9V 1A schematic needed

[John Fields]

On Sun, 20 Mar 2005 23:16:07 GMT, Fred Bloggs <nospam@nospam.com>
wrote:

Larry Brasfield wrote:

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.

---
I would appreciate seeing the waveform, and graphics (along with
supporting documentation) can be posted to
alt.binaries.schematics.electronic.
---

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.

---
That was my understanding as well, and his: "But if the mains had some
other waveform,"... smacks of: "If frogs had wings..."

--
John Fields

 

[Mac]

On Sun, 20 Mar 2005 11:21:13 -0800, 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.

[snip]

I could not read the whole thread. It got too tedious and personal for me.

I am noticing, however, that threads you actively participate in
sometimes get very long and eventually degenerate into accusations and
counter-accusations. Something to think about!

--Mac

 

[Mac]

On Sun, 20 Mar 2005 10:48:50 -0800, Larry Brasfield wrote:

"Mac" <foo@bar.net> wrote in message newsan.2005.03.20.17.45.41.593055@bar.net...
On Sat, 19 Mar 2005 07:52:21 -0800, Larry Brasfield wrote:

"John Woodgate" <jmw@jmwa.demon.contraspam.yuk> wrote
in message news:+vZqtVBIt9OCFw3u@jmwa.demon.co.uk...
I read in sci.electronics.design that Mac <foo@bar.net> wrote (in
pan.2005.03.19.07.11.50.951783@bar.net&gt about '9V 1A schematic
needed', on Sat, 19 Mar 2005:

Choosing capacitor values is a bit tricky, too. The bigger they are, the
more they force the transformer into a low duty-cycle, high-current mode
of operation.

In fact, the overall efficiency, measured as Idc/Irms, is higher at low
rectifier conduction angles, even though the peak diode current is
higher. A large filter capacitor doesn't increase transformer heating.


That statement really makes no sense to me.
How can you crowd the same charge transfer
into less time and get its RMS value to decrease?

[snip]

I'm with Larry on this one.
Thanks.

If we draw a box around the capacitor and
regulator, then the charge going into the box, integrated over one period
has to equal the charge coming out of the box, integrated over one period.

That's a fine way to approach it. The concept
you just outlined of some components grouped
together for simpler analysis is known as a "cutset".

...
The integral of the charge going out is fixed, because we have a linear
regulator. 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.

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 .

Don't put words in my mouth. I didn't say anything about the average value
of anything, except insofar as 'm' in 'rms' stands for 'mean.' And in the
context, I don't think I am incorrect, technically or otherwise.

I said, in effect, that the area under the 'I' curve for 'I' going into
the capacitor and regulator from the bridge is constant and equal to the
area of the curve for 'I' going out of the regulator into the load. And if
we shorten the duration where the integral is non-zero but keep the area
constant, we will never produce a lower RMS.

If you can come up with a counterexample *consistent with the context*, I
would like to see it so I can understand my mistake. You can't postulate
some wild waveform. This is just a bridge charging up a cap.

But for the signals in question, you are correct.

Then why not just leave it at that? ;-)

[snip]

--Mac

 

[J-vibe]

"Frank Bemelman" <f.bemelmanq@xs4all.invalid.nl> wrote in message
news:423ded74$0$146$e4fe514c@news.xs4all.nl...

"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.

Oh! A rogue processor on a spitting rampage!!!!

Let's see.....

1st: (3.69) x 833D06.77 = 307401.9813
2nd: 4236513.4080304 (you didn't spec D as a $H, so I'll ignore it)
3rd: 8165624.8347608 or 7C98F8$H
4th: 12094736.2614912 or B88D10
etc....

Do I get a pair of tickets to the show?

 

[Active8]

On Sun, 20 Mar 2005 13:30:25 -0600, John Fields wrote:

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.

I've had accounts closed for inactivity, but I don't remember the
amount left - not much if any, maybe zero. Thanks for the tip.

--
Best Regards,
Mike

 

[Active8]

On Sun, 20 Mar 2005 19:37:27 +0000 (UTC), Ken Smith wrote:

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?

