Chip with simple program for Toy

[Lord Garth]

John,

I've noticed that alt.binaries.schematics.electronic is now missing
from my news server. Apparently replaced by
free.binaries.schematics.electronic ... what has happened over the
last week?

 

[Lord Garth]

"Eeyore" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:487D20F6.E5C8C0F5@hotmail.com...

Lord Garth wrote:

John,

I've noticed that alt.binaries.schematics.electronic is now missing
from my news server. Apparently replaced by
free.binaries.schematics.electronic ... what has happened over the
last week?

American censorship. You know - free country and all that. Look up NYAG
+ Usenet.

I'm not interested in political drivel, I just don't wish to miss binary
postings!
I will search the terms you suggest however.

Interestingly, I've not seen my own post yet, only your reply.

 

[Lord Garth]

"John Fields" <jfields@austininstruments.com> wrote in message
news:ae9q74d9fuadh8162pd4rhqdn4hot53btu@4ax.com...

On Tue, 15 Jul 2008 23:13:10 +0100, Eeyore
rabbitsfriendsandrelations@hotmail.com> wrote:



Lord Garth wrote:

John,

I've noticed that alt.binaries.schematics.electronic is now missing
from my news server. Apparently replaced by
free.binaries.schematics.electronic ... what has happened over the
last week?

American censorship. You know - free country and all that. Look up NYAG
+ Usenet.

---
Here's what I found:

http://news.cnet.com/8301-13578_3-9964895-38.html

Note that a _Federal_ judge overturned the Pennsylvania State law for
being in violation of the first amendment of the US Constitution for
several reasons, so that's censorship overturned and a somewhat freer
environment restored.

Much like Larry Flynt's trip to the US Supreme Court which allowed him
to freely publish porn after lots of porn Nazis tried to shut him
up/down because of their own inhibitions and feelings of religious
guilt.

After all, if they didn't want to view or buy Flynt's stuff they
should have just left it alone, individually, and saved their own
souls instead of trying to shut him up/down, "for the good of all of
mankind." ,thinking that everyone else's souls were their business to
save by trying to erect legislation designed to curtail Flynt's
actions.

Hooray for the US Supreme Court's decision, don't you agree?

I often think that if a potential sexual predator gets himself/herself
off by masturbating to what they find on USENET or on the Internet
then that's well worth the alternative which might occur if that porn
wasn't there.

Cuomo's position, as a Roman Catholic, is interesting in that it would
cut off that avenue of expression and _require_ celibacy even though
celibacy might not be a real option.

So much for politicians...

JF

I was at UT when the Flynt case came to a head. I seem to recall him
publishing many pictures of mutilated US soldiers from the Vietnam
war. I wonder if that is what really was the root cause of the censorship
case?

 

[Mr.T]

"Richard Crowley" <rcrowley@xp7rt.net> wrote in message
news:6e480hF53dqtU1@mid.individual.net...

"Rich Grise" wrote ...
Mr.T wrote:
"Joerg" wrote
Some large Polynesian mussel shells supposedly do that. When you hold
them to your ear you hear the sounds of the sea.

Or the sound of your blood flow being acoustically amplified and
reflected back to your ear, at least.

I'm sure the ambient noise swamps the sound of your blood by a few
orders
of magnitude. ;-)

Similar to GWB being sure Iraq had nuclear weapons capability no doubt
You actually believe a sea shell stores sounds? Why only sea sounds then
after it has been far removed for years?
I think at least Joerg was joking... surely?

The sound of your heartbeat can seem deafening if you spend
much time in a good anechoic chamber. I tried it in ElectroVoice's
chamber in Buchannan many years ago.

Yep, and I can hear my blood pulsing in an isolation booth, doesn't even
require a good anechoic chamber.
Of course it's the blood flow in your head you hear, not your "heartbeat".

MrT.

 

[Phil Allison]

<emailaddress@insightbb.com>


I'm sorry if I did not clarify this, but I do not need an introductory
course in sizing transformers. The whole point was moving beyond the
generic overestimates towards the real science in what is actually
necessary. I lecture towards the end of saying, ok, but this is
common knowlege, what else do you know? Maybe it takes a village,
maybe no one person has the entire answer but it seems to get there we
need to have some lecture, some dispelling of comfort zones and get
right down to the actual criteria necessary and have that proven
through real worl examples of success or failure, not just saying "use
a bigger hammer", until it is proven to be needed.

