Let's consider npn transistor, when emitter-base is forward with

Outrageous

Let's consider npn transistor, when emitter-base is forward with voltage supplied to it, then the majority charge carrier(electron) will diffuse to base(p-type) ,
Then (in base-emitter) the minority charge carrier in p-type (electron) will that move close to the depletion layer of base-collector will drift to the collector(the flow of minority charge is spontaneous) so there will be current flow even without voltage supplied to base-collector circuit, right?
But why book say when no voltage supplied to the base-collector then there will not be current? Or this is because the current caused by minority charge in base is too small to be detected?

Outrageous

I was correct because that is why a transistor can act as a switch?

Kholdstare

First of all the majority carriers are injected over the barrier from n-emitter to p-base.

If you do not provide any base-collector bias, there wont be any collector current and the transistor will act similar to a diode.

uart

Yes this time outragous is basically correct (though I did have to edit it a bit before I could really make that claim ).

Certainly on the important point that collector current can continue to flow even when Vbc=0 you are correct. In fact, it (a positive collector current) can continue to flow even when the bc junction is lightly forward biased (collector voltage lower than the base voltage for an npn device). This operating mode is called saturation.

Outrageous

Thanks.
Do you mean that when it is heavily forward biased the minority charge carriers (hole) will not drift due to the decrease in depletion region of base-collector.

Saturation mode is two forward biased diode . For npn, n-p is forward, and p-n is also forward?

uart

We were talking about npn right? The minority carriers in the base are electrons.

If the base-collector junction is too heavily forward biased then you'll start to get significant injection of holes from base to collector, a component of current flow opposite to what you desire. (current exiting the collector of npn rather then entering).

If however the base collector junction is only lightly forward biased then little BC injection will occur, but the electrons injected at the BE junction will continue to diffuse across the base and to the collector.

Taking for convenience the emitter voltage as ground, for an npn that's lightly saturated the base voltage will be approx 0.7 volts and the collector voltage about 0.5 volts. This makes the collector (n) about 0.2 volts lower than the base (p) and hence the BC junction forward biased.

For an npn in heavy saturation the collector voltage could be a low as 0.3 volts (or less) and the BC junction forward biased by around 0.5 volts. At this point the transistor current gain is falling fast (with increasing saturation) preventing it getting much more deeply saturated.

BE is forward biased, and BC is also forward biased, but not as strongly as BE.

Outrageous

Then since there will be an injection of holes from base to collector ,then why do we need to forward biased BC with voltage supplied?

uart

- There is no (significant) injection of holes from base to collector. As I explained above, saturation doesn't go that far under normal forward (current entering the collector for an npn device) operation.

-In general we don't forward bias the BC junction with a supplied voltage. We normally reverse bias the BC junction.

Outrageous

I am sorry, ask wrong thing.
I mean if we don't forward the BC , there will still be an injection of minority charge carriers (electron) into the collector. Then why do we need to forward?
And if we don't put any voltage supplied to BC ,then there will be more current flow into the collector compare with forward biased BC , is this statement correct?

uart

Firstly, we don't normally forward bias the BC junction with the external supply, we in fact reverse bias it (eg collector more positive than base for an npn).

Now I'm not sure where this is leading, but it looks like you're thinking the you can achieve active mode operation without any external supply to bias it - you can't.

Yes saturation mode is extremely interesting, in the sense that we can maintain a current gain despite the external voltage between base and collector reversing slightly (so that this voltage actually opposes the direction of collector current flow). This is something that does confuse many students, and to understand it you need to realize that it is totally dependent upon base-emitter injection maintaining highly non thermal-equilibrium minority carrier levels in the base.

You can't look at whats happening at the BC junction in isolation. Looking at the device overall as a three terminal device you need a positive CE voltage (npn) to get a positive collector current. The transistor is NOT an energy source.

Anyway all of this is getting away from what I thought your initial question was about. It started out looking like an interesting question regarding some apparent contradictions of saturation mode, but has kind of gone off course since then. At this point I'm not entirely sure what you're asking or where your confusion is.

Enthalpy

You can operate a bipolar with zero base-collector voltage. It works almost normally, including the current gain.

In saturation mode, the base injects carriers in the epitaxial zone of the collector. This is where charge are stored in saturation and makes desaturation slow.

Beware some older books state wrongly that saturation injects carriers in the base. This was once with allied transistors, back in the germanium era, before epitaxy.

Studiot

This is nonsense.

If the base and emitter are connected to a circuit so that the base-emitter voltage is defined then the base-collector and emitter-collector voltages are also defined if the collector is also connected to some part of the same circuit.

If the collector is not connected then the collector voltage is undefined.

You cannot connect the collector without supplying some specified voltage (even zero).


Did you mean alloy transistor?

Enthalpy

If the transistor operates at or near zero collector-to-base voltage, about all emitter current is still swallowed in the collector, available for the external circuit.

That's because the depleted collector-base zone attracts the minority carriers injected in the base by the emitter.

It is the same process as in photovoltaic cells, except that the carriers are injected by the emitter, instead of being created by photons.

Just as in a photovoltaic cell, the current is available until the (collector-base) junction is directly polarized to a value where the corresponding direct current compensates the "photo"-current.

Correspondingly, a saturated bipolar transistor can have a collector current with Vce smaller than Vbe - can and does, if ohmic losses don't overshadow that.

Outrageous

Thanks for so many replies. Still have something uncertain.


V of emitter is 0V, V of base is 0.7V, V of collector is 0.3 V.
What is heavy saturation ? Is that mean current gain will be very small? What do you mean bu deeply saturated?

"Dependent upon base-emitter injection here mean that the
V of base minus V of emitter must be 0.7V, let's say V of emitter -0.7V, V of base is 0 , as long as V of collector is more positive than (-0.7V) then the electrons will definitely flow towards the collector?
What is highly non thermal equilibrium minority carrier levels in the base used for?

If V of base is -1V , V of collector is -1V , but V of emitter is-0.8V. Do you mean that I can still operate it? But why not when V of base and emitter are 0V ?
Is that because all the electrons in base will flow out from base instead of flowing into collector? Or
Is that because there won't be any change in the base-collector junction?

Charged are stored in saturation mean there are a lot of charge in the collector? What is desaturation?



In order to operate any mode of transistor, the collector terminal must be supplied with voltage, correct?

What do you mean by "swallowed" ? Disappear ? Or flow out from base?
Vce smaller than Vbe ?

Thank you .