chip swelling up and getting fried

[Peter Bennett]

On Mon, 16 Aug 2004 11:55:12 -0400, "Michael A. Covington"
<look@ai.uga.edu.for.address> wrote:

Voltage is the force that causes amps to flow. That's Ohm's Law
(high-school physics). You don't get to do any electronics without
mathematics.

What it means is that the number of amps actually flowing depends on the
resistance of the load. Try lighting a 10-amp headlight bulb with your
computer power supply -- if it can deliver 10 amps at 12 volts, it will do
it.

Now for the bad news. 12.0 volts is not enough to charge a lead-acid
battery fully. Fully charged, the battery itself would be at 12.6 volts.
Normal output of a float charger is 14.1 volts. Automotive battery
chargers drive a large current into the battery by putting out a much higher
voltage (something like 20 volts).

Perhaps a "charge a dead battery in 30 minutes" charger might deliver
20 volts, at great risk to the battery, but a more normal charge
routine will use a maximum of 14.4 volts, or so.

In any case, the 12 volts you get from a computer power supply is
definitely not adequate to charge a "12 volt" lead-acid battery.




--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[Jerry G.]

You'd be best off to get a proper charger for your 12 Volt battery. Chargers
are properly current controlled, and have the proper over-head voltage to
charge the battery. Chargers are properly designed for charging batteries.

Using a device to charge a battery that is not properly regulated and rated
for the application, and meeting proper standards for safety, can result in
a dangerous hazard.

--

Jerry G.
==========================


"ilaboo" <plener@earthlink.net> wrote in message
news:l%2Uc.27273$Jp6.3817@newsread3.news.atl.earthlink.net...
googling has provided some info but still confused maybe someone can help


i want to use a computer battery supply as a battery charger--lead acid
nicad, etc


it puts out 12v dc and 5 volts dc


advice talks about putting a load resister from 12v to ground to turn on
power supply ( power supply is a switching type0


confused at this point as there is 12 v available when i test it--i am
assuming at this point the power supply is on--a load across a 5 ohm
power resister in series gives 3 volts
using ohms law i get .600 amps--am i right so far?


power supply specs say i can get 10 amps at 212 v

what do i have to do to get this?
any help really is appreciated

i have no info such as schematic on this power supply

tia
pter

 

[Robert Baer]

Chad Waldman wrote:

I posted a message a week back but never received the answer I was
looking for.

I have an electromagnet hooked-up to a 9V battery. The magnet is
strongest when the battery is at full charge. I want to replace the
battery with an adapter to run the magnet from an outlet. What I am
wondering is, what is the max current the battery is outputting when it
is new? If you have an ammeter, can you please test the output of a new
battery.

Thank you,
Chad

I do not have the "magnet" that you have for testing, and it is not
possible for anyone else to do that testing without an accurate
description (manufactureer and model number for starters).
That *electro*magnet has a measureable resistance; one could calculate
the current drain at 9 volts.
Alternately, *you* could use a multimeter or a DVM to measure that
current.

Be advised, that most wall adaptors are unregulated, and so the output
voltage is *always* higher than the rating, even at full rated load.
That means, that your *electro*magnet will be stronger when you use a
9V adaptor.
An adaptor with even a 500mA rating would seem to be more than
adequate.

 

[Sam Goldwasser]

"Jon Davis" <jon@REMOVE.ME.jondavis.net> writes:

I was using my new emachines M6811 laptop, which has an A/C adapter with a
"90W" blue sticker on the end of it, with a 150W cigarette lighter DC/AC
inverter, Tripp Lite brand, in a 2002 Toyota Echo. The car has a new battery
and
the cig lighter had never before been used. The first day, with about
7 or 8 hours on the road, there were no problems. Eventually, however, one
day I plugged it in the inverter again and within about ten minutes the
inverter alarm started sounding. I kept resetting it and in another five
minutes the inverter's fuse blew, and the car's fuse blew, too, but it was
somehow delayed by a few hours I think for some reason because we didn't
notice the stereo wasn't working until hours later).

A few dumb questions:

- Is it possible that the laptop adapter is already going bad, drawing more
power than should be? (Every time I plug it into the wall while the laptop
is running I get a VERY visible and loud spark.)

This may not be anything. It it were really drawing over 150 W, something
would be getting very hot very quickly.