Yes. $30. They sent the next guy that came through the drive-thru
(they knew we were on the same job) to the hotel to get me to bring
it back The american bank we had in Germany used to nickle and
dime us to death and could make their adding machines show they were
right.
--
Best Regards,
Mike

 

[Active8]

On Sun, 20 Mar 2005 08:50:02 -0700, Jim Thompson wrote:

On Sun, 20 Mar 2005 06:12:20 -0500, Active8 <reply2group@ndbbm.net
wrote:


I thought it was 26 Feb.

WHAT was "26 Feb"? My birthday is February 29, 1940.

Ok

--
Best Regards,
Mike

 

[Active8]

On Sun, 20 Mar 2005 08:50:02 -0700, Jim Thompson wrote:

On Sun, 20 Mar 2005 06:12:20 -0500, Active8 <reply2group@ndbbm.net

I thought it was 26 Feb.

WHAT was "26 Feb"? My birthday is February 29, 1940.

Fortune teller said she could guess the state I was born in and how

many birthdays I've had for $1 ... something like that. I let her
try.

"You were born in the state of ignorance and you've only had one
birthday, the rest were anniversaries."

Sorry.
--
Best Regards,
Mike

 

[John Woodgate]

I read in sci.electronics.design that John Fields <jfields@austininstrum
ents.com> wrote (in <m43s31tk8mja1ntddeiclm3a1tu09dlc5o@4ax.com&gt about
'9V 1A schematic needed', on Sun, 20 Mar 2005:

That was my understanding as well, and his: "But if the mains had some
other waveform,"... smacks of: "If frogs had wings..."

It isn't the mains supply that matters, it's the waveforms of the
current through the diodes. If there is only resistance in the supply
voltage circuit, this waveform is nearly sinusoidal pulses, but of
duration corresponding to a higher frequency than the mains supply,
which is just another way of saying that the conduction angle is less
than 180 degrees (usually a lot less). However, the peak of this pulse
is not at the same time as the peak of the supply voltage waveform; the
effect of this is rather small but does exist.

If there is inductance in series with the rectifier, then current
waveform is no longer approximately sinusoidal, and algebraic analysis
is very difficult. It is necessary to use simulation.

The most-often cited modern papers on low-power single-phase rectifiers
appear to be:

Lieders A, Single-phase rectifiers with CR filters, Part 1 - Theory,
Electronic Components and Applications, Vol 1 No.3 pp. 153-163 (May
1979) ISSN 0141-6219

Lieders A, Single-phase rectifiers with CR filters, Part 2 - Design
procedure, Electronic Components and Applications, Vol 1 No.4 pp.
216-230 (August 1979) ISSN 0141-6219
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[lemonjuice]

On Sun, 20 Mar 2005 09:42:24 +0000, John Woodgate
<jmw@jmwa.demon.contraspam.yuk> wrote:

I read in sci.electronics.design that Larry Brasfield
donotspam_larry_b
rasfield@hotmail.com> wrote (in <QUX_d.14$te2.382@news.uswest.net&gt
about '9V 1A schematic needed', on Sat, 19 Mar 2005:
That statement really makes no sense to me. How can you crowd the
same
charge transfer into less time and get its RMS value to decrease?
(For
conceptual simplification, assume the DC current taken from the
bridge
is constant, so that the charge, (Idc * (2 f)), is held constant.

I re-examined my models. With a refinement of one, I can predict the
*small* increase reported by someone else. The original model assumed
that the output ripple was negligible, thus restricting the range of
filter capacitor values. If you include in the range of capacitor
values, those so small that very large amounts of ripple occur, and
the
conduction angle approaches 180 degrees, you get much more change of
r.m.s current, but so much ripple is not normally realistic.

Why should the range of filter values be restricted for negligible
ripple? Or you mean they'd be excessively high.


High ripple circuits due to their advantages are realistic. For power
maximisation and increase of dc current I've see many converters
connected directly to the ac mains purposely to increase the
conduction angle with rectifying capacitors below 1uF. At the output of
these though the large ripple removing capacitor(s) is placed.

