Chip with simple program for Toy

Bret Cahill · Feb 18, 2013

How many times can you flip a 30 Amp CB on and off before it wears
out?
Say you as gentle as possible with the switch.

Bret Cahill

There is no "gentle" for a normal circuit breaker: the toggle action is
so strong that the contacts are going to thump hard no matter what.

What Mr. Terrell said about switch-rated breakers: there was a breaker
box at my dad's shop that had the circuit breakers turned on and off
every working day for the ten years that I worked there, and we never
had a problem (and I would know -- it would have been my job to swap
the breaker out).

How would you know w/o a test, i.e., a short with a known resistance?
Maybe nothing ever went wrong in your shop and the CBs still worked
great as manual switches but had, over the years, been altered as far as
the amperage that shut them off?

A shop would be the last place to offer as evidence as the "do
everything within specs" mentality prevails. Much more compelling would
be the dtisy residential situation test.

Thank you both for the insult to my father's abilities to wisely select
appropriate switches, and for once again discouraging me to answer your
questions.

No insult was intended nor should be taken.

Some people do everything on spec and some people occasionally do
things off spec.

And some will try to turn off spec into a religion. The more gnarly
the better. President Obama is not one of those people.

Use switch-rated circuit breakers.

A retired electrician just told me there was no way to damage CBs by
flipping them like switches.

Bret Cahill

Michael A. Terrell · Feb 18, 2013

Tim Wescott wrote:

Stop being a troll.

He can't. YOU can killfile him.

Mr Ron · Feb 18, 2013

In article
<437210fd-6d88-47ba-a45b-01a454c59ca6@m4g2000pbc.googlegroups.com>,
Bret Cahill <Bret_E_Cahill@yahoo.com> wrote:

It seems like it was "over fused" or "over breakered" but how do you
know it was faulty if there was no test to determine if it was faulty?

Apparently you had _some_ test to determine if it was faulty.
They have a rated time/load curve. As the load increases the trip time is

reduced which is why you can load them to the max current rating. The cable
resistance is measured to to give an Earth Fault Current figure that would
trip the CB within 0.4 seconds. If the Live/Earth resistance is too high
for the breaker it won't trip fast enough.

Ron

--

MrRonMan@googlemail.com

George Herold · Feb 18, 2013

On Feb 17, 3:55 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Sun, 17 Feb 2013 14:33:28 -0500, Jamie

jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
Fred Abse wrote:

On Sat, 16 Feb 2013 22:05:03 -0800, Daniel Pitts wrote:

I went with 1MHz = 1/(2?*2mH*C) which results in C=50.6pF

It's f=1/(2*pi*sqrt(L*C))

That's not the easiest way to remember it:

Just remember that 2*pi*f*L=1/(2*pi*f*C) at resonance....(1)

From which you can derive everything you need for whatever circumstance.

There's a further trick. 2*pi*f crops up so many times, that we allocate a
symbol to it, actually the lower case Greek omega, which I will type as a
lower case "w", here. That means you only have to do a single pi
approximation for the whole calculation, which improves precision.
Ok, but since "w" omega is kind of not a constant, because you still
need to define it at the start, may I add the part there to even shorten it

w= ?*f

Most likely the tau symbol didn't come out but some where in my math
history it once represented 2 * PI.

for some reason I can't find the lower case Omega (? ) in my chart.

Jamie

EEs use tau to mean "time constant." We represent 2*pi as 2*pi.

I'm always just taking 2*pi = 10, (for approximations)
It's the metric system for frequencies... 1 cycle = 10 radians :^)

George H.
(I then remember there's a factor of ~1.6 floating around)

--

John Larkin Highland Technology Incwww..highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators- Hide quoted text -

- Show quoted text -

George Herold · Feb 18, 2013

On Feb 17, 6:24 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Sun, 17 Feb 2013 15:16:48 -0800, John Larkin

jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 17 Feb 2013 14:11:54 -0800, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:

On 2/17/13 2:17 AM, John Larkin wrote:
On Sat, 16 Feb 2013 22:05:03 -0800, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:

So, I have a circuit:

|-------------
___ |
D1 \,/ S
===== S
| S
VR ----+ S 2 mH inductor
| S
===== S
D1 /^\ S
--- S
| |
|----------+-+
|
-----
---
-

D1 and D2 are Varactors. I'm supposed to find the value of D1 and D2
for a resonance of 1MHz.

