NLDD dose influence on threshold voltage

Dear All

Anyone can help to give me some idea what will happen to Vtn if the
NLDD dose is increase or decrease?
ANy BSIM equation I can relate Vtn and NLDD dose?
I appreciate if you have any paper or book, you can refer me to.
Kindly help me to understand the physics behind for the relations.

Thank you all

best rgds
Jason

 

[Stephen]

Jason,

I will give you my best answer to your question, with quotations, but
hopefully someone else can verify that my answer is correct.

According to "Introduction to VLSI Circuits and Systems" by Uyemura, p.
137:

"A lightly doped drain MOSFET is designed to reduce the electric fields
in the channel region by providing n- (lightly doped) drain and source
regions instead of the usual n+ regions. Theoretically, this reduces
the maximum electric field intensity, which in turn increases the
reliability of the transistors."

Since you did not ask about why reliability is increased, I will not go
into that here.

However, since you are asking about the effect on Vtn, my answer is
that I'm not sure that there is an effect. If you look at the standard
equation for the threshold:

Vtn = Vto + lambda[ sq(-2*fermi_potential + Vsb) -
sq(2*fermi_potential) ]

Now if we assume that Vsb = 0, then we just have:

Vtn = Vto

Vto is a function of the work function difference between the gate
material and the silicon substrate. It seems to me that Vtn is simply
trying to create the channel between the source and drain that will
allow electrons (or holes) to flow through, and that the doping
concentration of the source/drain regions determines the magnitude of
the electric field. (In small channel devices, this electric field
causes velocity saturation - or the current to saturate very quickly,
and other frusturating design effects such as reliability problems,
which LDD helps solve by decreasing the doping concentration, and
thereby decreasing the electric field). I cannot find any explainable
links between Vtn and LDD, at least not directly.

Do others agree with this?

- zielstep

cheanglong@gmail.com wrote:

Dear All

Anyone can help to give me some idea what will happen to Vtn if the
NLDD dose is increase or decrease?
ANy BSIM equation I can relate Vtn and NLDD dose?
I appreciate if you have any paper or book, you can refer me to.
Kindly help me to understand the physics behind for the relations.

Thank you all

best rgds
Jason

 

Dear Stephen

Firstly I would like to thank you for posting to my message. It is my
honour to have people around here who is willing to discuss with me. It
is not easy to find people who willing to share ideas.

After going through your message, I got to know a bit more of the short
channel effects and the reason for using LDD.

Currently I am sorting out my thoughts on this subject. I will post
back within this week after I sort out my mind.

As of now, I totally agree with you.
But since the LDD will decrease the electric field at drain side, would
not the lower electric field will help in improving the threshold
voltage( not too low) ?
Any comments?

Thank you Stephen!

Anyone has idea, kindly share with us
Cheers


best rgds
Jason
Stephen wrote:

Jason,

I will give you my best answer to your question, with quotations, but
hopefully someone else can verify that my answer is correct.

According to "Introduction to VLSI Circuits and Systems" by Uyemura, p.
137:

"A lightly doped drain MOSFET is designed to reduce the electric fields
in the channel region by providing n- (lightly doped) drain and source
regions instead of the usual n+ regions. Theoretically, this reduces
the maximum electric field intensity, which in turn increases the
reliability of the transistors."

Since you did not ask about why reliability is increased, I will not go
into that here.

However, since you are asking about the effect on Vtn, my answer is
that I'm not sure that there is an effect. If you look at the standard
equation for the threshold:

Vtn = Vto + lambda[ sq(-2*fermi_potential + Vsb) -
sq(2*fermi_potential) ]

Now if we assume that Vsb = 0, then we just have:

Vtn = Vto

Vto is a function of the work function difference between the gate
material and the silicon substrate. It seems to me that Vtn is simply
trying to create the channel between the source and drain that will
allow electrons (or holes) to flow through, and that the doping
concentration of the source/drain regions determines the magnitude of
the electric field. (In small channel devices, this electric field
causes velocity saturation - or the current to saturate very quickly,
and other frusturating design effects such as reliability problems,
which LDD helps solve by decreasing the doping concentration, and
thereby decreasing the electric field). I cannot find any explainable
links between Vtn and LDD, at least not directly.

