# Another transistor question

Discussion in 'Electronic Basics' started by Martin Cote, Jun 30, 2007.

1. ### Martin CoteGuest

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

2. ### EeyoreGuest

Your use of the terminology open and closed is rather odd and seems to be back
to front.

Think of the transistor as a switch. It's either on or off. Try that and come
back if it still doesn't make sense.

Graham

3. ### Martin CoteGuest

Ok, make the following replacements in my original post:

open = on
closed = off

Martin

4. ### petrus bitbyterGuest

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

5. ### EeyoreGuest

That terminology is plain wrong.

When the transistor's ON (current supplied to the base) it's equivalent to a switch
being CLOSED.

Graham

6. ### Martin CoteGuest

It really make sense the way you phrased it. I think that part of my
confusion is caused by the fact that the current goes in the inverse
direction of the electrons flow.

My question is why there's no current left for the output when the
transistor is on? Why isn't the current split into both the 1K
transistor and the output? My guess is that anything connected to the
output is actually in parallel with the transistor (is this right?), and
since the transistor path has less resistance, the full current goes there.

Am I on the right track?
Martin

7. ### John PopelishGuest

The load on the output is assumed to be something like
another copy of this circuit. Or it might be a volt meter
connected between the output and 0v. In either case, any
current into the load must come through the 1k collector
resistor. The load and the transistor in this circuit are
in parallel, and divide the current that arrives through the
1k collector resistor. When the transistor is conducting
(when the input signal is positive) nearly all the current
the 1k resistor can supply is diverted to 0v through the
transistor. When the transistor is off (input signal less
positive than .6 volt), nearly all the current the 1k
resistor can provide goes to the load, and the output
voltage depends on the resistance of the load, compared to
the 1k collector resistor. If the load is another copy of
this inverter (10k resistor in series with a base to emitter
junction) more than 9/10ths of +Vs will appear at the output.
Because the signal voltage is measured with respect to 0v.
So the transistor can only short the output signal to 0v,
not create a positive output voltage.
Bingo. The transistor and the load form a current divider.

Got it!

Martin

9. ### EeyoreGuest

Almost all the current. Even when ON, a real transistor has a small voltage drop
across it.

Graham

10. ### Martin CoteGuest

Thanks for the confirmation!
Martin

11. ### petrus bitbyterGuest

An "on" transistor can be considered a very low resistance, almost a short.
And as the current takes the way of the lowest resistance all current goes
through the transistor.

As for the load, it depends. You can connect it between output and ground
but also between output and Vs. If you replace the 1k by a bulb, it *is* the

petrus bitbyter

12. ### JasenGuest

----+-- +5v
|
1K
|
+---- out
|
/
|/
in ---[10k]---|
|\|
~\
|
-------+----
"open" is a bad word to use in this context, an open switch is one
that's turned off, "on", or "conducting", or "saturated" are more
apropriate here.
nothing but the resistance or other associates voltage requirements
of whatever is connected to the output.

whemn the transistor is fully on, the voltage at the output will be
less than half a volt this is not enough to turn another transistor on
(needs atleast 0.6v) or to light a LED ( needs 1 to 3v), so for most
practical purposes the output is off

Bye.
Jasen

13. ### Fred AbseGuest

A bit like Shannon's MIT Master's thesis. He uses 0 for a closed contact
and 1 for an open contact. Makes the math easier to develop, but odd in a
modern context. Threw me completely until I got used to thinking
"negative" logic ;-)

14. ### Rich GriseGuest

Wouldn't we all! ;-)

They explain transistors a lot better than I can in a USENET newsgroup.

Have Fun!
Rich

15. ### neon

1,325
0
Oct 21, 2006
there are spes for all transistors about open base Ic leakage open emiters and so forth. if the transistor has a hi leakage curent and you put a 500k on the collector it might seem like it is on just by the voltage drop because of Ico leakage. a transistor is no a switch is a voltage amplifier. but if ever it saturates the current can and will conduct both way the two diode act lik a short.