- If I bought an inverter that supported more wattage--say, 250 watts--would
that decrease the chances of the *car* fuse from blowing out, or does the
car fuse not "care about" the power handling of the inverter? (I don't know
much about electricity.)

The original inverter might just have been defective.

- Would a direct DC-to-DC step-up converter be better? Any recommendations
on a model of such an adapter for a laptop marked on the bottom with "18.5
watts"?

No, you're probably better off with the DC to AC inverter.

--- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ: http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites: http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html

Note: These links are hopefully temporary until we can sort out the excessive
traffic on Repairfaq.org.

Important: Anything sent to the email address in the message header is ignored.
To contact me, please use the feedback form on the S.E.R FAQ Web sites.

 

[andy]

On Tue, 17 Aug 2004 23:11:39 -0700, Chad Waldman wrote:

I posted a message a week back but never received the answer I was
looking for.

I have an electromagnet hooked-up to a 9V battery. The magnet is
strongest when the battery is at full charge. I want to replace the
battery with an adapter to run the magnet from an outlet. What I am
wondering is, what is the max current the battery is outputting when it
is new? If you have an ammeter, can you please test the output of a new
battery.

Thank you,
Chad

it doesn't just depend on the battery - it depends on the resistance of
the magnet coils as well. The easiest way is to buy or borrow a meter and
measure it.

--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with

 or [attachment] in the subject line.

 

[Michael A. Covington]

"Chad Waldman" <chad.waldman@mail.com> wrote in message
news:42ef8598.0408172039.11565db6@posting.google.com...

I posted a message a week back but never received the answer I was
looking for.

I have an electromagnet hooked-up to a 9V battery. The magnet is
strongest when the battery is at full charge. I want to replace the
battery with an adapter to run the magnet from an outlet. What I am
wondering is, what is the max current the battery is outputting when it
is new? If you have an ammeter, can you please test the output of a new
battery.

What you need is to use an ammeter to measure the current flowing through
your electromagnet.

 

[Jack Edin]

Sam Goldwasser wrote:

"Jon Davis" <jon@REMOVE.ME.jondavis.net> writes:

- Would a direct DC-to-DC step-up converter be better? Any recommendations
on a model of such an adapter for a laptop marked on the bottom with "18.5
watts"?


No, you're probably better off with the DC to AC inverter.

I don't agree at all!

Buy an adapter specifically for you laptop, that plugs into a cigarette
lighter.

I have Compaq Armada series laptop(s). One I have mounted in the car.
And YES it uses a Compaq brand DC to DC converter. It outputs 18.5 volts
for my laptop.

To invert up to 120V AC, then back down is NOT efficient!

My two cents...


Jack

 

[Jamie]

replace the LapTop's AC power supply.


Jon Davis wrote:

I was using my new emachines M6811 laptop, which has an A/C adapter with a
"90W" blue sticker on the end of it, with a 150W cigarette lighter DC/AC
inverter, Tripp Lite brand, in a 2002 Toyota Echo. The car has a new battery
and
the cig lighter had never before been used. The first day, with about
7 or 8 hours on the road, there were no problems. Eventually, however, one
day I plugged it in the inverter again and within about ten minutes the
inverter alarm started sounding. I kept resetting it and in another five
minutes the inverter's fuse blew, and the car's fuse blew, too, but it was
somehow delayed by a few hours I think for some reason because we didn't
notice the stereo wasn't working until hours later).

A few dumb questions:

- Is it possible that the laptop adapter is already going bad, drawing more
power than should be? (Every time I plug it into the wall while the laptop
is running I get a VERY visible and loud spark.)

- If I bought an inverter that supported more wattage--say, 250 watts--would
that decrease the chances of the *car* fuse from blowing out, or does the
car fuse not "care about" the power handling of the inverter? (I don't know
much about electricity.)

- Would a direct DC-to-DC step-up converter be better? Any recommendations
on a model of such an adapter for a laptop marked on the bottom with "18.5
watts"?

Thanks,
Jon

 

[Jan-Erik Söderholm]

Was it out-of-stock from the Elektor "back issues" service ?`
Jan-Erik.

Trevor Campbell Davis wrote:

In case anyone can help, I need a copy of the March 2004 issue of Elektor
Electronics magazine (Nr 330), to complete a break in a library sequence.

I will of course pay for the copy plus postage.

Please help if you can.

