Brasfield wrote:

Actually, as you state this, (and I also suggested),
we are both a bit incorrect, hypertechnically.
The average value and the RMS value of a signal
taken over a fixed period are independent except
for the constraint |Savg| <= Srms .

Really? Sure about that? Call Im the average value, I the instantaneous
current, and Iac=I-Im or the zero average component of I. Then even you
can understand that Irms^2=<Iac>^2 + Im^2 where < > denotes cycle
average. Is that right? Im^2 is a component of Irms^2? That doesn't look
"independent" to me...and when you further particularize it to the
charging situation here where Im=Load Current, you are even more wildly
off the mark.
Hey! Link us to that learned treatise you posted in rec.audio- we want
to see more of your "wisdom". Or do you prefer to keep dodging on that
one too?

 

[Fred Bloggs]

Larry Brasfield wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:lqer31pv3l2b7828o4iatqeebcv674pr1f@4ax.com...

On Sun, 20 Mar 2005 08:38:38 -0800, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:


If you search my posts on rec.audio.tech, you could
find that analysis already done, if you're not doing
this for fun or education.

---
So, instead of referring the OP to a particular post or thread
containing the salient information, (if, indeed, such a post or thread
exists)

[More on this later.]

you would have him wallow in your offal. To what end?


I thought, since he is doing an analysis that I went
to some trouble awhile back to complete and put
into moderately clear form, that Mac might be
interested enough to do a simple Google news
search. It appears that you are unfamilier with
the power of that tool because I was able to
find that thread in less than a minute using the
obvious name and group search criteria plus
body containing ("transformer" and "capacitor"
and ("rectifier" or "bridge")). I had no need of
any special knowledge to spec that search.
Here is the link:
http://groups-beta.google.com/groups?as_q=capacitor+transformer&num=10&scoring=r&hl=en&ie=UTF-8&as_epq=&as_oq=rectifier+bridge&as_eq=&as_ugroup=rec.audio.tech&as_usubject=&as_uauthors=Larry+Brasfield&lr=&as_drrb=q&as_qdr=&as_mind=1&as_minm=1&as_miny=1981&as_maxd=20&as_maxm=3&as_maxy=2005&safe=off
(watch for word wrap)

You might want to take a look at it. There
are examples of both rational discourse and
dealing with a spew emitting crackpot who
has achieved some fame. There is even an
example of "ass handed" if you like.

[Insulting speculation cut.]
Just for your edification: I was not ready to
go spend the same minute Mac could spend
if he wanted without some reason to believe
he was interested. It was a pure efficiency
consideration, nothing as awful as you would
like to believe and proclaim.

And John, before you elect to insinuate that I
am a liar, would not common decency tempt
you to attempt a search?

Is this what you're talking about?

The model behind this analysis employs an ideal
transformer having an output resistance Ro and
open circuit peak output voltage Vp, rectifier
diodes that turn on at a fixed forward bias and
have no resistance [1], and an infinite output
filter capacitance [2]. An independent variable,
Vd, represents the difference between the filter
capacitor voltage limited by R and what it would
be if R was 0 Ohms. The transformer open circuit
output is assumed to be sinusoidal. The analysis
is for full wave rectification, but can be easily
adjusted for half wave.

[1. Diode resistance can be folded into the
transformer output resistance. The actual
P/N junction current/voltage characteristic
contributes negligible error because it only
affects the current flow when that current
is relatively close to zero anyway.]

[2. The infinite filter capacitor represents
the limiting (and worst) case for RMS/average
current ratio. The situation can improve for
lesser capacitors as the ripple becomes an
appreciable fraction of Vp.]

Get the rectifier turn-on level:
Vx := Vp - Vd
Substitute an angle, a, for the phase velocity:
a := 2 pi Fp t
(where Fp is line frequency and t is time)
The voltage charging the cap thru Ro:
v(a) = max(0, Vp cos(a) - Vx)
Derive (half of) the conduction angle
g = acos(Vx/Vp)
The squared charging voltage during conduction:
v(a)^2 = Vp^2 cos(a)^2 - 2 Vp Vx cos(a) + Vx^2
Mean charging voltage over a half cycle:
v_ = 1/pi Integral[from -g to +g] v(a) d(a)
= 1/pi (2 Vp sqrt(1 - (Vx/Vp)^2) - 2 Vx g)
Mean squared charging voltage over a half cycle:
vsq_ = 1/pi Integral[from -g to +g]
(Vp^2 cos(a)^2 - 2 Vp Vx cos(a) + Vx^2) d(a)
= 1/pi (Vp^2 g + Vp^2 sin(2 g) / 2
- 4 Vp Vx sqrt(1 - (Vx/Vp)^2) + 2 Vx^2 g)

Note that the integrals for finding means
are evaluated only within the conduction
angle and the interval over which the mean
is computed is a half cycle (pi radians).