 

[lemonjuice]

On Sun, 20 Mar 2005 17:45:42 GMT, Mac <foo@bar.net> wrote:

On Sat, 19 Mar 2005 07:52:21 -0800, Larry Brasfield wrote:

"John Woodgate" <jmw@jmwa.demon.contraspam.yuk> wrote
in message news:+vZqtVBIt9OCFw3u@jmwa.demon.co.uk...
I read in sci.electronics.design that Mac <foo@bar.net> wrote (in
pan.2005.03.19.07.11.50.951783@bar.net&gt about '9V 1A schematic
needed', on Sat, 19 Mar 2005:

Choosing capacitor values is a bit tricky, too. The bigger they
are, the
more they force the transformer into a low duty-cycle, high-current
mode
of operation.

In fact, the overall efficiency, measured as Idc/Irms, is higher at
low
rectifier conduction angles, even though the peak diode current is
higher. A large filter capacitor doesn't increase transformer
heating.


That statement really makes no sense to me.
How can you crowd the same charge transfer
into less time and get its RMS value to decrease?

[snip]

I'm with Larry on this one. If we draw a box around the capacitor and
regulator, then the charge going into the box, integrated over one
period
has to equal the charge coming out of the box, integrated over one
period.
This is just conservation of charge.

The integral of the charge going out is fixed, because we have a
linear
regulator. 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. What am I missing?

--Mac

John is right. You and Larry are missing the fact that by changing the
conduction angle you are actually changing the exact instance in time
when the capacitor starts conducting and thats going to change the
values of your charge and integration intervals in your equation
above.

 

[lemonjuice]

On Sun, 20 Mar 2005 06:45:25 GMT, Mac <foo@bar.net> wrote:

On Sat, 19 Mar 2005 07:56:56 +0000, John Woodgate wrote:

I read in sci.electronics.design that Mac <foo@bar.net> wrote (in
pan.2005.03.19.07.11.50.951783@bar.net&gt about '9V 1A schematic
needed', on Sat, 19 Mar 2005:

Choosing capacitor values is a bit tricky, too. The bigger they are,
the
more they force the transformer into a low duty-cycle, high-current
mode
of operation.

In fact, the overall efficiency, measured as Idc/Irms, is higher at
low
rectifier conduction angles, even though the peak diode current is
higher. A large filter capacitor doesn't increase transformer
heating.


I put together a spreadsheet to calculate Irms for a power supply.
This is
a bit like a simulation, I guess, and like any simulation, you have to
decide what to model and what to ignore. I chose to ignore resistance
in
the transformer and filter caps. I did this because I figured it would
just over-estimate the RMS current, which was the main thing I wanted
to find.

Anyway, unless I made some kind of mistake, the result of my
"simulation"
is that Irms goes up significantly with capacitor value. In my
simulation
I hold Idc constant, since the intent was to design a regulated power
supply. So the efficiency goes down with increasing capacitor value.

I am definitely interested in this topic, so if you can shed some
light on
it, I would be appreciative.

regards,
Mac

Its normal the that Irms goes up with increase in capacitance value.
Higher capacitance means less ripple which means higher average current
(Idc). As Idc is directly proportional to Irms and hence is DEPENDENT
(and not independent as someone claims) on Irms... Irms also increases.

Higher capacitance though means you're dissipating more power in the
there so your efficiency goes down.

 

[Active8]

On Sun, 20 Mar 2005 20:26:28 -0700, Jim Thompson wrote:

On Sun, 20 Mar 2005 22:14:04 -0500, Active8 <reply2group@ndbbm.net
wrote:

On Sun, 20 Mar 2005 19:37:27 +0000 (UTC), Ken Smith wrote:

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?

Yes. $30. They sent the next guy that came through the drive-thru
(they knew we were on the same job) to the hotel to get me to bring
it back The american bank we had in Germany used to nickle and
dime us to death and could make their adding machines show they were
right.

I had a bank in San Jose credit my account for over $400 that wasn't
mine, in 1968.

