# BJT - understanding its operation

Discussion in 'Electronic Basics' started by Seeker, Mar 15, 2006.

1. ### SeekerGuest

I'm an electric engineering student. I'm getting into electronics and I
can't figure out the bjt transistor.

The fact that it has two p-n contacts means that I'm confused every time I
see a schematic. Of course the book I'm reading does a pretty good job of
explaining but I still can't figure it out.

For example:
http://img518.imageshack.us/img518/7008/screenshotlineartechnologyltsp.png

This is a circuit I took from the book, it says it's used for sampling.
When V1=0 => Vo=Vi=V2
and when V1=5=>Vo=0,2V

V1 is generally a pulse signal generator and V2 is the signal we want to
sample. I understand that the bjt is used here as a switch, it operates at
core or cut-off mode etc.

What I don't understand is, isn't the V2 source affecting the circuit? We
control the Ib current with the V1 generator but why doesn't the V2 signal
affect our circuit whatsoever?

I can learn the circuits and study them on SwCAD but I don't want to have
to refer to a book everytime I see a new circuit so I can understand how
it works. Any help would be appreciated.

2. ### Charles SchulerGuest

V2 is the dc power supply.

3. ### Jonathan KirwanGuest

I'm still learning about more general models, myself.
Well, they don't show one of them. But it's there. It's just usually
not symmetrically built, so reversing the emitter and collector leads
won't get you the same behavior. It's supposed to go in one way. But
I understand that there are rare times when you might prefer to
reverse it and use it the other way. Not that I've done so for any
design purpose, though.
It's good to hear that the book is okay.
It does. It affects the voltage seen at the collector when the BJT is
"off" and there is effectively no Ib current. However, when the Ib
current is active and substantially driving the BJT into saturation,
the collector will be close to or less than a single diode drop above
the emitter. This is because, in saturation, the base-collector diode
is generally no longer reverse-biased and is starting to conduct in
the forward direction. Whatever collector current is required to
reach that point will flow, assuming the BJT's base drive is
sufficient and maintained.

How that takes place is another story. You might take a look at John
Popelish's recent contribution here to a thread titled, "transistors:
so confusing!!" on his 'drunken bum on a crazy street' theory.

Jon

4. ### Anthony FremontGuest

We

Yes it is. It certainly affects the current flowing out the emitter.
Like I said, it does.
Look at it like this: V1 will cause 5mA to flow thru Rb, into the base
and subsequently out the emitter to ground. Rc will allow up to 100mA
to flow thru it and into the collector and out the emitter to ground
(assuming no current is being drawn from Vo). As long as the DC beta
(Hfe) is at least 20, a 5V pulse from V1 will cause the transistor to
turn on and shunt all current available from V2 to ground. This has the
measurable effect of lowering the voltage at Vo to something very close
to ground (0.2V in your case). You should think of the BJT transistor
in terms of current and not voltages as that's what it is all about.
Transistors amplify the current applied to base by the beta of the
transistor and will allow that amount of current to flow thru the
collector. Something like a water valve allows you to apply a small
amount of force to regulate a much larger force.

Sorry if it's not a very clear explanation. I'm sure others will do
much better.

5. ### kellGuest

For example:
Yes indeed, that's about all you need to know to figure out how this
circuit works. Never mind about the internal pn stuff.
V1 controls the transistor. So with V1 high, the transistor on, the
transistor grounds out the node where you see Rc, the collector, and Vo
all connected and thus you see essentially zero volts (actually 0.2) at
Vo.

Now put V1 low. That turns the transistor off. Now with the
transistor's collector/emitter terminals looking like an open circuit,
the voltage at Vo equals V2 and Vi (why did they label it twice?), and
you have your sample. This assumes that whatever you are using to
detect the voltage at Vo has a high impedance, so it doesn't affect

to Vo and ground. This is your sample detector. Use a switch to turn
the 5 volts on and off (the circuit drawing omitted that part). Use
some low voltage at V2, like a twelve volt battery. The voltage on
your meter's screen will either equal V2 or read very low (just a few
tenths of a volt), depending on what's going on at the transistor's
base.

6. ### JamieGuest

V0 is what you want to sample.

7. ### redbellyGuest

Here's a link to John's drunken-bum description: