Effect of source to bulk voltage on Vth

Hi,

I am having a confusion in understanding increase of Vth voltage by increasing Vsb voltage. I am quoting from Behzad Razavi book [1].


"As Vb becomes more negative, more holes are connected to the substrate connection, leaving a large negative charge behind, i.e., as depicted in Fig. 2.22, the depletion region becomes wider. now recall from Eq. (2.1) that the threshold voltage is a function of total charge in depletion region because the gate charge must mirror Qd before an inversion layer is formed.Thus, as Vb drops and Qd increases, Vth also increases. This is called body effect".

My question is, if there are more electrons in the depletion region then less voltage at gate should be required in order to make the current flow between drain and source (i.e to create an inversion layer). This means Vth should decrease not increase. Can somebody explain this.


regards

There are not more electrons in the depletion region, see below.

We are talking about an NMOS in p-substrate, don't we? Leaving a large negative charge behind doesn't mean mobile electrons, but immobile negatively charged (Boron) ions and mobile positively charged holes - which in "reality" are electrons jumping from one (negatively charged) immobile Boron ion to another one (previously not yet negatively charged) into the opposite direction, leaving the former one uncharged, which "behaves" like a movement of a positive charge carrier to the counter-direction.

Now this greater amount of immobile negatively charged ions - as depicted in Fig. 2.22 - is compensated by the same amount of (positively charged) holes, thus generating a wider depletion zone - depleted of immobile electrons (!) - because, as long as Vgs < Vth - there's no inversion yet. So a larger Vth is needed for inversion, because some part of it is needed to compensate for the "mobile" holes' charge.

I admit that the description in Razavi's book - together with his figure 2.22 - is rather confusing, and I'm still not sure if I could make this body effect conceivable.

Hi erikl -

I think your explanation is equally confusing, sorry - there is no compensation of mobile hole charge by gate voltage - mobile hole charge is compensated by ionized impurities, in the neutral region.

I propose the following kind of intuitive explanation of the body voltage effect on Vt.
We are talking about nMOS device, with p-type substrate / body.

Suppose there is inversion layer of electrons, and depletion region is established in the substrate.
Negative charge of electron inversion layer and negative charge of depletion layer is compensated by positive gate voltage (actually - positive charge at the bottom of the gate).
Now, let's apply reverse body bias - i.e. applying negative body/source voltage Vbs.
Depletion region in the substrate becomes wider, containing more negative charge.
Voltage and charge on the gate remains the same.
Thus, electron density in the inversion layer should decrease - which is equivalent to increase of Vt (or decrease of gate overdrive Vg-Vt).

Max
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timof said:

Hi erikl -

I think your explanation is equally confusing, sorry - there is no compensation of mobile hole charge by gate voltage - mobile hole charge is compensated by ionized impurities, in the neutral region.

Click to expand...

You're right, Maxim! The depleted p region of course is depleted from the "mobile" holes - that's where from it gets is name.
The second part of my explanation above is nonsense, whereas the first part considers only the neutral p region, sorry!

Your explanation is intuitively better.

erikl

I am sorry but i can't understand both explanations....What i have got from above is something like this..

negative body voltage create wider depletion region. i.e. wider region of boron atoms where electrons have jumped from p-substrate. As there are now more negatively charged boron atoms, do these negatively charged boron atoms not cause ease in current flow? If not then why this depletion region condition is different than inversion layer....what happens when inversion layer is formed.

regards

Let me try to answer:

viperpaki007 said:

negative body voltage create wider depletion region. i.e. wider region of boron atoms where electrons have jumped from p-substrate.

Click to expand...

No electrons have jumped from p-substrate. Originally, the mobile charge carriers in the (neutral) p-region were holes. After depletion (of a part) of this p-region (because of the reverse voltage between source & bulk, resp. the mirror layer of electrons in silicon under the gate generated by the positive gate voltage, which both generate an n⁺p-junction, the n⁺ region being the inversion layer), those holes have been attracted to the electron mirror layer under the gate, and annihilated there (actually this means that some of these mirror electrons occupy all the boron atoms in the depletion layer and by this change them to ions, i.e. immobile, negatively charged boron ions. This creates an extremely thin depletion region also in the n⁺ region of the n⁺p junction).

viperpaki007 said:

As there are now more negatively charged boron atoms, do these negatively charged boron atoms not cause ease in current flow?

Click to expand...

No, these are boron ions fixed in the silicon lattice, immobile charges which cannot contribute to a charge flow (current).

viperpaki007 said:

If not then why this depletion region condition is different than inversion layer

Click to expand...

The depletion region is depleted from mobile charge carriers, both in the p and in the (very thin) n⁺ region of the n⁺p junction. The inversion layer is the n⁺ layer consisting of mobile electrons (minus this thin n⁺ depletion layer mentioned above, adjacent to the junction).

viperpaki007 said:

....what happens when inversion layer is formed.

Click to expand...

As soon as it is thicker than the thin n⁺ depletion region, it contains mobile electrons. These constitute the conducting channel.