Yes, I will argue with an answer when I ask for an equation and
someone tells me otherwise. I have built plenty of PSU over the
years, if I needed to know what I have already done successfully then
I would have asked a different question? No offense intended, but you
need to focus on what I asked, as do others. If they are ignorant of
the answer, there is no need to reply.


** Go get fucked.

You revolting brat.




....... Phil

 

[Paul E. Schoen]

"John Fields" <jfields@austininstruments.com> wrote in message
news:ninr74hjcudl6v8eaf69i3mpplk45rest9@4ax.com...

On Wed, 16 Jul 2008 02:55:03 -0700 (PDT), emailaddress@insightbb.com
wrote:


I'm sorry if I did not clarify this, but I do not need an introductory
course in sizing transformers. The whole point was moving beyond the
generic overestimates towards the real science in what is actually
necessary. I lecture towards the end of saying, ok, but this is
common knowlege, what else do you know? Maybe it takes a village,
maybe no one person has the entire answer but it seems to get there we
need to have some lecture, some dispelling of comfort zones and get
right down to the actual criteria necessary and have that proven
through real worl examples of success or failure, not just saying "use
a bigger hammer", until it is proven to be needed.

Yes, I will argue with an answer when I ask for an equation and
someone tells me otherwise. I have built plenty of PSU over the
years, if I needed to know what I have already done successfully then
I would have asked a different question? No offense intended, but you
need to focus on what I asked, as do others. If they are ignorant of
the answer, there is no need to reply.

---
You may have _built_ plenty of PSU over the years, but from your
question re. transformer sizing it appears that all you were doing was
grunt work, putting together someone else's design, with no real
insight as to the "why" of the transformer's capacity needing to be
larger than what was required to drive the load.

Instead of being an annoying little ass you might try thanking those
of us who have given you good advice instead of pretending that,
somehow, we did you a disservice.

Oh, and by the way, Fuck you.

I usually give someone the benefit of doubt and try to come up with various
ways of looking at a problem and proposing a reasonable solution, or
pointing out the need for more information. The OP has chosen not to
comment on those posts, but only those where he feels compelled to argue
and assume a self-righteous attitude. So I second your motion. All in
favor?

Paul

 

[Dan Coby]

<emailaddress@insightbb.com> wrote in message news:daad0917-cdf2-4ff5-8a75-

I'm sorry if I did not clarify this, but I do not need an introductory
course in sizing transformers. The whole point was moving beyond the
generic overestimates towards the real science in what is actually
necessary.

The usual limiting factor for the current output of a transformer is heat dissipation.

The factors that need to be considered are:
1) The amount of heat being produced.
2) How the heat is removed from where it is produced.
3) The maximum desired operating temperature.

Heat is produced in a transformer in several different ways. There are electrical
losses in the windings due to the currents in the windings. There are electrical losses
in the core materials dues to induced currents in the core There are probably other
factors related to frequency and materials.

The easiest one of these to handle quantitatively is the power (heat) lost in the
windings' resistance. As the current increases, so does the power that is lost in the
resistance of the windings. For more information (and equations) about the power loss in
a resistance see: http://en.wikipedia.org/wiki/Root_mean_square#Average_electrical_power

Short of giving you the equations for RMS current we cannot answer your question
since we do not know that the winding currents are for your application. (You mentioned
DC currents but that is pretty much meaningless for a transformer.) Please note that the
RMS current is not the same as the average current. There can be a big difference in
RMS current (and thus power lost) between a current waveform that consists of pulses
with a low duty cycle and high peak current versus a constant DC with the same
average current. If you are building power supplies, the RMS current will depend upon
(among other things) the output current, the filter capacitance, and the internal resistance
of the transformer. You have given us no information about any of the details of what
you are trying to accomplish and we really do not want to repeat the contents of
textbooks on basic electrical engineering.


How the heat is removed from where it is produced depends upon the specific
details of the design of the transformer and the environment in which is being used.
You have given us no details on your application and we really do not want to repeat
the contents of the textbooks on heat flow.


