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You use Outlook Express and top-post. It doesn't take an Occam...Actually, occams razor states that the easiest answer to this is that I'm a
idiot.
I was kind of suggesting that, in a dumb-post way.Zardoz wrote:
On Sun, 18 Jan 2004 19:12:34 GMT, "www.free-celebs.com"
hxqbabhi@hvobaahh.com> wrote:
Christina Aguilera naked
Why go through all of the trouble to post a "mime" format? Why the
hell not just posta jpg??? Could it be there is something wrong with
your post?
Its a virus. Where have you been for the last year or two?
You would be amazed at the number of otherwise intelligent people whoI was kind of suggesting that, in a dumb-post way.
Oh, and I have only been online since June 03, so for the last "year
or two" I've been out of it, but still smart enough to know not to
open any file that says it's a picture & not be .jpg or .gif.
No entry for "Crest Factor"?Thousands of Electronic Definitions - Must see
Literally thousands terms and definitions - must see:
http://www.electronicdefinitions.com
Sorry sir!
I'm a student now.I have a studying about Spam Mail issuse
case.Therefore I need more and more spam mail for this study.
Don't crosspost a useless post. And if you bother to research what theseI hate those pre-formatted, one size fits all, just add water, forum
formats. They are too boxy and those smilie faces and user logos make
me want to puke. This forum is the real deal. I can't believe how many
real experts hang here, always jumping in to help anyone who asks. The
3 to 9 hour posting lag bites but there is no equal. I do however miss
the pre-takeover Deja days!
There are several things in your circuit that need revision.As a beginner whose highest level of electronics is at GCSE, I'd
appreciate
some help on this circuit please.
It would help if I told you what I want the circuit to do:
Between the +12v rail and base of Q1 is a set of clips connected to paper.
I
want to be able to trigger the relay by drawing a pencil line between the
clips. The graphite would conduct a tiny current, and I want to amplify
that
to trigger the relay. SW1 and VR1 simulate the presence or absence of the
pencil line.
The resistance of the pencil line is high - anything from 4M to 12M and
above - so the current coming through is tiny.
So, I tested this circuit using New Wave Concept's "Livewire" software,
and
closing SW1 would in turn trigger the relay.
However, I made the circuit up on stripboard, and as soon as I connect the
power to the circuit, the relay closes immediately. Q2 becomes very hot.
I just measured a thick 2cm graphite line to be about 100k ohms. Maybe itDimitrij,
Thanks so much for your very kind help - the lesson I've learnt is that
electronics from a text book is far too ideal! I've revised the circuit
with
your suggestions (see the attachment).
THe circuit appears to work well at VR1 of up to 12M, but my multimeter
shows that the resistance of a thick line of graphite from a pencil drawn
on
paper is actually up to 15-18M. Am I asking too much to have a simple
circuit that can cope with such a variance in resistance?
Are the values for C1 and C2 enough to attenuate any AC?
And you will need a resistor in the B(NPN) circuitI just measured a thick 2cm graphite line to be about 100k ohms. Maybe it
depends on how thick the line is?
With the circuit you have (the original one, not the 'improved' one) the
current through the relay is max about 10,000 times the current into the
base. If you have 18M across 12-1.4V=10.6V, then the current will be just
shy of a microamp. Thus, the current through the relay will be 10,000
times
that, or 10mA, which might not be enough to close the relay.
Instead of using a darlington connected pair, you could use the following
circuit:
(view with courier font)
12V
-+-------------------+
| | |
| .-. |
| | | |
| | |1MEG |
| '-' |
o | |
+-------G||-+S
o | ||-> P-MOSFET
| | ||-+D
| |c |
'----b| NPN |
|e Relay Coil Here
| |
| |
GND------+-----------+
When there is no connection between the 'o's, the transistor is off, and
so
the P-MOSFET base is at 12V, turning it off.
When there is a connection through a resistance of 100MEG or less, the NPN
will turn on fully, causing the gate of the P-MOSFET to go near ground.
This
turns it on, thus closing the relay.
Any P-MOSFET where Vgs(th) is < 8V should work. You don't need to parallel
a
diode, because the mosfet has a substrate diode which should prevent any
damage when the coil is turned off. If the relay isn't a 12V relay (say,
its
a 5V relay), you should add in a resistor between the P-MOSFET and the
coil.
Oh, and you may need a 0.2uF cap _before_ your 7812, between in and gnd.
Are they used on the Starship Enterprise ?Oh not just resistors, NEXT generation resitors!