These identities are were substituted:
sin(+acos(x)) = +sqrt(1-x^2)
cos(x)^2 = 1/2 + cos(2 x)/2

The mean and RMS output current will be:
I_ = v_ / Ro
irms = sqrt(vsq_) / Ro

Because the RMS and average charging voltages
are not analytically invertable (at least not
by me), I wrote a Perl program to tabulate
their normalized values versus Vx normalized
by Vp, and show a few other interesting values.

================================================================
# List normalized Vx values for which to get RMS and mean.
@vx = (
0.0, 0.5, 0.6, 0.7, 0.75,
0.80, 0.8125, 0.825, 0.8375, 0.85, 0.8625, 0.875,
0.90, 0.9125, 0.925, 0.9375,
0.95, 0.955, 0.96, 0.965, 0.97,
0.975, 0.98, 0.985, 0.99, 0.995, 1.0
);
# Synthesize inverse cosine.
sub acos {
return atan2( sqrt(1.0 - $_[0] * $_[0]), $_[0] );
}

$pi = 4.0 * atan2(1.0, 1.0);
# Evaluate the mean normalized charging voltage at normalized Vx.
sub vm { local($d) = @_;
local($r);
$r = (1.0/$pi) * (2.0 * sqrt(1.0-$d*$d) - 2.0 * $d * &acos($d));
return $r;
}

# Evaluate the RMS normalized charging voltage at normalized Vx.
sub rms { local($d) = @_;
local($r,$a);
$a = &acos($d);
$r = $a + sin(2.0 * $a)/2.0 - 4.0 * $d * sqrt(1.0 - $d*$d) + 2.0 * $d*
$d * $a;
return sqrt($r/$pi);
}

# Tabulate conduction time and angle, mean, RMS, ratio, and Vd.
print "mS(60Hz) angle mean RMS RMS/mean drop\n";
foreach $d (@vx) {
$a = &acos($d);
$vavg = &vm($d);
$vrms = &rms($d);
$rmr = ($vavg > 0.0)? sprintf("%4.2f", $vrms/$vavg) : ' -- ';
# Compute conduction time for 60 Hz power.
$t60 = 1000.0 * $a * (2.0/$pi) / 120.0;
# Convert to conduction angle in degrees.
$a = 2.0 * $a * (180.0 / $pi);
# Show independent variable Vx as Vd (drop).
$d = 1.0 - $d;
printf(" %4.2f %6.2f %7.5f %7.5f %s %5.3f\n",
$t60, $a, $vavg, $vrms, $rmr, $d);
}

================================================================

Running the above program produces this table:

mS(60Hz) angle mean RMS RMS/mean drop
8.33 180.00 0.63662 0.70711 1.11 1.000
5.56 120.00 0.21800 0.29411 1.35 0.500
4.92 106.26 0.15510 0.22208 1.43 0.400
4.22 91.15 0.10018 0.15470 1.54 0.300
3.83 82.82 0.07600 0.12306 1.62 0.250
3.41 73.74 0.05424 0.09302 1.71 0.200
3.30 71.32 0.04920 0.08579 1.74 0.188
3.19 68.82 0.04434 0.07868 1.77 0.175
3.07 66.25 0.03965 0.07170 1.81 0.162
2.94 63.58 0.03514 0.06486 1.85 0.150
2.81 60.80 0.03082 0.05816 1.89 0.137
2.68 57.91 0.02670 0.05162 1.93 0.125
2.39 51.68 0.01908 0.03904 2.05 0.100
2.24 48.29 0.01560 0.03303 2.12 0.088
2.07 44.66 0.01237 0.02723 2.20 0.075
1.89 40.73 0.00941 0.02168 2.30 0.063
1.68 36.39 0.00673 0.01640 2.44 0.050
1.60 34.51 0.00574 0.01437 2.50 0.045
1.51 32.52 0.00481 0.01240 2.58 0.040
1.41 30.41 0.00394 0.01050 2.67 0.035
1.30 28.14 0.00312 0.00866 2.77 0.030
1.19 25.68 0.00238 0.00689 2.90 0.025
1.06 22.96 0.00170 0.00521 3.07 0.020
0.92 19.87 0.00110 0.00364 3.30 0.015
0.75 16.22 0.00060 0.00219 3.65 0.010
0.53 11.46 0.00021 0.00092 4.34 0.005
0.00 0.00 0.00000 0.00000 -- 0.000

