Basic question in Transistors

Why Transistor is built with 3 terminals NPN.

why not 4 terminals like NPNP.

Thanks,
novice_vlsi

Go find a good basic electronics book that introduces transistors; there is a LOT of information there that would answer your question.

By controlling the current through the base-emitter P-N junction, you can control the current flow across the collector-emitter path. A transistor is, in simple terms, a current-controlled current source, and is typically operated in two modes: as a switch, or as an amplifier.

Why would you expect a transistor to need two PN junctions?

For example read this Ebook:

Power Electronics and Its Applications - Google Books

It explains also the NPNP and PNPN (usually: thyristor) structures ...

i think you are referring to thyristors...

PN junctioned devices...

hey friend
first understand the basic working priciple of diode, then transistor, and then thyrister. You will find the difference between their operation. You need to find the answer to this question.....

enjunear said:

A transistor is, in simple terms, a current-controlled current source,

Click to expand...

Actually, it is a voltage controlled current source.

The collector current is a function of the base - emitter voltage.

However, for many applications. it can be considered to be current controlled.

But you can't explain circuits such as the current mirror if you assume current control.

ljcox said:

Actually, it is a voltage controlled current source.
The collector current is a function of the base - emitter voltage.
However, for many applications. it can be considered to be current controlled.
But you can't explain circuits such as the current mirror if you assume current control.

Click to expand...

Yes! Good comment. Fazit: Don`t blindly rely on (few) textbooks.

ljcox said:

Actually, it is a voltage controlled current source.

The collector current is a function of the base - emitter voltage.

However, for many applications. it can be considered to be current controlled.

But you can't explain circuits such as the current mirror if you assume current control.

Click to expand...

Just asking....

as Ic=(beta)*Ib thus output current depends upon the input current.....so it should be current controlled current source....
although we can control the input current(Ib for common emitter) and thus it indirectly becomes voltage controlled current source...

please correct if i am wrong !! ;-)

nikhilsigma said:

Just asking....

as Ic=(beta)*Ib thus output current depends upon the input current.....so it should be current controlled current source....
although we can control the input current(Ib for common emitter) and thus it indirectly becomes voltage controlled current source...

please correct if i am wrong !! ;-)

Click to expand...

Yes, Ic = β Ib is correct. That is a very useful relationship.

But if you study how a bipolar transistor works, you will find that the collector current is determined by the charge in the base - emitter region.

Since that charge is proportional to the B-E voltage, then it is voltage controlled.

Consider the case where the transistor is used as a switch & is in saturation.

If it were current controlled, then Ic would stop immediately Ib was switched off.

But in fact, there is a delay before Ic drops to zero since the charge has to dissipate.

This is called the storage time.

There are techniques for reducing the storage time which involve discharging the B-E region as quickly as possible.

EDIT.
I have attached a simulation of a Bipolar Transistor characteristic.

It shows Ic versus Vbe. Note that it is an expotential function.

ljcox said:

Actually, it is a voltage controlled current source.

The collector current is a function of the base - emitter voltage.

However, for many applications. it can be considered to be current controlled.

But you can't explain circuits such as the current mirror if you assume current control.

Click to expand...

Yes, once you get out of the linear region, that is the more correct way of analyzing the device. For a beginner doing simple circuits, using Ic=β*Ib will be much easier to do a quick analysis to get him/her moving forward.

YMMV.

enjunear said:

Yes, once you get out of the linear region, that is the more correct way of analyzing the device. For a beginner doing simple circuits, using Ic=β*Ib will be much easier to do a quick analysis to get him/her moving forward.

YMMV.

Click to expand...

I agree that Ic=β*Ib is very useful for linear and switching circuits.

However, the expotential function is also necessary in the linear region in some cases, eg. the current mirror.

okay i got it....

so in a nut shell....

Ic depends on Ib and Ib depends on Vbe

so we get.... that Ic depends on Vbe

so it becomes... Ic = β*Io(exp(-Vbe/Vt)-1)

Thus it's an Voltage controlled current source.... :-D ;-)
Hence proved....

nikhilsigma said:

okay i got it....

so in a nut shell....

Ic depends on Ib and Ib depends on Vbe

so we get.... that Ic depends on Vbe

so it becomes... Ic = β*Io(exp(-Vbe/Vt)-1)

Thus it's an Voltage controlled current source.... :-D ;-)
Hence proved....

Click to expand...

No, it is approximately

Ic = Io(exp(-Vbe/Vt) - 1)

Ib is the recombination current.

See the attachment.

It shows that for both cases Ic is a function of Vbe, not Ib.