Do others agree with this?

- zielstep

cheanglong@gmail.com wrote:
Dear All

Anyone can help to give me some idea what will happen to Vtn if the
NLDD dose is increase or decrease?
ANy BSIM equation I can relate Vtn and NLDD dose?
I appreciate if you have any paper or book, you can refer me to.
Kindly help me to understand the physics behind for the relations.

Thank you all

best rgds
Jason

 

[Stephen]

Jason,

I've throught it through some more, and here's my new conclusion:

At very low channel lengths, as the length continues to decrease we are
seeing that Vth is actually decreasing! This is because the
source/drain regions actually have a depletion region of their own, and
this subtracts from the channel area where the Vgs has to overcome Vth.
Therefore, at larger values of L, the source/drain depletion regions
are very small and have no real effect, but at small values of L, they
cause Vth to lower.

In the same way, if we lightly dope part of the drain region, then this
will cause the (rather small) depletion region to shrink just a little.
Therefore, Vth will rise for very low L (it will stay the same
otherwise, because the change will not be noticeable comparibly
speaking). However, the trend of Vth decreasing will continue with
smaller values of L, but by having a lightly doped drain it just won't
decrease by as much.

I hope this helps.

- zielstep

cheanglong@gmail.com wrote:

Dear Stephen

Firstly I would like to thank you for posting to my message. It is my
honour to have people around here who is willing to discuss with me. It
is not easy to find people who willing to share ideas.

After going through your message, I got to know a bit more of the short
channel effects and the reason for using LDD.

Currently I am sorting out my thoughts on this subject. I will post
back within this week after I sort out my mind.

As of now, I totally agree with you.
But since the LDD will decrease the electric field at drain side, would
not the lower electric field will help in improving the threshold
voltage( not too low) ?
Any comments?

Thank you Stephen!

Anyone has idea, kindly share with us
Cheers


best rgds
Jason
Stephen wrote:
Jason,

I will give you my best answer to your question, with quotations, but
hopefully someone else can verify that my answer is correct.

According to "Introduction to VLSI Circuits and Systems" by Uyemura, p.
137:

"A lightly doped drain MOSFET is designed to reduce the electric fields
in the channel region by providing n- (lightly doped) drain and source
regions instead of the usual n+ regions. Theoretically, this reduces
the maximum electric field intensity, which in turn increases the
reliability of the transistors."

Since you did not ask about why reliability is increased, I will not go
into that here.

However, since you are asking about the effect on Vtn, my answer is
that I'm not sure that there is an effect. If you look at the standard
equation for the threshold:

Vtn = Vto + lambda[ sq(-2*fermi_potential + Vsb) -
sq(2*fermi_potential) ]

Now if we assume that Vsb = 0, then we just have:

Vtn = Vto

Vto is a function of the work function difference between the gate
material and the silicon substrate. It seems to me that Vtn is simply
trying to create the channel between the source and drain that will
allow electrons (or holes) to flow through, and that the doping
concentration of the source/drain regions determines the magnitude of
the electric field. (In small channel devices, this electric field
causes velocity saturation - or the current to saturate very quickly,
and other frusturating design effects such as reliability problems,
which LDD helps solve by decreasing the doping concentration, and
thereby decreasing the electric field). I cannot find any explainable
links between Vtn and LDD, at least not directly.

Do others agree with this?

- zielstep

cheanglong@gmail.com wrote:
Dear All

Anyone can help to give me some idea what will happen to Vtn if the
NLDD dose is increase or decrease?
ANy BSIM equation I can relate Vtn and NLDD dose?
I appreciate if you have any paper or book, you can refer me to.
Kindly help me to understand the physics behind for the relations.

Thank you all

best rgds
Jason

 

Dear Stephen

I totally agree with what you said. I have tried to link short channel
effect to the Nldd and have the similar thought as you have shared.
So whenever channel is very short, Vt will decrease. But when the Nldd
is lighter, it will compensate the Vt roll off up to some degree which
is unknown to me.
Therefore, the Vt will stay the same or increase very slightly.
I will write back if I read about some new stuff ...
Thank you so much Stephen

best rgds
Cheang Long