Now, given the definition of load regulation,
(which is the percentage drop in output from
no load to full rated load (sometimes known
as "100% load regulation")), the transformer
output resistance can be calculated as
Ro = Vp * (LR/100) / Imax
where Imax is the rated RMS output current
and LR is the load regulation in percent.
If necessary, LR can be adjusted from the
load regulation specified by the transformer
manufacturer to accomodate rectifier series
resistance and the filter capacitor's ESR.

To find where in the above table a transformer
will operate when delivering its rated output
current (Imax), you can find (or interpolate)
the row where the RMS value is equal to LR/100.
The deliverable load current can be found by
dividing Imax by the row's "RMS/mean" value.
The output drop under that full load can be
found by multiplying Vp by the "drop" value.

Notice that the RMS/mean value is 1.62 when
the load regulation is 12.3%. For devices
with better load regulation, the RMS/mean
value is higher. This means that the 1.62
factor underestimates required transformer
RMS output current rating for a given DC
load current for transformers that happen
to have load regulation tighter than 12.3%.

To assuage doubt as to whether transformers
might have better than 12% load regulation,
look at what some manufacturers say. For an
example of one making efficient devices:
http://www.plitron.com/pages/Products/Std/selectio.htm
http://www.plitron.com/pages/Products/Std/ratings.htm
Their torroidal power transformers vary
from 20% to 4% load regulation, with all
500 VA and larger devices at 4%. For those
devices, the safe RMS/mean current ratio is
about 2.04 (interpolated from the table).

(If my name-calling pal is reading, I am not
willing to call that factor "several times".)

As Mr. Strand has observed, the safe ratio
may be affected by unusual conditions, such
as degraded cooling. Likewise for ambient
temperatures above where the transformer was
rated. These effects can be accounted for
by derating LR in proportion to whatever RMS
current derating is appropriate.

 

[John Fields]

On Sun, 20 Mar 2005 18:02:34 +0000, John Woodgate
<jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that John Fields <jfields@austininstrum
ents.com> wrote (in <t8cr31dq63l0asne3h84n7ds67g503gmtp@4ax.com&gt about
'9V 1A schematic needed', on Sun, 20 Mar 2005:

It's a PITA, but what's worked for me in the past is to make sure that
the last check drawn on the account takes it down to _precisely_ $0.00.

Again, in UK, asking to 'close the account' gives the bank the problem
of making sure it goes to exactly 0.00, even if at 23:59:59 they add
bank charges.

---
OK, but my point was that here, IME, (and, apparently,in JT's) it's
not necessary to _ask_ to close the account, merely drawing it down to
precisely $0.00 will _automatically_ close it.

--
John Fields

 

[Fred Bloggs]

Fred Bloggs wrote:

Larry Brasfield wrote:


Actually, as you state this, (and I also suggested),
we are both a bit incorrect, hypertechnically.
The average value and the RMS value of a signal
taken over a fixed period are independent except
for the constraint |Savg| <= Srms .


Really? Sure about that? Call Im the average value, I the instantaneous
current, and Iac=I-Im or the zero average component of I. Then even you
can understand that Irms^2=<Iac>^2 + Im^2 where < >...

....should be <Iac^2> in that Irms^2 expression.

 

[Ken Smith]

In article <369%d.341$H06.77@newsread3.news.pas.earthlink.net>,
J-vibe <nospam@nospam.com> wrote:

"Ken Smith" <kensmith@green.rahul.net> wrote in message
news:d1imv7$bqm$2@blue.rahul.net...
In article <c62%d.14861$cN6.7359@newsread1.news.pas.earthlink.net>,
J-vibe <nospam@nospam.com> wrote:
I forgot, I'm using modern social lingustic terminology, I'll rephrase
it:
1FAA 0010 , 11F1A2 021101, 12F111A122 0121120111, What the next
sequence?