Told them about it, they said they'd correct their error.

This went on for months with no correction.

When I went back to Motorola I closed the account and walked with the
money ;-)

I got a check for over $500 from an insurance company. It was meant
for someone else. They sent me a promisory note to pay it back which
I never signed. That was like a free premium payment.
--
Best Regards,
Mike

 

[Frank Bemelman]

"Active8" <reply2group@ndbbm.net> schreef in bericht
news:1f2zhdw85vty6$.dlg@ID-222894.news.individual.net...

On Sun, 20 Mar 2005 20:26:28 -0700, Jim Thompson wrote:

On Sun, 20 Mar 2005 22:14:04 -0500, Active8 <reply2group@ndbbm.net
wrote:

On Sun, 20 Mar 2005 19:37:27 +0000 (UTC), Ken Smith wrote:

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?

Yes. $30. They sent the next guy that came through the drive-thru
(they knew we were on the same job) to the hotel to get me to bring
it back The american bank we had in Germany used to nickle and
dime us to death and could make their adding machines show they were
right.

I had a bank in San Jose credit my account for over $400 that wasn't
mine, in 1968.

Told them about it, they said they'd correct their error.

This went on for months with no correction.

When I went back to Motorola I closed the account and walked with the
money ;-)

I got a check for over $500 from an insurance company. It was meant
for someone else. They sent me a promisory note to pay it back which
I never signed. That was like a free premium payment.
--

So you and Jim Thompson are both confessed thieves. Jim in particular,
because $400 in 1968 is not what I would call pocket change.

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

 

[Larry Brasfield]

"lemonjuice" <exskimos@anonymous.to> wrote in message
news:1111406725.887396.127330@g14g2000cwa.googlegroups.com...

On Sun, 20 Mar 2005 06:45:25 GMT, Mac <foo@bar.net> wrote:
....
Anyway, unless I made some kind of mistake, the result of my "simulation"
is that Irms goes up significantly with capacitor value. In my simulation
I hold Idc constant, since the intent was to design a regulated power
supply. So the efficiency goes down with increasing capacitor value.

I am definitely interested in this topic, so if you can shed some light on
it, I would be appreciative.

Its normal the that Irms goes up with increase in capacitance value.

That is true, in the strictest sense. But people who have studied
this issue or observed real instances of the circuit for that effect
know that the Irms value approaches a definite limit as the
capacitance value approaches infinity.

Higher capacitance means less ripple which means higher average current
(Idc).

Huh? If you hold everything else constant, (including the
DC current taken from the cap by the regulator/load),
then Idc stays constant and so does the amount of
charge transferred per line cycle.

As Idc is directly proportional to Irms and hence is DEPENDENT
(and not independent as someone claims) on Irms... Irms also increases.

Whatever makes you think Idc = K * Irms where K
is a constant? Anybody who analyzes this correctly
can see that, if Idc is held constant, Irms increases
as the conduction angle decreases. This is all very
fundamental.

As for "(and not independent as someone claims)", you
have (apparently) misunderstood and misconstrued my
remarks. Since they make sense in their original context,
I won't explain that here. You need to learn that there is
a real difference between RMS and mean. If you knew
the difference, you would have a much better chance of
understanding the independence concept.

Higher capacitance though means you're dissipating more power in the
there so your efficiency goes down.

Yes, but how could that happen when Idc is constant
and Irms is "directly proportional to" Irms, holding Irms
constant? You are writing riddles here. Please try to
get those sorted out before adding new ones.

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

 

[John Woodgate]

I read in sci.electronics.design that Frank Bemelman
<f.bemelmanq@xs4all.invalid.nl> wrote (in
<423ef180$0$140$e4fe514c@news.xs4all.nl&gt about '9V 1A schematic
needed', on Mon, 21 Mar 2005:

So you and Jim Thompson are both confessed thieves. Jim in particular,
because $400 in 1968 is not what I would call pocket change.

