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 This should be easy ??? 
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Post This should be easy ???
Hello:

I need advise on what kind of capacitor I could use to store a 12v
signal in the following application. I have been a lifelong tinkerer
but know nothing about electronics.

I have a piece of audio video equipment (let's call it "A") which
outputs a 12 vdc signal to another piece of AV equipment (called "B").
When "B" senses the power signal it turns on.

During the normal startup sequence for "A", the 12 v signal is
interrupted for ten to fifteen seconds causing "B" to cycle on - off -
back on again. It would seem to me I ought to be able to insert a
capacitor (something from Radio Shack) into the 12vdc positive to
maintain the signal for 20 seconds or so. I doubt we're talking even
1/10th of an amp. I have no problem if there is a slight delay for
capacitor to build up charge initially or a delay later when "A"
disconnects the signal before "B" shuts down.

Any suggestions would be most welcome. Thank you.

Tim W., Wheaton, IL Timsw1@hotmail.com


29 Dec 2007, 17:01
Post Re: This should be easy ???
On Mon, 13 Aug 2007 11:29:05 -0700, TimW <timsw1@hotmail.com> wrote:

>Hello:
>
>I need advise on what kind of capacitor I could use to store a 12v
>signal in the following application. I have been a lifelong tinkerer
>but know nothing about electronics.
>
>I have a piece of audio video equipment (let's call it "A") which
>outputs a 12 vdc signal to another piece of AV equipment (called "B").
>When "B" senses the power signal it turns on.
>
>During the normal startup sequence for "A", the 12 v signal is
>interrupted for ten to fifteen seconds causing "B" to cycle on - off -
>back on again. It would seem to me I ought to be able to insert a
>capacitor (something from Radio Shack) into the 12vdc positive to
>maintain the signal for 20 seconds or so. I doubt we're talking even
>1/10th of an amp. I have no problem if there is a slight delay for
>capacitor to build up charge initially or a delay later when "A"
>disconnects the signal before "B" shuts down.

---
View in Courier:

The simplest way to accomplish your goal would be to do this:


.. +---[DIODE>]--+------+
.. |+ |+ |+
.. [A] [BFC] [B]
.. | | |
.. +-------------+------+


The problem that arises is that for the BFC to supply 100 mA to "B"
during "A"'s 15 second shutdown and to keep the input voltage droop
at "B" from falling more than 1V during that 15 seconds, the BFC
needs to be:


Idt 0.1A * 15s
C = ---- = ------------ = 1.5F
dv 1V

which, depending on how long the first "ON" of the ON-OFF-ON
sequence lasts and the amount of current "A" can supply during that
time might make it impossible to fill the BFC up with enough charge
to last for 15 seconds.

BUT...

It may be that if the ON_OFF_ON sequence is signal only (not power)
the 100mA is way overkill. Why don't you measure it and post back
with what you find and that way we may be able to come up with
something more sensible.

Some spec's on your AV equipment (or links to data sheets) would
also be helpful.


--
JF


29 Dec 2007, 17:01
Post Re: This should be easy ???
TimW wrote:

> It would seem to me I ought to be able to insert a
> capacitor (something from Radio Shack) into the 12vdc positive to
> maintain the signal for 20 seconds or so.

And you'll probably need a diode to stop any 'reverse flow' too.

> I doubt we're talking even 1/10th of an amp.

Even 100mA results in a fairly unrealistic value of storage cap. You need to
measure the current drawn.

Graham


29 Dec 2007, 17:01
Post Re: This should be easy ???
On Aug 13, 3:01 pm, Eeyore <rabbitsfriendsandrelati...@hotmail.com>
wrote:
> TimW wrote:
> > It would seem to me I ought to be able to insert a
> > capacitor (something from Radio Shack) into the 12vdc positive to
> > maintain the signal for 20 seconds or so.
>
> And you'll probably need a diode to stop any 'reverse flow' too.
>
> > I doubt we're talking even 1/10th of an amp.
>
> Even 100mA results in a fairly unrealistic value of storage cap. You need to
> measure the current drawn.
>
> Graham

Mikkel/JF/Graham

Thank you all for replying to my post. I will measure the current draw
and get back to you. I need to retrieve my meter and it's not at my
house so it may take me a few days to get back to you.

I know the signal is definitly not power. Per the AV equipment specs
(and there's not much to them) unit "B" requires a 5vdc input. Unit
"A" is quoted as delivering a 12vdc trigger signal. So even if the 12v
were to drop off during the 15 seconds below a nominal 12v the signal
should still be seen by "B" as long as it remains above 5v. Both units
are of course 120vac plug-in for actual power.

Hope this info helps and like I said I will try to get actual current
draw measurement. The wiring going into unit "B" is very small guage,
perhaps #28 or even smaller. Thanks again for your collective
assistance.

Tim


29 Dec 2007, 17:02
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