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'[EE]: Prolonging life-time of alkaline cell'
2002\01\28@184744 by 859-1?Q?Alexandre_Guimar=E3es?=

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Hi, Jinx

> Some info here that I think fits the bill
>
> http://www.rampower.co.nz/technical.html
>
> It's not envisioned or desirable to have the cell flat. Even when
> called on to perform, it need supply only 85pA
>
> A super cap would have been a possibility, but the cap leakage
> and the necessary associated regulator circuitry consumption
> rules it out for long-term (a day or two ?) use

   Be VERY carefull... I tried rechargeable alkalines like that one and in
about 4 months inside the equipament it started to put that "blue thing"
out. They leak exactly like the non rechageable alkaline cells ! They are
great for toys but not for equipament that needs backup. The super cap might
be a better idea. It has been some years since I tried them, they may have
changed, but check it out before. The "blue stuff" corrodes everything
inside the equipament.

Best regards,
Alexandre Guimaraes

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2002\01\28@211418 by Jinx

face picon face
I'm nosing around for a definitive answer to the mechanism
of cell leakage. So far no luck with the local Energiser and
Duracell reps, but they're e-mailing o'seas for me. I'll pass
it on if/when I get it

Took a look at http://www.howstuffworks.com  for any info, and came
across these gems of examples of the consumer society. Part
of some instructions on how to make your own battery -

"Another simple experiment you can try involves a baby food
jar (if you don't have a baby around the house, go buy a few jars
of food at the grocery and empty them out)......."

Not "look for an empty one" or "give the contents to a poor
baby"

The page on Power Paper said the developers were looking
for backers to make "a range of disposable appliances using
their flexible batteries"

Look forward more poison in a landfill near you

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2002\01\28@213804 by 859-1?Q?Alexandre_Guimar=E3es?=

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Hi, Jinx

> I'm nosing around for a definitive answer to the mechanism
> of cell leakage. So far no luck with the local Energiser and
> Duracell reps, but they're e-mailing o'seas for me. I'll pass
> it on if/when I get it

   The RAM cells are a canadian technology, you can look at more details at
http://www.bti.ca/ram.htm .

   They seem to be a nice option for many situations but I would never use
them inside an expensive equipament, at least until they solve the leakage
problem. I was able to talk to a tech guy over the phone that really knew
how the cells work and how they have to be recharged. You have to do it with
individual control for each cell if they are on a pack. They have some
simple ways of doing it but it is still a problem for some applications. You
may want to try a call directly to Canada. The leakage problem is something
they swore to me that would not happen ! It has happened :-( I sent them the
batteries to show it. Those were the only 2 problems I saw on them, charging
in packs and leaking if not used and left in charge. It is a nice and cheap
technology.

Best regards,
Alexandre Guimaraes

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2002\01\29@055137 by Russell McMahon

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I am using rechargeable alkalines (RA's) in a consumer product that is in
moderate volume production so far.
I don't like them but they were expressly required by the client.
You can make them leak enthusiastically if you abuse them.
At very low trickle rates such as Jinx originally proposed I would suspect,
but only suspect, that they would potentially behave much better.

It is important to note that not all RA's are created equal. My client tried
a number of brands and reported inferior lifetime performance from some
brands. While the patents are held by a Canadian firm they are made by a
number of manufacturers including Rayovac.
The brand my customer has currently chosen are Alcava made in Korea.

A good site with a reasonable amount of simple detail including performance
graphs under various operating conditions.

       http://www.alcava.ch/Allemand/Data_d.html


While these cells have a high open circuit voltage when new they droop very
rapidly in use and have very poor recharge lifetimes if deep cycled or
rapidly discharged. Rayovac publish an excellent document covering all
aspects of their product "The High Performance Alkaline Battery in a
Rechargeable Alkaline System" . 430 kB PDF - well worth downloading by
anyone ever liable to use them in any capacity.

While the literature usually proposes a charge/pause & test / charge
regimen, a simple charger that limits peak charge voltage and decreases
current as the terminal voltage rises will meet many requirements. This does
not violate the spec sheet charging requirements - despite the use of pulse
type charging in most applications. A simple way of achieving this is a
series resistor from a higher voltage supply and a zener diode clamp. A
slightly more complex scheme with a zener and 2 resistors provide a dual
slope charging curve and max terminal voltage control. This has a faster
charging time than simple 1 resistor chargers but less complexity than pulse
chargers (work out how to do it yourself :-) ).

I am not of the opinion that monitoring of individual cells is required for
typical applications. This may be desirable in high performance apps.

Rayovac target terminal voltage is 1.65v/cell.