One posible answer is:


1121F113A11122 012112210113

Correct! You are the man!
Fred could not even understand where to begin. He was busy counting his
toes.

I wouldn't bet on that. Evidence is that Fred knows a lot is not any sort
of dummy when he stays on his meds. There also seems to be some evidence
that he doesn't like you.



--
--
kensmith@rahul.net forging knowledge

 

[Ken Smith]

In article <1tqlrx73frjuz.dlg@ID-222894.news.individual.net>,
Active8 <reply2group@ndbbm.net> wrote:
[...]

I'm curious. Who are the morons that write the sofware that does
this:

I've left out of town jobs and just withdrew whatever money was left
in the bank or wrote a check to my permanent account. I may leave a
few pennies. The bank decides to charge a service fee each month for
being below the minimun balance. The account goes negative and the
program charges a fee for non-sufficient funds. Then I can't open an
account elsewhere until I wake them up.

I quote "Errors in our favour are not errors"

Have you ever had a bank give you extra money?

--
--
kensmith@rahul.net forging knowledge

 

[Clarence_A]

"Active8" wrote

fuclarence had nothing to say:
Clarence_A wrote:
"John Larkin" wrote
Fred Bloggs wrote:

In the case of the OP292 thread, the motivation
was to make a dead certain diagnosis of
snipped Boggs speak
Clarence and his problem- and that is what
I did- also ordered some OP292's and played
around with it some more.
snipped Boggs speak

You actually ordered parts and built circuits
to prove somebody from a newsgroup to be wrong?
That's pushing the top range of the getalifeometer!

Yes that is pretty strange for even Fred
and he is as strange as they come.
The problem with the amplifier circuit
has been verified by Analog devices and
the alternate circuit with diodes to
assure equal amplitude are in place.
The Brass board is in the hands of the
customer and the final PCB layout is
complete for the first build.
The unit now works as intended.

BTW, I killfiled FRED ages ago,
so I never see him unless someone
replies to his inane rambling.

Once again- <snipped Boggs speak> came
on the NG lying about breadboarding
something when in fact all you had
done was run a half assed simulation
on a SPICE freebie.

"SABER" is not free! BUT LTspice is! same result.
ALSO: Three other Engineers verified my finding.
<snipped Boggs speak>

Looks like he can't spec a 3-term regulator either.

I do not use three terminal regulators,
they do not work well with 100 volt plus inputs
or from -55C to 125C!

Besides, No problem with the power supply.
Did replace the POR though,
it oscillated at high temp! (Not my design)

 

[Fred Bloggs]

Clarence_A wrote:

"fuclarence" <nobody@wateoftime.com> wrote in message
news:423D610C.6080805@wateoftime.com...

PLONK Again!

Ooooh my- am I reeling from the insult of being P-LONKED by such a
personage....

 

[John Fields]

On Sun, 20 Mar 2005 11:21:13 -0800, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

And John, before you elect to insinuate that I
am a liar, would not common decency tempt
you to attempt a search?

---
Decency has nothing to do with it, and since I'm not interested in
this particular subject, reading what you have to say about it holds
no interest for me. OTOH, the motivation behind your "style" does,
and I will continue to comment on that as I see fit.

--
John Fields

 

[Larry Brasfield]

"Fred Bloggs" <nospam@nospam.com> wrote in
message news:423DD239.9010105@nospam.com...

Larry Brasfield wrote:
Actually, as you state this, (and I also suggested),
we are both a bit incorrect, hypertechnically.
The average value and the RMS value of a signal
taken over a fixed period are independent except
for the constraint |Savg| <= Srms .

Fred Bloggs wrote, (edited only for clarity):
Really? Sure about that? Call Im the average value, I the instantaneous
current, and Iac=I-Im or the zero average component of I. Then even you
can understand that Irms^2=<Iac^2> + Im^2 where < >...

