i.e. E and b
how does it work ? what is the reason for C=0.1uf
How? Well, simplified, the transistor requires a voltage of less than
about 3V to conduct and light the LED. The three resistor voltage
divider puts more than 3V across the 680k when the phone is on hook,
and the voltage drops below 3V when the phone is off hook, and the LED
lights up.
You could also look at it as a current summing point at the junction
of the transistor's base and the 680k. When the current from the
phone line is equal to or greater than the current thru the 680k, the
base sees no forward bias and the LED stays off. When the current is
less than the current thru the 680k, that current from the 680k
forward biases the transistor and the LED lights.
And yes, you have to take into account the .6V drop across the base to
emitter junction.
The 0.1 uF is to filter out RFI or other noise, even ringing current,
from the phone line, so all you see is the DC voltage.
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###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it:
http://physics.nist.gov/cuu/Units/binary.html
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