I suspect that Jinx's idea of using one such cell with a very light trickle
current would be an acceptable backup solution.

I would query the claimed 10 year alkaline lifetime - this is certainly
longer than most domestic cells "use by" lives (about 5 years) although I
suppose they want these to have some useful life when actually used !.

Note that the ability to recharge STANDARD alkaline cells was widely
reported for some years before the rechargeable cells appeared and a recent
report suggests that a modern standard alkaline will in fact recharge every
bit as well (or as badly) as the custom built rechargeable alkalines.

I consider that they are not a nice battery for serious secondary cell use
unless the application heavily targets some of their advantages. This is the
case in my customers application where long shelf life behind charges and a
high terminal voltage per cell and lowish cost are very desirable.


regards,

                 Russell McMahon

_____________________________

{Original Message removed}

2002\01\29@060800 by Jinx

face picon face
> I would query the claimed 10 year alkaline lifetime

My typo - 10 years for Lithium, 5-6 for alkaline

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2002\01\29@061844 by Jinx

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> A good site with a reasonable amount of simple detail
> including performance graphs under various operating
> conditions

       http://www.alcava.ch/Allemand/Data_d.html

Thanks for passing the info on. Not familiar with RAM
batteries as yet, maybe I'll pick up a couple

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2002\01\29@064411 by Russell McMahon

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> > A good site with a reasonable amount of simple detail
> > including performance graphs under various operating
> > conditions
>
>         http://www.alcava.ch/Allemand/Data_d.html
>
> Thanks for passing the info on. Not familiar with RAM
> batteries as yet, maybe I'll pick up a couple


I can give you a few of a less preferred brand. OK for initial playing.


       Russell McMahon

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2002\01\29@161641 by Tim McDonough

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> I've been following this thread, which I find very interesting,
especially
> since I am working on a project that is battery powered. I have done
web
> searches without specific results, and I have yet to come up with a
good
> power management/saving scheme to apply. Are there any good sources
that
> will help me do my homework on this? I don't mind doing the work, I
just
> don't know where to find info on this subject....

The January 2002 issue of Embedded Systems Programming
(http://www.embedded.com) has an article by Mike Willey and Kris Stafford
called "Low-Power Design" on Page 25. It's a pretty good starting
point and includes some material on battery types in addition to
hardware and software concerns.

Tim

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2002\01\30@210606 by Kathy Quinlan

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----- Original Message -----
From: "Jinx" <spam_OUTjoecolquittTakeThisOuTspamCLEAR.NET.NZ>
> If anyone knows what current (if any) a 1N4148
>uses at 85nA through-current I'd like to know.


Ummm Jinx,

I think I am a little rusty on my EE theory ;o) but if I put a battery in
serries with 4 diodes then the current in to 4 diodes is the same as the
current out ;o) IE 85nA in = 85nA out :o)

the reverse current IIRC (not 100% as i slept through semi theory ;o)) is
the current the diode will leak ie if the voltage applied to the Cathode is
grater than the anode, this is the current you would expect to flow, I do
not know if the Li cell will handle a 25nA current (my guess is it should)

Regards,

Kat.
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2002\01\30@222241 by Jinx

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> > If anyone knows what current (if any) a 1N4148
> >uses at 85nA through-current I'd like to know.
>
> I think I am a little rusty on my EE theory ;o) but if I put a
> battery in serries with 4 diodes then the current in to 4
> diodes is the same as the current out ;o) IE 85nA in =
> 85nA out :o)

I wasn't sure if a diode is a passive or active component or
something in between. In normal circumstances that probably
wouldn't be an issue, but as the currents are very small I had
to think twice and check

> the reverse current IIRC (not 100% as i slept through semi
> theory ;o)) is the current the diode will leak ie if the voltage
> applied to the Cathode is grater than the anode, this is the
> current you would expect to flow

I knew that - I think it was the mere mention of currents close
to what I'd be using that got me spooked

> I do not know if the Li cell will handle a 25nA current (my guess
> is it should)

In what regard ? You mean supply ?

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2002\01\31@012040 by Jinx

face picon face
> I think I am a little rusty on my EE theory ;o) but if I put a battery
> in serries with 4 diodes then the current in to 4 diodes is the
> same as the current out ;o) IE 85nA in = 85nA out :o)

I know it sounded like a bunny question, but it wasn't as clear-
cut for me. For example, diodes and transistors heat up, so
there is some resistance (resistivity). Which had me assuming
that the p-n junction is not just simply a transparent lossless
migration of ions. Sound reasonable ?

