BJT under saturation condition??

Under saturation condition base current is very large that voltage across Base-emitter junction of  silicon based BJT is more than 0.7 volt. In this condition the voltage at collector (Vc) is 0.2 volt.

By kirchoff's voltage law Vc=Vce( imagine emitter is directly grounded)
Vce =Vcb+Vbe.

If Vbe>0.7volt how Vce=0.2volt?
As Vce includes Vbe, it must also be more than 0.7 volt.
But it is 0.2 volt wich is less that 0.7 volt.

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You can see he's looking at Vbe(sat) at 0.7 V, and wondering why Vce(sat) is only 0.2V. He's missed the bit about this being a three terminal device.....
Doesn't understand how the transistor - or perhaps resistance/current and voltage work together.

Most people want to look at BJT's as voltage operated devices because that's what their used to.

If the BJT is saturated then it's current Ice will be maximized (that's what saturated means) - so it's apparent resistance is low. (high current = low resistance - otherwise there would be a lot of heat)

So the voltage Vce will be low

(You can't measure a voltage across a short circuit)

(See 2nd down in the list on the right.)
deepakmurali (author)  rickharris5 years ago
Consider 3 points C,B,E in a straight line such that distance between C and E is same sum of the distance between C anb B +distance between B and E.
(i.e) CE=CB+BE

where CE is distance between C and E
BE is distance between B and E
CB is distance between C and B

If CE =0.2 metre , how BE=0.7 metre. BE must be less than CE na?

will CB be negative????????
And herein lies your problem. You are confined to straight line thinking, and this isn't that system. You aren't tapping a resistor. This is a 4 terminal device, not the three you are seeing in your head.
Oh no I don't do homework for students - You either have to justify the problem in real terms ie it exists, or work it out for your self.
It depends where your measuring the voltage.

As the "apparent" resistance of the transistor decreases the voltage across the transistor collector to emitter will also decrease.

Homework again?

(edited because of typo.)
You're confusing Kirchoff's voltage law with his There are two independent loops to consider for the device, Vbe, and Vce. Ie=Ic+Ib as it should do.