I went with 1MHz = 1/(2?*2mH*C) which results in C=50.6pF
Oops I wrote the formula wrong in here.

I get 12.7 pF.
Running through it the second time, I get the same answer you do. Not
sure where I messed up, but such is arithmatic.

So, I'm not entirely sure if that is the entire capacitance of the
circuit, or the capacitance of of each of D1 and D2.

The varicaps are in series, so each one needs to be about 25 pF *if* VR is
applied through a big resistor. If VR is literally a voltage source, the lower
diode is shorted for AC signals and only the upper one participates in the
resonance. So it's a trick question or a poorly stated problem.
VR provides the reverse bias on the diodes which "sets" the capacitance,
no? Or am I not understanding?

Yes; futzing VR will tune the resonant frequency. But it matters whether VR is a
weak drive (like, through a 1M resistor) or a stiff voltage source.

http://en.wikipedia.org/wiki/Varicap

There's a mistake in the wiki article. Does anybody see it?

Only one?
I didn't read the whole thing, and I don't know much about
varactors.
But I'd rather it read 'at DC only a very small reverse bias current
flows' rather than 'no current flows'.

What didn't you like?

George H

And they left out parametric amplifiers and mixers.

--

John Larkin Highland Technology Incwww..highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators- Hide quoted text -

- Show quoted text -

default · Feb 18, 2013

On Sun, 17 Feb 2013 10:51:18 -0500, Michael Black <et472@ncf.ca>
wrote:

It's only good if there's enough capacitance, and only good if you
actually want to bypass there.

Yup.

Those old radios with the "spark plates" (probably named by some auto
mechanic with a screw driver in the wrong place) where just a 1"
square of copper with the power lead soldered to it and a 1-1/4" piece
of mica for a dielectric. Presumably that keeps the ~100 khz ignition
ringing out of the AM band radio.

Tim Wescott · Feb 18, 2013

On Mon, 18 Feb 2013 02:05:43 -0500, Michael A. Terrell wrote:

Tim Wescott wrote:

Stop being a troll.

He can't. YOU can killfile him.

I let my killfile expire every once in a while to give people second
chances, particularly when they show some signs of competence.

He's earned his way back in.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

John Larkin · Feb 19, 2013

On Mon, 18 Feb 2013 08:00:18 -0800 (PST), George Herold
<gherold@teachspin.com> wrote:

On Feb 17, 6:24 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 17 Feb 2013 15:16:48 -0800, John Larkin

jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 17 Feb 2013 14:11:54 -0800, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:

On 2/17/13 2:17 AM, John Larkin wrote:
On Sat, 16 Feb 2013 22:05:03 -0800, Daniel Pitts
newsgroup.nos...@virtualinfinity.net> wrote:

So, I have a circuit:

|-------------
___ |
D1 \,/ S
===== S
| S
VR ----+ S 2 mH inductor
| S
===== S
D1 /^\ S
--- S
| |
|----------+-+
|
-----
---
-

D1 and D2 are Varactors. I'm supposed to find the value of D1 and D2
for a resonance of 1MHz.

I went with 1MHz = 1/(2?*2mH*C) which results in C=50.6pF
Oops I wrote the formula wrong in here.

I get 12.7 pF.
Running through it the second time, I get the same answer you do. Not
sure where I messed up, but such is arithmatic.

So, I'm not entirely sure if that is the entire capacitance of the
circuit, or the capacitance of of each of D1 and D2.

The varicaps are in series, so each one needs to be about 25 pF *if* VR is
applied through a big resistor. If VR is literally a voltage source, the lower
diode is shorted for AC signals and only the upper one participates in the
resonance. So it's a trick question or a poorly stated problem.
VR provides the reverse bias on the diodes which "sets" the capacitance,
no? Or am I not understanding?

Yes; futzing VR will tune the resonant frequency. But it matters whether VR is a
weak drive (like, through a 1M resistor) or a stiff voltage source.

http://en.wikipedia.org/wiki/Varicap

There's a mistake in the wiki article. Does anybody see it?

Only one?
I didn't read the whole thing, and I don't know much about
varactors.
But I'd rather it read 'at DC only a very small reverse bias current
flows' rather than 'no current flows'.

What didn't you like?

The claim of octave tunability based on a 2:1 capacitance range.

