Since the current through the transistor is controllable, it means that the internal resistance of the transistor is changing. The fact that when its base-emitter resistance changes it reflects the change in its collector-emitter resistance.
Hence it is termed as transfer of resistance-transistor.

Inverter is defined as the circuit that converts DC voltage source into AC voltage at required frequency.
Now CMOS-FET inverter is a device which is actually a FET (Field Effect Transistor) in CMOS technology.
CMOS-FET inverter is a simply a push-pull type FET circuit as shown below, which amplifies both halves of the AC cycle and give you reliable output of AC source.

thank you for your immediate response.

sir I would like to know working of CMOSFET inverter.

It is always the case, that when voltage across base-emitter junction of NPN transistor is greater than or equal to 0.7V, then it is forward biased.

When this voltage is quite large, say around 3V or more (upto certain limit), then total the number of free electrons in emitter layer start participating in conduction of electric current through the transistor.
This condition is called as saturation of transistor and in this condition we can use it just like a switch which is ON.
And when the base voltage is quickly reduced to zero, this process of conducting current suddenly stops and we can use the transistor just like a switch which is OFF…!

What type of CMOS info you need?

I have doubt when npn transistor is on or off i.e is it right if the voltage at the base is more than the emiter volage applied the transistor is on.
I would also like to know when does the transistor act as switch as well as CMOS working.

You may also refer other articles on configurations of transistor, to clear your doubts.

Refer this article…
http://vsagar.com/2012/10/20/simple-details-common-collector-configuration-npn-transistor-2/

My dear friend,

• The BE junction is forward biased. So the physical barrier within it is more easy to surmount for electrons coming from emitter layer.
• Once then enter base layer, they have strong mutual repulsion among them.
• So some of them have the tendancy to get attracted towards the +ve terminal of Vbe.
• Hence base current (conventional current) flows INTO base, rather than coming out…!
• But don’t consider the -ve terminal of Vcb battery, since it is NOT CONNECTED to BE junction, it is connected to CB junction.
• So you are not allowed to consider the ELECTRICAL CONNECTIVITY of Vcb with this process.

Please also read the explanation given on your next comment.
I like the way, you are asking the queries. It increases the wisdom of understanding this process in deep.

The reason that the Vcb >> Vbe, is as follows….

• The CB junction is reverse biased.
• So it has large physical gap than BE junction, as this junction is forward biased.
• When electrons come from emitter layer into base layer, they are crowded in small area of base.
• So they acquire sufficient repulsive force among them to go into collector layer.
• But this force is still slightly less than the required force to overcome CB junction barrier.
• So a strong attractive force from collector side is required to pull these electrons towards collector layer and then towards +ve terminal of battery.
• Hence the Vcb >> Vbe.

Please explain ………

I don’t understand this basic concept …..

THANKS…….

The electrons repel each other due to mutual repulsion…like charges, as they have…

from fwd bias-> f) Now large number of electrons in emitter layer is repelled by negative terminal of Vbe and they flow towards b-e junction.

why are the electrons in emitter layer is repelled when it’s connected to negative junction of the battery which is also full of electrons??

Your queries can be explained in short as follows…

• Here we have to consider the minority charge carriers. When a large reverse biased PD is applied across the BC junction, these minority charges acquire strong repulsive force and they CAN NOW penetrate through the junction.
• The reason that the base is small in size and lightly doped are as follows…
• When transistor is forward biased, a large number of negative charges flow towards BE junction.
• When they enter in base region, due to small space, they acquire further mutual repulsive force among them.
• Due to this they have sufficient KE to cross the reverse biased CB junction.
• Also in base layer we don’t want to “waste” electrons, within holes.
• So the doping concentration of base is less.

Also, I thank you for putting this query, as me too, was thinking of writing down a short note on these points in particular.
Thanks also for visiting the site.
Please don’t hesitate to point out the “points” missing…by sparing out of your valuable time….

May I please know what sort of more description you need further?
Please reply, so that I can make necessary updates and enrich my website.
Thanks for visiting the site…