73

Trevor Campbell Davis
G3YMM

tcd@eurocom.demon.co.uk
_____________________

 

[Sam Goldwasser]

Jack Edin <nospam@logicunlimited.com> writes:

Sam Goldwasser wrote:
"Jon Davis" <jon@REMOVE.ME.jondavis.net> writes:

- Would a direct DC-to-DC step-up converter be better? Any recommendations
on a model of such an adapter for a laptop marked on the bottom with "18.5
watts"?
No, you're probably better off with the DC to AC inverter.

I don't agree at all!

Buy an adapter specifically for you laptop, that plugs into a
cigarette lighter.

That I agree with. What I don't agree with is just buying any old DC-DC
converter to power your laptop. That would probably void the warranty.

--- sam | Sci.Electronics.Repair FAQ Mirror: http://repairfaq.ece.drexel.edu/
Repair | Main Table of Contents: http://repairfaq.ece.drexel.edu/REPAIR/
+Lasers | Sam's Laser FAQ: http://repairfaq.ece.drexel.edu/sam/lasersam.htm
| Mirror Sites: http://repairfaq.ece.drexel.edu/REPAIR/F_mirror.html

Note: These links are hopefully temporary until we can sort out the excessive
traffic on Repairfaq.org.

Important: Anything sent to the email address in the message header is ignored.
To contact me, please use the feedback form on the S.E.R FAQ Web sites.

I have Compaq Armada series laptop(s). One I have mounted in the
car. And YES it uses a Compaq brand DC to DC converter. It outputs
18.5 volts for my laptop.

To invert up to 120V AC, then back down is NOT efficient!

My two cents...


Jack

 

[Jon Davis]

"Sam Goldwasser" <sam@saul.cis.upenn.edu> wrote in message
news:6w4qn0avbz.fsf@saul.cis.upenn.edu...

Jack Edin <nospam@logicunlimited.com> writes:

Sam Goldwasser wrote:
"Jon Davis" <jon@REMOVE.ME.jondavis.net> writes:

- Would a direct DC-to-DC step-up converter be better? Any
recommendations
on a model of such an adapter for a laptop marked on the bottom with
"18.5
watts"?
No, you're probably better off with the DC to AC inverter.

I don't agree at all!

Buy an adapter specifically for you laptop, that plugs into a
cigarette lighter.

That I agree with. What I don't agree with is just buying any old DC-DC
converter to power your laptop. That would probably void the warranty.

Well I don't see how plugging the AC adapter into a step-up AC/DC inverter
is any better. You're making TWO voltage transformations (stepping up to
120V, then stepping down again), plus your going from direct current to
alternating current and then back to direct current which is undoubtedly
going to be more likely to cause problems than just stepping up the voltage
of a direct current.

I am sure a DC-DC step-up adapter will be the best solution for me. It's
electrically IDENTICAL to the universal external laptop batteries you can
buy to place under a laptop and hook directly to the power input.

Jon

 

[John Larkin]

On 17 Aug 2004 23:11:39 -0700, chad.waldman@mail.com (Chad Waldman)
wrote:

I posted a message a week back but never received the answer I was
looking for.

I have an electromagnet hooked-up to a 9V battery. The magnet is
strongest when the battery is at full charge. I want to replace the
battery with an adapter to run the magnet from an outlet. What I am
wondering is, what is the max current the battery is outputting when it
is new? If you have an ammeter, can you please test the output of a new
battery.

Thank you,
Chad

RatShack 9v alkaline, I got 6 amps into a dead-short ammeter for the
first second or so, and then it dropped rapidly. After about a minute,
it was down to 1 amp and the battery was getting pretty hot, so I
quit. A rechargable would likely behave differently.

We'd have to know the magnet's resistance to estimate the actual
current. How long does the battery last, by your standards?

You should probably buy a cheap DVM and measure actual coil resistance
and the voltage that makes it work OK.

John

 

[Jonathan Kirwan]

On Wed, 18 Aug 2004 16:39:22 -0700, John Larkin
<jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

RatShack 9v alkaline, I got 6 amps into a dead-short ammeter for the
first second or so, and then it dropped rapidly.

I just checked the Energizer datasheet for it and they list a starting internal
resistance of just under 2 Ohms, so that 6A seems about right.

After about a minute,
it was down to 1 amp and the battery was getting pretty hot,

The curve suggests that this should happen at about 80-90% discharge.

so I quit.

And scratch one ratshack battery.

