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John Fields...
Posted: Sat Jul 12, 2008 1:02 pm
Guest
On Sat, 12 Jul 2008 07:56:53 -0700 (PDT), emailaddress at (no spam) insightbb.com
wrote:

Quote:
On Jul 11, 2:34 pm, ehsjr
e.h.s.j.r.removethespampunctuat... at (no spam) bellatlantic.net> wrote:

I don't want to argue - that seems your intent.
I know you want a magic formula - you have made that
clear.  There is no single formula for you.  Those who
have replied have made that clear.

We've tried to show you some of the factors that are
involved.  You insist on ignoring them.  So, we can't
help you, no matter how hard we try.  Sorry.

A formula yes, but no it's not magic. It simply requires considering
all the significant variables which I'd hoped others would assist
with, but obviously nobody else wants to do more than argue instead of
putting thought into what such an equation would look like. If you
say "it depends", then that should be in an equation. If you say some
other thing depends too, then that too can be put into the equation.

There is an equation that could be made. How accurate the answer from
it would depend on how complete it was.

---
Funny, meteorologists feel the same way, and yet it seems that
detecting that last elusive flap of butterfly wings has been
perpetually beyond their grasp.

JF
Paul E. Schoen...
Posted: Sat Jul 12, 2008 1:56 pm
Guest
<emailaddress at (no spam) insightbb.com> wrote in message
news:da3d3fd0-1853-4851-adaa-d062241579be at (no spam) x41g2000hsb.googlegroups.com...
On Jul 11, 2:34 pm, ehsjr
<e.h.s.j.r.removethespampunctuat... at (no spam) bellatlantic.net> wrote:

Quote:
I don't want to argue - that seems your intent.
I know you want a magic formula - you have made that
clear. There is no single formula for you. Those who
have replied have made that clear.

We've tried to show you some of the factors that are
involved. You insist on ignoring them. So, we can't
help you, no matter how hard we try. Sorry.

"A formula yes, but no it's not magic. It simply requires considering
all the significant variables which I'd hoped others would assist
with, but obviously nobody else wants to do more than argue instead of
putting thought into what such an equation would look like. If you
say "it depends", then that should be in an equation. If you say some
other thing depends too, then that too can be put into the equation."

"There is an equation that could be made. How accurate the answer from
it would depend on how complete it was. I came here looking to make
as complete an equation as possible but it seems everyone else is
apathetic about the idea and only wants to tell me I don't "need" to
know or that it can't be done."

"It's not magic. A caveman looking at a bic lighter would think that
is magic but is it? Your great great grandfather, if he were alive
today, might think a computer is magic, but is it? Just because
someone doesn't know something it doesn't become magic. It would just
require someone bothering to do so, which I was attempting but it
seems I'm going it alone because others think it's too hard or they're
too lazy or whatever the reason. That's fine, nobody is compelled to
do anything but if they had no assistance (some replying did!) then as
always they should've just moved on to the next thread."

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

I find it curious that you reply only to those who either disagree with you
or do not give you a formula that automagically takes into account every
design parameter and spits out the answers you demand. And who is the lazy
one? You are asking for free advice, and you are getting a lot of bang for
your buck. Contract a design engineer at $100/hour and they will not be
"lazy".

The factors involved in your design are numerous:

Input voltage range
Frequency range
Physical size constraints
Efficiency
Costs (NRE, materials, labor)
Environmental conditions (temp, humidity, altitude, etc.)
Safety (UL, NEMA, Medical, etc.)
Isolation/insulation/leakage current
Surge withstand
EMI/RFI
Load variations
Duty cycle
Overcurrent and short circuit protection
Reverse connection protection
Charging current/voltage/time/temperature profile
Acceptable MTBF

You have constrained the parameters by saying that the charger circuit has
been designed and you are not willing or able to change that. So you are
only looking for a cheap and dirty AC to DC power supply that will meet
your needs. Your charger circuit must have specifications for what it can
accept as raw DC input, and what sort of current it will draw under all
conditions. IOW, you need to model this circuit with a "formula" or SPICE
model. Then you must decide if you want to use an off-the-shelf transformer
or design one specifically to meet your needs. An OTS solution will be
initially cheaper, but you will sacrifice something, like efficiency or
physical size that will not be optimized. You will need to choose one that
meets (but probably exceeds) your specifications.