Well, not thieves under English law, since, as it happens, 1968. In both
cases, the owner took no effective steps to re-possess. I'm quite sure
that if they had, the pay-back would have occurred.
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Active8]

On Mon, 21 Mar 2005 16:25:54 +0000, John Woodgate wrote:

I read in sci.electronics.design that Frank Bemelman
f.bemelmanq@xs4all.invalid.nl> wrote (in
423ef180$0$140$e4fe514c@news.xs4all.nl&gt about '9V 1A schematic
needed', on Mon, 21 Mar 2005:

So you and Jim Thompson are both confessed thieves. Jim in particular,
because $400 in 1968 is not what I would call pocket change.

Well, not thieves under English law, since, as it happens, 1968. In both
cases, the owner took no effective steps to re-possess. I'm quite sure
that if they had, the pay-back would have occurred.

In my case the owner was no longer the owner because they gave me
the money - IMO at the time. If they felt they had legal precedent
to collect the money, they wouldn't have needed to try to get me to
sign a promissory note which would have been an admission of debt.
--
Best Regards,
Mike

 

[Frank Bemelman]

"John Woodgate" <jmw@jmwa.demon.contraspam.yuk> schreef in bericht
news:gLcis+ESWvPCFw+j@jmwa.demon.co.uk...

I read in sci.electronics.design that Frank Bemelman
f.bemelmanq@xs4all.invalid.nl> wrote (in
423ef180$0$140$e4fe514c@news.xs4all.nl&gt about '9V 1A schematic
needed', on Mon, 21 Mar 2005:

So you and Jim Thompson are both confessed thieves. Jim in particular,
because $400 in 1968 is not what I would call pocket change.

Well, not thieves under English law, since, as it happens, 1968. In both
cases, the owner took no effective steps to re-possess. I'm quite sure
that if they had, the pay-back would have occurred.

Yes, that is my definition of an effective step too. Anyway, I found
it remarkable to read this 'aneckdote' from a person that insinuated
the other day that someone purchased a few opamps "at his EMPLOYERS
expense".

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

 

[Fred Bloggs]

Larry Brasfield wrote:

"lemonjuice" <exskimos@anonymous.to> wrote in message
news:1111406725.887396.127330@g14g2000cwa.googlegroups.com...

On Sun, 20 Mar 2005 06:45:25 GMT, Mac <foo@bar.net> wrote:

...

Anyway, unless I made some kind of mistake, the result of my "simulation"
is that Irms goes up significantly with capacitor value. In my simulation
I hold Idc constant, since the intent was to design a regulated power
supply. So the efficiency goes down with increasing capacitor value.

I am definitely interested in this topic, so if you can shed some light on
it, I would be appreciative.


Its normal the that Irms goes up with increase in capacitance value.


That is true, in the strictest sense.

What strictest sense is that you pesuod-intellectual trash (PIT from now
on out)?

But people who have studied
this issue or observed real instances of the circuit for that effect
know that the Irms value approaches a definite limit as the
capacitance value approaches infinity.

No shit , PIT?

Higher capacitance means less ripple which means higher average current
(Idc).


Huh? If you hold everything else constant, (including the
DC current taken from the cap by the regulator/load),
then Idc stays constant and so does the amount of
charge transferred per line cycle.

No shit, PIT? Too dumb to figure out that Idc is the load current yet?
And you apparently don't even know what transformer regulation is...

As Idc is directly proportional to Irms and hence is DEPENDENT
(and not independent as someone claims) on Irms... Irms also increases.


Whatever makes you think Idc = K * Irms where K
is a constant? Anybody who analyzes this correctly
can see that,

-which you didn't do, PIT- your Perl program is a joke- and you're too
damned dumb to know that your little grade school crap computation
becomes invalid because of ridiculously high harmonic content.

if Idc is held constant, Irms increases
as the conduction angle decreases. This is all very
fundamental.

Here's where your stupid beyond words , PIT- can't have your large
conduction angle with constant Idc and infinite capacitance. You have so
many ratios and normalizations in there it has made you confused ,PIT.
This also explains why you hesitated to link us to your bullshit
"analysis". You're quite the PIT, PIT.