--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro acquisition and simulation

John Larkin · Feb 19, 2013

On Mon, 18 Feb 2013 07:51:27 -0800 (PST), George Herold
<gherold@teachspin.com> wrote:

On Feb 17, 3:55 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 17 Feb 2013 14:33:28 -0500, Jamie

jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
Fred Abse wrote:

On Sat, 16 Feb 2013 22:05:03 -0800, Daniel Pitts wrote:

I went with 1MHz = 1/(2?*2mH*C) which results in C=50.6pF

It's f=1/(2*pi*sqrt(L*C))

That's not the easiest way to remember it:

Just remember that 2*pi*f*L=1/(2*pi*f*C) at resonance....(1)

From which you can derive everything you need for whatever circumstance.

There's a further trick. 2*pi*f crops up so many times, that we allocate a
symbol to it, actually the lower case Greek omega, which I will type as a
lower case "w", here. That means you only have to do a single pi
approximation for the whole calculation, which improves precision.
Ok, but since "w" omega is kind of not a constant, because you still
need to define it at the start, may I add the part there to even shorten it

w= ?*f

Most likely the tau symbol didn't come out but some where in my math
history it once represented 2 * PI.

for some reason I can't find the lower case Omega (? ) in my chart.

Jamie

EEs use tau to mean "time constant." We represent 2*pi as 2*pi.

I'm always just taking 2*pi = 10, (for approximations)
It's the metric system for frequencies... 1 cycle = 10 radians :^)

Physicists are barbarians. Everybody knows that 2*pi = 6.

--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro acquisition and simulation

Phil Hobbs · Feb 20, 2013

On 2/19/2013 5:31 PM, John Larkin wrote:

On Mon, 18 Feb 2013 07:51:27 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Feb 17, 3:55 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 17 Feb 2013 14:33:28 -0500, Jamie

jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
Fred Abse wrote:

On Sat, 16 Feb 2013 22:05:03 -0800, Daniel Pitts wrote:

I went with 1MHz = 1/(2?*2mH*C) which results in C=50.6pF

It's f=1/(2*pi*sqrt(L*C))

That's not the easiest way to remember it:

Just remember that 2*pi*f*L=1/(2*pi*f*C) at resonance....(1)

From which you can derive everything you need for whatever circumstance.

There's a further trick. 2*pi*f crops up so many times, that we allocate a
symbol to it, actually the lower case Greek omega, which I will type as a
lower case "w", here. That means you only have to do a single pi
approximation for the whole calculation, which improves precision.
Ok, but since "w" omega is kind of not a constant, because you still
need to define it at the start, may I add the part there to even shorten it

w= ?*f

Most likely the tau symbol didn't come out but some where in my math
history it once represented 2 * PI.

for some reason I can't find the lower case Omega (? ) in my chart.

Jamie

EEs use tau to mean "time constant." We represent 2*pi as 2*pi.

I'm always just taking 2*pi = 10, (for approximations)
It's the metric system for frequencies... 1 cycle = 10 radians :^)

Physicists are barbarians. Everybody knows that 2*pi = 6.

George is suffering from Fortran dyslexia--I think he meant pi**2 not 2*pi.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot net
http://electrooptical.net

George Herold · Feb 20, 2013

On Feb 19, 8:07 pm, Phil Hobbs
<pcdhSpamMeSensel...@electrooptical.net> wrote:

On 2/19/2013 5:31 PM, John Larkin wrote:

On Mon, 18 Feb 2013 07:51:27 -0800 (PST), George Herold
gher...@teachspin.com> wrote:

On Feb 17, 3:55 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 17 Feb 2013 14:33:28 -0500, Jamie

jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
Fred Abse wrote:

On Sat, 16 Feb 2013 22:05:03 -0800, Daniel Pitts wrote:

I went with 1MHz = 1/(2?*2mH*C) which results in C=50.6pF

It's f=1/(2*pi*sqrt(L*C))

That's not the easiest way to remember it:

Just remember that 2*pi*f*L=1/(2*pi*f*C) at resonance....(1)

From which you can derive everything you need for whatever circumstance.

There's a further trick. 2*pi*f crops up so many times, that we allocate a
symbol to it, actually the lower case Greek omega, which I will type as a
lower case "w", here. That means you only have to do a single pi
approximation for the whole calculation, which improves precision.
Ok, but since "w" omega is kind of not a constant, because you still
need to define it at the start, may I add the part there to even shorten it

w= ?*f

Most likely the tau symbol didn't come out but some where in my math
history it once represented 2 * PI.

for some reason I can't find the lower case Omega (? ) in my chart.