The maximum operating temperature for a transformer depends upon the materials
and upon operating considerations. As the operating temperature increases the
rates of break down of the insulation increase. In general, the higher the operating
temperature of a transformer, the shorter its life will be. There can also be other
operating considerations. For instance, the materials that are used to build a
transformer, may allow for a long life while operating at 100 C but I would not like
to have that transformer in close proximity to my skin at that temperature.

 

[Tom Biasi]

"W. eWatson" <notvalid2@sbcglobal.net> wrote in message
news:xZrfk.16520$89.2021@nlpi069.nbdc.sbc.com...

I have a RS small Optimus speaker (stereo pair). It operates on four AA
batteries. I plugged the powered speaker into my CC Crane radio's headphone
jack, and it easily played the music at various volumes. I then took it to
an elder neighbor's house to attach it to his TV. He's hard of hearing, and
the plan is to put the speaker close to where he sits. That way he doesn't
have to boom the TV sound when others are present.

He has a Sharper Image (now defunct store) device that plugs into the
audio RCA jacks of his TV. The remote headphones are placed in the
device's box to recharge the batteries in them. The set up works, but he
manages to leave the headphones off it and often forgets to turn the power
off. That drains the headphone power. He often puts the headphones in the
cradle the wrong way, and they don't get charged up.

The speakers should be the answer, but when I plugged them into the RCA
audio cable from the TV, single plug into the speaker, nothing happens.
That male audio plug looks identical to the one I used with the radio.
What's wrong here?
Hard to say from here. maybe the TV needs its audio directed to the external

jacks. Hit the program mode and look at the audio out settings.

Tom

 

[ian field]

"aleksa" <aleksaZR@gmail.com> wrote in message
news:29f3968d-491c-4257-8ad8-d059f77253bd@t54g2000hsg.googlegroups.com...

I want to solder one TQFP144, but do not have
any experience with anything more than DIP40.

This is what I'm planning to do. I will:

1. apply some tinnol on all the pins on the PCB
2. remove excess tinnol with dump cloth
3. place the chip and solder it pin by pin with solder iron

step 3 will certanly solder several pins together which I plan to
fix with solder braid. The question is: what width should I buy?

Desolder braid risks pulling up pins if you don't keep it fully and
uniformly heated.

Hold the board upside down but slanted with the ends of the pins downwards
and tease the excess solder away with the tip of the iron, you need enough
flux to prevent a skin of oxide forming.

 

[Paul E. Schoen]

"whit3rd" <whit3rd@gmail.com> wrote in message
news:6a58988a-5bd9-4c16-af89-69652a6d8f6c@k30g2000hse.googlegroups.com...

On Jul 13, 4:35 pm, John Fields <jfie...@austininstruments.com> wrote:

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

Well, not completely, no. Consider a 1A (RMS)
rated transformer, with a diode/capacitor output.

When attached to a 1A (DC) load, the transformer
actually forward-biases the diode for a brief
time at the peak voltage, and otherwise current
from the transformer is nil. If the conduction
period is one-tenth of the full cycle period, that
means 10A current from the transformer during
the active time.

Here's where it gets mathematical: the power dissipated
in a 1A transformer with (for instance) 1 ohm winding

resistance is 1 watt, when the load is taking a simple
1A (AC) current.

Heat = 1A **2 * 1 ohm = 1 watt

When the same transformer feeds the DC load as described above, the
power is

Heat = (0.9 * 0) + (0.1 * 10A**2 * 1 ohm) = 10 watt

So a perfectly good transformer can burn up feeding a
DC rectifier and load, when a similar AC load wouldn't

bother it. That isn't always covered by a 'factor of two'
or any other rule of thumb. I've seen manufacturers offer
tables of the permissible DC output ratings for their
transformers, but you can't count on that. The AC rating
looks better, so salesmen will feed you that info first.
-----------------------------------------------------------

I beg to differ with your analysis, if you are talking about an ordinary
rectifier and capacitor circuit. As an example, I simulated a FWB with a 12
VAC nominal output transformer with 1 ohm series resistance, and a load of
12 ohms, and a capacitor of 100,000 uF, which should produce the highest
possible current peaks. The simulation shows peak currents of 3.7 amps.
With 1000 uF, the peaks are 3.2 amps. Now, during the charging period, with
100,000 uF, the peaks start at 14.6 amps and then diminish to 4.3 amps at
0.5 seconds. In the first 200 mSec, the tranny is supplying 34 watts, but
then settles down to 15.8 watts when the capacitor is fully charged. At
that time, the load is essentially pure DC, and the resistor dissipates
13.9 watts. So only about 2 watts is left, and that is shared among the
rectifiers (305 mW each), and the tranny (about 0.8 watts).