If I first substitute your 'Im' and 'Irms' with 'Savg' and 'Srms', I get:
Srms^2=<Iac^2> + Savg^2 where < >...
A brief application of algebra and notation unnoting yields:
Savg^2 = Srms^2 - <Iac^2>
Now, square each side of my stated constraint to get:
Savg^2 <= Srms^2
Since the term <Iac^2> can never be negative, I believe
that my statement is correct and consistent with yours.
And, to answer your question: Yes, I am sure about it.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Fred Bloggs]

Larry Brasfield wrote:

"Fred Bloggs" <nospam@nospam.com> wrote in
message news:423DD239.9010105@nospam.com...

Larry Brasfield wrote:

Actually, as you state this, (and I also suggested),
we are both a bit incorrect, hypertechnically.
The average value and the RMS value of a signal
taken over a fixed period are independent except
for the constraint |Savg| <= Srms .


Fred Bloggs wrote, (edited only for clarity):
Really? Sure about that? Call Im the average value, I the instantaneous
current, and Iac=I-Im or the zero average component of I. Then even you
can understand that Irms^2=<Iac^2> + Im^2 where < >...

If I first substitute your 'Im' and 'Irms' with 'Savg' and 'Srms', I get:
Srms^2=<Iac^2> + Savg^2 where < >...
A brief application of algebra and notation unnoting yields:
Savg^2 = Srms^2 - <Iac^2
Now, square each side of my stated constraint to get:
Savg^2 <= Srms^2
Since the term <Iac^2> can never be negative, I believe
that my statement is correct and consistent with yours.
And, to answer your question: Yes, I am sure about it.

Big damn difference between your weak Savg<=Srms and the exact
expression Irms=sqrt(<Iac^2>+Im^2). And your assertion that Irms and Im
are independent is total manure when you can see that they are
algebraically related. What is more, in the case of the filtering
circuit under discussion , Im is one independent variable that defines
Irms through that formula- making Irms a dependent quantity. Your brain
is defective, you have no sense of cause and effect, and you are NO
damned analyst- all you do is spew a bunch pretentious crap designed to
make you look sophisticated- and damned if you don't hate it when
someone who actually knows what they're doing comes along and blows your
ship out of the water. I will dig into that pathetic Perl kluge soon-

 

[Frank Bemelman]

"Fred Bloggs" <nospam@nospam.com> schreef in bericht
news:423D6513.4080304@nospam.com...

PS I wasted no time at all working on your puzzle, pussy boy. Ha Ha Ha
.....

Let's see if he can solve the new puzzle. It's easier than the
first. No, don't waste your time, it isn't worth it

--
Thanks, Frank.
(remove 'q' and 'invalid' when replying by email)

 

[Frank Bemelman]

"J-vibe" <nospam@nospam.com> schreef in bericht
news:369%d.341$H06.77@newsread3.news.pas.earthlink.net...

LADIES AND GENTLEMEN, WELCOME TO AMATUER COMEDY NIGHT. TONIGHT WE
HAVE FRED BLOGGER, SO GIVE HIM A BIG HAND: (clap, clap, clap....):

Yeah. I have a puzzle for you. After all, the 'AMATUER COMEDY NIGHT'
show needs a few more 'AMATUERS'.

Our mysterious processor spits out:
369%d.341$H06.77, 423D6513.4080304,

What is the next sequence? I guarantee there is a next
sequence and you will be able to check it.


--
Thanks, Frank.
(remove 'q' and 'invalid' when replying by email)

 

[Frank Bemelman]

"Frank Bemelman" <f.bemelmanq@xs4all.invalid.nl> schreef in bericht
news:423ded1c$0$144$e4fe514c@news.xs4all.nl...

"Fred Bloggs" <nospam@nospam.com> schreef in bericht
news:423D6513.4080304@nospam.com...

PS I wasted no time at all working on your puzzle, pussy boy. Ha Ha Ha
....

Let's see if he can solve the new puzzle. It's easier than the
first. No, don't waste your time, it isn't worth it

And I expect an answer in less than an hour of course

--
Thanks, Frank.
(remove 'q' and 'invalid' when replying by email)

 

[Jonathan Kirwan]

On Sat, 19 Mar 2005 03:26:15 GMT, "J-vibe" <nospam@nospam.com> wrote:

I have found this schematic online (not schematic symbol) and I think
I could adapt it to supply a generic/home electronics 9 volts I need to
run a generic floppy drive. I'll also be checking out that book you
mentioned for future references.
http://www.bonavolta.ch/hobby/en/audio/reg_sup.htm

There are several examples in there, some of them boiling down to an
open-loop emitter follower tethered to a zener reference. One of the
designs there, the 53V version, uses feedback and I like that one as a
basis for a 9V supply.