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2002\01\31@022025 by David Duffy

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At 05:59 PM 31/01/2002 +1300, you wrote:
> > I think I am a little rusty on my EE theory ;o) but if I put a battery
> > in serries with 4 diodes then the current in to 4 diodes is the
> > same as the current out ;o) IE 85nA in = 85nA out :o)
>
>I know it sounded like a bunny question, but it wasn't as clear-
>cut for me. For example, diodes and transistors heat up, so
>there is some resistance (resistivity). Which had me assuming
>that the p-n junction is not just simply a transparent lossless
>migration of ions. Sound reasonable ?

If it's just in series you can't have a different current on the "in"
and "out" legs of it. A bit like the quote from the Electrical Officer
that said there was no current in the 240Vac Neutral line because
the appliance used it all up! Hmmmm...
Regards...

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2002\01\31@044244 by Jinx

face picon face
> If it's just in series you can't have a different current on the
> "in" and "out" legs of it.

Without trying to be difficult  (trust me, I don't have to try
very hard ;-)  )

If you have a diode passing a large current it will get hot.
Wouldn't this thermal energy be from the "in" current, so
the "out" will be lower than the "in" ? I understand some
junction electro-chemistry, but not that much. Surely a
diode is not a perfect conductor

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2002\01\31@044909 by Michael Rigby-Jones

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> -----Original Message-----
> From: Jinx [SMTP:@spam@joecolquittKILLspamspamCLEAR.NET.NZ]
> Sent: Thursday, January 31, 2002 9:42 AM
> To:   KILLspamPICLISTKILLspamspamMITVMA.MIT.EDU
> Subject:      Re: [EE]: Prolonging life-time of alkaline cell
>
> > If it's just in series you can't have a different current on the
> > "in" and "out" legs of it.
>
> Without trying to be difficult  (trust me, I don't have to try
> very hard ;-)  )
>
> If you have a diode passing a large current it will get hot.
> Wouldn't this thermal energy be from the "in" current, so
> the "out" will be lower than the "in" ? I understand some
> junction electro-chemistry, but not that much. Surely a
> diode is not a perfect conductor
>
A diode however is a two terminal device.  The current flowing into the
device has nowhere else to go but out the other side.  It is the voltage
drop accross the junction that causes the heat to be generated (P=IV).

Mike

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2002\01\31@050223 by Jinx

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> It is the voltage drop accross the junction that causes
> the heat to be generated (P=IV)

Got it now. VA (in) = VA (out) + thermal energy

So, the answer to the original question is - A diode in
series with a battery consumes no current of its own

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2002\01\31@053744 by Russell McMahon

picon face
> > It is the voltage drop across the junction that causes
> > the heat to be generated (P=IV)
>
> Got it now. VA (in) = VA (out) + thermal energy
>
> So, the answer to the original question is - A diode in
> series with a battery consumes no current of its own

Two useful basic theorems of electrical circuitry are --

- Mean current flow into any circuit node is zero

- The sum of potentillas around a closed loop equals zero

The latter is, AFAIR after all these years, Kirchoffs theorum and the former
is, um err, brain fade, it's, oh bother ....
Cheats - looks in text book - no wonder I can't remember the name - it's the
same guy - it's Kirchoffs current theorum (the 2nd being his voltage theorum
of course). Very useful (indeed essential) for all sorts of circuit
analysis.



       RM

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2002\01\31@054954 by steve

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> - The sum of potentillas around a closed loop equals zero

Shouldn't that be a flock of potentillii ?
:-)

Steve.

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2002\01\31@055016 by Jinx

face picon face
> Two useful basic theorems of electrical circuitry are --
>
> - Mean current flow into any circuit node is zero
>
> - The sum of potentillas around a closed loop equals zero

Just having a "bad diode" day

Bye-bye January, I'll miss you. Not

The suggestion to use a higher V cell with dropping diodes
still isn't an option though.

The circuit takes 1.6V at 85nA, which I think is ideal for
alkalines. The 4 diodes in series with a lithium cell will use
just as much power (0.136uW) as the circuit itself

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2002\01\31@091408 by Alan B. Pearce

face picon face
>The specs say the reverse current at
>20C is 25nA, but I don't know enough about diode data to
>say whether those 4 diodes are consuming 100nA. If they
>are, then that has implications for the cell. If anyone knows
>what current (if any) a 1N4148 uses at 85nA through-current
>I'd like to know

The diodes don't "use" any current, as they are 2 terminal devices.
The current going in one terminal comes out the other.

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2002\01\31@124500 by David Koski

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But it does consume power.