If you can show me a circuit where you will get these 10 ampere peaks at
10% duty cycle, then I will agree that the tranny will be overloaded. But
you will probably need to use some sort of PWM, and there will also be a
lot more power being dumped into the load. If you are talking about AC to
DC rectifier circuits, it's a safe bet to design the circuit so that the DC
output voltage under load is about the same as the nominal RMS AC voltage
of the transformer, and in this case the RMS input current is 1.81/1.08 =
less than twice the output current. So the 2:1 ratio that John proposed is
very reasonable.

Paul

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

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SYMBOL polcap 400 192 R0
SYMATTR InstName C2
SYMATTR Value 100000ľ
SYMATTR Description Capacitor
SYMATTR Type cap
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SYMBOL voltage 128 176 R0
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SYMATTR Value SINE(0 18 60 0 0 0 100)
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SYMBOL res 512 144 R0
SYMATTR InstName R1
SYMATTR Value 12
SYMBOL schottky 352 272 M270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D3
SYMATTR Value 1N5818
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL schottky 288 384 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D1
SYMATTR Value 1N5818
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL schottky 352 496 M270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D4
SYMATTR Value 1N5818
SYMATTR Description Diode
SYMATTR Type diode
TEXT 184 528 Left 0 !.tran 1

 

[Phil Allison]

"trickyrick" <rickj@sympatico.ca>

I think that the AC in Europe is 220. Can't I just wire up something
from the outlet like oneside and the ground for 110 VAC.
Thanks

** No.

That will still give you 220 volts - in Europe.


....... Phil

 

[Ben Jackson]

On 2008-07-16, aleksa <aleksaZR@gmail.com> wrote:

1. apply some tinnol on all the pins on the PCB
2. remove excess tinnol with dump cloth
3. place the chip and solder it pin by pin with solder iron

I'm not sure what 'tinnol' is, but assuming it's some kind of liquid
flux: Skip step 2. You can readily solder fine pitch parts like that
in a puddle of flux. Fine pitch + lots of flux = easy. In fact, the
first remedy I would suggest for almost any problem (including bridged
pins) is to add flux and reheat it. You can often "pick up" the excess
solder with the iron and remove it or redistribute it. With lots of
flux, the solder will strongly prefer to stick to the pins rather than
itself which can cure the bridge.

If you get braid, make it very fine, and NEVER PULL IT OFF THE BOARD
if it resists. If you do, you will remove traces, bend pins, etc. It
can help to tin the braid as well, but it will corrode over time.

--
Ben Jackson AD7GD
<ben@ben.com>
http://www.ben.com/

 

[Tony Burch]

"aleksa" <aleksaZR@gmail.com> wrote in message
news:29f3968d-491c-4257-8ad8-d059f77253bd@t54g2000hsg.googlegroups.com...

I want to solder one TQFP144, but do not have
any experience with anything more than DIP40.

This is what I'm planning to do. I will:

1. apply some tinnol on all the pins on the PCB
2. remove excess tinnol with dump cloth
3. place the chip and solder it pin by pin with solder iron

step 3 will certanly solder several pins together which I plan to
fix with solder braid. The question is: what width should I buy?

Hi,

I would not recommend using solder braid. Please have a look at the video at
http://www.supersolderingsecrets.com for a reliable technique for had
soldering TQFPs.

If you don't have a "reservoir tip" for your iron, you can just use a broad
chisel tip.

Kind regards,
Anthony Burch

 

[Phil Allison]

"The Phantom"

I have a transformer with a nominal 24 volt secondary, rated at 8 amps.
It
has a measured series resistance (secondary plus reflected primary) of
about .125 ohms.