I'm not a designer, just a hobbyist, but here is how I'd proceed from
that 53V design:

(1) I like the idea of using the Darlington topology, with a cheap
and readily available 2N3055 for the power pass transistor. I might
also like this because I still have a few laying around.

(2) The other transistor can be a 2N3904 or 2N2222. At Ic=1A for the
2N3055, I'd expect no better than beta=30, so the Ic on this other
transistor would need to be about 33mA. Since Vce might be anywhere
from 2.5V to perhaps 6V or so, this suggests that the transistor
should be able to dissipate at least 1/4 watt, itself. I think both
of those mentioned above can do this ... but not much more. So watch
this.

(3) 1A is a lot of current. Radio Shack sells only two kinds of
transformers (12.6V and 25.2V) and its their 12.6V ones that come in
1.2A and 3.0A versions. With the 12.6V transformers, you'll need to
use a bridge rectifier and with an average current of 1A, you'd be
better to select the 3A version since the peak currents will probably
be in that area.

(4) At 1A average load, a 12.6V transformer is going to need a pretty
seriously sized filter capacitor to get up to and hold a supply rail
for the circuit. The peak voltage would be about sqrt(2) *12.6V or
17.8V, less a pair of diode drops at high current, or about 2V less to
say 16V or so. There are books with charts to help you figure out
exactly what size the capacitor should be for some desired ripple (you
can work it out with trig) but my rough guess takes about 1/3rd of the
60Hz time (say, 5ms or so) and then takes some desired ripple voltage
(let's say 1V here) and calculates C=dt*I/dV, which would be 5ms*1A/1V
or 5000uF. Radio Shack sells 4,700uF at 35V. That should be okay.

(5) The feedback method shown in the 53V supply on the web page looks
reasonable to me. It's basically a zener being compared to a portion
of the output voltage by way of a transistor that then pulls the
Darlington base lower (more current) if the voltage climbs much above
the reference point. However, I'd just use a zener from 6.2V to 9.1V
or so, not the 15V (of course.) Call it a 6.2V zener for now. Here
again, the transistor can be a 2N3904 or 2N2222, I believe.

(6) The zener needs a current through it. I'd go with 1mA. The data
sheets often spec things at 20mA (Izt) but 1mA has worked well in the
past for me. You can redesign for larger Izt, if you want. This
current establishes the Ic for this transistor comparator and that,
together with the minimum estimated difference between Vin and the
base of the Darlington pair provides the size of the collector
resistor. With a ripple of about 1V, going let's say from 15V to 16V,
the minimum Vin will then be 15V. The Darlington pair will require
some 1.25V or so, so the base should be at 9V+1.25V or about 10.25V.
The difference is then 4.75V. At 1mA, this suggests a 4.7k resistor
in the collector. I'd go with even less... a 3.3k here, just to be
absolutely certain that at least 1mA was available for the zener.

(7) The base of the comparator transistor needs to divide the output
voltage down. And it would be nice to have some slight adjustment
range, so I'd recommend a potentiometer in here, as well. With a beta
of about 200 or so for the transistor, I'd estimate that a current
through this divider that is about equal to the zener current should
be okay... another 1mA or so. With 9V on the output, this suggests
about 9k ohms total. A little less would be okay. In any case, this
seems kind of convenient for a 1k potentiometer (common value) to give
us some variable range for adjustment. The base voltage should be
0.7V above the zener voltage, so about 6.2+.7 or about 6.9V.
Actually, because of the light base current, it will probably be a
little less than this. This suggests that we want the grounded side
resistor to be 6900 ohms (assuming 9k total) and the high side to be
the remainder, or 2100 ohms. However, with a 1k pot in the middle,
and set to the midpoint, we'd take about 500 ohms off both sides, so
we'd like 6400 and 1600, perhaps. Using 5% resistor values, I'd
select 6200 ohms and 1500 ohms and put the 1k pot in the middle of the
two.

(8) As shown in the 52V supply on that web page, a small cap would
also be nice across the zener.