Jamie

EEs use tau to mean "time constant." We represent 2*pi as 2*pi.

I'm always just taking 2*pi = 10, (for approximations)
It's the metric system for frequencies... 1 cycle = 10 radians :^)

Physicists are barbarians. Everybody knows that 2*pi = 6.

George is suffering from Fortran dyslexia--I think he meant pi**2 not 2*pi.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot nethttp://electrooptical.net- Hide quoted text -

- Show quoted text -

You must have heard about the dyslexic agnostic with insomnia...

He stayed up all night wondering if there really is a dog.
George H.

Feb 20, 2013

Same thing in pesotum illinois...or it moves where I do. Darla Brown11 chestnut pesotum illinois 61863. 2174186680 when it works

Daniel Pitts · Feb 20, 2013

On 2/17/13 2:17 AM, John Larkin wrote:

On Sat, 16 Feb 2013 22:05:03 -0800, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

So, I have a circuit:

|-------------
___ |
D1 \,/ S
===== S
| S
VR ----+ S 2 mH inductor
| S
===== S
D1 /^\ S
--- S
| |
|----------+-+
|
-----
---
-

D1 and D2 are Varactors. I'm supposed to find the value of D1 and D2
for a resonance of 1MHz.

I went with 1MHz = 1/(2?*2mH*C) which results in C=50.6pF

I get 12.7 pF.
Yup, after I reworked it, that is exactly what I got.

So, I'm not entirely sure if that is the entire capacitance of the
circuit, or the capacitance of of each of D1 and D2.

The varicaps are in series, so each one needs to be about 25 pF *if* VR is
applied through a big resistor. If VR is literally a voltage source, the lower
diode is shorted for AC signals and only the upper one participates in the
resonance. So it's a trick question or a poorly stated problem.
"About 25 pF" was the correct answer for this problem. I'm not sure how

the lower diode is shorted, since it is reverse biased.

And there are two D1s. Maybe it's a dual varactor.
My bad, D1 is the upper, I meant to label the bottom diode D2.

John Larkin · Feb 20, 2013

On Wed, 20 Feb 2013 10:10:39 -0800, Daniel Pitts
<newsgroup.nospam@virtualinfinity.net> wrote:

On 2/17/13 2:17 AM, John Larkin wrote:
On Sat, 16 Feb 2013 22:05:03 -0800, Daniel Pitts
newsgroup.nospam@virtualinfinity.net> wrote:

So, I have a circuit:

|-------------
___ |
D1 \,/ S
===== S
| S
VR ----+ S 2 mH inductor
| S
===== S
D1 /^\ S
--- S
| |
|----------+-+
|
-----
---
-

D1 and D2 are Varactors. I'm supposed to find the value of D1 and D2
for a resonance of 1MHz.

I went with 1MHz = 1/(2?*2mH*C) which results in C=50.6pF

I get 12.7 pF.
Yup, after I reworked it, that is exactly what I got.

So, I'm not entirely sure if that is the entire capacitance of the
circuit, or the capacitance of of each of D1 and D2.

The varicaps are in series, so each one needs to be about 25 pF *if* VR is
applied through a big resistor. If VR is literally a voltage source, the lower
diode is shorted for AC signals and only the upper one participates in the
resonance. So it's a trick question or a poorly stated problem.
"About 25 pF" was the correct answer for this problem. I'm not sure how
the lower diode is shorted, since it is reverse biased.

Both diodes are presumably reverse biased. But if VR is a voltage
source, the VR node is AC ground, so the lower diode is effectively
shorted, and the upper should be 12 pF. Give your instructor an
"incomplete."

--

John Larkin Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro acquisition and simulation

Feb 21, 2013

On Thu, 21 Feb 2013 06:41:32 -0800 (PST), George Herold
<gherold@teachspin.com> wrote:

On Feb 21, 2:06 am, DaveC <inva...@invalid.net> wrote:
This is one conductor in a cable from some iPod earphones:

http://www.tinyuploads.com/images/mizPu4.jpg

What is the best way to deal with the fibre strands and to tin the wire?

Thanks,
Dave

I did something like that... I just teased the wire away from the
fibers, cut the fibers, and the enamel on the wire burned off from the
heat of the soldering iron.