On Thu, 31 Jan 2002 23:01:01 +1300
Jinx <RemoveMEjoecolquittEraseMEspamEraseMECLEAR.NET.NZ> wrote:

{Quote hidden}

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2002\01\31@155017 by Michael Vinson

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Russell McMahon wrote, in part:
>- Mean current flow into any circuit node is zero

i.e. TOTAL current flow

>- The sum of potentials around a closed loop equals zero

As a physics note, the first is equivalent to the physical
principle of conservation of charge (if it weren't true,
charge would have to be created or destroyed, or at least
build up (which would rule out steady-state behavior), at
the node).

The second is equivalent to conservation of energy, if you
just remember the basic definition of electric potential:
it is the potential energy per unit charge.

Michael V

Thank you for reading my little posting.


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2002\01\31@175408 by 859-1?Q?Alexandre_Guimar=E3es?=

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Hi, Jinx

> I know it sounded like a bunny question, but it wasn't as clear-
> cut for me. For example, diodes and transistors heat up, so
> there is some resistance (resistivity). Which had me assuming
> that the p-n junction is not just simply a transparent lossless
> migration of ions. Sound reasonable ?


   It is reasonable and you indeed have heat there to loose energy. It is
proportional to the diode drop voltage and current. You get close to 0.6 x
85na watts for each diode. It is not lossless but at the currents involved
it will probably be negletable.

Best regards,
Alexandre Guimaraes

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2002\01\31@183837 by Russell McMahon

picon face
> The suggestion to use a higher V cell with dropping diodes
> still isn't an option though.
>
> The circuit takes 1.6V at 85nA, which I think is ideal for
> alkalines. The 4 diodes in series with a lithium cell will use
> just as much power (0.136uW) as the circuit itself


The milliamp hour rating of 85 nA  is 85E-9A * 8765 Hour/year *1000  mA/A
mAH
= 0.75 mAH/year
10 years = 7.5 mAH !

This will give you a full 10 year shelf life off all but the very smallest
Lithium cell.
Same for an Alkaline cell but with a reduced shelf life (5 years?)
Working backwards from an incomplete battery table, even a tiny "230" cell =
6mm x 4mm probably has a capacity of 10 to 30 mAH depending on chemistry.

A similar charging current is not liable to do any damage (or quite
possibly, I suspect,  have any significant charging action) on practical
batteries.


       Russell McMahon

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2002\01\31@184336 by Jinx

face picon face
> 85na watts for each diode. It is not lossless but at the
> currents involved it will probably be negletable

However in this case, the losses from the dropping
diodes are the same as the circuit consumes, so 50%
of the battery's power is gone as heat

That would be true for any supply that's reduced to half
by diodes. Perhaps not important for mains-powered
supplies, but a definite consideration for low voltage, ultra-
low power battery circuits

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2002\01\31@184654 by Jinx

face picon face
> 10 years = 7.5 mAH !
>
> This will give you a full 10 year shelf life off all but the very
> smallest Lithium cell.
> Same for an Alkaline cell but with a reduced shelf life (5 years?)

Not bad when you do the sums is it

> A similar charging current is not liable to do any damage (or
> quite possibly, I suspect,  have any significant charging action)
> on practical batteries

Still waiting for an answer from the US about that

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2002\01\31@203522 by Kathy Quinlan

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----- Original Message -----
From: "Jinx" <EraseMEjoecolquittspamEraseMECLEAR.NET.NZ>
> I know it sounded like a bunny question, but it wasn't as clear-
> cut for me. For example, diodes and transistors heat up, so
> there is some resistance (resistivity). Which had me assuming
> that the p-n junction is not just simply a transparent lossless
> migration of ions. Sound reasonable ?
>

Sure does, if I had not done an experiment on it in Uni, I would have
thought otherwise, I know that the voltage drop across the diode changes
with temperature, but without looking at the circuit, I could not say if the
change of voltage drop will cause a change in current :o)

Regards,

Kat.

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'[EE]: Prolonging life-time of alkaline cell'
2002\02\01@083844 by Peter L. Peres
picon face
> If you have a diode passing a large current it will get hot.
> Wouldn't this thermal energy be from the "in" current, so
> the "out" will be lower than the "in" ? I understand some
> junction electro-chemistry, but not that much. Surely a
> diode is not a perfect conductor

The heat does not involve any electrons, only photons, thus any electrons
entering on one leg also exit on the other, therefore the same current
flows in both legs. The heat is caused by 'collisions' in the junction
material (and some of it probably by IR quanta emission from electrons
falling between stable states). The energy loss comes from the potential
difference across the device times the current. If you want to picture it
graphically you throw balls into a long horizontal pipe. All the balls
come out the other end but they are somewhat slowed down. The electrical
equivalent of slowing down is potential loss (voltage). The balls are
equivalent to unit charges (electrons, current = charge * time). The pipe
heats up if you throw enough balls into it.