(9) On the output, we should have another filter capacitor. I'd use
a large value, perhaps 100uF or so.

All this arrives at a design looking like:

,------,
,-----| |-------+----------+-------------+------------,
| | |(+) | | | |
| | | --- 4700uF \ | |
12.6V | AC | --- 35V / 3300 | |
AC | | | \ |/c NPN |
| |bridge|(-) | +-----------| 2N3904 |
'-----| rect |--, --- | |\e |
'------' | /// |/c NPN | |/c NPN
| ,-| 2N3904 '----------| 2N3055
--- | |\e |\e
/// Vout | --+-----, |
| | | | |
| | /---/ | +---- Vout
\ | / \ --- 10uF |
1500 / | --- --- --- 100uF
\ | |6.2V | ---
| | | | |
| | --- --- ---
\ | /// /// ///
1000 /<----'
\
|
|
\
6200 /
\
|
|
---
///


(View this in a fixed-spaced font.)

Keep in mind that this is a design group and I'm not a designer. I've
done absolutely NO work here on looking at the sensitivities for
oscillation, for example, over various load types (capacitive,
inductive, resistive, or otherwise non-linear.) My example should
probably be in the basics group, really. But I'm posting it here
because that's where your question was at.

As a side point, regarding Fred Bloggs, I've learned some from both
reading his posts and, at times, watching him as he evolves them. And
on the math side, he's posted alternative viewpoints to my own ways of
thinking. By doing so, he's expanded my own thinking tools. He
expects __quantitative__ argument from designers and I've no problem
with that. In any case, I respect what he says on the subject of
electronic design and I'm glad he's shared his time here.

Jon

 

[Larry Brasfield]

"Fred Bloggs" <nospam@nospam.com> wrote in
message news:423DED25.9050401@nospam.com...

Larry Brasfield wrote:
"Fred Bloggs" <nospam@nospam.com> wrote in
message news:423DD239.9010105@nospam.com...

Larry Brasfield wrote:

Actually, as you state this, (and I also suggested),
we are both a bit incorrect, hypertechnically.
The average value and the RMS value of a signal
taken over a fixed period are independent except
for the constraint |Savg| <= Srms .


Fred Bloggs wrote, (edited only for clarity):
Really? Sure about that? Call Im the average value, I the instantaneous
current, and Iac=I-Im or the zero average component of I. Then even you
can understand that Irms^2=<Iac^2> + Im^2 where < >...

If I first substitute your 'Im' and 'Irms' with 'Savg' and 'Srms', I get:
Srms^2=<Iac^2> + Savg^2 where < >...
A brief application of algebra and notation unnoting yields:
Savg^2 = Srms^2 - <Iac^2
Now, square each side of my stated constraint to get:
Savg^2 <= Srms^2
Since the term <Iac^2> can never be negative, I believe
that my statement is correct and consistent with yours.
And, to answer your question: Yes, I am sure about it.


Big damn difference between your weak Savg<=Srms and the exact expression Irms=sqrt(<Iac^2>+Im^2). And your assertion that Irms
and Im are independent is total manure when you can see that they are algebraically related.

But Fred, they are related by a number that can
have any positive value. This means that the 2
quantities I mentioned in the constraint have no
a priori relationship other than that they must
adhere to the constraint. I never claimed that
those numbers were independent of the actual
signal, since, as you and I both know, they are
related as you have shown.

What is more, in the case of the filtering circuit under discussion , Im is one independent variable that defines Irms through
that formula- making Irms a dependent quantity.

You may recall that I introduced this little subject thusly:
Actually, as you state this, (and I also suggested),
we are both a bit incorrect, hypertechnically.
The average value and the RMS value of a signal
taken over a fixed period are independent except
for the constraint |Savg| <= Srms .
and made clear that it would be a sidetrack:
But for the signals in question, you are correct.

That was offered in response to Mac's assertion:
So as we reduce the conduction angle, we must have a larger
current for a shorter time, and because of the nature of the RMS
calculation, it seems to me that it is impossible for Irms to go down when
we do this.

Given sinusoidal mains voltage, Mac is surely right.
And if that assumption is embedded in your formula,
then I would agree that Irms is dependent as you say.