George H.
I don't even bother to try to tease the fibers away. I just get a good

blob of solder on the tip of the iron and push the wire into it. In my
experience everything burns away except the wire.
Eric

Jim Thompson · Feb 21, 2013

On Thu, 21 Feb 2013 09:05:25 -0800, etpm@whidbey.com wrote:

On Thu, 21 Feb 2013 06:41:32 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Feb 21, 2:06 am, DaveC <inva...@invalid.net> wrote:
This is one conductor in a cable from some iPod earphones:

http://www.tinyuploads.com/images/mizPu4.jpg

What is the best way to deal with the fibre strands and to tin the wire?

Thanks,
Dave

I did something like that... I just teased the wire away from the
fibers, cut the fibers, and the enamel on the wire burned off from the
heat of the soldering iron.

George H.
I don't even bother to try to tease the fibers away. I just get a good
blob of solder on the tip of the iron and push the wire into it. In my
experience everything burns away except the wire.
Eric

DaveC · Feb 21, 2013

Regular Formvar doesn't
dissolve in solder.
...Jim Thompson

For which you would recommend...?

John Fields · Feb 22, 2013

On Thu, 21 Feb 2013 10:16:08 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On Thu, 21 Feb 2013 09:05:25 -0800, etpm@whidbey.com wrote:

On Thu, 21 Feb 2013 06:41:32 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Feb 21, 2:06 am, DaveC <inva...@invalid.net> wrote:
This is one conductor in a cable from some iPod earphones:

http://www.tinyuploads.com/images/mizPu4.jpg

What is the best way to deal with the fibre strands and to tin the wire?

Thanks,
Dave

I did something like that... I just teased the wire away from the
fibers, cut the fibers, and the enamel on the wire burned off from the
heat of the soldering iron.

George H.
I don't even bother to try to tease the fibers away. I just get a good
blob of solder on the tip of the iron and push the wire into it. In my
experience everything burns away except the wire.
Eric

That works for "SolderEze" coated wire. Regular Formvar doesn't
dissolve in solder.

...Jim Thompson

---
None of it dissolves in solder, it just ablates when the temp gets
high enough.

--
JF

Jim Thompson · Feb 22, 2013

On Thu, 21 Feb 2013 19:45:42 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Thu, 21 Feb 2013 10:16:08 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On Thu, 21 Feb 2013 09:05:25 -0800, etpm@whidbey.com wrote:

On Thu, 21 Feb 2013 06:41:32 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Feb 21, 2:06 am, DaveC <inva...@invalid.net> wrote:
This is one conductor in a cable from some iPod earphones:

http://www.tinyuploads.com/images/mizPu4.jpg

What is the best way to deal with the fibre strands and to tin the wire?

Thanks,
Dave

I did something like that... I just teased the wire away from the
fibers, cut the fibers, and the enamel on the wire burned off from the
heat of the soldering iron.

George H.
I don't even bother to try to tease the fibers away. I just get a good
blob of solder on the tip of the iron and push the wire into it. In my
experience everything burns away except the wire.
Eric

That works for "SolderEze" coated wire. Regular Formvar doesn't
dissolve in solder.

...Jim Thompson

---
None of it dissolves in solder, it just ablates when the temp gets
high enough.

John Fields · Feb 22, 2013

On Thu, 21 Feb 2013 19:34:57 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On Thu, 21 Feb 2013 19:45:42 -0600, John Fields
jfields@austininstruments.com> wrote:

On Thu, 21 Feb 2013 10:16:08 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On Thu, 21 Feb 2013 09:05:25 -0800, etpm@whidbey.com wrote:

On Thu, 21 Feb 2013 06:41:32 -0800 (PST), George Herold
gherold@teachspin.com> wrote:

On Feb 21, 2:06 am, DaveC <inva...@invalid.net> wrote:
This is one conductor in a cable from some iPod earphones:

http://www.tinyuploads.com/images/mizPu4.jpg

What is the best way to deal with the fibre strands and to tin the wire?

Thanks,
Dave

I did something like that... I just teased the wire away from the
fibers, cut the fibers, and the enamel on the wire burned off from the
heat of the soldering iron.

George H.
I don't even bother to try to tease the fibers away. I just get a good
blob of solder on the tip of the iron and push the wire into it. In my
experience everything burns away except the wire.
Eric

That works for "SolderEze" coated wire. Regular Formvar doesn't
dissolve in solder.

...Jim Thompson

---
None of it dissolves in solder, it just ablates when the temp gets
high enough.

What does "ablating" look like ?>:-|

...Jim Thompson
---

http://en.wikipedia.org/wiki/Ablation

--
JF

Chip with simple program for Toy

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