# a doubt regarding saturation mode of BJT

#### devanandiamin7

Joined Apr 14, 2007
5
Hi all,
I have a doubt regarding saturation mode of BJT. Saturation occurs when both emitter-base junction and base-collector junction gets forward biased. For forward biasing a silicon p-n junction only 0.7V is needed. So is for emitter-base junction. But not for base-collector junction. Because we know at saturation voltage across collector and emitter Vce is 0.2V. Using Kirchoff's voltage laws the voltage at base-collector p-n junction is less than 0.7V. Why is it lesser?

#### Ron H

Joined Apr 14, 2005
7,014
First of all, 0.7V is just an approximation to be used in first-order calculations. The forward voltage is actually a function of current and current density (current per unit junction area).
In a BJT, the collector-base junction is much larger than the emitter-base junction. This means that, if junction currents are equal (they probably won't be), Vbc will be less than Vbe.

Joined Feb 6, 2012
11
Electrons flowing from the emitter to the base must recombine with the holes in the P material (NPN transistor). Therefore the Vbe is pretty close to a standard diode drop. However, once the base emitter is forward biased, the electrons travelling from the emitter to the collector do not recombine in the base, they pass right on through to the collector. With no diode junction effect, the Vce saturation voltage drop is due to mostly to the bulk resistance of the material.

#### Ron H

Joined Apr 14, 2005
7,014
Electrons flowing from the emitter to the base must recombine with the holes in the P material (NPN transistor). Therefore the Vbe is pretty close to a standard diode drop. However, once the base emitter is forward biased, the electrons travelling from the emitter to the collector do not recombine in the base, they pass right on through to the collector. With no diode junction effect, the Vce saturation voltage drop is due to mostly to the bulk resistance of the material.
But there is still a diode junction from base to collector that is forward biased. Our OP is wondering how the forward voltage drops of the B-C junction and the B-E junction can be different.
Hopefully, I explained that in my previous post.

#### cabraham

Joined Oct 29, 2011
82
The forward drops of the c-b and b-e junctions are indeed different due to differing doping levels. Emitter region is doped much heavier than the collector region. Base region is doped lightly. So the "n-p-n" structure is asymmetric. Also, if the base region is doped in a "heterogeneous" fashion, the density of acceptors at the collector edge of the base is heavier than that at the emitter edge of the base. This helps reduce c-b reverse leakage current.

So the c-b junction has a different scale current than the b-e junction. Recall for a p-n junction diode the following I-V relation:

Id = Is*exp((Vd/Vt) - 1), known as Shockley's equation.

The "Is" scale current factor, aka "reverse saturation current", is determined by dimensions, doping level/profile, temperature, etc. For a bjt structure, n-p-n, we have 2 junctions. So these 2 junctions have differing "Is" values due to doping density/profile, geometry differences. We distinguish these 2 Is values as "Ics" for the c-b junction, and "Ies" for the b-e junction.

When the bjt is in saturation, the c-b junction is forward biased with the base being positive, hence the collector is 1 forward drop below the base. But the base is 1 forward drop above the emitter. The "forward drop differs for the 2 junctions, such that if Vbe = 0.7V, while Vbc = 0.6V, then Vce = 0.1V.

A good bjt reference text can elaborate. The details can be found in books by Millman, Taub, Schilling, Sze, Muller, Kamens, and others. Search for books from these authors and you should find one that helps.

Claude