But if the mains had some other waveform, it could
happen that a reduced conduction angle with higher
mean current (holding charge constant) could reduce
the RMS current. For that to be true, the portion
of waveform excluded by the conduction angle
reduction would itself have to have a higher RMS
to mean ratio than the ratio within the conduction.
This can be seen from your own restatement (and
admittedly more comprehensive formulation) of
the constraint I stated by breaking each term
into a pair, members of which represent the
excluded portion and the rescaled kept portion.

Do I need to (somehow) post a waveform to
permit you to see this? I hope not, because I
think you are smart enough to do it yourself.

[ Load of "... crap ..." cut.]

I'm sorry we could not have one calm chat.
But it is nice of you to segregate the crap.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Fred Bloggs]

Larry Brasfield wrote:

"Fred Bloggs" <nospam@nospam.com> wrote in
message news:423DED25.9050401@nospam.com...

Larry Brasfield wrote:

"Fred Bloggs" <nospam@nospam.com> wrote in
message news:423DD239.9010105@nospam.com...


Larry Brasfield wrote:


Actually, as you state this, (and I also suggested),
we are both a bit incorrect, hypertechnically.
The average value and the RMS value of a signal
taken over a fixed period are independent except
for the constraint |Savg| <= Srms .

Fred Bloggs wrote, (edited only for clarity):
Really? Sure about that? Call Im the average value, I the instantaneous
current, and Iac=I-Im or the zero average component of I. Then even you
can understand that Irms^2=<Iac^2> + Im^2 where < >...

If I first substitute your 'Im' and 'Irms' with 'Savg' and 'Srms', I get:
Srms^2=<Iac^2> + Savg^2 where < >...
A brief application of algebra and notation unnoting yields:
Savg^2 = Srms^2 - <Iac^2
Now, square each side of my stated constraint to get:
Savg^2 <= Srms^2
Since the term <Iac^2> can never be negative, I believe
that my statement is correct and consistent with yours.
And, to answer your question: Yes, I am sure about it.


Big damn difference between your weak Savg<=Srms and the exact expression Irms=sqrt(<Iac^2>+Im^2). And your assertion that Irms
and Im are independent is total manure when you can see that they are algebraically related.


But Fred, they are related by a number that can
have any positive value. This means that the 2
quantities I mentioned in the constraint have no
a priori relationship other than that they must
adhere to the constraint.

That so-called "constraint" makes the one variable *dependant* on the
other. Iac is a function of the other circuit parameters as well as Im
too in some complicated way.

I never claimed that
those numbers were independent of the actual
signal, since, as you and I both know, they are
related as you have shown.

Well let's see if they're not independent of the original signal then
that means that are dependent on the original signal and therefore
dependent on each other mathematically known as implicitly.

What is more, in the case of the filtering circuit under discussion , Im is one independent variable that defines Irms through
that formula- making Irms a dependent quantity.


You may recall that I introduced this little subject thusly:
Actually, as you state this, (and I also suggested),
we are both a bit incorrect, hypertechnically.
The average value and the RMS value of a signal
taken over a fixed period are independent except
for the constraint |Savg| <= Srms .
and made clear that it would be a sidetrack:
But for the signals in question, you are correct.

That was offered in response to Mac's assertion:
So as we reduce the conduction angle, we must have a larger
current for a shorter time, and because of the nature of the RMS
calculation, it seems to me that it is impossible for Irms to go down when
we do this.

Given sinusoidal mains voltage, Mac is surely right.
And if that assumption is embedded in your formula,
then I would agree that Irms is dependent as you say.

But if the mains had some other waveform, it could
happen that a reduced conduction angle with higher
mean current (holding charge constant) could reduce
the RMS current. For that to be true, the portion
of waveform excluded by the conduction angle
reduction would itself have to have a higher RMS
to mean ratio than the ratio within the conduction.
This can be seen from your own restatement (and
admittedly more comprehensive formulation) of
the constraint I stated by breaking each term
into a pair, members of which represent the
excluded portion and the rescaled kept portion.

Do I need to (somehow) post a waveform to
permit you to see this? I hope not, because I
think you are smart enough to do it yourself.

[ Load of "... crap ..." cut.]

I'm sorry we could not have one calm chat.
But it is nice of you to segregate the crap.

It is understood that we are talking about a rectified ac- with a
sinusoidal variation with time.