There are many simplified formulas that have been given to you, that will
give results close enough to choose from several OTS transformers that
should meet your specs. You can also model the transformer pretty well in
SPICE. But you will, sooner or later, need to build a prototype and test
the hell out of it. After a few hours of testing, you can pretty well
characterize the transformer, and then fine tune your selection. Or you can
use the results to design a custom transformer that exactly meets your
needs. But tranny design gets rather complex when you are pushing the
limits of efficiency, cost, size, regulation, and other factors.

You have all the information you need to design, build, and simulate your
DC front-end circuit. We do not have all the information we need to model
your charger load, and even if we did, it is up to you to actually design
the thing. As I said before, SPICE is essentially a mathematical formula
that you can modify with all the parameters you want, and run step and
sweep analyses to make sure your initial design is OK. Then you can tweak
it ad nauseum to get it as close to perfect as it will ever be.

Paul
John Larkin...
Posted: Sat Jul 12, 2008 2:21 pm
Guest
On Sun, 13 Jul 2008 01:02:47 +1000, "Phil Allison"
<philallison at (no spam) tpg.com.au> wrote:

Quote:

emailaddress at (no spam) insightbb.com


I am not designing a charging circuit.


** You are NOT designing ANYTHING AT ALL

YOU STINKING CRIMINAL LIAR !!!!!!!!!!


FOAD - you anonymous pile of shit.




..... Phil




http://www.sfgate.com/cgi-bin/object/article?o=6&f=/c/a/2008/07/12/DDCOMICS12.DTL

John
Baron...
Posted: Sat Jul 12, 2008 3:14 pm
Guest
John Larkin wrote:

Quote:
On Sun, 13 Jul 2008 01:02:47 +1000, "Phil Allison"
philallison at (no spam) tpg.com.au> wrote:


emailaddress at (no spam) insightbb.com


I am not designing a charging circuit.


** You are NOT designing ANYTHING AT ALL

YOU STINKING CRIMINAL LIAR !!!!!!!!!!


FOAD - you anonymous pile of shit.




..... Phil





http://www.sfgate.com/cgi-bin/object/article?o=6&f=/c/a/2008/07/12/DDCOMICS12.DTL

John

LOL. Nice ! Thats a keeper. :-)

--
Best Regards:
Baron.
Paul E. Schoen...
Posted: Sat Jul 12, 2008 4:29 pm
Guest
<mrdarrett at (no spam) gmail.com> wrote in message
news:0631a23f-0a93-4a34-a703-b3bd4e22ba06 at (no spam) j22g2000hsf.googlegroups.com...
Quote:
On Jul 12, 7:40 am, emailaddr... at (no spam) insightbb.com wrote:
..snip..

However, it is fairly beside the point that I am looking for
a universal equation that ignores all of this. Many people seem to be
saying the rest matters and yes of course it does - but it is still
theoretically expressable in an equation with the factors that change
as variables.


Universal equation. Here you go.

http://en.wikipedia.org/wiki/Heat_equation
http://en.wikipedia.org/wiki/Newton%27s_law_of_cooling#Newton.27s_law_of_cooling

Also check out "Introduction to Heat Transfer" by Incropera and
DeWitt.

The game is to keep the transformer wires at only a moderate
temperature higher than ambient, or you will melt the wiring
insulation.

Knock yourself out.

