Another transistor question

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

Ok, make the following replacements in my original post:

open = on closed = off

Martin

"Martin Cote" schreef in bericht news:9Lyhi.265750$ snipped-for-privacy@fe09.news.easynews.com...

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

That terminology is plain wrong.

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

That's where your problem lies.

Graham

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

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.

Yes.

Bingo. The transistor and the load form a current divider.

Got it!

That was helpful! Thanks!

Martin

Yes

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

Graham

drop

Thanks for the confirmation! Martin

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 load.

petrus bitbyter

----+-- +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

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 ;-)

--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
durable; the other is a cheaper thing, but the moths get into it."
                                             (Stephen Leacock)

Wouldn't we all! ;-)

Try some of these links:

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

Have Fun! Rich