Jon

 

[Terran Melconian]

In article <l%2Uc.27273$Jp6.3817@newsread3.news.atl.earthlink.net>,
ilaboo <plener@earthlink.net> wrote:

i want to use a computer battery supply as a battery charger--lead acid
nicad, etc

You have several problems.

First, you must sort out the computer power supply. Depending on the
power supply, it may turn on with a switch (AT) or when you short a
certain wire to ground (ATX). Then, it may require a certain amount of
load on the 5v line in order to actually turn on, or it may require a
certain load on the 5v line in order to regulate properly, or it may not
require any load at all.

First figure out which type of power supply you have, then look up the
correct wiring colors on the web.

The computer power supply is regulated, but this doesn't help you much,
because neither 5V nor 12V are useful for charging batteries. To charge
NiCDs, you'll need to control the current flowing into the battery. In
the simplest case, you can do this with a resistor and a large voltage
drop. Let's suppose you have a 1.2V NiCD or NiMH (they charge very
similarly) battery. These are usually slow-charged at C/10, which means
that you take the capacity of the battery, divide by 10 in appropriate
units, and charge with this current. For example, if you had a 2
amp-hour battery, you'd charge it at 200 mA (and then set your alarm
clock for about 12 hours later and take it off the charger). Look up a
battery datasheet for more information about this.

Now, suppose you don't want to build any hard circuits. You take your
12v supply, subtract 1.2V for the battery, and have a 10.8v drop.
10.8/0.2 = 54, so you want a resistor close to 54 ohms, with higher
being safer than lower. You could do two batteries in series by redoing
the calculation, but I wouldn't go to four; at that point, the change in
voltage of the batteries as they charge (they'll go from about 1V to
1.5V as they charge) becomes significant and the resistor no longer
approximate a current source very well.

This resistor will be dissipating a good bit of power, so get a big one.
(0.2 A * 11 V = 2.2 watts)

The situation for lead-acid is rather worse. Lead acid batteries are
charged with a constant voltage somewhere in the range of 13.8 to 14.6,
depending on whether it's a float charge (left on constantly) or a
cyclic or top-up charge (can't be left on permanently without damaging
the battery). The only way you're going to get these voltages from your
computer power supply is to use the +12 and -5 (or +5 and -12) outputs,
and the negative outputs are usually limited to about 500 mA. If you
wanted to be cheap and approximate like we were with the "constant
current" charge for the NiCDs, you could get into the right ballpark
with four 1n400x diodes in series. However, you would then *still* have
the problem of making sure you don't load your computer power supply
beyond what it can handle and make it give up the magic smoke, and given
that even a smallish lead-acid battery is going to draw considerably
more than 500 mA to charge, I think you should forget about using the
computer power supply to charge lead acids unless you have a 6V one.

 

[Jonathan Kirwan]

On Wed, 18 Aug 2004 18:20:16 -0700, John Larkin
<jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

A 9v alkaline is good for maybe 1 a-h, so I probably didn't use it up
much by shorting it for a minute. Now that it's cooled off and rested
for a couple of hours, it's back up to 5.8 amps initial current. I
think there's some polarization that goes on at high current (tiny
bubbles or something) but it's temporary (some sort of catalytic
reaction depolarizes it?)

An accumulation of opposing charge on those bubbles, perhaps?

Jon

 

[Robert Baer]

John Larkin wrote:

On 17 Aug 2004 23:11:39 -0700, chad.waldman@mail.com (Chad Waldman)
wrote:

I posted a message a week back but never received the answer I was
looking for.

I have an electromagnet hooked-up to a 9V battery. The magnet is
strongest when the battery is at full charge. I want to replace the
battery with an adapter to run the magnet from an outlet. What I am
wondering is, what is the max current the battery is outputting when it
is new? If you have an ammeter, can you please test the output of a new
battery.

Thank you,
Chad

RatShack 9v alkaline, I got 6 amps into a dead-short ammeter for the
first second or so, and then it dropped rapidly. After about a minute,
it was down to 1 amp and the battery was getting pretty hot, so I
quit. A rechargable would likely behave differently.

We'd have to know the magnet's resistance to estimate the actual
current. How long does the battery last, by your standards?

You should probably buy a cheap DVM and measure actual coil resistance
and the voltage that makes it work OK.

John

The short circuit current of the battery is not relevant, period.
In fact, it is *stupid* to make such a useless measurement.