Michael

These equations are only part of the overall solution. Transformers are
complex physical entities with many materials having different thermal
characteristics, and the temperature of any spot in the transformer depends
on the amount of power being applied, time, and the way in which heat is
radiated, conducted, or convected away from the source of the heat. The
winding insulation is rated at a certain temperature according to the
insulation class, and may vary from about 105 C (Class A) to 180 C (Class
H).

http://www.engineeringtoolbox.com/nema-insulation-classes-d_734.html

That is for motors (and I think it applies also to transformers), but here
is information more specific to trannies, and allows for hot spots:
http://www.jeffersonelectric.com/cgi-bin/site.pl?3208&dwContent_contentID=12

This shows a wider range of classes up to "S", which is 250 C. Now, that's
hot!
http://www.pleo.com/ulsystem/eis_apply.htm

This is a detailed document with more formulas for the OP to use when he
designs his transformer:
http://www.superioressex.com/uploadedFiles/News/White_Papers/emcwa-nema_magnet-thermal-class-ratings.pdf

Paul
ehsjr...
Posted: Sat Jul 12, 2008 5:00 pm
Guest
John Fields wrote:
Quote:
On Sat, 12 Jul 2008 07:56:53 -0700 (PDT), emailaddress at (no spam) insightbb.com
wrote:


On Jul 11, 2:34 pm, ehsjr
e.h.s.j.r.removethespampunctuat... at (no spam) bellatlantic.net> wrote:


I don't want to argue - that seems your intent.
I know you want a magic formula - you have made that
clear. There is no single formula for you. Those who
have replied have made that clear.

We've tried to show you some of the factors that are
involved. You insist on ignoring them. So, we can't
help you, no matter how hard we try. Sorry.

A formula yes, but no it's not magic. It simply requires considering
all the significant variables which I'd hoped others would assist
with, but obviously nobody else wants to do more than argue instead of
putting thought into what such an equation would look like. If you
say "it depends", then that should be in an equation. If you say some
other thing depends too, then that too can be put into the equation.

There is an equation that could be made. How accurate the answer from
it would depend on how complete it was.


---
Funny, meteorologists feel the same way, and yet it seems that
detecting that last elusive flap of butterfly wings has been
perpetually beyond their grasp.

JF

Smile Nice analogy, wonder if he'll get it.

What is interesting to me is the apparent contradiction in
the op's thinking. He chooses to lecture respondents with
his thinking, rather than attempt to understand what they
are saying. Or, if not lecture, argue with points that are
made. I've seen this a few times here - someone comes here,
states that they don't know about X, and asks a question.
But then they argue with the answer! What kind of thinking
is that? Is it the "new way of thinking" or ???

Ed
Jamie...
Posted: Sat Jul 12, 2008 5:33 pm
Guest
emailaddress at (no spam) insightbb.com wrote:

Quote:
On Jul 11, 4:07 pm, mrdarr... at (no spam) gmail.com wrote:

On Jul 9, 6:39 pm, emailaddr... at (no spam) insightbb.com wrote:






On Jul 9, 9:26 pm, emailaddr... at (no spam) insightbb.com wrote:

I have an application in which I need to determine if a transformer is
suitable.
Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply.

Another thing I am wondering is if a basic, typical small brick AC-DC
switching supply is stable powering this kind of load. Stripping down
the circuit to the basic topology, what if it had for example:

AC-DC SMPS -> LM317 -> 1.25 Ohm resistor (LM317 in current regulating
config) -> 1A into Battery Pack + LM317 feedback

Since the SMPS is trying to regulate to it's spec'd voltage, let's say
that is 13V even if an uncommon value, will it run stable doing so and
if so, is that at the same constant output wattage from the SMPS the
whole time, with the linear regulator simply dropping more voltage,
creating more heat at the beginning of a battery recharge cycle since
it's suppling constant current to batteries that are at a lower
initial, low state-of-charge voltage?

Why so many LM317s? They will generate heat too, and waste energy.


First, the example is not something I intend to build at all, it is
just providing an example, a load.