I believe 20,000 ohms/volt is (or was) considered a "pretty goodHello all,
As I am a newbie in the world of theoretical and practical
electronics I have come across quite a few ares that I have questions
in. One subject that I am not clear on is the level needed for
accurate measurements via an analog voltmeter and/or a multimeter.
I understand that, as per the mathematics, the higher the impedence
value the more accurate the measurement from the meter. However I was
wondering what would be the necessary level of ohms/volt that an
analog voltmeter and/or a multimeter would need to operate at in order
for the measurement(s) being displayed to be considered accurate for
the testing of compuers/arcade PCB's, as well as general trouble
shooting of other common electrical devices (phone, PDA, etc.). Is
20,000 ohms/volt generally satisfactory, or does one really need a
50,000 ohms/volt meter?
Chances are that the meter will not even indicate a spikes unless it lastsHello all,
As I am a newbie in the world of theoretical and practical
electronics I have come across quite a few ares that I have questions
in. One subject that I am not clear on is the level needed for
accurate measurements via an analog voltmeter and/or a multimeter.
I understand that, as per the mathematics, the higher the impedence
value the more accurate the measurement from the meter. However I was
wondering what would be the necessary level of ohms/volt that an
analog voltmeter and/or a multimeter would need to operate at in order
for the measurement(s) being displayed to be considered accurate for
the testing of compuers/arcade PCB's, as well as general trouble
shooting of other common electrical devices (phone, PDA, etc.). Is
20,000 ohms/volt generally satisfactory, or does one really need a
50,000 ohms/volt meter?
As a note, I have a digital multimeter that is rated at 4 megaohms,
however, I am interested in the analog multimeter so that I can see
any spikes that may be produced by the electrical device.
----------
Thank you all for your time and advice!
Respectfully,
Sam
Spikes are too fast to move the needle.I am interested in the analog multimeter
so that I can see any spikes that may be produced by the electrical device.
Sam (jacobson98)
A 20,000 ohms/volt meter, when set to a 5 volt rangewhat would be the necessary level of ohms/volt
that an analog voltmeter...would need to operate at...
for the testing of compuers/arcade PCB's ?
"Mr Anon" <anon@anon.co.uk> wrote in message
news:IXv_b.9718$h44.1089599@stones.force9.net...
Dimitrij,
Thanks so much for your very kind help - the lesson I've learnt is that
electronics from a text book is far too ideal! I've revised the circuit
with
your suggestions (see the attachment).
THe circuit appears to work well at VR1 of up to 12M, but my multimeter
shows that the resistance of a thick line of graphite from a pencil
drawn
on
paper is actually up to 15-18M. Am I asking too much to have a simple
circuit that can cope with such a variance in resistance?
Are the values for C1 and C2 enough to attenuate any AC?
I just measured a thick 2cm graphite line to be about 100k ohms. Maybe it
depends on how thick the line is?
With the circuit you have (the original one, not the 'improved' one) the
current through the relay is max about 10,000 times the current into the
base. If you have 18M across 12-1.4V=10.6V, then the current will be just
shy of a microamp. Thus, the current through the relay will be 10,000
times
that, or 10mA, which might not be enough to close the relay.
Instead of using a darlington connected pair, you could use the following
circuit:
(view with courier font)
12V
-+-------------------+
| | |
| .-. |
| | | |
| | |1MEG |
| '-' |
o | |
+-------G||-+S
o | ||-> P-MOSFET
| | ||-+D
| |c |
'----b| NPN |
|e Relay Coil Here
| |
| |
GND------+-----------+
When there is no connection between the 'o's, the transistor is off, and
so
the P-MOSFET base is at 12V, turning it off.
When there is a connection through a resistance of 100MEG or less, the NPN
will turn on fully, causing the gate of the P-MOSFET to go near ground.
This
turns it on, thus closing the relay.
Any P-MOSFET where Vgs(th) is < 8V should work. You don't need to parallel
a
diode, because the mosfet has a substrate diode which should prevent any
damage when the coil is turned off. If the relay isn't a 12V relay (say,
its
a 5V relay), you should add in a resistor between the P-MOSFET and the
coil.
Oh, and you may need a 0.2uF cap _before_ your 7812, between in and gnd.
Regards,
Bob Monsen
I am looking for a circuit that will trigger a pulse after waiting for
like
60 minutes.
You're not stupid, you're schizophrenic, get help.Stupid fuck? Well pindick my IQ is over 130 what is yours? I obviously know
[]
Death to psychotronic weaponry. Religion is fraud. James M. Vierling Jr.
----------------