 

[Jerry G.]

Did you check the voltages at the tuner module to see if there is anything
supplying it? This includes checking if the AGC bias to the tuner is also
correct.

Did you check that in the TV mode, that the source switching circuits are
infact working correctly to pass the video signal from the decoding (output
of the IF detection) through its proper path?

If the above complies, then check to see if the IF module is getting the
proper supply. If so, then you will have to feed a known signal in to the
tuner section, and use a scope to see if there is any output from the tuner.
Then follow through the IF stages. Following the schematic, and careful
evaluation of the bias voltages, and signal paths would be the key to
finding the fault. There is no simple answer for this, unless it is just a
missing DC voltage supply to the IF or tuner sections.

--

Jerry G.
==========================


"Lee" <lee@fireserve.net> wrote in message
news:c438d194.0408182055.5c17ec50@posting.google.com...
I replaced a burned up D409 and vertical IC, got it to stay on and
picture is back fine now (menu works fine, picture is perfect).
However, nothing seems to be coming out of the tuner, or at least the
RF input section. I see the channel on the screen, but it finds no
channels. No snow, no noise, nothing. Any thoughts? Thx!

 

[Qwerty Keyboard]

"Robert Baer" <robertbaer@earthlink.net> wrote in message
news:4122F7B8.2EF329A9@earthlink.net...

The short circuit current of the battery is not relevant, period.
In fact, it is *stupid* to make such a useless measurement.

WTF? The only one who is *stoopid* here is you. Short circuit testing a
battery provides valuable information about how the battery will behave
under heavy loads and short circuit fault conditions. The original poster
wasn't very specific about what kind of electromagnet was in use, but smart
money bets it was a hobbyist's homebrew electromagnet with far too few turns
with far to large wire and the 9V battery impedance provides most of the
current limiting.

You are clearly so stoopid to make such a comment you evidently belong in
the lower two quartiles of this study.

http://www.apa.org/journals/psp/psp7761121.html

Even pond scum has more potential than you.

http://www.acfnewsource.org/environment/pond_scum.html

And your momma wears combat boots.

Bitch.




-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

 

[John Larkin]

On Thu, 19 Aug 2004 06:34:20 GMT, Robert Baer
<robertbaer@earthlink.net> wrote:

The short circuit current of the battery is not relevant, period.
In fact, it is *stupid* to make such a useless measurement.

How is learning something stupid? Oh, yeah, for some people, it is a
waste of time.

John

 

[Mjolinor]

"Andre" <testing_h@yahoo.com> wrote in message
news:2c2cf14c.0408191002.5138004e@posting.google.com...

Hi group.

With all the computers being scrapped, there must be literally
thousands of useless HDD's dumped every week in the UK alone.

So, things to do with them.

1) Maglev (take off heads, use one of the actuator magnets)
2) Hard drive speakers (see Afrotech's site for this one)
3) Novelty paperweight (take off lid, fill with casting resin)
4) Novelty keyring (old Microdrives work well for this)
5) Exhibit in computer museum
6) Drive failure mode analysis
7) Send back to manufacturer and get new drive under warranty
8) Strip down for useful connectors and SMD tantalums
9) Bulk sell the PCB's on Ebay for data recovery purposes
10) Sell them on Ebay for Ł0.99 each as "Faulty"
11) Fix them with Spinrite 6.0 and sell as refurbished drives
12) Stress relief (spin up to insane RPM and hit with hammer)
13) Novelty doorbell (put button in centre of platter)
14) Doorstop
15) Pack with Thermite and light it to see how much is left of drive
16) Maglev train (same principle as *1 but a row of them)
17) Miniature centrifuge (take off platters, add tube carrier unit)
18) Time capsule (write data, seal in durable case then bury in
concrete)
19) Laser scanner (glue mirrors to the spindle motor/actuator arm)
20) Tesla Turbine (nice flat platters- might work!)
21) Spin-coating machines
22) Gyroscopes for small satellites
23) Ultra-high RPM sanding machines
24) CD Destroyer (fix CD clamp to spindle motor and pin to head arm)
25) Strip spindle motors out, and use for small R/C helicopters
26) High efficiency motors for solar water pumps
27) Attach propellors and use as wind turbines

Please add to this

I made a "winder" for the kids beyblades from a laptop disc and two laptop
Ni-MH cells.