Second, in the example there is only one LM317, when I mentioned
"LM317" the 2nd and third time I was only describing the typical
topology of using one LM317 in the context of it being in current
limiting mode. I used an example that is straight from National's
LM317 datasheet since many would presumably be familiar with it.


What are you charging? What supply voltage do you need?



I am not building that example. The first example or something fairly
similar I did build a few years ago but now I have no need to consider
the charging or other aspects of the circuit beyond the supply to it,
because it used to be a stationary charger with capability of charging
several different sized packs but it has been reused now for a single
set of packs all with 6 cells. Therefore the most desirable supply
has changed from what it used formerly to something more closely
matched. However, it is fairly beside the point that I am looking for
a universal equation that ignores all of this. Many people seem to be
saying the rest matters and yes of course it does - but it is still
theoretically expressable in an equation with the factors that change
as variables.

if it has any meaning at all for the subject..


If you're working with an existing xformer for example. What I have
done when not having all the spec's for the xformer was to measure the
primary and secondary DC resistance of the xformer. the primary side
can be scaled by the reference of the ratio of the transformer..

For example, lets assume you have 120 volt service into a wallwart,
the unit outputs 12 volts, this would be a 10:1 ratio. So if we were to
measure lets say 200 ohms on the primary side, we would thus scale that
down by 10 = 20 ohms. The DC resistance on the secondary side is simply
added to this to give you a sum R..
You can look at this as an imaginary resistor in the xformer..

Not knowing the factors of the xformer makes efficiency unknown here,
we can only guess an average and we shouldn't assume it to be near 1.0

I would say 0.8 to be safe.

When calculating the voltage output, you use the R in series
with the load to help determine the voltage drop on the output.
Of course, with a full bride you must also consider the .7 or 1.3
voltage lost depending on what type you use.

A 2 diode full wave rectifier works the best how ever, you'd need the
proper output on the xformer for this..

THen there is the ripple voltage at the filter caps which gets
influenced from the R in the xformer..

In many cheap designs, this R in the xformer is actually used as part
of the filtering to help give a more stable voltage at lower current
demand, the same as putting in a real resistor in series to the caps for
the same reason.

That's about as basic I can describe it.
http://webpages.charter.net/jamie_5"
John Fields...
Posted: Sat Jul 12, 2008 5:54 pm
Guest
On Sat, 12 Jul 2008 22:00:45 GMT, ehsjr
<e.h.s.j.r.removethespampunctuation at (no spam) bellatlantic.net> wrote:

Quote:
John Fields wrote:
On Sat, 12 Jul 2008 07:56:53 -0700 (PDT), emailaddress at (no spam) insightbb.com
wrote:


On Jul 11, 2:34 pm, ehsjr
e.h.s.j.r.removethespampunctuat... at (no spam) bellatlantic.net> wrote:


I don't want to argue - that seems your intent.
I know you want a magic formula - you have made that
clear. There is no single formula for you. Those who
have replied have made that clear.

We've tried to show you some of the factors that are
involved. You insist on ignoring them. So, we can't
help you, no matter how hard we try. Sorry.

A formula yes, but no it's not magic. It simply requires considering
all the significant variables which I'd hoped others would assist
with, but obviously nobody else wants to do more than argue instead of
putting thought into what such an equation would look like. If you
say "it depends", then that should be in an equation. If you say some
other thing depends too, then that too can be put into the equation.

There is an equation that could be made. How accurate the answer from
it would depend on how complete it was.


---
Funny, meteorologists feel the same way, and yet it seems that
detecting that last elusive flap of butterfly wings has been
perpetually beyond their grasp.

JF

:-) Nice analogy, wonder if he'll get it.

What is interesting to me is the apparent contradiction in
the op's thinking. He chooses to lecture respondents with
his thinking, rather than attempt to understand what they
are saying. Or, if not lecture, argue with points that are
made. I've seen this a few times here - someone comes here,
states that they don't know about X, and asks a question.
But then they argue with the answer! What kind of thinking
is that? Is it the "new way of thinking" or ???

