Re: Another transistor question
- From: "petrus bitbyter" <pieterkraltlaatditweg@xxxxxxxxxxxxxxxxx>
- Date: Sat, 30 Jun 2007 22:42:20 +0200
"Martin Cote" <mentalray@xxxxxxxxxxxx> schreef in bericht
news:9Lyhi.265750$f82.220242@xxxxxxxxxxxxxxxxxxxxxxxxx
Hi,
I was trying to understand a NOT gate implementation at the following
page:
http://www.kpsec.freeuk.com/trancirc.htm#inverter
AFAIU, if the transistor is open (i.e., there is voltage at the input),
the current from +Vs is sink into the ground, so there is no voltage for
the output. If the transistor is closed, the only path for the current is
to go to through the output. Is this correct?
When the transistor is open, what does prevent the current to go through
both the transistor and the output? Does the transistor has full priority
on the current? If so, why?
I would appreciate some help!
Thanks,
Martin
You start at the wrong side.
The base-emitter junction of a Si transistor requires a voltage of over
about 0.6V before it will pass current. Below this so called knee-voltage
there will flow no current into the base. So when the input voltage is low -
let's say <0.5V - the base will not pass current so there will be no
collector current either. The transistor is said to be off and the output
voltage will be high.
When the input voltage is high - let's say about Vs - a current of about
Vs/1k[mA] wil flow into the base. In a general purpose transistor this
should cause a collector current of some 100 times the base current... if
that current was not limited by the 1k resistor. The collector current now
causes a voltage over the 1k resistor that will almost be as high as Vs. The
collector voltage will lower accordingly to some 0.2V. So the output voltage
is low and the transistor is said to be on or saturated.
petrus bitbyter
.
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