As for "(and not independent as someone claims)", you
have (apparently) misunderstood and misconstrued my
remarks. Since they make sense in their original context,

No it doesn't PIT- your original context is a bunch of
pseudo-intellectual trash. You may impress others with your
cut-and-paste of elementary integral table junk- but as usual with
mediocre little pretenders like you- you can't make sense of it.

I won't explain that here. You need to learn that there is
a real difference between RMS and mean. If you knew
the difference, you would have a much better chance of
understanding the independence concept.

Not from you he won't- you're too damned to get cause and effect
straight, and it all you can do to RMS and mean straight formula- but
since you have no understanding, you completely miss quite a few vital
points.

Higher capacitance though means you're dissipating more power in the
there so your efficiency goes down.


Yes, but how could that happen when Idc is constant
and Irms is "directly proportional to" Irms, holding Irms
constant? You are writing riddles here. Please try to
get those sorted out before adding new ones.

Don't you need to write another Perl routine that produces a bs table
that you can only index into for maximum Irms rating of transformer and
no other value? Yeah- that's useful retard and PIT.

>

 

[lemonjuice]

On Mon, 21 Mar 2005 08:29:44 -0800, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

"lemonjuice" <exskimos@anonymous.to> wrote in message
news:1111406725.887396.127330@g14g2000cwa.googlegroups.com...
On Sun, 20 Mar 2005 06:45:25 GMT, Mac <foo@bar.net> wrote:
...
Anyway, unless I made some kind of mistake, the result of my
"simulation"
is that Irms goes up significantly with capacitor value. In my
simulation
I hold Idc constant, since the intent was to design a regulated
power
supply. So the efficiency goes down with increasing capacitor value.

I am definitely interested in this topic, so if you can shed some
light on
it, I would be appreciative.

Its normal the that Irms goes up with increase in capacitance value.

That is true, in the strictest sense. But people who have studied
this issue or observed real instances of the circuit for that effect
know that the Irms value approaches a definite limit as the
capacitance value approaches infinity.

Mr.Brasfield must be joking. Now tell me who said Irms doesn't have a
limit? Its obvious to anyone that in any finite power circuit a
current has a limit. Were we dealing with infinite power? No.

Higher capacitance means less ripple which means higher average
current
(Idc).

Huh? If you hold everything else constant, (including the
DC current taken from the cap by the regulator/load),
then Idc stays constant and so does the amount of
charge transferred per line cycle.

Mr Brasfield. I'm changing the Capacitance which is not going to hold

everything constant. I guess you know that I inst = V inst * jwC => I
change C => I change the instantaneous current I inst. Idc is the mean
of that .. so Idc HAS TO change. All very simple.

As Idc is directly proportional to Irms and hence is DEPENDENT
(and not independent as someone claims) on Irms... Irms also
increases.

Whatever makes you think Idc = K * Irms where K
is a constant? Anybody who analyzes this correctly
can see that, if Idc is held constant, Irms increases
as the conduction angle decreases. This is all very
fundamental.

Mr. Brasfield . You can actually see it using common sense but as you

seem so stubborn here goes.
the definition of Idc is assuming f = 60Hz
Idc = 1/T* Integral( Ipeak*Sin 2*PI*60*t) dt
the definition of Irms is
Irms = sqrt (1/T Integral ( Ipeak^2 * Sin^ 2*PI*60*t) dt)
Carry out the integration and tell me if you don't get my result.

As for "(and not independent as someone claims)", you
have (apparently) misunderstood and misconstrued my
remarks. Since they make sense in their original context,
I won't explain that here. You need to learn that there is
a real difference between RMS and mean. If you knew
the difference, you would have a much better chance of
understanding the independence concept.

I just defined it for you above. I never said the Irms = Idc but that

Irms = constant * Idc.

Higher capacitance though means you're dissipating more power in the
there so your efficiency goes down.

Yes, but how could that happen when Idc is constant
and Irms is "directly proportional to" Irms, holding Irms
constant?

You are writing riddles here. Please try to
get those sorted out before adding new ones.

Who said I was holding I rms constant?