---
I've seen it a few times also, and I think it's a way of trying to
save face by not admitting that the gift that was given incurred
emotional indebtedness.

That is, if you give me a gift of knowledge and I dis it, I'll still
have the gift but in my mind I won't owe you anything for it.

Make sense?

JF
Phil Allison...
Posted: Sat Jul 12, 2008 9:19 pm
Guest
"Jamie" = Maynard A. Philbrook = radio ham KA1LPA

WARNING: Ignore everything this total moron posts.


Quote:

For example, lets assume you have 120 volt service into a wallwart, the
unit outputs 12 volts, this would be a 10:1 ratio. So if we were to
measure lets say 200 ohms on the primary side, we would thus scale that
down by 10 = 20 ohms. The DC resistance on the secondary side is simply
added to this to give you a sum R..


** WRONG !!!!!!!!

Primary resistance appears at the secondary divided by the SQUARE of the
transformer's actual turns ratio.

The turns ratio is always less than the nominal voltage ratio, to allow for
voltage drop under load, so - for the above case, ( very small
transformer) the turns ratio is about 8 and reflected resistance is
therefore 200 /64 = 3.13 ohms.



....... Phil
Phil Allison...
Posted: Sat Jul 12, 2008 9:21 pm
Guest
"John Fields"

Quote:
I've seen it a few times also, and I think it's a way of trying to
save face by not admitting that the gift that was given incurred
emotional indebtedness.

That is, if you give me a gift of knowledge and I dis it, I'll still
have the gift but in my mind I won't owe you anything for it.

Make sense?


** If you imagine some little a brat throwing a toy that is not the one he
wanted back at Santa.



...... Phil
Paul E. Schoen...
Posted: Sat Jul 12, 2008 11:47 pm
Guest
"Phil Allison" <philallison at (no spam) tpg.com.au> wrote in message
news:6dt714F47182U1 at (no spam) mid.individual.net...
Quote:

"Jamie" = Maynard A. Philbrook = radio ham KA1LPA

WARNING: Ignore everything this total moron posts.



For example, lets assume you have 120 volt service into a wallwart,
the unit outputs 12 volts, this would be a 10:1 ratio. So if we were to
measure lets say 200 ohms on the primary side, we would thus scale that
down by 10 = 20 ohms. The DC resistance on the secondary side is simply
added to this to give you a sum R..


** WRONG !!!!!!!!

Primary resistance appears at the secondary divided by the SQUARE of the
transformer's actual turns ratio.

The turns ratio is always less than the nominal voltage ratio, to allow
for voltage drop under load, so - for the above case, ( very small
transformer) the turns ratio is about 8 and reflected resistance is
therefore 200 /64 = 3.13 ohms.

You're right. I just tried a SPICE model with a tranny that has 400 mH
primary and 25 mH secondary, which is a 4:1 transformer with an impedance
ratio of 16:1. Adding 16 ohms to the primary series resistance, or 1 ohm to
the secondary, gives the same result into a load. It makes sense, because
you have four times the current and 1/4 the voltage in the secondary, so
the effective resistance (or impedance) ratio is a square function.

If you try the simulation, you will need to keep at least about 1 uOhm of
series R in the primary.