Peter

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2002\02\01@101059 by Roman Black

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Michael Vinson wrote:
>
> Russell McMahon wrote, in part:
> >- Mean current flow into any circuit node is zero
>
> i.e. TOTAL current flow
>
> >- The sum of potentials around a closed loop equals zero

And in regards to AC current? Ever touched
415v AC when wearing rubber soled shoes? One
connection only, no closed loop and "ouchy"
current.
;o)
-Roman

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2002\02\01@102335 by Roman Black

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Jinx wrote:
>
> > 85na watts for each diode. It is not lossless but at the
> > currents involved it will probably be negletable
>
> However in this case, the losses from the dropping
> diodes are the same as the circuit consumes, so 50%
> of the battery's power is gone as heat
>
> That would be true for any supply that's reduced to half
> by diodes. Perhaps not important for mains-powered
> supplies, but a definite consideration for low voltage, ultra-
> low power battery circuits


Maybe you should be thinking cost per mWh
more than being concerned with waste? Or even
simpler, if you can be sure of a few years use
before failure, it becomes an issue of total
manufacturing cost. I don't think all that
heat from the "wasted" nW is going to annoy
your user too much. ;o)
-Roman

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2002\02\01@153405 by Jinx

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> The heat does not involve any electrons

Thanks for the explanation Peter

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2002\02\01@154234 by Jinx

face picon face
> Maybe you should be thinking cost per mWh
> more than being concerned with waste? Or even

I think what this made me realise is that reducing the
battery voltage by using diodes can be more wasteful
than is desirable, and that it's better to find a more
appropriate battery. However, your point about about
whether the customer will be truly inconvenienced by
a "cost v performance" issue is noted

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2002\02\01@191237 by Aaron Lahman

flavicon
face
Roman Black wrote:
>
>Michael Vinson wrote:
>>
>> Russell McMahon wrote, in part:
>> >- Mean current flow into any circuit node is zero
>>
>> i.e. TOTAL current flow
>>
>> >- The sum of potentials around a closed loop equals zero
>
>
>And in regards to AC current? Ever touched
>415v AC when wearing rubber soled shoes? One
>connection only, no closed loop and "ouchy"
>current.
>;o)
>-Roman

Still the sum = 0.  At the peak, there is a drop of 587 volts (for 415v rms)
across your finger|terminal|ground, and almost the same across your
grount|boot|foot capacitor (as it is still being 'charged') which leads to a
voltage drop in your body.  Current flows and you scream :)


diagram:       \|/     <---- hair standing up
              ( )
       587v   ,-,
---------. +-- | \   <----- voltage drop across body.  owwww
 device  |     | /
         |    / \   577v <-- boots not fully charged
_____________/___\_____
  0v   ground

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2002\02\02@054151 by Roman Black

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face
Aaron Lahman wrote:
{Quote hidden}

So we are in agreeance. :o)
I wasn't talking about the sum of potentials but
the previous discussion of current in vs current
out in a closed loop (as Jinx had suggested)...
-Roman

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2002\02\02@182914 by Aaron Lahman

flavicon
face
Roman Black wrote:
>
>Aaron Lahman wrote:
>> diagram:       \|/     <---- hair standing up
>>                ( )
>>         587v   ,-,
>>  ---------. +-- | \   <----- voltage drop across body.  owwww
>>   device  |     | /
>>           |    / \   577v <-- boots not fully charged
>>  _____________/___\_____
>>    0v   ground
>
>
>So we are in agreeance. :o)
>I wasn't talking about the sum of potentials but
>the previous discussion of current in vs current
>out in a closed loop (as Jinx had suggested)...
>-Roman
>

I just liked the chance to be artistic :)

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2002\02\02@183751 by Roman Black

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Aaron Lahman wrote:
{Quote hidden}

Ha ha! Yeah, and it does look a little like me.
;o)
-Roman

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2002\02\02@190937 by Jinx

face picon face
> >> diagram:       \|/     <---- hair standing up
> >>                ( )
> >>         587v   ,-,
> >>  ---------. +-- | \   <----- voltage drop across body.  owwww
> >>   device  |     | /
> >>           |    / \   577v <-- boots not fully charged
> >>  _____________/___\_____
> >>    0v   ground
> >
>
> I just liked the chance to be artistic :)

Looks like Pollock and Picasso had a fight over the brush ;-)

Only got two eyes though !!

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