** This is a fabricated falsehood - the numbers simply do not add up.

A 192VA rated transformer does not have 4% regulation - correctly rated
it has 8%.

Look up makers data if you doubt this.

The oft quoted ratio of 1.6 for AC amps to DC amps applies ONLY at full
load and for a correctly rated transformer.

BTW:

Percent voltage regulation and percent power loss with resistive load are
virtually the same numbers (ignoring I mag loss).


...... Phil

 

[Phil Allison]

"The Phantom"
"Phil Allison"

I have a transformer with a nominal 24 volt secondary, rated at 8 amps.
It has a measured series resistance (secondary plus reflected primary)
of about .125 ohms.


** This is a fabricated falsehood - the numbers simply do not add up.

A 192VA rated transformer does not have 4% regulation - correctly
rated
it has 8%.

This one does.

** Cos it is not correctly VA rated.

Look up makers data if you doubt this.

This transformer has no maker's identification, so I can't look it up.

** So you also have no idea what its VA rating is.

However, the primary (it's a 60 Hz transformer) says 120 VAC and it
measures .961 ohms, cold.

** So it is a circa 360 VA transformer.

The *correctly* rated secondary load is not 8 amps - but more like 15.

The oft quoted ratio of 1.6 for AC amps to DC amps applies ONLY at full
load and for a correctly rated transformer.



...... Phil

 

[Ben Jackson]

On 2008-07-16, Tim Wescott <tim@seemywebsite.com> wrote:

aleksa wrote:
I want to solder one TQFP144, but do not have

Contrary to what Ian and Ben have said (and they're both good guys),
when I have to do this I use the following procedure:

1. Locate the part and tack down opposing pins (i.e. pins
1 and 73) so it sits where it belongs.
2. Solder all the pins on one side. Just drag the iron and
make a big glob of solder across all the pins. Think of
it as the Golden Gate of solder bridges.
3. Wick up the solder (carefully) with some honking big (1/8")
solder wick. Just lay it sideways on that Golden Gate of
solder and suck it all up.

If you make a mistake and have a less-capable iron, it's easy to end
up with the solder wick attached to so much thermal mass you can't
remove it.

I basically do your steps 1, 2 except I use a more moderate amount of
solder in 2. I end up with small groups of pins bridged but not the
whole lot.

And for the OP, liquid flux is not solder paste or solder. It's a
substance that cleans the pins and pads and makes it easier for solder
to "wet". Make sure you get some designed for electronics, not plumbing.
And if you can, choose water soluble.

--
Ben Jackson AD7GD
<ben@ben.com>
http://www.ben.com/

 

[Paul E. Schoen]

"The Phantom" <phantom@aol.com> wrote in message
news:8t1u74d5s811flfu5bj3s4q36lnhp86d2l@4ax.com...

On Wed, 16 Jul 2008 18:27:11 -0400, "Paul E. Schoen" <pstech@smart.net
wrote:


I beg to differ with your analysis, if you are talking about an ordinary
rectifier and capacitor circuit. As an example, I simulated a FWB with a
12
VAC nominal output transformer with 1 ohm series resistance, and a load
of
12 ohms, and a capacitor of 100,000 uF, which should produce the highest
possible current peaks. The simulation shows peak currents of 3.7 amps.
With 1000 uF, the peaks are 3.2 amps. Now, during the charging period,
with
100,000 uF, the peaks start at 14.6 amps and then diminish to 4.3 amps at
0.5 seconds. In the first 200 mSec, the tranny is supplying 34 watts, but
then settles down to 15.8 watts when the capacitor is fully charged. At
that time, the load is essentially pure DC, and the resistor dissipates
13.9 watts. So only about 2 watts is left, and that is shared among the
rectifiers (305 mW each), and the tranny (about 0.8 watts).

I have a transformer with a nominal 24 volt secondary, rated at 8 amps.
It
has a measured series resistance (secondary plus reflected primary) of
about .125 ohms. I connected a bridge rectifier consisting of 4 80 amp
Schottky diodes, and a real 100,000 uF capacitor.