Paul

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

Version 4
SHEET 1 880 680
WIRE 128 128 16 128
WIRE 336 128 224 128
WIRE 128 208 16 208
WIRE 336 208 224 208
WIRE 16 224 16 208
WIRE 336 224 336 208
FLAG 16 224 0
FLAG 336 224 0
SYMBOL ind2 112 112 R0
WINDOW 0 4 -10 Left 0
WINDOW 3 -12 124 Left 0
SYMATTR InstName L1
SYMATTR Value 400m
SYMATTR Type ind
SYMATTR SpiceLine Rser=16
SYMBOL ind2 240 224 R180
WINDOW 0 2 122 Left 0
WINDOW 3 -8 -11 Left 0
SYMATTR InstName L2
SYMATTR Value 25m
SYMATTR Type ind
SYMATTR SpiceLine Rser=0
SYMBOL voltage 16 112 R0
WINDOW 0 -12 -2 Left 0
WINDOW 3 -34 192 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(0 2.828 60 0 0 0 100)
SYMBOL res 320 112 R0
SYMATTR InstName R1
SYMATTR Value 1
TEXT 112 272 Left 0 !K1 L1 L2 1
TEXT -20 268 Left 0 !.tran 1
John Fields...
Posted: Sun Jul 13, 2008 7:51 am
Guest
On Sun, 13 Jul 2008 12:21:58 +1000, "Phil Allison"
<philallison at (no spam) tpg.com.au> wrote:

Quote:

"John Fields"

I've seen it a few times also, and I think it's a way of trying to
save face by not admitting that the gift that was given incurred
emotional indebtedness.

That is, if you give me a gift of knowledge and I dis it, I'll still
have the gift but in my mind I won't owe you anything for it.

Make sense?


** If you imagine some little a brat throwing a toy that is not the one he
wanted back at Santa.

---

Yeah, good one. :-)

JF
default...
Posted: Sun Jul 13, 2008 8:15 am
Guest
On Sun, 13 Jul 2008 01:02:47 +1000, "Phil Allison"
<philallison at (no spam) tpg.com.au> wrote:

Quote:

emailaddress at (no spam) insightbb.com


I am not designing a charging circuit.


** You are NOT designing ANYTHING AT ALL

YOU STINKING CRIMINAL LIAR !!!!!!!!!!


FOAD - you anonymous pile of shit.

Can't you just feel the love?

--


----== Posted via Pronews.Com - Unlimited-Unrestricted-Secure Usenet News==----
http://www.pronews.com The #1 Newsgroup Service in the World! >100,000 Newsgroups
---= - Total Privacy via Encryption =---
Michael A. Terrell...
Posted: Sun Jul 13, 2008 11:23 am
Guest
default wrote:
Quote:

On Sun, 13 Jul 2008 01:02:47 +1000, "Phil Allison"
philallison at (no spam) tpg.com.au> wrote:


emailaddress at (no spam) insightbb.com


I am not designing a charging circuit.


** You are NOT designing ANYTHING AT ALL

YOU STINKING CRIMINAL LIAR !!!!!!!!!!


FOAD - you anonymous pile of shit.

Can't you just feel the love?


Phil must be out of toasters to molest. :(


--
http://improve-usenet.org/index.html

If you have broadband, your ISP may have a NNTP news server included in
your account: http://www.usenettools.net/ISP.htm

Sporadic E is the Earth's aluminum foil beanie for the 'global warming'
sheep.
John Fields...
Posted: Sun Jul 13, 2008 6:35 pm
Guest
On Sat, 12 Jul 2008 07:44:00 -0700 (PDT), emailaddress at (no spam) insightbb.com
wrote:

Quote:
On Jul 11, 4:37 pm, John Fields <jfie... at (no spam) austininstruments.com> wrote:
On Wed, 9 Jul 2008 18:26:12 -0700 (PDT), emailaddr... at (no spam) insightbb.com
wrote:





I have an application in which I need to determine if a transformer is
suitable.

Transformer is rated for 12VAC, 1.67A but I need DC.  I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?

I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?

Perhaps a little more info about the project is in order.  Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries.  

Why can't you just used the raw unfiltered DC and current limit the
input to the battery?


I am not designing a charging circuit. The charging circuit is fixed,
done years ago, works fine. It's just being mated with a different
supply because the need for it has changed. I am looking for an
equation to qualify transformers per their AC spec for DC supply when
used with a basic bridge rectifier following them.

---
Use Irms = 2.0 Idc and you'll always be safe .

JF
 
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