If you simulate this, a DC load which gives 8 amps RMS in the secondary
may
give a DC current of less than 4 amps. The ratio of secondary RMS
current
to DC load current will exceed 2 to 1 if the transformer is much larger
than this, with a series resistance less than this transformer has.

I posted a partial analysis over on ABSE in which I indicate that the
grid
waveform has a large effect on the RMS to DC current ratio in these
rectifier circuits.

I have not looked at the analysis, but I did find an error in my analysis
as stated above, although it does not change the essential fact that the
transformer will not be overloaded if you keep the DC current out to about
50% of the AC current rating.

My error was that I used the voltage and current out of the transformer as
a measure of the power it was delivering, and that is correct in a sense,
but the internal resistance sees an RMS current of about 1.8 amps, for a
power dissipation of 3.24 watts, and not 0.8. I found it easier to use an
external resistance for the simulation. This model would be for a 12 VAC
transformer rated at 2 amps (24 VA) with 2/12 = 16.7% regulation. Larger
transformers will generally have better regulation, partly because they do
not have as much surface area to volume, and cannot as easily get rid of
internal heat by convection.

Simulating your circuit with a 3.3 ohm load, I get Pin = 142W, Pout = 127W,
Iin = 8.14A, Iout=4.39A. The internal resistance of the tranny dissipates
8.7 watts, and the diodes 1.7 watts each. The peak current is 19.8 amps.
The Iin/Iout is 1.85. Using a transformer with less internal resistance, or
better regulation, will give a ratio over 2:1, but it will then be a
transformer with a much higher rating, or rated much more conservatively
than normal (as even this one seems to be). New ASCII file follows:

Paul

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

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SHEET 1 880 680
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WIRE 288 144 240 144
WIRE 384 144 352 144
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WIRE 416 192 416 144
WIRE 240 256 240 144
WIRE 288 256 240 256
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WIRE 416 304 416 256
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WIRE 640 336 640 304
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FLAG 512 144 V+
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SYMATTR Description Diode
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SYMATTR Value 100000ľ
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SYMATTR Type cap
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SYMBOL schottky 288 384 R270
WINDOW 0 32 32 VTop 0
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SYMATTR InstName D1
SYMATTR Value MBR745
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL schottky 352 496 M270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D4
SYMATTR Value MBR745
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL res 128 160 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R2
SYMATTR Value .13125
TEXT 184 528 Left 0 !.tran 3

 

[Phil Allison]

"The Phantom"
"Phil Allison"

I have a transformer with a nominal 24 volt secondary, rated at 8
amps.
It has a measured series resistance (secondary plus reflected primary)
of about .125 ohms.


** This is a fabricated falsehood - the numbers simply do not add up.

A 192VA rated transformer does not have 4% regulation - correctly
rated it has 8%.

This one does.


** Cos it is not correctly VA rated.

The manufacturer rated it,

** Nevertheless, it is not correctly rated.

** So you also have no idea what its VA rating is.

I am taking the manufacturer's word for it. Printed on the transformer is
the
designation 24 volts, 8 amps.

** Nevertheless, it is not correctly rated.

Your argument is entirely false.

However, the primary (it's a 60 Hz transformer) says 120 VAC and it
measures .961 ohms, cold.


** So it is a circa 360 VA transformer.

The *correctly* rated secondary load is not 8 amps - but more like 15.

That would depend on the insulation system the manufacturer used, and the
resultant allowable temperature rise, wouldn't it?

** The 8% figure is for the lowest temp grade insulation in common use.

Using higher temp grade will only increase the figure.




....... Phil

 

[Phil Allison]

"John Popelish"

If all transformers were manufactured to a single regulation and
temperature rise standard,

** The vast majority on offer do.

The oft quoted ratio of circa 1.6 applies to stock lines transformers.



...... Phil

 

[Phil Allison]

"The Phantom"

For example, for the 24 volt, 8 amp transformer I've mentioned in this
thread,

** An incorrectly rated example.

Plucked out of his arse.

The
recommendation that John Fields made, to assume Irms/Idc = 2, which he
says will
always be safe, may not be safe if you are using a transformer with good
regulation and if your grid waveform is a good sinusoid.

** A transformer with unusually good regulation ALSO has unusually LOW temp
rise.

Which wipes you asinine case out.

Piss off.


....... Phil